C 字符串初始值设定项不包括终止符? [英] C string initializer doesn't include terminator?

问题描述

我对以下 C 代码片段感到有些困惑:

I am a little confused by the following C code snippets:

printf("Peter string is %d bytes\n", sizeof("Peter")); // Peter string is 6 bytes

这告诉我，当 C 编译双引号中的字符串时，它会自动为空终止符添加一个额外的字节.

This tells me that when C compiles a string in double quotes, it will automatically add an extra byte for the null terminator.

printf("Hello '%s'\n", "Peter");

printf 函数知道什么时候停止读取字符串Peter"，因为它到达了空终止符，所以......

The printf function knows when to stop reading the string "Peter" because it reaches the null terminator, so ...

char myString[2][9] = {"123456789", "123456789" };
printf("myString: %s\n", myString[0]);

此处，printf 打印所有 18 个字符，因为没有空终止符(如果不去掉 9，它们将不适合).C 不会在变量定义中添加空终止符吗?

Here, printf prints all 18 characters because there's no null terminators (and they wouldn't fit without taking out the 9's). Does C not add the null terminator in a variable definition?

推荐答案

你的字符串是 [2][9].那些 [9] 是 ['1', '2', etc... '8', '9'].因为您在第一个数组维度中只给了它 9 个字符的空间，并且因为您使用了所有 9 个字符，所以它没有空间放置 '\0' 字符.重新定义你的字符数组:

Your string is [2][9]. Those [9] are ['1', '2', etc... '8', '9']. Because you only gave it room for 9 chars in the first array dimension, and because you used all 9, it has no room to place a '\0' character. redefine your char array:

char string[2][10] = {"123456789", "123456789"};

它应该可以工作.

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