char 数组声明中的字符串文字周围的大括号有效吗?(例如 char s[] = {“Hello World"})

问题描述

偶然我发现行 char s[] = {"Hello World"}; 被正确编译并且似乎被视为与 char s[] = "Hello世界";.不是第一个 ({"Hello World"}) 一个包含一个字符数组元素的数组，因此 s 的声明应为 char *s[]?事实上，如果我将其更改为 char *s[] = {"Hello World"}; 编译器也会按预期接受它.

By accident I found that the line char s[] = {"Hello World"}; is properly compiled and seems to be treated the same as char s[] = "Hello World";. Isn't the first ({"Hello World"}) an array containing one element that is an array of char, so the declaration for s should read char *s[]? In fact if I change it to char *s[] = {"Hello World"}; the compiler accepts it as well, as expected.

正在寻找答案，我找到的唯一提到这个的地方是 这个 但没有引用标准.

Searching for an answer, the only place I found which mentioned this is this one but there is no citing of the standard.

所以我的问题是，为什么行 char s[] = {"Hello World"}; 被编译，尽管左侧是 array of char 和右边是char数组的array?

So my question is, why the line char s[] = {"Hello World"}; is compiled although the left side is of type array of char and the right side is of type array of array of char?

以下是一个工作程序:

#include<stdio.h>
int main() {
    char s[] = {"Hello World"};
    printf("%s", s); // Same output if line above is char s[] = "Hello World";
    return 0;
}

感谢您的澄清.

附言我的编译器是 gcc-4.3.4.

P.S. My compiler is gcc-4.3.4.

推荐答案

这是允许的，因为标准是这样说的:C99 section 6.7.8, §14:

It's allowed because the standard says so: C99 section 6.7.8, §14:

一个字符类型的数组可以由一个字符串字面量初始化，可选括在大括号中.字符串字面量的连续字符(包括如果有空间或数组大小未知，则终止空字符)初始化数组的元素.

这意味着两者

char s[] = { "Hello World" };

和

char s[] = "Hello World";

只不过是语法糖

char s[] = { 'H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd', 0 };

在相关注释(同一部分，第 11 节)中，C 还允许在标量初始值设定项周围使用大括号，例如

On a related note (same section, §11), C also allows braces around scalar initializers like

int foo = { 42 };

顺便说一句，它非常适合复合文字的语法

which, incidentally, fits nicely with the syntax for compound literals

(int){ 42 }

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