Question 1 |
What is the output of the following program?
#include < stdio.h >
int main()
{
int a=10, b=20;
int *p;
p=&a;
*p=b;
a=35;
*p+=5;
b=*p;
p=&b;
*p=a;
printf("%d %d",a,b);
return 0;
}
10 20 | |
10 40 | |
40 40 | |
40 10 |
Question 1 Explanation:
OUTPUT : 40 40
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Question 2 |
What is the output of the following program?
#include < stdio.h >
int main()
{
int a[]={1,2,3,4,5};
int b[]={1,2,3,4,5};
if(a==b)
{
printf("IITB");
}
else
{
printf("IISC");
}
return 0;
}
IITB | |
IISC | |
Compiler error | |
Output depends on Compiler |
Question 2 Explanation:
If statement will compare the base address of two arrays a and b,and they are not same.
So condition becomes false and program prints no.
OUTPUT : IISC
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Question 3 |
What is the output of the following program?
#include < stdio.h >
int main()
{
int a[5]={1,2,3,4,5};
int *p;
p=a;
printf("%d",*p++);
printf("%d",(*p)++);
printf("%d",*p);
printf("%d",*++p);
printf("%d",++*p);
return 0;
}
1 2 3 3 4 | |
1 2 2 3 4 | |
1 2 3 4 4 | |
Compiler error |
Question 3 Explanation:
OUTPUT : 12334
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Question 4 |
What is the output of the following program?
#include < stdio.h >
int main()
{
int a[5]={1,2,3,4,5};
int *p;
p=a;
printf("%d",*a++);
printf("%d",(*a)++);
printf("%d",*a);
printf("%d",*++a);
printf("%d",++*a);
return 0;
}
1 2 3 3 4 | |
1 1 2 3 4 | |
1 2 3 4 5 | |
Compiler error |
Question 4 Explanation:
When we declare int a[5]={1,2,3,4,5}; in c program. "a" becomes the constant pointer. It means that, we can not modify the "a" in program.
In the above program two statements
printf("%d",*a++);
printf("%d",*++a);
tries to modify the "a".
Hence, it's a compiler error.
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In the above program two statements
printf("%d",*a++);
printf("%d",*++a);
tries to modify the "a".
Hence, it's a compiler error.
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Question 5 |
What is the output of the following program?
#include < stdio.h >
int main()
{
int a[5]={1,2,3,4,5};
int *p;
p=a;
printf("%d",*(p+2));
printf("%d",*(a+2));
p++;
printf("%d",*p);
printf("%d",*a);
return 0;
}
3 3 2 2 | |
3 3 2 1 | |
3 3 Gabage 1 | |
3 3 Gabage 2 |
Question 5 Explanation:
OUTPUT : 3321
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There are 5 questions to complete.
Q3. Doubt.
*(1008) gives 3.
*(1004) gives 3.
*(1008) has value 4. But gives 3!
*(1008) is 3 as per the answer. At last line ++*p it becomes 4.
Never mind.
Question 7.
Explanation doubt in last but line.
p=p+3; //last update of variable p e.i base address is 1002+3*sizeof(char) e.i 1005.
printf(“%d”,p+2); //does not update p, I guess.
so, printf(“%c”,p[-3]) will be the base address 1005-3*sizeof(char)=1002. So *(1002) gives “a”.
Thank You Manoj,
We have updated the answer.