在c中的snprintf中使用const char数组

Question

I'm very new comer to C. I've this type of code and when I tried to execute it this warning message showed up "passing argument 1 of 'snprintf' discards 'const' qualifier from pointer target type" and nothing happened.

What I did wrong? Thank you

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

int main()
{
  int i;
  const char *msg[3] = {"Hello", "Good Morning", "Hello World"};
  const char *strings[];


  for(i=0; i<3; i++)
  snprintf(strings[i], 20, "%s %d", msg[i], i);

  for(i=0; i<3; i++)
  printf("strings[%d]: %s\n", i, strings[i]);

  return 0;
}

Answer 1

snprintf(strings[i], 20, "%s %d", msg[i], i);

that tries to write into strings[i] . Since it's been declared as constant, compiler just refuses to do that because it violates the contract.

But here, it's even more serious: strings[i] doesn't have any allocated memory for the strings or even the pointers (!), so removing the const qualifier would result in undefined behaviour when running your program.

Answer 2

You need to allocate space for each string you want to print, like this

char *strings[3];
for (i = 0; i < 3 ; ++i) {
    size_t length = snprintf(NULL, 0, "%s %d", msg[i], i);
    strings[i] = malloc(length + 1);
    if (string[i] != NULL) {
        snprintf(strings[i], length, "%s %d", msg[i], i);
    }
}

for (i = 0; i < 3 ; ++i) {
    if (string[i] != NULL) {
        printf("string[%d]: %s\n", i, strings[i]);
    }
}

In the first part, we allocate 3 poitners on the stack. And then, for every string we first compute the length, and then snprintf() into the successfuly allocated memory.