sizeof always returns 4.

[Ploe]

I'm trying to get to grips with malloc however the sizeof function - no matter what I do, always returns 4. Can someone explain to me what is up?

#include <stdlib.h>
#include <stdio.h>

int main()
{
    int *ptr = malloc(sizeof(int) * 10);
    int len = sizeof(ptr);
    printf("Len: %i\n", len);
}

[useruseruser]

You are getting the size of the pointer, most likely.

[Adak]

Why don't you check out a char, a char array with 20 elements, a double, a long int, etc. ?

Two quantities being the same, does not an "always" make.

And as noted above, you're only showing the size of the pointer.

[zacs7]

I'm trying to get to grips with malloc however the sizeof function - no matter what I do, always returns 4. Can someone explain to me what is up?

sizeof is an operator, not a function. Secondly, C does not automatically maintain any size information at runtime for you. sizeof is done at compile time, it's not going to de-reference your pointer and work out how many bytes you asked to be allocated.

All that it has told you is, sizeof(int *) is 4 for your machine.

[Ploe]

So at runtime I'd need some form of variable as a reference for the size of the array? I didn't realise that sizeof was an operator, however that would explain why Code::Blocks was all neat and changed its colour.

Thanks.

[laserlight]

So at runtime I'd need some form of variable as a reference for the size of the array?

Yes. Alternatively, you designate some special value to serve as the "last" value, e.g., how the null character in a string is used. Passing the size tends to be less error prone though.