c - switch-case 中有效但毫无值(value)的语法？

Question

通过一个小错别字，我无意中发现了这个构造:

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 似乎是有效的，但也完全无法访问。

我得到了一个干净的编译，甚至没有关于无法访问代码的警告，但这似乎毫无意义。

编译器是否应该将此标记为无法访问的代码？
这有什么用吗？

Answer 1

也许不是最有用的，但并非完全毫无值(value)。您可以使用它来声明一个在 switch 范围内可用的局部变量。

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

标准 (N1579 6.8.4.2/7) 有以下示例:

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);
}

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，示例不是有效的 C++ 代码。在那种情况下(N4140 6.7/3，强调我的):

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

因此将 int i = 4; 替换为 int i; 使其成为有效的 C++。