not able to get the length of an integer array

Question

Possible Duplicate:
length of array in function argument

I am trying to get the length of an integer array but i am not getting the right answer

void main()
{
    int x[]={33,55,77};
    printf("%d",getLength(x));//outputs 1
    printf("%d",sizeof(x)/sizeof(int));//outputs 3
}

int getLength(int *inp)
{
    return sizeof(inp)/sizeof(int);
}

So why is getLength returning value 1 instead of 3 ?

This question or one like it has been asked hundreds of times at SO. I wonder where people think the size of a C array is stored in memory. — Jim Balter, Jan 15 '13 at 16:19
@dasblinkenlight Please read my comment again ... I wrote "where", not "why". If people do not understand the C storage model, they have no business writing C code; they certainly will fail. — Jim Balter, Jan 18 '13 at 04:09

score 3 · Answer 1 · answered Jan 15 '13 at 16:16

3

Because arrays aren't pointers, and pointers aren't arrays. sizeof(pointer) is not the same as sizeof(the array which the pointer points to).

answered Jan 15 '13 at 16:16

score 2 · Accepted Answer · answered Jan 15 '13 at 16:16

Since arrays decay to pointers, you cannot perform length calculation in a function: the function gets a pointer, not an array. You need to do the computations inline the way your function does when it prints 3, or use a macro to compute array length:

#define GET_LENGTH(inp) (sizeof(inp)/sizeof(*inp))

Here is a link to a demo on ideone.

score 1 · Answer 3 · answered Jan 15 '13 at 16:16

1

The sizeof(inp) here is returning the size of pointer and not the size of array and the size of pointer = 4

So

sizeof(inp)/sizeof(int)
 ^__ 4        ^__ 4      = 1

answered Jan 15 '13 at 16:16

MOHAMED

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not able to get the length of an integer array

3 Answers3