C++ – Pass the current state of a function into another function in C/C++

c++calling-conventionfunction

Is there a way to pass the current state of a function into another function in C/C++? I mean all the parameters and local variables by current state. For example:

void funcA (int a, int b)
{
    char c;
    int d, e;
    // Do something with the variables.
    // ...
    funcB();
    // Do something more.
}

void funcB()
{
    // funcB() should be able to access variables a,b,c,d & e
    // and any change in these variables reflect into funcA().
}

The code is in bad shape if there is a need for funcB() kind of functions. But can it be achieved?

This can help if someone is starting to re-factor a long method with multiple parameters.

Best Solution

Introduce a common struct.

struct State {
    char c;
    int d,e;
};

void funcA(int a, int b){
    State s;
    s.d = 1234; // ...
    // ...
    funcB(s);
}

void funcB(State& s)
{
    //...
}

Related Solutions

How to pass a function as a parameter in C

Declaration

A prototype for a function which takes a function parameter looks like the following:

void func ( void (*f)(int) );

This states that the parameter f will be a pointer to a function which has a void return type and which takes a single int parameter. The following function (print) is an example of a function which could be passed to func as a parameter because it is the proper type:

void print ( int x ) {
  printf("%d\n", x);
}

Function Call

When calling a function with a function parameter, the value passed must be a pointer to a function. Use the function's name (without parentheses) for this:

func(print);

would call func, passing the print function to it.

Function Body

As with any parameter, func can now use the parameter's name in the function body to access the value of the parameter. Let's say that func will apply the function it is passed to the numbers 0-4. Consider, first, what the loop would look like to call print directly:

for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
  print(ctr);
}

Since func's parameter declaration says that f is the name for a pointer to the desired function, we recall first that if f is a pointer then *f is the thing that f points to (i.e. the function print in this case). As a result, just replace every occurrence of print in the loop above with *f:

void func ( void (*f)(int) ) {
  for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
    (*f)(ctr);
  }
}

Source

What’s the difference between a method and a function

A function is a piece of code that is called by name. It can be passed data to operate on (i.e. the parameters) and can optionally return data (the return value). All data that is passed to a function is explicitly passed.

A method is a piece of code that is called by a name that is associated with an object. In most respects it is identical to a function except for two key differences:

A method is implicitly passed the object on which it was called.
A method is able to operate on data that is contained within the class (remembering that an object is an instance of a class - the class is the definition, the object is an instance of that data).

(this is a simplified explanation, ignoring issues of scope etc.)

Best Solution

Related Solutions

How to pass a function as a parameter in C

What’s the difference between a method and a function

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