为什么带有初始化程序的 C++17 if 语句不能按预期工作？

Question

struct A
{
    auto g1()
    {
        return true;
    }

    void f()
    {
        if (auto b = g1(); b) // ok
        {
            return;
        }

        if (auto b = g2(); b) // error: use of 'auto A::g2()' before deduction of 'auto'
        {
            return;
        }
    }    
    
    auto g2()
    {
        return true;
    }
};

为什么带有初始化器的 C++17 if 语句不能按预期工作？

Answer 1

因为标准是这么说的（引用最新草案）：

[dcl.spec.auto.general]

If a variable or function with an undeduced placeholder type is named by an expression ([basic.def.odr]), the program is ill-formed. Once a non-discarded return statement has been seen in a function, however, the return type deduced from that statement can be used in the rest of the function, including in other return statements.

[Example 4:
auto n = n;                     // error: n's initializer refers to n
auto f();
void g() { &f; }                // error: f's return type is unknown

为了澄清一点，g2 的“声明”是“可见的”，因为 g1 的定义在完整的 class 上下文中。但这并没有扩展到看到 g2.

的定义

为什么带有初始化程序的 C++17 if 语句不能按预期工作？

Why does C++17 if statement with initializer not work as expected?

c++

c++17

language-lawyer

auto

type-deduction