我该如何解决这个问题code创建一个字符串数组？ [英] How do I fix this code to create an array of strings?

问题描述

我想创建一个字符串数组。这里是code：

I want to create an array of strings. Here is the code:

#include <stdio.h>
int main(void){
  char str1[] = {'f','i'};
  char str2[] = {'s','e'};
  char str3[] = {'t','h'};
  char arry_of_string[] = {str1,str2,str3};
  printf("%s\n",arry_of_string[1]);
  return 0;
}

这是行不通的行：

char arry_of_string[] = {str1,str2,str3}; 

我如何改正？

推荐答案

如果你想创建一个数组的字符串的，你缺少一个星号，并终止零：

If you would like to create an array of strings, you are missing an asterisk, and terminating zeros:

char str1[] = {'f','i','\0'};
char str2[] = {'s','e','\0'};
char str3[] = {'t','h','\0'};
char *arry_of_string[] = {str1,str2,str3};

有是做个体串的更简单的方法，也：

There is an easier way of doing the individual strings, too:

char str1[] = "fi";
char str2[] = "se";
char str3[] = "th";
char *arry_of_string[] = {str1,str2,str3};

当您使用字符X [] =...构造，您的字符串的内容（包括终止零）复制到内存你可以写，产生相同的效果字符x [] = {''，''，......'\\ 0'} 结构。

When you use the char x[] = "..." construct, the content of your string literal (which includes a terminating zero) is copied into memory that you are allowed to write, producing the same effect as char x[] = {'.', '.', ... '\0'} construct.

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