发生在C中声明，未初始化的变量是什么？它有一个价值？ [英] What happens to a declared, uninitialized variable in C? Does it have a value?

问题描述

如果用C我写：

int num;

在我分配什么 NUM ，是价值 NUM 不确定？

Before I assign anything to num, is the value of num indeterminate?

推荐答案

静态变量（文件范围和功能静态）初始化为零：

Static variables (file scope and function static) are initialized to zero:

int x; // zero
int y = 0; // also zero

void foo() {
    static int x; // also zero
}

非静态变量（局部变量）的不确定的。前阅读他们在不确定的行为赋值结果。

Non-static variables (local variables) are indeterminate. Reading them prior to assigning a value results in undefined behavior.

void foo() {
    int x;
    printf("%d", x); // the compiler is free to crash here
}

在实践中，他们往往只是在那里开始有些无厘头的价值 - 一些编译器甚至可放入特定的固定值，以寻找在调试器，当它显而易见的 - 但严格来说，编译器是免费的，从做任何事轰然召唤通过你的鼻腔恶魔的。

In practice, they tend to just have some nonsensical value in there initially - some compilers may even put in specific, fixed values to make it obvious when looking in a debugger - but strictly speaking, the compiler is free to do anything from crashing to summoning demons through your nasal passages.

至于为什么它是不确定的行为，而不是简单的未定义/任意值，也有一些有额外的标志位在其再presentation各类CPU架构。一个现代的例子是安腾，该公司在其注册到不是东西位;当然，C标准起草者考虑一些较旧的架构。

As for why it's undefined behavior instead of simply "undefined/arbitrary value", there are a number of CPU architectures that have additional flag bits in their representation for various types. A modern example would be the Itanium, which has a "Not a Thing" bit in its registers; of course, the C standard drafters were considering some older architectures.

尝试与设定会导致在的操作的CPU的例外，这些标志位的值的工作确实的不应该失败（例如，整数加法，或分配给另一变量）。如果你去，留下一个未初始化的变量，编译器可能会选择与设置这些标志位一些随机的垃圾 - 这意味着触摸的未初始化的变量可能是致命的。

Attempting to work with a value with these flag bits set can result in a CPU exception in an operation that really shouldn't fail (eg, integer addition, or assigning to another variable). And if you go and leave a variable uninitialized, the compiler might pick up some random garbage with these flag bits set - meaning touching that uninitialized variable may be deadly.

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