Can we have functions inside functions in C++?

Question

I mean something like:

int main() 
{
  void a() 
  {
      // code
  }
  a();

  return 0;
}

Answer 1

You can't have local functions in C++. However, C++11 has lambdas. Lambdas are basically variables that work like functions.

A lambda has the type std::function (actually that's not quite true, but in most cases you can suppose it is). To use this type, you need to #include . std::function is a template, taking as template argument the return type and the argument types, with the syntax std::function. For example, std::function is a lambda returning an int and taking two arguments, one std::string and one float. The most common one is std::function, which returns nothing and takes no arguments.

Once a lambda is declared, it is called just like a normal function, using the syntax lambda(arguments).

To define a lambda, use the syntax [captures](arguments){code} (there are other ways of doing it, but I won't mention them here). arguments is what arguments the lambda takes, and code is the code that should be run when the lambda is called. Usually you put [=] or [&] as captures. [=] means that you capture all variables in the scope in which the value is defined by value, which means that they will keep the value that they had when the lambda was declared. [&] means that you capture all variables in the scope by reference, which means that they will always have their current value, but if they are erased from memory the program will crash. Here are some examples:

#include 
#include 

int main(){
    int x = 1;

    std::function lambda1 = [=](){
        std::cout << x << std::endl;
    };
    std::function lambda2 = [&](){
        std::cout << x << std::endl;
    };

    x = 2;
    lambda1();    //Prints 1 since that was the value of x when it was captured and x was captured by value with [=]
    lambda2();    //Prints 2 since that's the current value of x and x was captured by reference with [&]

    std::function lambda3 = [](){}, lambda4 = [](){};    //I prefer to initialize these since calling an uninitialized lambda is undefined behavior.
                                                                 //[](){} is the empty lambda.

    {
        int y = 3;    //y will be deleted from the memory at the end of this scope
        lambda3 = [=](){
            std::cout << y << endl;
        };
        lambda4 = [&](){
            std::cout << y << endl;
        };
    }

    lambda3();    //Prints 3, since that's the value y had when it was captured

    lambda4();    //Causes the program to crash, since y was captured by reference and y doesn't exist anymore.
                  //This is a bit like if you had a pointer to y which now points nowhere because y has been deleted from the memory.
                  //This is why you should be careful when capturing by reference.

    return 0;
}

You can also capture specific variables by specifying their names. Just specifying their name will capture them by value, specifying their name with a & before will capture them by reference. For example, [=, &foo] will capture all variables by value except foo which will be captured by reference, and [&, foo] will capture all variables by reference except foo which will be captured by value. You can also capture only specific variables, for example [&foo] will capture foo by reference and will capture no other variables. You can also capture no variables at all by using []. If you try to use a variable in a lambda that you didn't capture, it won't compile. Here is an example:

#include 

int main(){
    int x = 4, y = 5;

    std::function myLambda = [y](int z){
        int xSquare = x * x;    //Compiler error because x wasn't captured
        int ySquare = y * y;    //OK because y was captured
        int zSquare = z * z;    //OK because z is an argument of the lambda
    };

    return 0;
}

You can't change the value of a variable that was captured by value inside a lambda (variables captured by value have a const type inside the lambda). To do so, you need to capture the variable by reference. Here is an exampmle:

#include 

int main(){
    int x = 3, y = 5;
    std::function myLambda = [x, &y](){
        x = 2;    //Compiler error because x is captured by value and so it's of type const int inside the lambda
        y = 2;    //OK because y is captured by reference
    };
    x = 2;    //This is of course OK because we're not inside the lambda
    return 0;
}

Also, calling uninitialized lambdas is undefined behavior and will usually cause the program to crash. For example, never do this:

std::function lambda;
lambda();    //Undefined behavior because lambda is uninitialized

---

Examples

Here is the code for what you wanted to do in your question using lambdas:

#include     //Don't forget this, otherwise you won't be able to use the std::function type

int main(){
    std::function a = [](){
        // code
    }
    a();
    return 0;
}

Here is a more advanced example of a lambda:

#include     //For std::function
#include       //For std::cout

int main(){
    int x = 4;
    std::function divideByX = [x](int y){
        return (float)y / (float)x;    //x is a captured variable, y is an argument
    }
    std::cout << divideByX(3) << std::endl;    //Prints 0.75
    return 0;
}