Return from calling function inside lambda - c++

Question

Lambdas are an awesome way to create reusable code inside a function/method without polluting the parent class. They're a very functional replacement for C-style macros most of the time.
However, there's one bit of syntactic sugar from macros that I can't seem to replicate with a lambda, and that's the ability to exit from the containing function. For example, if I need to return while checking the range of a series of ints, I can do that easily with a macro:
const int xmin(1), xmax(5);
#define CHECK_RANGE(x) { if((x) < xmin || (x) > xmax) return false; }
bool myFunc(int myint) {
CHECK_RANGE(myint);
int anotherint = myint + 2;
CHECK_RANGE(anotherint);
return true;
}
Obviously this is an oversimplified example, but the basic premise is that I'm performing the same check over and over on different variables, and I think it's more readable to encapsulate the check and related exits. Still, I know that macros aren't very safe, especially when they get really complex. However, as far as I can tell, trying to do the equivalent lambda requires awkward additional checks like so:
const int xmin(1), xmax(5);
auto check_range = [&](int x) -> bool { return !(x < xmin || x > xmax); };
bool myFunc(int myint) {
if(!check_range(myint)) return false;
int anotherint = myint + 2;
if(!check_range(anotherint)) return false;
return true;
}
Is there a way to do this with a lambda? Or am I missing some alternative solution?
Edit: I recognize that returning from inside a macro is generally a bad idea unless significant precautions are taken. I'm just wondering if it's possible.

Answer 1

You are correct--there's no way to return from the caller from inside a lambda. Since a lambda can be captured and stored to be called later, from inside an arbitrary caller, doing so would result in unpredictable behavior.
class Foo
{
Foo(std::function<void(int)> const& callMeLater) : func(callMeLater) {}
void CallIt(int* arr, int count)
{
for (index = count; index--;)
func(count);
// do other stuff here.
}
std::function<void(int)> func;
};
int main()
{
auto find3 = [](int arr)
{
if (arr == 3)
return_from_caller; // making up syntax here.
};
Foo foo(find3);
};

Answer 2

Is there a way to do this with a lambda?
Not exactly like the macro but your lambda, instead of returning a bool, can throw a special exception (of type bool, by example)
auto check_range
= [](int x) { if ( (x < xmin) || (x > xmax) ) throw bool{false}; };
and the function myFunc() can intercept this special type
bool myFunc (int myint)
{
try
{
check_range(myint);
int anotherint = myint + 2;
check_range(anotherint);
return true;
}
catch ( bool e )
{ return e; }
}
For a single check_range() call, this is (I suppose) a bad idea; if you have a lot of calls, I suppose can be interesting.
The following is a full working example
#include <iostream>
constexpr int xmin{1}, xmax{5};
auto check_range
= [](int x) { if ( (x < xmin) || (x > xmax) ) throw bool{false}; };
bool myFunc (int myint)
{
try
{
check_range(myint);
int anotherint = myint + 2;
check_range(anotherint);
return true;
}
catch ( bool e )
{ return e; }
}
int main ()
{
std::cout << myFunc(0) << std::endl; // print 0
std::cout << myFunc(3) << std::endl; // print 1
std::cout << myFunc(7) << std::endl; // print 0
}

Answer 3

No better way to do this than just to use the return value of the lambda and then return from the calling function. Macros are ew for this.
As it stands in C++, that is the idiomatic way to exit from a function that uses another condition to determine whether or not to exit.

Answer 4

Not C++11, but people have hacked C++2a coroutines to basically do this.
It would look a bit like:
co_await check_range(foo);
where the co_await keyword indicates that in some cases, this coroutine could return early with an incomplete result. In your cases, this incomplete result would be non-resumabable error.
The playing around I saw was with optionals, and required using a shared ptr, but things may improve before it is standardized.