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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
17 lines
224 B
C
17 lines
224 B
C
// PR c++/86429
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// { dg-do compile { target c++14 } }
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struct A
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{
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int i;
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constexpr int f(const int&) const { return i; }
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};
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void g()
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{
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constexpr A a = { 42 };
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[&](auto x) {
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constexpr auto y = a.f(x);
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}(24);
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}
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