在C语言中，是初始化数组只有一个元素经过特殊处理的？ [英] In C, is array initialization with only one element treated specially?

问题描述

在阅读<一href=\"http://stackoverflow.com/questions/8060931/append-to-the-beginning-of-an-array/8061034#8061034\">this问题我想测试在GCC的输入，看到错误会是什么输出。令我惊讶的下面一行：

 字符数组[] = {的};
 

编译没有错误或警告，导致含有S \\ 0 2大小的数组。我本来期望一个编译器错误，因为前pression右侧的类型为的char * [] 。

是一个数组初始化与不被视为在这种情况下一个数组，为什么？

只有一个元素

解决方案

 字符数组[] = {的};
 

是一样的：

 字符数组[] =S;
 

下面 {} 在此情况下可选的，因为s为字符串文字。

或者

 字符数组[] = {'S'，'\\ 0'};
 

在这种情况下， {} 是必要的初始化数组

While reading this question I wanted to test the input in GCC to see what errors would be output. To my surprise the following line:

char array[] = {"s"};

compiles without error or warning, resulting in an array of size 2 containing "s\0". I would have expected a compiler error because the right side of the expression is of type char*[].

Is an array initialization with only one element not treated as an array in this case, and why?

解决方案

char array[] = {"s"};

is same as:

char array[] = "s";

Here { } are optional in this case because "s" is string literal.

Or,

char array[] = {'s', '\0'};

In this case, { } are necessary to initialize the array.

这篇关于在C语言中，是初始化数组只有一个元素经过特殊处理的？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！