Can I define functions in C++ inline? I am talking about lambda functions, not the inline keyword that causes a compiler optimization.
4 Answers
C++11 added lambda functions to the language. The previous versions of the language (C++98 and C++03), as well as all current versions of the C language (C89, C99, and C11) do not support this feature. The syntax looks like:
[capture](parameters)->return-type{body} For example, to compute the sum of all of the elements in a vector:
std::vector<int> some_list; int total = 0; for (int i=0;i<5;i++) some_list.push_back(i); std::for_each(begin(some_list), end(some_list), [&total](int x) { total += x; }); 
answered Sep 18, 2012 at 19:37
Adam Rosenfield
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7 Comments
Xeo
int total = std::accumulate(begin(some_list), end(some_list), 0); :)
Qaz
Any particular reason this was downvoted? It answers the question just fine.
Grizzly
Isn't the code missing a
mutable? IIRC lambdas take captures as const by default, so it should be [&total](int x) mutable.Xeo
@Grizzly: That's only interesting for by-value captures, since
constness doesn't propagate through indirection.
ildjarn
@Desolator : They're in namespace
std, found by ADL. And yes, they wrap some_list.begin() and some_list.end() in this case. |
In C++11, you can use closures:
void foo() { auto f = [](int a, int b) -> int { return a + b; }; auto n = f(1, 2); } Prior to that, you can use local classes:
void bar() { struct LocalClass { int operator()(int a, int b) const { return a + b; } } f; int n = f(1, 2); } Both versions can be made to refer to ambient variables: In the local class, you can add a reference member and bind it in the constructor; and for the closure you can add a capture list to the lambda expression.
5 Comments
Xeo
Remember that you can't use local types in templates prior to C++11.
Kerrek SB
@Xeo: That's true, but I didn't think that was a requirement... the OP is welcome to clarify.
Xeo
Oh, I didn't mean it that way. It was a general reminder, since you didn't mention that restriction.
Thomas Eding
Neither of your examples form closures. The first example does form an anonymous function though.
Kerrek SB
@ThomasEding: That depends on your personal disposition towards the empty set, I suppose.
i dont know if i understand you well, but you want a lambda function?
http://en.cppreference.com/w/cpp/language/lambda
#include <vector> #include <iostream> #include <algorithm> #include <functional> int main() { std::vector<int> c { 1,2,3,4,5,6,7 }; int x = 5; c.erase(std::remove_if(c.begin(), c.end(), [x](int n) { return n < x; } ), c.end()); std::cout << "c: "; for (auto i: c) { std::cout << i << ' '; } std::cout << '\n'; std::function<int (int)> func = [](int i) { return i+4; }; std::cout << "func: " << func(6) << '\n'; } if you dont have c++11x then try:
Comments
Pre C++11, if you want to localize a function to a function, that can be done:
int foo () { struct Local { static int bar () { return 1; } }; return Local::bar(); } or if you want something more complicated:
int foo (int x) { struct Local { int & x; Local (int & x) : x(x) {} int bar (int y) { return x * x + y; } }; return Local(x).bar(44); } But if you want a true function literal in pre C++11, that is not possible.
[](int num1, int num2) {return num1 + num2);}.int n = 1==2?function(){return 10;} : -1;something like that?