Using noexcept as a lambda modifier or parameter constraint

Question

Can the noexcept modifier be applied to a lambda expression? If so, how?

Can noexcept be made a constraint on a function argument? For example, something like in the following code, where the meaning is that the callback function must be noexcept?

//probably not valid code - I'm just trying to express the idea void f_async(std::function<void (int) noexcept> callback) noexcept {     ... } 

This can almost be accomplished with the following code, but I'm wondering if there is a way to use something like the above alternative.

void f_async(std::function<void (int)> callback)     noexcept(callback(std::declval<int>())) {     ... } 

The problem here of course is that f_async can be noexcept(false) if the callback is noexcept(false) - I want to make a stronger statement that f_async is always noexcept, meaning it's only callable if you use a noexcept callback.

Answer 1

Can the noexcept modifier be applied to a lambda expression? If so, how?

Add noexcept after the parenthesis:

[](Args args) noexcept { ... } 

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Can noexcept be made a constraint on a function argument?

Yes, use enable_if:

template <typename F> auto f_async(const F& func) noexcept          -> typename std::enable_if<noexcept(func(0))>::type {     func(0); }  int main() {     f_async([](int x) noexcept {});     f_async([](int x) {}); // <- this line won't compile } 

However, this method cannot work directly in g++ 4.7 (it does work in clang++ 3.2), because it cannot mangle noexcept expression yet:

3.cpp:5:6: sorry, unimplemented: mangling noexcept_expr

You could workaround it using a wrapper struct:

template <typename F, typename... Args> struct EnableIfNoexcept          : std::enable_if<noexcept(std::declval<F>()(std::declval<Args>()...))> {};  template <typename F> auto f_async(const F& func) noexcept -> typename EnableIfNoexcept<F, int>::type {     func(0); }