Question

char* temp;

temp = (char*) malloc (strlen(window->entry.value)+1);
//strncpy( temp, window->entry.value, sizeof(temp) ); DOESN"T WORK
memcpy (temp, window->entry.value, strlen(window->entry.value) + 1); //WORKS

(where window->entry.value is a string.)

Thanks.

Answer 1

sizeof(temp) doesn’t do what you think it does. It tells you the size, in bytes, of a pointer to char, which is what temp is. Most likely on your system either 4 or 8.

This has nothing to do with the length of any particular string, or the size of any particular buffer returned from malloc.

It’s pure fluke that you’ve passed the right length when using memcpy, and the wrong length when using strncpy. memcpy would also not work if passed sizeof(temp) as the length, and strncpy would work if passed the right length.