How to pass array of strings as parameter in C

[bcz]

Trying to get C code to handle the passing an array of 80 character strings to a subroutine



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

static int i;
static char r[5][81]; // pascal array [1..5] of string;

void sub(char *r[5][81]){
fprintf(stdout,"%s\n",*r[2]);
}

main(int argc, char **argv) {
fprintf(stdout,"Program tto5 running\n");
strncpy(r[2],"Test OK",80);
sub(&r);
fprintf(stdout,"Program tto5 exiting\n");
}


Note strings are NULL terminated, rather than length in byte 0. Also by default 80 characters long

Any help appreciated

Rgds Bill

[Bachstelze]

Why are you making a pointer to the array? You don't need to, this works:


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

void sub(char r[5][81])
{
printf("%s\n", r[2]);
}

int main(void)
{
char r[5][81];
strncpy(r[2], "Test.", 80);
sub(r);
return 0;
}


If for some reason you really want to use a pointer, you must add some parentheses:


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

void sub(char (*r)[5][81])
{
printf("%s\n", (*r)[2]);
}

int main(void)
{
char r[5][81];
strncpy(r[2], "Test.", 80);
sub(&r);
return 0;
}


char *foo[] is an array of char*s, char (*foo)[] is a pointer to an array of chars.

[bcz]

Thanks for example code and explanation. Just what I needed

The trivial code is simplified. The subroutine needs to setup 3 arrays of strings, so seems I actually need a pointer

Rgds Bill