Question

char**总是让我困惑。以下代码生成分段错误。请解释一下......

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

int main()
{
    char** nameList;
    nameList = malloc(4*sizeof(char*));
    nameList[0] = malloc(12); //not sure if needed but doesn't work either
    nameList[0] = "Hello "; 
    printf("%s  ",nameList[0]);// even this statement isn't executed
    strcat(nameList[0], "World");
    printf("%s ",nameList[0]);
    return 0;
}

Answer 1

nameList = malloc(4*sizeof(char*));之后你有： nameList [0] =垃圾 nameList [1] =垃圾 nameList [2] =垃圾 nameList [3] =垃圾

你有nameList[0] = "Hello ";之后 nameList [0] =“你好” nameList [1] =垃圾 nameList [2] =垃圾 nameList [3] =垃圾

所以，当你执行strcat(nameList[1], "World");时，你很可能会得到一个段错误，因为nameList [1]可以指向内存中的任何位置。

Answer 2

您的代码通过写入只读存储并尝试写入其末尾来显示未定义的行为。

您的malloc想法是朝着正确方向迈出的一步。但是，您应该使用strcpy将"Hello"复制到新分配的内存中。此外，您需要考虑计划追加的字符串的大小，以及计算动态分配大小时的空终止符。

显然，您还需要在程序结束时释放所有已分配的内存：

char** nameList;
nameList = malloc(4*sizeof(char*));
nameList[0] = malloc(12);
strcpy(nameList[0], "Hello ");
printf("%s  ",nameList[0]);
strcat(nameList[0], "World"); // You were strcat-ing into a wrong element
printf("%s ",nameList[0]);
free(nameList[0]);
free(nameList);

Demo on ideone

Answer 3

在使用双ptrs之前，获取使用单个ptr的代码。此外，您不希望“过度编程”简单的代码。但是，如果要编写double ptr的用法，请从此代码开始并修改为使用双ptrs。

int main()
{
        char *nameList;

        nameList = malloc(12);   // point nameList to 12 bytes of storage

        strncpy(nameList, "Hello \0", 7);
        printf("%s\n",nameList);   // notice no space, its already after hello

        strncat(nameList, "World", 5);
        printf("%s\n",nameList);

        free(nameList);

        return 0;
}

这个简单的代码使用字符串，malloc和strcat有什么问题？

3 个答案: