get constexpr variable from a lambda function is fine , but compile fail (Visual C++) and fine (gcc) when such statement is in a new lambda

Question

This code compiles fine in gcc, but fail in Visual C++.

MCVE = https://wandbox.org/permlink/SqNI85EospSrwm5T

int main() {
    auto func_do1Pass2=[&]() {
        return 8;
    };
    constexpr int g=func_do1Pass2();
    auto sLambda=[&]() {
        constexpr int g2=func_do1Pass2();  //<== error at Visual C++
    };
}

Error C2131 expression did not evaluate to a constant

1. Who is wrong? Why? Please give reference.

2. Here is my more real example. What is a workaround for the below code in Visual C++ case?

Code:

void testtt() {
    auto func_do1Pass2=[&]<int s>() {
        return 8;
    };
    [&]<int s2>() {
        ///vvv can edit only here
        constexpr int g2=func_do1Pass2.operator()<s2>();
        ///^^^ can edit only here
    }.operator()<2>();
}

Answer 1

Just make func_do1Pass2 constexpr as shown below. The problem is that in your given code func_do1Pass2 isn't constexpr so it can't be used in constexpr context.

void testtt() {
//--vvvvvvvvv-------------------------------->added constexpr here
    constexpr auto func_do1Pass2=[&]<int s>()  {
        return 8;
    };
    [&]<int s2>() {
        
        constexpr int g2=func_do1Pass2.operator()<s2>();
        
    }.operator()<2>();
}

Working demo