如何将char指针数组的索引设置为char指针？

Question

I'm tokenizing by commas, which gives me char * as an output in a while loop. How do I assign each of these char pointers in the while loop to the index of a char pointer []?

Pseudocode:

char * p;
char * args[30];
int i = 0;
while(p!=NULL){
    p = strtok(NULL,",");
    args[i] = p; //attempt 1
    *(args + i) = p; //attempt 2
    strcpy(p,args[i]); //attempt 3
    i++;
}

Error: I print out the value of p and after printing index 0, it fails. Here is my code for printing it out:

 for(int j=0; j<i; j++){
      printf("%s \n",args[j]);
 }

Here is my error: "0 g" when my input is "gmn" and it prints out Segmentation fault: 11.

Answer 1

Your program is mostly correct, but I think your problem is that you're using strtok() incorrectly. Upon its first call, strtok() expects a string and delimiters. Subsequent calls expect NULL and delimiters.

I modified your C "pseudo-code" to a working program.

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

void main(int argc, char* argv[]) {
    char* p;
    char* args[30];
    int i = 0;
    int j;

    char input[30];
    puts("Please enter a string:");
    scanf("%29s", &input); /* get a string to break up */

    p = args[i++] = strtok(input, ",");  /* first call to strtok() requires the input */

    while(p!=NULL && i < 30) { /* added bounds check to avoid buffer overruns */
        p = strtok(NULL,","); /* subsequent calls expect NULL */
        args[i] = p; /* this is the best way to assign values to args, but it's equivalent to your attempt 2*/
        i++;
    }

    for(j = 0; j < i; j++){
            printf("%s \n",args[j]);
    }
}

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Edit: I just realized that my original code used an uninitialized pointer p . This is undefined behavior, and I have corrected the code.