C++ : how to define a function in main and then pass it as an argument

Question

Firstly, I want to create a function inside main. Secondly, I want to pass this function as an argument to another function. Is it possible please?

For the first point, I know that it is possible using lambda functions or structures, as explained in the following link: Can we have functions inside functions in C++?

However, I did not manage to pass this function as an argument to another function.

More precisely, I want something like:

int main()
{
    void add(int x, int y){cout<<x+y<<endl;}
    int a = 1;
    int b = 2;
    apply_operator(add,a,b);
}

where apply_operator is defined in another file by:

void apply_operator(void (*operator)(int,int),int x,int y)
{
    operator(x,y);
}

I'm sure that `apply_operator` is not defined as you say, since `operator` is a C++ keyword. — Drew Dormann, Jun 16 '21 at 16:45
One cannot declare or define a function inside another function. The best you can do is lambda. — ALX23z, Jun 16 '21 at 16:46

score 1 · Answer 1 · answered Jun 16 '21 at 16:47

1

Using lambdas and refraining from using the operator keyword we get:

#include <iostream>

void apply_operator(void (*op)(int, int), int x, int y)
{
    op(x, y);
}

int main()
{
    auto add = [](int x, int y) { std::cout << (x + y) << std::endl; };
    int a = 1;
    int b = 2;
    apply_operator(add, a, b);
}

Which prints 3 as expected.

answered Jun 16 '21 at 16:47

Nathan Pierson

5,461
1
12
30

Thank you very much, it worked very well. I tried the same thing with "apply_operator(&add, a, b);" instead and it did not work. – ZZZ Jun 16 '21 at 17:01
That's because a lambda is already convertible to a function pointer. Taking its address gives you something different. – Nathan Pierson Jun 16 '21 at 17:03

score 1 · Answer 2 · answered Jun 16 '21 at 16:47

1

You can certainly use a lambda.

int main()
{
    auto add = [](int x, int y){cout<<x+y<<endl;};
    int a = 1;
    int b = 2;
    apply_operator(add,a,b);
}

See it work!

answered Jun 16 '21 at 16:47

Drew Dormann

59,987
13
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180

Thank you very much, it worked very well. I tried the same thing with "apply_operator(&add, a, b);" instead and it did not work. – ZZZ Jun 16 '21 at 17:02
@ZZZ that is because a captureless lambda will implicitly convert to a function pointer. Your address-of-a-lambda will not. – Drew Dormann Jun 16 '21 at 17:07

score 0 · Answer 3 · answered Jun 16 '21 at 16:47

You may be interested in using a lambda in this case

int main()
{
    auto add = [](int x, int y){ cout<<x+y<<endl; };
    int a = 1;
    int b = 2;
    apply_operator(add,a,b);
}

As long as the lambda does not capture it can decay into a function pointer.

score 0 · Answer 4 · answered Jun 16 '21 at 16:53

You could use function pointers. I have pasted the code below. I have declared a the pointer "f" to a fucntion which takes in two integers and returs nothing in return and assigned the address of the fucntion add to it.

#include <iostream>

using namespace std;

void add(int x, int y)
{
    cout<<x+y<<endl;
}

void apply_operator(void (*f)(int, int),int a, int b)
{
    f(a, b);
}
int main()
{
    void (*f)(int, int);
    f = &add;
    int a = 1;
    int b = 2;
    apply_operator(f,a,b);
}

score -1 · Answer 5 · answered Jun 16 '21 at 17:02

-1

# include<iostream>

# include<conio.h>
using namespace std;
int fun(int a, int b)
{
    if(a>b)
    {
    cout<<"1st number is larger";
    }
    else
    {   
    cout<<"2nd number is larger";
    }
}       

int main()
{
    int a,b;
    cout<<"enter the two numbers";
    cin>>a>>b;
    fun(a,b);
    getch();

   return 0;

    getch();

}

answered Jun 16 '21 at 17:02

pushplata

1
1

I don't see what relevance this has to the question at all. – Nathan Pierson Jun 16 '21 at 17:03

C++ : how to define a function in main and then pass it as an argument

5 Answers5