所有 char 数组都自动以空字符结尾吗? [英] Are all char arrays automatically null-terminated?

问题描述

可能我对谷歌搜索太垃圾了，但我一直认为 char 数组只能通过文字初始化 (char x[]="asdf";) 终止，并且有点惊讶当我看到情况似乎并非如此.

Probably I'm just too dump for googling, but I always thought char arrays get only null terminated by an literal initialization (char x[]="asdf";) and got a bit surprised when I saw that this seems not to be the case.

int main()
{
    char x[2];
    printf("%d", x[2]);
    return 0;
}

输出:0

声明为 size=2*char 的数组不应该实际获得 2 个字符的大小吗?或者我在这里做错了什么?我的意思是将 char 数组用作简单的 char 数组而不是字符串并不少见，或者是吗?

Shouldn't an array declared as size=2*char actually get the size of 2 chars? Or am I doing something wrong here? I mean it isn't uncommon to use a char array as a simple char array and not as a string, or is it?

推荐答案

您正在访问超出其边界的未初始化数组.这是双重未定义行为，任何事情都可能发生，甚至将 0 作为输出.

You are accessing an uninitialized array outside its bounds. That's double undefined behavior, anything could happen, even getting 0 as output.

回答你真正的问题:只有字符串文字得到空终止，这意味着 char x[]="asdf" 是一个 5 个元素.

In answer to your real question: Only string literals get null-terminated, and that means that char x[]="asdf" is an array of 5 elements.

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