Menu
How following recursive lambda call ends/terminates ?
#include <cstdio> auto terminal = [](auto term) // <---------+ { // | return [=] (auto func) // | ??? { // | return terminal(func(term)); // >---------+ }; }; auto main() -> int { auto hello =[](auto s){ fprintf(s,"Hello\n"); return s; }; auto world =[](auto s){ fprintf(s,"World\n"); return s; }; terminal(stdout) (hello) (world) ; return 0; } What am I missing over here ?
Running code
411
Visual Studio 2017 version 15.3 and later (available in /std:c++17 mode and later): A lambda expression may be declared as constexpr or used in a constant expression when the initialization of each data member that it captures or introduces is allowed within a constant expression.
Permalink. All the alternatives to passing a lambda by value actually capture a lambda's address, be it by const l-value reference, by non-const l-value reference, by universal reference, or by pointer.
Lambdas can both capture variables and accept input parameters. A parameter list (lambda declarator in the Standard syntax) is optional and in most aspects resembles the parameter list for a function. auto y = [] (int first, int second) { return first + second; };
Generic lambdas were introduced in C++14 . Simply, the closure type defined by the lambda expression will have a templated call operator rather than the regular, non-template call operator of C++11 's lambdas (of course, when auto appears at least once in the parameter list).
It's not a recursive function call, look at it step-by-step:
terminal(stdout) - this simply returns a lambda which has captured stdout hello, which executes the lambda (func(term)), the result of which is passed to terminal(), which simply returns a lambda as in 1.world, which does the same as 2, this time the return value is discarded...
66
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
