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Career Registration Links for Top IT companies in india

Hello Jobseekers, Do Upload Your Resume And Ask Our Friends Too...
Use below links to register and upload your resume for Assenture, CTS, HCL, HP (Hewlett-Packard), IBM,  Infosys,  L&T infotech,  Mahendira satyam,  Microsoft, Oracle,  TCS,  Tech Mahindira, Wipro.
final year students, and others can also register your profile and submit your Resume through the given links.





        Accenture Registration Link:


·         CTS Registration Link :


·         HCL Registration Link :


·         HP Online Registration Link :


·         IBM Registration Link :


·         Infosys Registration Link :


·         L&T Infotech Registration Link :


·         Mahindra Satyam Registration Link :


·         Microsoft Registration Link :


·         Oracle Registration Link :

           
·         Tech Mahindra Registration Link :


·         TCS Registration Link :

      

·         Wipro Registration Link :


·         Wipro WASE Registration Link :( Only For BCA, BCM and BSc Final Year Students)

                  http://careers.wipro.com/it/campus/india/wase.htm


 This is small list only. we will upload other companies registration links soon.
if you know other companies registration links please specify in comments.. we will upload soon.

C Programming Interview Questions Part Six

C Programming Interview Questions with answers, interview questions and answers, c programming interview question part six.learn c programming online, c programs, online c questions
 




76)

 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.


77)

 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




78)

 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.



79)

 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 datatypes.



80)

 main()
{
 char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
 x=convert(c);
 printf("%c",x);
}
 convert(z)
 {
return z32;
}

Answer:

Compiler error

Explanation:
declaration of convert and format of getc() are wrong.




81)

 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.


82)

 # include <stdio.h>
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.




83)
 # include<stdio.h>
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.


85)

#include<stdio.h>
 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.



86)

 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.




87) 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.




88)     

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:

40 31 22

Explanation:

Let us assume some x= scanf("%d",&i)t the values during
execution will be, t ix 4 04 3 12 220




89)

 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.



90)
 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.



91)

 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.



92)

            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





93)

main()
 {
 char *p;
p="%d\n";
p++; p++;
printf(p2,300);
 }

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.



94)

 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.




95)

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'.

96) void main()
{
 static int i=5;
if(i)
{
main();
 printf("%d ",i);
}
 }

Answer:

0000

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.




97)

 void main()
{
 int k=ret(sizeof(float));
 printf("\n here value is %d",++k);
}
int ret(int ret) {
ret += 2.5;
return(ret);
 }

Answer:

Here value is 7

Explanation:

The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.




98)

void main()
{
char a[]="12345\0";
 int i=strlen(a);
 printf("here in 3 %d\n",++i);
}

Answer:

here in 3 6

Explanation:

The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the preincrement in the printf statement, the 6 will be printed.



99)

 void main()

{

 unsigned giveit=1;
 int gotit;
 printf("%u ",++giveit);
 printf("%u \n",gotit=giveit);

}

Answer:

0 65535


100)

void main()
{

 int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
 else
printf("Forget it\n");
 }

Answer:

Ok here

Explanation:

Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.


C Programming Interview Questions Part Five

Note : All the programs are tested under Turbo C/C++ compilers.






61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since
void is an empty type. In the second line you are creating variable
vptr of type void * and v of type void hence an error.




62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of
the pointer variable. In second sizeof the name str2 indicates the
name of the array whose size is 5 (including the '\0' termination
character). The third sizeof is similar to the second one.




63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the
boolean value FALSE, and any non-zero value is considered to be
the boolean value TRUE. Here 2 is a non-zero value so TRUE.
!TRUE is FALSE (0) so it prints 0.



64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the
preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
21 6/26/2004
21
Preprocessor doesn't replace the values given inside the double
quotes. The check by if condition is boolean value false so it goes
to else. In second if -1 is boolean value true hence "TRUE" is
printed.



65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by whitespace)
they are concatenated (this is called as "stringization"
operation). So the string is as if it is given as "%d==1 is %s". The
conditional operator( ?: ) evaluates to "TRUE".



66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.



67) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be
used to declare the variable name of the type arr2. But it is not the
case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are
used for declaring new types.




68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost
block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf
prints 30. In the next block, i has value 20 and so printf prints 20.
In the outermost block, i is declared as extern, so no storage space
is allocated for it. After compilation is over the linker resolves it to
global variable i (since it is the only variable visible there). So it
prints i's value as 10.


69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside
that block only. But the lifetime of i is lifetime of the function so it
lives upto the exit of main function. Since the i is still allocated
space, *j prints the value stored in i since j points i.


70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In
printf first you just print the value of i. After that the value of the
expression -i = -(-1) is printed.
71) #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant


72) #include<stdio.h>
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:
garbagevalue..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
meAnswer:it will print first element of 3D array.


73) #include<stdio.h>
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer
and it will take integer value. i value may be stored either in
register or in memory.


74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)

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