C中的变量定义忽略 [英] Variable Definition Ignore in C

问题描述

代码:

int main()  
{
  int a=1;
  switch(a)
  {
    int b=20;

    case 1:
    printf("b is %d\n",b);
    break;

    default:
    printf("b is %d\n",b);
    break;
  }
  return 0;
}

输出:它为 b 打印一些垃圾值b 的声明何时发生在这里为什么这里 b 没有用 20 初始化???

Output: It prints some garbage value for b when does the declaration of b takes place here Why b is not initialized with 20 here???

推荐答案

因为内存会分配给 int b 但是当应用程序运行时 "b = 20"永远不会被评估.

Because memory will be allocated for int b but when the application is run "b = 20" will never be evaluated.

这是因为您的 switch 语句将跳转到 case 1:¹ 或 default:, 跳过有问题的语句 - 因此 b 将未初始化并调用未定义的行为.

This is because your switch-statement will jump down to either case 1:¹ or default:, skipping the statement in question - thus b will be uninitialized and undefined behavior is invoked.

以下两个问题(以及他们接受的答案)将进一步帮助您寻找答案:

The following two questions (with their accepted answers) will be of even further aid in your quest searching for answers:

• 绕过定义时如何使用变量? ²

为什么不能变量在 switch 语句中声明?

将编译器警告/错误提高到更高级别有望在您尝试编译源代码时为您提供此信息.

Turning your compiler warnings/errors to a higher level will hopefully provide you with this information when trying to compile your source.

以下是gcc 对此事的说明；

Below is what gcc says about the matter;

foo.cpp:6:10: error: jump to case label [-fpermissive]
foo.cpp:5:9: error:   crosses initialization of 'int b'

<小时>

^{1 因为 int a 永远是 1(一)所以它永远跳到这里.}

---

^{1 since int a will always be 1 (one) it will always jump here.}

^{2 两个链接中最相关的，由我回答.}

^{2 most relevant out of the two links, answered by me.}

这篇关于C中的变量定义忽略的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！