在c中声明和复制char字符串数组

Question

I made ac program that attempts to add the values of one string array to another using a separate method:

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

void charConv(char *example[])
{
  example= (char* )malloc(sizeof(char[4])*6);
  char *y[] = {"cat", "dog", "ate", "RIP", "CSS", "sun"};

  printf("flag\n");

  int i;
  i=0;

  for(i=0; i<6; i++){
    strcpy(example[i], y[i]);
  }
}

int main() {
  char *x[6];
  charConv( *x[6]);

  printf("%s\n", x[0]);
}

However it keeps returning a segmentation fault. I'm only beginning to learn how to use malloc and c in general and its been puzzeling me to find a solution.

Answer 1

To pin-point your problem: you send *x[6] (here - charConv( *x[6]); ) which is the first char of the 7 'th (!!!) string (Remember, C is Zero-Base-Indexed ) inside an array of 6 string you didn't malloc -> using memory you don't own -> UB.

Another thing I should note is char[] vs char * [] . Using the former, you can strcpy into it strings. It would look like this:

'c' | 'a' | 't' | '\0' | 'd' | 'o' | 'g' | ... [each cell here is a `char`]

The latter ( what you used ) is not a contiguous block of char s but a array of char * , hence what you should have done is to allocate memory for each pointer inside your array and copy into it. That would look like:

 0x100 | 0x200 | 0x300... [each cell is address you should malloc and then you would copy string into]

But, you also have several problems in your code. Below is a fixed version with explanations:

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

void charConv(char *example[])
{
  // example= (char* )malloc(sizeof(char[4])*6); // remove this! you don't want to reallocate example! When entering this function, example points to address A but after this, it will point to address B (could be NULL), thus, accessing it from main (the function caller) would be UB ( bad )

  for (int i = 0; i < 6; i++)
  {
    example[i] = malloc(4); // instead, malloc each string inside the array of string. This can be done from main, or from here, whatever you prefer
  }


  char *y[] = {"cat", "dog", "ate", "RIP", "CSS", "sun"};

  printf("flag\n");

 /* remove this - move it inside the for loop
  int i;
  i=0;
  */

  for(int i=0; i<6; i++){
    printf("%s\t", y[i]); // simple debug check - remove it 
    strcpy(example[i], y[i]);
    printf("%s\n", example[i]); // simple debug check - remove it 
  }
}

int main() {
  char *x[6];
  charConv( x); // pass x, not *x[6] !!

  for (int i = 0; i < 6; i++)
  {
    printf("%s\n", x[i]);  // simple debug check - remove it 
  } 
}

As @MichaelWalz mentioned, using hard-coded values is not a good practice. I left them here since it's a small snippet and I think they are obvious. Still, try to avoid them

Answer 2

You need to start by understanding the pointers and some other topics as well like how to pass an array of strings to a function in C etc. In your program, you are passing *x[6] in charConv() which is a character.

Made corrections in your program -

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

void charConv(char *example[], int num)
{
  int i;

  for (i = 0; i < num; i++){
      example[i] = (char* )malloc(sizeof(char)*4);
  }

  const char *y[] = {"cat", "dog", "ate", "RIP", "CSS", "sun"};

  for(i = 0; i < num; i++){
      strcpy(example[i], y[i]);
  }
}

int main() {
  char *x[6];
  int i = 0;

  charConv(x, 6);

  /* Print the values of string array x*/
  for(i = 0; i < 6; i++){
      printf("%s\n", x[i]);
  }

  return 0;
}