c++ assign a class member function a lambda function for computation efficiency [duplicate]

Question

UPDATED: (Rephrased). I'm looking to boost the computation efficiency of my code by make an run-time assignment of a class member function to one of many functions conditional on other class members.

One recommended solution uses #include <functional> and function<void()>, as shown in the simple test example:

#include <iostream>
using namespace std;

struct Number {
  int n;
  function(void()) doIt;

  Number(int i):n(i) {};

  void makeFunc() {

      auto _odd  = [this]() { /* op specific to odd */ };
      auto _even = [this]() { /* op specific to even */ };

    // compiles but produces bloated code, not computatinally efficient
      if (n%2) doIt = _odd;   
      else     doIt = _even;  
  };
};

int main() {
  int i;
  cin >> i;
  Number e(i);
  e.makeFunc();
  e.doIt();
};

I'm finding that the compiled code (i.e. debug assembly) is grotesquely complicated and presumably NOT computationally efficient (the desired goal).

Does someone have an alternative construct that would achieve the end goal of a computationally efficient means of conditionally defining, at run-time, a class member function.

Did you try to replace void (*doIt)(); with std::function<void> doIt;? — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Oct 28, 2015 at 16:30
If you have a new question, post a new question. This one cannot get any new answers. — n. m. could be an AI
– n. m. could be an AI, Commented Oct 28, 2015 at 18:09
When you're saying it's not efficient, what was the baseline code you were comparing the performance against? It's a rhetorical question, I know you didn't. What I'm saying: you should. — Karoly Horvath
– Karoly Horvath, Commented Oct 28, 2015 at 18:31
"I'm finding that the compiled code (i.e. debug assembly) is grotesquely complicated and presumably NOT computationally efficient" Don't presume, benchmark. And bloated assembly code does not imply inefficiency; generally quite the opposite. — Ian Kemp - SO dead by AI greed
– Ian Kemp - SO dead by AI greed, Commented Oct 28, 2015 at 20:32

Piotr Skotnicki · Accepted Answer · 2015-10-28 16:31:05Z

3

A capturing lambda expression cannot be assigned to a regular function pointer like you have.

I suggest using

std::function<void()> doIt;

instead of

void (*doIt)();

edited Oct 28, 2015 at 16:31

Piotr Skotnicki

48.8k7 gold badges122 silver badges167 bronze badges

answered Oct 28, 2015 at 16:30

R Sahu

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1 Comment

Karoly Horvath Over a year ago

"cannot be assigned" - An that's because the type of a lambda is not utterable, so you cannot assign it to something which has a type. You need a construct that does type erasure.

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