size of allocated array

This is a discussion on size of allocated array within the C Programming forums, part of the General Programming Boards category; When I try to check the size of an array that is allocated, it always tells me the size is ...

  1. #1
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    size of allocated array

    When I try to check the size of an array that is allocated, it always tells me the size is 8, no matter how many bytes I allocate. When I check the size of an array that is declared the normal way, it gives me the correct size. I figure that it just checks the size of the pointer and not of the array? How can i fix this? Running on a macbook pro

    Code:
    #include <stdio.h>
    #include <math.h>
    #include <stdlib.h>
    
    int main()
    {
    	int *a,b[] = {1,2,3};
    	a = (int *)malloc(48);
    	a[0] = 1;
    	a[1] = 2;
    	a[2] = 3;
    	printf("size of b: %li\n", sizeof(b)); -> returns 12
    	printf("size of a: %li\n", sizeof(a)); -> returns 8
    	return 0;
    }

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    Yep! that's the thing with dynamic storage; workaround would be to put a magic no. at the top so you don't need to calculate it everytime
    Code:
    #define SIZEOF_A 48

  3. #3
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    Short answer: You can't.
    Long answer: The best thing to do is to keep track of the size. It's the only portable way.
    (Meaning there may be other ways, but they aren't portable.)
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    Quote Originally Posted by boxden
    I figure that it just checks the size of the pointer and not of the array?
    You figured correctly.

    Quote Originally Posted by boxden
    How can i fix this?
    You already know the size of the dynamic array in order to allocate space for it... trying to find out the size again is kind of like asking, "what's your name, Bob?"

    In this case, the value that you want is equal to 48 / sizeof(int). But you should not be writing:
    Code:
    a = (int *)malloc(48);
    instead, you should write, say:
    Code:
    size_t a_size = 12;
    a = malloc(a_size * sizeof(*a));
    Now you can write:
    Code:
    printf("size of a: %lu\n", a_size);
    EDIT:
    A drawback with itCbitC's suggestion is that you then might as well not use a dynamic array.
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    You can't get the size of an array directly using the sizeof() function in C. sizeof() returns the size of the data type you give it in bytes.

    There are macros you can define to get around this but they are not always useful. Best way to deal with this sort of thing is to define a variable to keep track to the length of arrays when you declare them.

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    Quote Originally Posted by laserlight View Post
    instead, you should write, say:
    Code:
    size_t a_size = 12;
    a = malloc(a_size * sizeof(*a));
    Now you can write:
    Code:
    printf("size of a: %lu\n", a_size);
    Except it would return 96 instead of 48 on the o/p's machine.
    Quote Originally Posted by laserlight View Post
    A drawback with itCbitC's suggestion is that you then might as well not use a dynamic array.
    Yep! that's why it's a workaround

  7. #7
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    Quote Originally Posted by itCbitC
    Except it would return 96 instead of 48 on the o/p's machine.
    Maybe, maybe not. That is why I wrote "say".

    EDIT:
    That said, sizeof(int) == 8 seems rare at the moment. Does MacBook Pro really have 8-byte ints? What integer type corresponds to int32_t from C99's <stdint.h> for MacBook Pro?
    Last edited by laserlight; 06-28-2010 at 03:52 PM.
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    Quote Originally Posted by laserlight View Post
    Does MacBook Pro really have 8-byte ints?
    8-byte pointers not ints, to be precise.

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    It's not so much whether a particular hardware supports 64-bit integers natively. As long as the compiler has generated code to do appropriate multiple-register math where needed, any architecture can support any size integer.

  10. #10
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    Quote Originally Posted by itCbitC
    8-byte pointers not ints, to be precise.
    My guess is that you managed to misread sizeof(*a) as sizeof(int*).
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    Quote Originally Posted by laserlight View Post
    My guess is that you managed to misread sizeof(*a) as sizeof(int*).
    Yep! got that backwards

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