Question

Please, tell me, what is the correct way of copying allocated char array to “normal” char array?
I have attempted to do the following, but it fails :

char * buf = (char *) malloc (BUFSIZ * sizeof(char));
// filling up the allocated array with stuff...
char temp[BUFSIZ];
strcpy(temp, buf); // strcpy doesn't work

Answer 1

First things first, you should not cast the return value of malloc (in C anyway) since it can hide errors.

Secondly, you never need to multiply by sizeof(char) since it’s always guaranteed to be one – doing so clogs up your code.

And, as to the actual question, you can use:

memcpy (temp, buff, BUFFSZ);

to copy the entire character array.

I’m assuming that’s what you want because you make no mention of handling C “strings”, only a character array.

If indeed you are handling C strings, strcpy will work fine in this case, provided:

• you have room at the end of the buffer for the terminating zero-byte; and
• you’ve actually put the zero-byte in there.

For example, this little snippet works fine:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (void) {
    // Two buffers.

    char buff1[4];
    char *buff2 = malloc (4);
    if (buff2 == NULL) {
        puts ("No memory!");
        return 1;
    }

    // Populate first buffer.

    buff2[0] = 'p';
    buff2[1] = 'a';
    buff2[2] = 'x';
    buff2[3] = '\0';

    // Transfer and print.

    strcpy (buff1, buff2);
    puts (buff1);

    // Free and exit.

    free (buff2);
    return 0;
}