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main()
{
int ptr[] = {1,2,23,6,5,6};
printf("%d",&ptr[3]-&ptr[0]);
}
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Answer / sandeep
3
because aray pointer arthematic considers positions(index) of aray
| Is This Answer Correct ? | 46 Yes | 10 No |
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 |
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 |
Answer / mitesh mahera
I need a answer aboutthis question..if any can ?!
| Is This Answer Correct ? | 4 Yes | 6 No |
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 |
write the program for prime numbers?
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