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