I got this warning: sizeof on array function parameter will return size of 'const char *' instead of 'const char []'

Question

Possible Duplicate:
Why sizeof(param_array) is the size of pointer?

I'm new to C, I got an warning from clang when compiling my code:

#include<stdio.h>

char *strcpy (char destination[],const char source[]);
int main(void) {
    char str1[] = "this is a very long string";
    char str2[] = "this is a short string";
    strcpy(str2, str1);
    puts(str2);
    return 0;
}
char *strcpy (char destination[], const char source[]) {
    int size_of_array = sizeof source / sizeof source[0];
    for (int i = 0; i < size_of_array; i++) {
        destination[i] = source[i];
    }
    return destination;
}

I don't know what does the following warning mean:

string_copy_withou_pointer.c:12:29: warning: sizeof on array function parameter
      will return size of 'const char *' instead of 'const char []'
      [-Wsizeof-array-argument]
        int size_of_array = sizeof source / sizeof source[0];
                                   ^
string_copy_withou_pointer.c:11:46: note: declared here
char *strcpy (char destination[], const char source[]) {

Any idea?

Other than all the other answers you've got that point out the problems, I'd recommend you to not redefine functions from the standard library. Invent new names, it's safer. — Art, Oct 22 '12 at 07:30
for the sake of this duplicate question while i don't think its duplicate but concludes the same basic idea, the answer should be as many has mentioned this array will eventually happen to be a pointer which you don't want to get the size of directly instead you need to reference it using sizeof &source on that line. — Waheed, Jun 28 '19 at 07:46

score 11 · Answer 1 · answered Oct 22 '12 at 07:24

11

This warning is telling you that if you call sizeof(char[]) you won't get the size of the array but the size of a char* pointer.

This means that your variable size_of_array will be wrong because it won't represent the size of the real array.

answered Oct 22 '12 at 07:24

alestanis

21,519
4
48
67

2

hwo to fix it then? – MartianMartian Sep 21 '18 at 21:28
1

To fix it, you can use the [strlen](http://www.cplusplus.com/reference/cstring/strlen/) function. – john Feb 25 '21 at 00:03

score 9 · Accepted Answer · answered Oct 22 '12 at 07:23

That's because const char source[] in argument position is just syntactic sugar for const char *source. See, e.g., Steven Summit's C notes.

In this particular case, you'll want to call strlen. When not dealing with strings, pass the size of the array as a separate argument.

score 1 · Answer 3 · answered Oct 22 '12 at 07:25

1

The size of the array doesn't follow when you pass it to the function. Actually it's passed as a pointer which is why the warning message mentions

will return size of 'const char *'

answered Oct 22 '12 at 07:25

Some programmer dude

400,186
35
402
621

I got this warning: sizeof on array function parameter will return size of 'const char *' instead of 'const char []'

3 Answers3

Linked