Clang rejects constexpr variable initialiation from constepxr function template parameter

[ranaanoop]

The following program is rejected by clang but accepted by gcc and msvc. Demo

template <typename T>
constexpr auto func(T c) {
    constexpr auto k = c(); 
    return k;
};


int main() {
    
    constexpr int i = 0;
    
    constexpr auto lambda = []()constexpr { constexpr int j = i; return j; };  //compiles
    constexpr auto lambda2 = [i]()constexpr { constexpr int j = i; return j; };//compiles
   
    constexpr auto a = func(lambda); //clang and others compiles this

    
    constexpr auto b = func(lambda2); //clang doesn't compile this
    
}   

Clang says:

<source>:3:20: error: constexpr variable 'k' must be initialized by a constant expression
    3 |     constexpr auto k = c(); 
      |                    ^   ~~~
<source>:18:24: note: in instantiation of function template specialization 'func<(lambda at <source>:13:30)>' requested here
   18 |     constexpr auto b = func(lambda2); //clang doesn't compile this
      |                        ^
<source>:13:65: note: function parameter 'c' with unknown value cannot be used in a constant expression
   13 |     constexpr auto lambda2 = [i]()constexpr { constexpr int j = i; return j; };//compiles
      |                                                                 ^
<source>:3:24: note: in call to 'c.operator()()'
    3 |     constexpr auto k = c(); 
      |                        ^~~
<source>:2:23: note: declared here
    2 | constexpr auto func(T c) {
      |                       ^
1 error generated.
Compiler returned: 1

[llvmbot]

@llvm/issue-subscribers-clang-frontend

Author: Anoop Rana (ranaanoop)

The following program is rejected by clang but accepted by gcc and msvc. [Demo](https://godbolt.org/z/95Pfrqrex)

template <typename T>
constexpr auto func(T c) {
    constexpr auto k = c(); 
    return k;
};


int main() {
    
    constexpr int i = 0;
    
    constexpr auto lambda = []()constexpr { constexpr int j = i; return j; };  //compiles
    constexpr auto lambda2 = [i]()constexpr { constexpr int j = i; return j; };//compiles
   
    constexpr auto a = func(lambda); //clang and others compiles this

    
    constexpr auto b = func(lambda2); //clang doesn't compile this
    
}   

Clang says:

<source>:3:20: error: constexpr variable 'k' must be initialized by a constant expression
    3 |     constexpr auto k = c(); 
      |                    ^   ~~~
<source>:18:24: note: in instantiation of function template specialization 'func<(lambda at <source>:13:30)>' requested here
   18 |     constexpr auto b = func(lambda2); //clang doesn't compile this
      |                        ^
<source>:13:65: note: function parameter 'c' with unknown value cannot be used in a constant expression
   13 |     constexpr auto lambda2 = [i]()constexpr { constexpr int j = i; return j; };//compiles
      |                                                                 ^
<source>:3:24: note: in call to 'c.operator()()'
    3 |     constexpr auto k = c(); 
      |                        ^~~
<source>:2:23: note: declared here
    2 | constexpr auto func(T c) {
      |                       ^
1 error generated.
Compiler returned: 1

[shafik]

I believe this is: #63139

[zygoloid]

Looks like clang is correct to me. The lambda reads from its capture when evaluating constexpr int j = i;, but the value of the capture is unknown. Making that variable static should fix it.

But this case is weird. It's not obvious to me whether a non-odr-use access should constitute a read. Probably worth checking that with CWG.

[zygoloid]

Oh, no, [expr.prim.lambda.capture]/11 says we don't rewrite i to name the capture here. If we're getting that wrong, that'd explain why lambda works but lambda2 does not.