Airports from all around the world To Improve Your GK

Wednesday, September 30, 2009

Airports
An airport is a location where aircraft such as fixed-wing aircraft, helicopters, and blimps take off and land. Aircraft may be stored or maintained at an airport. An airport consists of at least one surface such as a runway, a helipad, or water for takeoffs and landings, and often includes buildings such as hangars and terminal buildings.

Larger airports may have fixed base operator services, seaplane docks and ramps, air traffic control, passenger facilities such as restaurants and lounges, and emergency services. A military airport is known as an airbase or air station. The terms airfield, airstrip, and aerodrome may also be used to refer to airports, and the terms heliport, seaplane base, and STOLport refer to airports dedicated exclusively to helicopters, seaplanes, or short takeoff and landing aircraft. In some jurisdictions, the term airport is used where the facility is licensed as such by the relevant government organization. Elsewhere the distinction is one of general appearance. Other jurisdictions define an airport by its having the customs offices etc expected of a port, though the more general term is airport of entry.







































C Program Aptiude Questions

Friday, September 18, 2009

1)struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.

2) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}

Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes

3) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

4) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.

5) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.

6) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

7) # include
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

8) # include
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

9) #include
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.

10) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

11) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

12) int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t i x
4 0 -4
3 1 -2
2 2 0

13) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

14) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

15) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.

16) What are the following notations of defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation

17) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

18) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.

19) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

20) void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

C++ Aptitude Questions and Answers

Thursday, September 10, 2009

Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived object, calling of that virtual method will result in which method being called?
a. Base method
b. Derived method..

Ans. b

For the following C program
#define AREA(x)(3.14*x*x)
main()
{float r1=6.25,r2=2.5,a;
a=AREA(r1);
printf("\n Area of the circle is %f", a);
a=AREA(r2);
printf("\n Area of the circle is %f", a);
}
What is the output?

Ans. Area of the circle is 122.656250
Area of the circle is 19.625000

void main()
{
int d=5;
printf("%f",d);
}

Ans: Undefined

void main()
{
int i;
for(i=1;i<4,i++)
switch(i)
case 1: printf("%d",i);break;
{
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}

Ans: 1,2,3,4

void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}

Ans: 6

void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(i
printf("less");
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}

Ans: less

void main()
{
float j;
j=1000*1000;
printf("%f",j);
}

1. 1000000
2. Overflow
3. Error
4. None

Ans: 4

How do you declare an array of N pointers to functions returning pointers to functions returning pointers to characters?

Ans: The first part of this question can be answered in at least
three ways:

1. char *(*(*a[N])())();

2. Build the declaration up incrementally, using typedefs:

typedef char *pc; /* pointer to char */
typedef pc fpc(); /* function returning pointer to char */
typedef fpc *pfpc; /* pointer to above */
typedef pfpc fpfpc(); /* function returning... */
typedef fpfpc *pfpfpc; /* pointer to... */
pfpfpc a[N]; /* array of... */

3. Use the cdecl program, which turns English into C and vice
versa:

cdecl> declare a as array of pointer to function returning
pointer to function returning pointer to char
char *(*(*a[])())()

cdecl can also explain complicated declarations, help with
casts, and indicate which set of parentheses the arguments
go in (for complicated function definitions, like the one
above).
Any good book on C should explain how to read these complicated
C declarations "inside out" to understand them ("declaration
mimics use").
The pointer-to-function declarations in the examples above have
not included parameter type information. When the parameters
have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)
What are the number of syntax errors ?
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}

Ans: None.

void main()
{
int i=7;
printf("%d",i++*i++);
}

Ans: 56

#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");

Ans: "one is defined"

void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}

Ans: 20 20 20

What is alloca()
Ans : It allocates and frees memory after use/after getting out of scope

main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}

Ans: 321

char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}

Ans: anything is good.

void main()
{
char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}

Ans: "harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
"ewlett-packard"

Output of the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}

a) 0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error

Ans. (d)

What is the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c
printf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(i
printf("pass2");
else
printf("Fail2")
}

a) Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these

Ans. (c)

What will the following program do?
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//

a) Swap contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1

In the following code segment what will be the result of the function,
value of x , value of y
{unsigned int x=-1;
int y;
y = ~0;
if(x == y)
printf("same");
else
printf("not same");
}

a) same, MAXINT, -1
b) not same, MAXINT, -MAXINT
c) same , MAXUNIT, -1
d) same, MAXUNIT, MAXUNIT
e) not same, MAXINT, MAXUNIT

Ans. (a)

What will be the result of the following program ?
char *gxxx()
{static char xxx[1024];
return xxx;
}
main()
{char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is : %s",gxxx());
}

a) The string is : string
b) The string is :Oldstring
c) Run time error/Core dump
d) Syntax error during compilation
e) None of these

Ans. (b)

Find the output for the following C program
main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf("%s\n",p2);
}

Ans. An empty string

Find the output for the following C program
main()
{
int x=20,y=35;
x = y++ + x++;
y = ++y + ++x;
printf("%d %d\n",x,y);
}

Ans. 57 94

Find the output for the following C program
main()
{
int x=5;
printf("%d %d %d\n",x,x<<2,x>>2);
}

Ans. 5 20 1

Find the output for the following C program
#define swap1(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5,y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}
int swap2(int a,int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}

Ans. 10 5

Find the output for the following C program
main()
{
char *ptr = "Ramco Systems";
(*ptr)++;
printf("%s\n",ptr);
ptr++;
printf("%s\n",ptr);
}

Ans. Samco Systems

Find the output for the following C program
#include
main()
{
char s1[]="Ramco";
char s2[]="Systems";
s1=s2;
printf("%s",s1);
}

Ans. Compilation error giving it cannot be an modifiable 'lvalue'

Find the output for the following C program
#include
main()
{
char *p1;
char *p2;
p1=(char *) malloc(25);
p2=(char *) malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}

Ans. RamcoSystems

Find the output for the following C program
# define TRUE 0
some code
while(TRUE)
{
some code
}

Ans. This won't go into the loop as TRUE is defined as 0

struct list{
int x;
struct list *next;
}*head;

the struct head.x =100

Is the above assignment to pointer is correct or wrong ?

Ans. Wrong

What is the output of the following ?

int i;
i=1;
i=i+2*i++;
printf(%d,i);

Ans. 4

FILE *fp1,*fp2;

fp1=fopen("one","w")
fp2=fopen("one","w")
fputc('A',fp1)
fputc('B',fp2)
fclose(fp1)
fclose(fp2)
}

Find the Error, If Any?

Ans. no error. But It will over writes on same file.

What are the output(s) for the following ?
#define MAN(x,y) (x)>(y)?(x):(y)
{int i=10;
j=5;
k=0;
k=MAX(i++,++j);
printf(%d %d %d %d,i,j,k);
}

Ans. 10 5 0

void main()
{
int i=7;
printf("%d",i++*i++);
}

Ans: 56

C Aptitude Questions With Answer

Predict the output or error(s) for the following:
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}

Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}

Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}

Answer:
I hate U

Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}

Answer:
5 4 3 2 1

Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) j="0;j<5;j++){">}

Answer:
2 2 2 2 2 2 3 4 6 5

Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

Predict the output or error(s) for the following:
7
. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

Answer:
0 0 1 3 1

Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
1 2

Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}

Answer :
three

Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()
{
printf("%x",-1<<4);>
}

Answer:
fff0

Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}

Answer:
Compiler Error : Type mismatch in redeclaration of function display

Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12. main()
{
int c=- -2;
printf("c=%d",c);
}

Answer:
c=2;

Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}

Answer:
sizeof(i)=1

Explanation:
Since the #define replaces the string int by the macro char

14. main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}

Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

Predict the output or error(s) for the following:
15. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
77

Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

Answer:
SomeGarbageValue---1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
You should not initialize variables in declaration

18. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

Answer:
Compiler Error

Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}

Answer:
hai

Explanation:
\n - newline
\b - backspace
\r - linefeed

20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:
45545

Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:
64

Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}

Answer:
ibj!gsjfoet

Explanation:
++*p++ will be parse in the given order
Ø *p that is value at the location currently pointed by p will be taken
Ø ++*p the retrieved value will be incremented
Ø when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23. #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}

Answer:
50

Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}

Answer:
100

Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem

Predict the output or error(s) for the following:
25. main()
{
printf("%p",main);
}

Answer:
Some address will be printed.

Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

26. main()
{
clrscr();
}
clrscr();

Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

27. enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}

Answer:
0..1..2

Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

28. void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Answer:
4..2

Explanation:
the second pointer is of char type and not a far pointer

29. main()
{
int i=400,j=300;
printf("%d..%d");
}

Answer:
400..300

Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.

30. main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}

Answer:
H

Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

31. main()
{
int i=1;
while (i<=5) { printf("%d",i); if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}

Answer:
Compiler error: Undefined label 'here' in function main

Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

32. main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}

Answer:
Compiler error: Lvalue required in function main

Explanation:
Array names are pointer constants. So it cannot be modified.

33. void main()
{
int i=5;
printf("%d",i++ + ++i);
}

Answer:
Output Cannot be predicted exactly.

Explanation:
Side effects are involved in the evaluation of i

34. void main()
{
int i=5;
printf("%d",i+++++i);
}

Answer:
Compiler Error

Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

35. #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}

Answer:
Compiler Error: Constant expression required in function main.

Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

Predict the output or error(s) for the following:
36. main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}

Answer:
1

Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

37. #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}


Answer:
100

38. main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}

Answer:
1

Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

39. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
M

Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");

40. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
Initialization should not be done for structure members inside the structure declaration

41. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}


Answer:
Compiler Error

Explanation:
in the end of nested structure yy a member have to be declared

42. main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}

Answer:
Linker error: undefined symbol '_i'.

Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

43. main()
{
printf("%d", out);
}

int out=100;
Answer:
Compiler error: undefined symbol out in function main.

Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

Predict the output or error(s) for the following:
44. main()
{
extern out;
printf("%d", out);
}
int out=100;

Answer:
100

Explanation:
This is the correct way of writing the previous program.

45. main()
{
show();
}
void show()
{
printf("I'm the greatest");
}

Answer:
Compier error: Type mismatch in redeclaration of show.

Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

46. main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}

Answer:
100, 100, 100, 2
114, 104, 102, 3

47. main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; p =" a;" j="0;">
Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

48. main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}

Answer:
111
222
333
344

49. main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}

Answer:
g20fy

Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

50. main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}

Answer:
ck

Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.