c++ - 调用函数时切换 "transfer of control bypasses initialization of:"

Question

当我尝试构建以下开关时，出现“控制转移绕过初始化:”错误:

switch (retrycancel)
{
    case 4:    //The user pressed RETRY
        //Enumerate all visible windows and store handle and caption in "windows"
        std::vector<MainHandles::window_data> windows = MainHandles().enum_windows().get_results(); 
        break;

    case 2: 
        //code
}

这与我调用枚举函数有关。如果不允许在 switch 内调用函数，是否有解决此类问题的方法？

Answer 1

C++ 标准第 6.6.4 节:

The goto statement unconditionally transfers control to the statement labeled by the identifier. The identifier shall be a label (6.1) located in the current function.

C++ 标准第 6.7 节:

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps from a point where a local variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has POD type (3.9) and is declared without an initializer

重点由我添加。由于 switch 实际上是 goto 变相的，因此您会遇到这种行为。要解决这个问题，如果必须使用 switch

，请添加大括号

switch (retrycancel)
    {
    case 4:
    {
        const std::vector<MainHandles::window_data> windows(
            MainHandles().enum_windows().get_results()
        );
        break;
    }
    case 2: 
        //code
    }

或重构为if/else

if (retrycancel == 4) {
    const std::vector<MainHandles::window_data> windows(
        MainHandles().enum_windows().get_results()
    );
} else if (retrycancel == 2)
    // code
} else {
    ...
}

虽然在 switch 中创建 windows vector 对我来说并不明显，但你可能想要重新思考你的设计。 注意我在 windows 中添加了一个 const 限定符，因为它在您的示例中没有被修改。