读字符串不起作用

Question

I have this little code and I really can't figure out why it always gives me SIGSEGVs.

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

int read_strs(char **a, int *len) {

  int i;
  scanf("%d", len);
  if(*len <= 0) return 1;
  a = (char **) malloc(sizeof(char*) * (*len));
  if(a == NULL) return 1;
  for( i = 0; i < *len; i++ ) 
  {
      a[i] = malloc(sizeof(char) * 1000);
      scanf("%s", a[i]);
  }

  return 0;

}


int main(void) {
  int i, n;
  char **array;

  read_strs(array, &n);
  for( i = 0; i < n; i++ ) 
    printf("%s\n", array[i]);
    return 0;
}

It seems it doesn't even alloc the memory.

Answer 1

The problem is that char ** array; in your main function never gets loaded with the char ** array you malloc in test. You send a copy of the pointer to read_strs - it won't get updated when you assign to it here:

// a is local to test_str
a = (char **) malloc(sizeof(char*) * (*len));

You would need to pass a pointer to your char ** array variable if you want it to work in your current scheme. You could also simply return a from read_strs .

edit : note, if you pass a pointer to char ** array , you'll also have to do work to make read_strs operate on a pointer to a char ** , whereas now it's operating on a simple char ** .

Answer 2

You need a couple fixes, but basically Walter is right

 int read_strs(char***ax, *len){

then

char **a;
a = *ax = (char**) malloc(sizeof(char*) * (*len));

then in the call, pass the location of array so that it can be filled by the function:

read_strs(&array, &n);