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When we refer to a captured variable from a constant-expression context inside a lambda, the closure (like any function parameter) is not constant because we aren't in a call, so we don't have an argument. So the capture is non-constant. But if the captured variable is constant, we might be able to use it directly in constexpr evaluation. PR c++/82643 PR c++/87327 * constexpr.c (cxx_eval_constant_expression): In a lambda function, try evaluating the captured variable directly. From-SVN: r269951
24 lines
317 B
C
24 lines
317 B
C
// PR c++/87327
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// { dg-do compile { target c++17 } }
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template <int N>
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struct Foo {
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constexpr auto size() const {
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return N;
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}
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};
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constexpr int foo() {
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constexpr auto a = Foo<5>{};
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[&] {
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Foo<a.size()> it = {};
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return it;
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}();
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return 42;
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}
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constexpr int i = foo();
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