有效，但在转换情况下毫无价值的语法吗? [英] Valid, but worthless syntax in switch-case?

问题描述

通过一些拼写错误，我偶然发现了这个结构:

Through a little typo, I accidentally found this construct:

int main(void) {
    char foo = 'c';

    switch(foo)
    {
        printf("Cant Touch This\n");   // This line is Unreachable

        case 'a': printf("A\n"); break;
        case 'b': printf("B\n"); break;
        case 'c': printf("C\n"); break;
        case 'd': printf("D\n"); break;
    }

    return 0;
}

switch语句顶部的printf似乎有效，但也完全不可访问.

It seems that the printf at the top of the switch statement is valid, but also completely unreachable.

我得到了干净的编译，甚至没有关于无法访问代码的警告，但这似乎毫无意义.

I got a clean compile, without even a warning about unreachable code, but this seems pointless.

编译器是否应将其标记为无法访问的代码?
这有什么用吗?

Should a compiler flag this as unreachable code?
Does this serve any purpose at all?

推荐答案

也许不是最有用，但并非完全没用.您可以使用它来声明在switch范围内可用的局部变量.

Perhaps not the most useful, but not completely worthless. You may use it to declare a local variable available within switch scope.

switch (foo)
{
    int i;
case 0:
    i = 0;
    //....
case 1:
    i = 1;
    //....
}

标准(N1579 6.8.4.2/7)具有以下示例:

The standard (N1579 6.8.4.2/7) has the following sample:

在人工程序片段中的示例

EXAMPLE    In the artificial program fragment

switch (expr)
{
    int i = 4;
    f(i);
case 0:
    i = 17;
    /* falls through into default code */
default:
    printf("%d\n", i);
}

其标识符为i的对象以自动存储持续时间存在(在块内)，但从不存在 初始化，因此如果控制表达式的值非零，则对printf函数的调用将 访问一个不确定的值.同样，无法调用函数f.

the object whose identifier is i exists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will access an indeterminate value. Similarly, the call to the function f cannot be reached.

P.S.顺便说一句，该示例不是有效的C ++代码.在这种情况下(N4140 6.7/3，重点是我的):

P.S. BTW, the sample is not valid C++ code. In that case (N4140 6.7/3, emphasis mine):

从具有自动存储持续时间的变量不在范围内的点跳转 ⁹⁰ 的程序 除非变量具有标量类型 ，否则它在范围内的格式不正确 构造函数和琐碎的析构函数，这些类型之一的cv限定版本或以下类型之一的数组 之前的 类型，并且声明时没有初始化程序 (8.5).

A program that jumps⁹⁰ from a point where a variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has scalar type, class type with a trivial default constructor and a trivial destructor, a cv-qualified version of one of these types, or an array of one of the preceding types and is declared without an initializer (8.5).

^{90)从switch语句的条件到案例标签的转移在这方面被认为是跳跃.}

^{90) The transfer from the condition of a switch statement to a case label is considered a jump in this respect.}

因此，用int i;替换int i = 4;使其成为有效的C ++.

So replacing int i = 4; with int i; makes it a valid C++.

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