C 字符数组初始化 [英] C char array initialization

问题描述

我不确定通过以下方式初始化后char数组中会有什么.

I'm not sure what will be in the char array after initialization in the following ways.

1.char buf[10] = "";
2. char buf[10] = " ";
3. char buf[10] = "a";

对于情况 2，我认为 buf[0] 应该是 ' '，buf[1] 应该是 '\0'，从 buf[2] 到 buf[9] 将是随机内容.对于情况 3，我认为 buf[0] 应该是 'a'，buf[1] 应该是 '\0'，并且从buf[2] 到 buf[9] 将是随机内容.

For case 2, I think buf[0] should be ' ', buf[1] should be '\0', and from buf[2] to buf[9] will be random content. For case 3, I think buf[0] should be 'a', buf[1] should be '\0', and from buf[2] to buf[9] will be random content.

正确吗?

对于情况 1，buf 中会有什么?buf[0] == '\0' 并且从 buf[1] 到 buf[9] 会是随机内容吗?

And for the case 1, what will be in the buf? buf[0] == '\0' and from buf[1] to buf[9] will be random content?

推荐答案

这不是初始化数组的方式，而是用于:

This is not how you initialize an array, but for:

1. 第一个声明:

1. The first declaration:

char buf[10] = "";

相当于

char buf[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};

第二个声明:

The second declaration:

char buf[10] = " ";

相当于

char buf[10] = {' ', 0, 0, 0, 0, 0, 0, 0, 0, 0};

第三个声明:

The third declaration:

char buf[10] = "a";

相当于

char buf[10] = {'a', 0, 0, 0, 0, 0, 0, 0, 0, 0};

如您所见，没有随机内容:如果初始化器较少，则数组的剩余部分使用 0 进行初始化.即使在函数内部声明数组也是如此.

As you can see, no random content: if there are fewer initializers, the remaining of the array is initialized with 0. This the case even if the array is declared inside a function.

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