以下代码是我的较大程序的简化版本，用于演示问题。

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

const int m = 5;
char **words1;
char **words2;

void f(void *v) {
    printf("v: %pn", v);

    int len = v==words1 ? 3 : 4;
    printf("len: %dn", len);

    for (int i=0; i<m; i++) {
        // What goes here?
        /*
        char *c = malloc(sizeof(char) * len);
        c = (char*)v; // Something _like_ this?!
        printf("%sn", c);
        */
    }
}

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

    words1 = malloc(m * sizeof(char*));
    printf("%pn", words1);
    words2 = malloc(m * sizeof(char*));
    printf("%pn", words2);

    for (int i=0; i<m; i++) {
        words1[i] = malloc(sizeof(char) * 3);
        words2[i] = malloc(sizeof(char) * 4);
        strcpy(words1[i], "22");
        strcpy(words2[i], "333");
    }

    f(words1);
    f(words2);

    for (int i=0; i<m; i++) {
        free(words1[i]);
        free(words2[i]);
    }
    free(words1);
    free(words2);
}

我有两个全局**char我认为它们在堆上（因为malloc）。

的签名f()不能改变，即只能接受void *。

重要的是，在实际程序**char中，堆栈中的数据太大了。

简而言之：我如何**char从*void?

回答

   char **f_words = v;
    for (int i=0; i<m; i++) {
        printf("%sn", f_words[i]);
    }