尝试使用双指针复制字符串时出现分段错误

Question

Jus started learning about pointers and im stuck with this program outputting a segmentation fault. Its supposed to copy the first 10 Characters of a string to the location pointed by the double pointer using gdb ive found that **pt=*s; produces the seg fault

#include <stdio.h>
#include <stdlib.h>
void str1(char *s, char **pt);
void str1(char *s, char **pt){
    for(int i=0;i<10;i++){
        **pt=*s;
        pt++;
        s++;

    }
}
int main (void) {
   char str[30] = "223This is test";
   char *ptr;
   str1(str, &ptr);
   printf("%s", ptr);
   return 0;
}

Answer 1

First of all ptr is not initialized, you can't really use it until you reserve space for it or store a valid memory address in it, ie make it point to some valid variable.

char *ptr = malloc(11);

Then you need to increment it properly in the function:

(*pt)++;

Once the copy is completed you need to null terminate the char array so it can be treatead as a string, aka a null terminated char array.

**pt = '\0';

Now as ptr was passed as a pointer to pointer, the increment is known by the caller, main in this case, so when you try to print it, it prints nothing because it's pointing to the end of the char array, we need to bring it back to the beggining.

*pt -= 10;

Corrected code with comments taking yours as base:

Live demo

#include <stdio.h>
#include <stdlib.h>
#define SIZE 10

void str1(char *s, char **pt) {
    for (int i = 0; i < SIZE; i++) {
        **pt = *s;
        (*pt)++; //properly increment pt
        s++;
    }
    **pt = '\0'; //null terminate copied string

    //since ptr was passed as **, the increment is known by the caller
    //now ptr will be pointing to the end of the string
    //we have to bring it back to the beginning
    *pt -= SIZE;
}

int main(void) {

    char str[] = "223This is test";  
    char *ptr = malloc(SIZE + 1); //space for 10 character + null-terminator

    //check for allocation errors
    if(ptr == NULL){
        perror("malloc");
        return EXIT_FAILURE;
    }

    str1(str, &ptr);
    printf("%s", ptr); 
    free(ptr);
    return EXIT_SUCCESS;
}

Answer 2

You probably want this:

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

void str1(char* s, char** pt) {
  char *p = malloc(100);           // allocate memory for destination
  *pt = p;                         // store it for the caller

  for (int i = 0; i < 10; i++) {
    *p = *s;
    p++;
    s++;
  }

  *p = 0;     // put string terminator, otherwise printf won't work correctly
}

int main(void) {
  char str[30] = "223This is test";
  char *ptr;                       // for now p points nowhere
  str1(str, &ptr);                 // now p points to the memory allocated in str1
  printf("%s", ptr);
  free(ptr);                       // free memory for completeness
  return 0;
}