C -Implementing strcpy() But Segfault

Question

I have made a strcpy() function in C, and I am copying words from one array to other not just letters, but when I run it I am getting Segmentation fault what to do?

#include <stdio.h>

void strcpy1(char *dest[], char source[])
{
    while ((*dest++ = *source++));
}

int main()
{
    char source[3][20] = { "I", "made", "this" };
    char dest[3][20];

    strcpy1(&dest, source);
    
    //printing destination array contents   
    for (int i = 0; i < 3; i++) {
        printf("%s\n", dest[i][20]);
    }

    return 0;
}

Answer 1

There are multiple problems in your code:

• the prototype for your custom strcpy1 function should be:

  void strcpy1(char *dest[], char *source[]);
  

• the arrays source and dest are 2D char arrays: a very different type from what strcpy1 expects, which are arrays of pointers. Change the definition to:

   char *source[4] = { "I", "made", "this" };
   char *dest[4];
  

• you should pass the destination array as dest instead of &dest

• the source array should have a NULL pointer terminator: it should be defined with a length of at least 4. Same for the destination array.

• in the print loop dest[i][20] refers to a character beyond the end of the i-th string. You should just pass the string as dest[i].

Here is a modified version:

#include <stdio.h>

void strcpy1(char *dest[], char *source[])
{
    while ((*dest++ = *source++));
}

int main()
{
    char *source[4] = { "I", "made", "this" };
    char *dest[4];

    strcpy1(dest, source);
    
    //printing destination array contents   
    for (int i = 0; dest[i]; i++) {
        printf("%s\n", dest[i]);
    }

    return 0;
}

Note that it is somewhat confusing to name strcpy1 a function that has very different semantics from the standard function strcpy().

Answer 2

The %s specifier is for strings, eg, a char* referring to the first character of a string.

When you pass dest[i][20] to the printf function, it is not a char*. It is a single char, the 21st char (valid indices are 0-19, for a total of 20 elements).

So it is an array-out-of-bounds-index, and also, not a char* as printf expects.

printf("%s\n", dest[i][20]);