当我们在C的字符串末尾不包含'\ 0'时会发生什么? [英] What happened when we do not include '\0' at the end of string in C?

问题描述

在C中，当我以这种方式初始化数组时:

In C, when I initialize my array this way:

char full_name[] = {
    't', 'o', 'a', 'n'
};

并用printf("%s", full_name);

并使用 valgrind 运行它

未初始化的值是通过堆栈分配创建的

Uninitialised value was create by stack allocation

为什么会这样?

推荐答案

由于%s格式说明符期望以空字符结尾的字符串，因此代码的结果行为是不确定的.您的程序被认为是格式错误的，可以完全不产生任何输出，不产生任何输出，崩溃等.简短地说，不要那样做.

Since %s format specifier expects a null-terminated string, the resulting behavior of your code is undefined. Your program is considered ill-formed, and can produce any output at all, produce no output, crash, and so on. To put this shortly, don't do that.

这并不是说所有字符数组都必须以空字符结尾:该规则仅适用于打算用作C字符串的字符数组，例如可以传递给%s格式说明符上的printf，或者传递给strlen或Standard C库的其他字符串函数.

This is not to say that all arrays of characters must be null-terminated: the rule applies only to arrays of characters intended to use as C strings, e.g. to be passed to printf on %s format specifier, or to be passed to strlen or other string functions of the Standard C library.

如果打算将char数组用于其他用途，则无需将其终止为null.例如，此用法已完全定义:

If you are intended to use your char array for something else, it does not need to be null terminated. For example, this use is fully defined:

char full_name[] = {
    't', 'o', 'a', 'n'
};
for (size_t i = 0 ; i != sizeof(full_name) ; i++) {
    printf("%c", full_name[i]);
}

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