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main()
{
int ptr[] = {1,2,23,6,5,6};
printf("%d",&ptr[3]-&ptr[0]);
}

Answers were Sorted based on User's Feedback



main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / sandeep

3
because aray pointer arthematic considers positions(index) of aray

Is This Answer Correct ?    46 Yes 10 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / vivek

ans - none

Is This Answer Correct ?    18 Yes 12 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / shivam jindal

That should print a 3. It's really the same as

printf("%d", 3-0);

...since:

ptr[3] is the 4th element in the ptr[] array.

&ptr[3] is a pointer to the 4th element in the ptr[] array.

&ptr[0] is similarly a pointer to the first element in ptr[].

&ptr[3] - &ptr[0] is a subtraction of two pointers. That's only defined (in standard C/C++) for pointers to elements in the same array, like in this case, and it's defined as the difference between the index values. That's where the 3-0 comes from.

The result of a pointer difference is an int. &ptr[0] - &ptr[3] results in 0-3 which is -3.

Is This Answer Correct ?    5 Yes 3 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / dipak

In int at least 2 bytes are used for size and we know that &ptr[] gives the address of ptr .
&ptr[3]-&ptr[0]
Size of &ptr[3] is 3*2=6 times greater than Size of &ptr[0] is 1*2=2
ptr[0] also have any value that's why I consider it 1
So 6-2=4

Is This Answer Correct ?    4 Yes 3 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / mitesh mahera

I need a answer aboutthis question..if any can ?!

Is This Answer Correct ?    4 Yes 6 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / deveshdashora

6

Is This Answer Correct ?    6 Yes 10 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / jithin ramakrishnan

2

Is This Answer Correct ?    0 Yes 4 No

main() { int ptr[] = {1,2,23,6,5,6}; printf("%d",&ptr[3]-&ptr[0]); }..

Answer / baba

Ans: 12

The expression in printf evaluates the difference of the memory addresses of ptr[3] and ptr[0]

Is This Answer Correct ?    5 Yes 10 No

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