二维字符指针分段错误

Question

I'm taking arguments for my C program in the linux terminal and am trying to pass them into another 2d array pointer, but I'm getting a segmentation fault where I attempt to assign the content in the argv 2d array to the strings 2d array. How can I successfully do this without getting an error?

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

int main(int argc, char *argv[]){

    char **strings = malloc((argc-1) * 100 * sizeof(char));

    int j = 1;
    while(*(argv + j) != NULL){
        for(int i = 0; i < strlen(*(argv + j)); i++){
            *(*(strings + j) + i) = *(*(argv + j) + i);
            printf("%c", *(*(strings + j) + i));
        }
        printf("\n");
        j++;
    }

    return 0;
}

Answer 1

It looks like you're trying to copy the contents of argv into memory and then print the contents. I wish I could draw a picture of the memory contents for you as it would be easy to see what's going on. This uses a jagged array, so just google that to see some pictures. In a nutshell, char** is an array of char* . Your code is not allocating the correct size of memory. You know exactly how many elements you have through argc so you can allocate strings . Then for each char* of strings , you allocate enough memory to fit the data taken from argv . The length comes from strlen as you have, then the string is copied into the memory pointed to by elements of strings .

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

int main(int argc, char* argv[]){
    char** strings = malloc((argc - 1) * sizeof(char*));

    for (int i = 0; i < argc - 1; ++i) {
        size_t len = strlen(argv[i + 1]);
        strings[i] = malloc(len + 1);
        strcpy(strings[i], argv[i + 1]);
        printf("%s\n", strings[i]);
    }

    // do what you would like with strings

    for (int i = 0; i < argc - 1; ++i)
        free(strings[i]);

    free(strings);
    return 0;
}