Array size in functions

Question

I'm working with some arrays in plain C99 and I need to know their size. Does anyone knows why this is happening:

int function(char* m){
    return sizeof(m) / sizeof(m[0]);
}

int main(){
    char p[100];
    int s = sizeof(p) / sizeof(p[0]);
    printf("Size main: %d\nSize function: %d\n",s,function(p));
    return 0;
}

Output:

Size main: 100
Size function: 8

Thanks!

When you pass it as a pointer to the function, sizeof returns sizeof char pointer, not length of the array. — 001, Apr 24 '14 at 14:07
this is why the length of an array is often passed as an additional arg in c functions — bph, Apr 24 '14 at 14:11

score 1 · Answer 1 · answered Apr 24 '14 at 14:07

1

This is happening because sizeof(m) == sizeof(char*), which is 8 on your platform.

When you call function(p), p decays into a pointer and loses its size information.

answered Apr 24 '14 at 14:07

NPE

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score 1 · Accepted Answer · edited Apr 24 '14 at 14:10

1

Arrays decay to pointers at the first chance they get, thus in function all you have is a pointer, which is 8 bytes in your case.

edited Apr 24 '14 at 14:10

jamessan

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answered Apr 24 '14 at 14:09

Casper Beyer

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Array size in functions

2 Answers2