memset + whitespace + memcpy

Question

How can i set a character array say of size 100 to whitespace and then copy 10 charters to that same string from other. For example:

there is one char array a[100] To do : set all of it to whitespace

Now there is another array : b[10] (suppose this is filled with some string of length 9) To do : copy this array to the previous one

What iam doing is : memset(a, ' ', sizeof(a));
350         memcpy(a,b, strlen(b))

But iam not getting space that i had set after 10 chars has been copied .

Answer 1

Try the following:

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

#define LENGTH 100

int main(int argc, char *argv[])
{
    char *a = NULL;
    char b[10] = "abcdefghi"; /* note that this has a trailing null character */

    a = malloc(LENGTH + 1);
    if (a) {
        *(a + LENGTH) = '\0';
        memset(a, ' ', LENGTH);
        fprintf(stdout, "a (before):\t[%s]\n", a);
        memcpy(a, b, sizeof(b) - 1); /* strip trailing null */
        fprintf(stdout, "a (after):\t[%s]\n", a);
        free(a);
    }

    return EXIT_SUCCESS;
}

Running this:

$ gcc -Wall test.c
$ ./a.out
a (before):     [...100 spaces...........]
a (after):      [abcdefghi...91 spaces...]                                                                                           ]

Answer 2

You are missing the null \0 character at the last position of the array.

Answer 3

You'll see. Try the following code:

for (int i = 0; i <= 99; i++)
    printf("%c", a[i]);
printf("\n");

The reason is when you memcpy the string to a, a NULL character is placed in a[10]. If you only output a, the output will stop just before the NULL.

Answer 4

don't you think that you need to type cast to (char *)

a = (char *) malloc(LENGTH + 1);