How to initialize items in a range in a captureless lambda, C++ - c++

Question

My code is below. This works, It allows me to have exactly one range in my lambda.
So I guess what my question is, is how do I achieve the same results without using
"if(LOOP > 2 && LOOP < 5){int THERANGEVALUE = 2; FUNC[THERANGEVALUE]();}"?
And instead initialize an item in my captureless lambda as being ranged instead. aka, item_2 being item_range(2,4). And then also being able to continue my lambda normally, whereas Item_3 will equate to item_5.
Thank you for any help in advance, I will gladly add more input if requested.
#include <iostream>
using namespace std;
void (*FUNC[3])(void) = { //captureless lambda.
/*ITEM_0*/[](){ cout << "x" << endl;},
/*ITEM_1*/[](){cout << "y" << endl;},
/*ITEM_2->ITEM_4*/[](){cout<<"z";}
};
/*Here the [](){code;} lambda is acting as a stand-in for void FUNC() so it shouldn't touch anything outside of its scope*/
int LOOP = 4;
int main()
{
if(LOOP > 2 && LOOP < 5){int THERANGEVALUE = 2; FUNC[THERANGEVALUE]();}
FUNC[LOOP]();
return 0;
}
Adding on to this, below is the solution I came up with after asking a friend. To my surprise it was actually a lot simpler than I expected. While I couldn't initialize each item in the lambda in a range easily, I could pass it into an array and set a range inside of the array instead. So while it's not quite what I was looking for, it's...good enough for my purposes. Thanks Jaime if you see this. Otherwise I'd use PilouPili's answer below.
#include <iostream>
using namespace std;
void (*FUNC[4])(void) = { //captureless lambda.
/*ITEM_0*/ [](){ cout << "x" << endl;},
/*ITEM_1*/ [](){cout << "y" << endl;},
/*ITEM_2->ITEM_4*/[](){cout<<"z";},
/*ITEM_5*/ [](){cout<<"z";}
};
int LOOP = 4;
int main()
{
int ARR[5]={};
for(int I = 0; I < 6;I=I+1){//handling of ranged values.
if(I>2 && I<5){ARR[I]=2;} else {ARR[I]=I;}
}
FUNC[ARR[LOOP]]();
return 0;
}

Answer 1

I only see to way :
either extend your function array -> That's FUNC1 in the next example
change the value given in operator [] -> That's FUNC2 in the next example
#include <iostream>
#include <vector>
using namespace std;
std::vector<void (*)(void)> init_FUNC()
{
std::vector<void (*)(void)> func(5, [](){cout<<"z";});
func[0]=[](){ cout << "x" << endl;};
func[1]=[](){ cout << "y" << endl;};
return func;
}
std::vector<void (*)(void)> FUNC1= init_FUNC();
class FUNC_MAP
{
void (*_FUNC[3])(void) = { //captureless lambda.
/*ITEM_0*/[](){ cout << "x" << endl;},
/*ITEM_1*/[](){cout << "y" << endl;},
/*ITEM_2->ITEM_4*/[](){cout<<"z";}
};
typedef void (*FUNC_MAP_OUT)(void);
public:
FUNC_MAP_OUT operator[](int i)
{
if(i>2 && i<5)
{return _FUNC[2];}
else
{return _FUNC[i];}
}
};
FUNC_MAP FUNC2;
/*Here the [](){code;} lambda is acting as a stand-in for void FUNC() so it shouldn't touch anything outside of its scope*/
int LOOP = 1;
int main()
{
FUNC1[LOOP]();
FUNC2[LOOP]();
return 0;
}