c - c中函数内部的Malloc用法

Question

我试图传递一个指向函数的指针。在这个函数中我使用 malloc 来保留空间。问题是当我在 main 中返回时程序没有响应。这是我的简化代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int def(char **B){
    int i;
    B = malloc(3 * sizeof( char ));
    for(i = 0; i < 2 ; i++){
        B[i] = malloc(5 * sizeof(char));
    }
    for(i = 0; i < 2 ; i++){
        scanf("%s" , B[i]);
    }
    for(i = 0; i < 2 ; i++){
        printf("%s\n" , B[i]);
    }
    return 0;
}

int main(int argc, char *argv[]) {
    char **B;
    int i;
    def(B);
    for(i = 0; i < 2 ; i++){
        printf("%s\n" , B[i]);
    }
    return 0;
}

Answer 1

int def(char **B)

应该是

char** def(char **B) 

它的返回值应该是

return B; 
 /* else the memory allocated inside the function will be freed at
  * the end and by accessing it later you have undefined behavior for the 
  * rest of the program
  */ 

B = malloc(3 * sizeof( char ));

应该是

B = malloc(3 * sizeof( char*)); // you have two levels of indirection. so char* first

for(i = 0; i < 2 ; i++) // similary with the other for loops

应该是

 for(i = 0; i < 3 ; i++) // you used 3 in the above step

def(B);

应该是

B=def(B);

使用 free() 释放分配的内存是一个很好的做法，尽管它会在程序结束时自动释放