Why can't I use constexpr with lambda function?

Question

I need an explanation for this. How does lambda differ from others function to it don't be allowed? Is this a nonsense to be not part of C++ standard? for example, I wanted to write a "nested" function in C++ but like it isn't possible to do directly I do use lambda instead of. This function does a very small job: take a single int parameter and multiply with some values and returns. This values are locals to where lambda function is defined and all them are constexpr and know at run-time for this reason I do want to let the compiler put a const value instead of call the function. It's very often used function (that's why I write this as function) and I don't want this beging computed at run-time.

n2859 says "N2859 notes that severe implementation difficulties would arise if lambdas were permitted in contexts which require their contents to be part of a mangled name, and the prohibition on lambdas in constant expressions form part of the resolution to those difficulties." But this seems like a self-reference, and the linked PDF doesn't contain anything about constexpr or constant expressions as far as I can see. — dyp
– dyp, Commented Mar 29, 2014 at 20:59
Oops, I got tricked by the address bar of my browser.. The quote is from N3597. — dyp
– dyp, Commented Mar 29, 2014 at 21:14

galop1n · Accepted Answer · 2014-03-29 21:04:46Z

It is because the standard list lambda as non constant expression : "5.19 Constant expressions" … "is a core constant expression unless it involves one of the following as a potentially evaluated subexpression" … "— a lambda-expression (5.1.2);"

That is enough for any compiler to reject constexpr function with a lambda involved.

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