C 中已声明但未初始化的变量会发生什么情况？它有价值吗？

如果我在 C 中写：

int num;

在我分配任何东西之前num，是值num不定？

静态变量（文件作用域和函数静态）初始化为零：

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

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

非静态变量（局部变量）是不定。在赋值之前读取它们会导致未定义的行为.

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

实际上，它们最初往往只是有一些无意义的值 - 一些编译器甚至可能会放入特定的固定值，以便在调试器中查看时变得明显 - 但严格来说，编译器可以自由地执行从崩溃到召唤的任何操作恶魔通过你的鼻道 http://catb.org/jargon/html/N/nasal-demons.html.

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 http://blogs.msdn.com/oldnewthing/archive/2004/01/19/60162.aspx the Itanium, which has a "Not a Thing" bit in its registers https://web.archive.org/web/20190127204712/https://blogs.msdn.microsoft.com/oldnewthing/20040119-00/?p=41003; of course, the C standard drafters were considering some older architectures.

尝试使用设置了这些标志位的值可能会导致 CPU 操作异常really不应失败（例如，整数加法或分配给另一个变量）。如果你留下一个未初始化的变量，编译器可能会在设置了这些标志位的情况下拾取一些随机垃圾 - 这意味着触摸该未初始化的变量可能是致命的。