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Pointers to Functions

 

Introduction

Imagine you are writing a program to process cylinder-related calculations, that is, to get its diameter, its circumference, its areas, and volume. You could start your program as follows:

using namespace System;

int main()
{
    double Radius;

    Radius = 25.55;

    Console::WriteLine("Cylinder Summary");
    Console::WriteLine("\nRadius:   {0}", Radius);
    
    return 0;
}

This would produce:

Cylinder Summary
Radius:   25.55
Press any key to continue...

When we studied functions that return a value, we saw that the result of such a function could be assigned to a value locally declared in the calling function:

using namespace System;

static double CalculateDiameter(double R);
int main()
{
    double Radius, Diameter;

    Radius = 25.52;
    Diameter = CalculateDiameter(Radius);

    Console::WriteLine("Cylinder Summary";
    Console::WriteLine("\nRadius:   {0}", Radius);
    Console::WriteLine("\nDiameter: {0}", Diameter);
    
    return 0;
}

double CalculateDiameter(double Rad)
{
    return Rad * 2;
}

At this time, we know that when a function returns a value, the calling of the function is a complete value that can be assigned to a variable. In fact, when calling a function that takes an argument, if that argument itself is gotten from a value returned by a function, the calling of the second function can be done directly when calling the first function. This seemingly complicated scenario can be easily demonstrated as follows:

using namespace System;
    
static double  CalculateDiameter(double R);
static double  CalculateCircumference(double D);

int main()
{
    double Radius, Circumference;

    Radius = 25.52;

    // Instead of calling the CalculateDiameter() function first and
    // assign it to another, locally declared variable, such as in
    // "double Diameter = CalculateDiameter(Radius)", we can call the
    // CalculateDiameter(Radius) directly when we are calling the
    // CalculateCircumference() function.
    // This is possible because the CalculateCircumference() function
    // takes an argument that is the result of calling the
    // CalculateDiameter() function. As long as we only need the
    // circumference and we don't need the diameter, we don't have
    // to explicitly call the CalculateDiameter() function.
    Circumference = CalculateCircumference(CalculateDiameter(Radius));

    Console::WriteLine("Cylinder Summary";
    Console::WriteLine("\nRadius:   {0}", Radius);
    Console::WriteLine("\nCircumference: {0}\n", Circumference);
    
    return 0;
}

double CalculateDiameter(double Rad)
{
    return Rad * 2;
}

double CalculateCircumference(double Diam)
{
    const double PI = 3.14159;
    return Diam * PI;
}

In some circumstances, such as this one, we may find out that the value we want to process in a function is in fact a value gotten from an intermediary function. Unfortunately, a regular function cannot be passed to a function like a regular variable. In reality, the C++ language allows this but the function must be passed as a pointer.

Declaring a Pointer to Function

The concept of a callback function is highly used in Microsoft Windows application programming. For this reason, you should know how callback functions work and how to use them. They are present on the Win32 API library. In Win32, they allow creating procedures.

A callback function is a pointer to a function. A pointer to a function is a special function that is declared as a pointer. Its name by itself is considered a variable. As such, and unlike a regular variable, the name of this function can be assigned a regular function. This allows the function to be passed as an argument. The function itself is not implemented but its name is used as a programmer type-defined object.

The reason a function can be passed as argument is because the name of a function is itself a constant pointer. The basic syntax to declare a pointer to a function is:

DataType (*FunctionName)();

The DataType can be any of the data types we have used so far. The FunctionName must be a valid name for a function. The name of the function must be preceded by an asterisk operator. To actually make this declaration a pointer to a function, the asterisk and the name of the pointer must be included between parentheses. If you omit the parentheses, the compiler would think that you are declaring a function that would return a pointer, which changes everything.

Because this is a pointer, you must use parentheses, required for every function declared. If this function will not take any argument, you can leave the parentheses empty or type void. Based on this, you can declare a pointer to a function as follows:

int main()
{
    void (*SomethingToSay)(void);
    
    return 0;
}

After declaring a pointer to a function, keep in mind that this declaration only creates a pointer, not an actual function. In order to use it, you must define the actual function that would carry the assignment the function is supposed to perform. That function must have the same return type and the same (number of) argument(s), if any. For example, the above declared pointer to function is of type void and it does not take any argument. you can define a function as follows:

void MovieQuote()
{
    Console::WriteLine("We went through a lot of trouble because of you!");
    Console::WriteLine("You owe us\n");
    Console::WriteLine("  From \"Disorganized Crime\"");
}

With such an associated function defined, you can assign it to the name of the pointer to function as follows

SomethingToSay = MovieQuote;

This assignment gives life to the function declared as pointer. The function can then be called as if it had actually been defined. Here is an example:

using namespace System;

void MovieQuote()
{
    Console::WriteLine("We went through a lot of trouble because of you");
    Console::WriteLine("You owe us");
    Console::WriteLine("  From \"Disorganized Crime\"");
}

int main()
{
    void (*SomethingToSay)();

    // Assign the MovieQuote() function to the pointer to function
    SomethingToSay = MovieQuote;

    // Call the pointer to function as if it had been defined already
    SomethingToSay();
    
    return 0;
}

This would produce:

We went through a lot of trouble because of you
You owe us
  From "Disorganized Crime"

Press any key to continue...

A Function Pointer that Returns a Value

You can also declare a pointer to function for a function that returns a value. Remember that both functions must return the same type of value. Here is an example:

using namespace System;

int  Addition()
{
    int a = 16, b = 442;
    return a + b;
}

int main()
{
    int  (*SomeNumber)();

    // Assign the MovieQuote() function to the pointer to function
    SomeNumber = Addition;

    // Call the pointer to function as if it had been defined already
    Console::WriteLine("The number is {0}", SomeNumber());
    
    return 0;
}

A Function Pointer With Arguments

If you want to use a function that takes arguments, when declaring the pointer to function, provide the return type and an optional name for each argument. Here is an example:

int (*SomeNumber)(int x, int y);

When defining the associated function, besides returning the same type of value, make sure that the function takes the same number of arguments. Here is an example:

using namespace System;

int Addition(int a, int b)
{
    return a + b;
}

int main()
{
    int (*SomeNumber)(int x, int y);
    int x = 128, y = 5055;

    // Assign the MovieQuote() function to the pointer to function
    SomeNumber = Addition;

    // Call the pointer to function as if it had been defined already
    Console::WriteLine("{0} + {1} = {2}\n", x, y, SomeNumber(x, y));
    
    return 0;
}

Type-Defining a Function Pointer

You can create a programmer-defined type as a pointer to function. Here is the syntax to use:

typedef (*TypeName)(Arguments);

The typedef keyword must be used.

The asterisk and the TypeName must be enclosed in parentheses. The name must follow the rules applied to functions so far.

The TypeName must be followed by parentheses. If the pointer to function will take arguments, provide its type or their types between parentheses. Otherwise, you can leave the parentheses empty (but you must provide the parentheses).

After creating such a custom type, the name of the type would be used as an alias to a pointer to function. Consequently, it can be used to declare a pointer to function. Here is an example:

using namespace System;

int  Addition(int a, int b)
{
    return a + b;
}

int main()
{
    // Creating a programmer-defined type
    typedef int  (*AddsTwoIntegers)(int x, int y);
    // Now, the AddsTwoIntegers name is a pointer to function
    // that can take two integers. It can be used for a declaration

    AddsTwoIntegers TwoNumbers;
    int x = 128, y = 5055;

    TwoNumbers = Addition;

    // Call the pointer to function as if it had been defined already
    Console::WriteLine("{0} + {1} = {2}\n", x, y, TwoNumbers(x, y));
    
    return 0;
}

This would produce:

128 + 5055 = 5183
Press any key to continue...

A Pointer to a Function as Argument

Using pointer to functions, a function can be passed as argument to another function. The function must be passed as a pointer. The argument is declared in a complete format as if you were declaring a function. Here is an example of a function that is passed a function as argument.

double Circumference(double (*FDiam)(double R))

This Circumference() function takes one argument. The argument itself is a pointer to function. This argument takes a double-precision number as argument and it returns a double-precision value. The Circumference() function returns a double-precision number.

It is important to know that the pointer to function that is passed as argument is declared completely, in this case as

double (*FDiam)(double R)

Although the FDiam declaration is accompanied by an argument, in this case R, this argument allows the compiler to know that FDiam takes an argument. This argument actually will not be processed by the Circumference() function when the Circumference() function is defined because the R argument does not belong to the Circumference() function.

When calling the Circumference() function, you will use the FDiam argument as a variable in its own right, using its name, as in

Circumference(Diameter)

When defining the Circumference() function, you must process the pointer to function that it takes as argument. If this argument is an alias to a function that returns a value, you can call it and pass it the argument as we studied in the last section. If you want to involve the FDiam argument in any operation, you can declare a local variable to the Circumference() function. If the FDiam argument must be involved in an operation that involves a value external to the Circumference() function, you must pass that type of value as argument to the Circumference() function, unless you are using a global variable. This means that, in most circumstances, the pointer to function passed as argument may be accompanied by at least one other argument. For example, if you want to use the FDiam as a diameter value to calculate the circumference (Circumference = Diameter * PI), you may have to declare it with an argument for the radius. It would be declared as follows:

double Circumference(double  (*FDiam)(double R), double Rad);

The function can then be implemented as follows:

double Circumference(double (*FDiam)(double R), double Rad)
{
    double Circf;
    const double PI = 3.14159;
    Circf = (*FDiam)(Rad);

    return Circf * PI;
}

Remember that, when declaring a function, the compiler does not care about the name(s) of the argument(s). If the function takes any, what the compiler cares about are the return type of the function, its name, and the type(s) of its argument(s), if any. Therefore, the above function could as well be declared as follows:

double Circumference(double (*)(double R), double);

This indicates that the Circumference() function takes two arguments whose names are not known. The first argument is a pointer to a function that takes one double-precision number as argument and returns a double. The second argument of the Circumference() function is also a double-precision number. The Circumference() function returns a double-precision number. This is what the program at this time would look like:

using namespace System;

double Diameter(double);
double Circumference(double  (*D)(double R), double r);

int main()
{
    double Radius;

    Radius = 25.52;

    Console::WriteLine("Cylinder Summary");
    Console::WriteLine("Radius:   {0}", Radius);
    Console::WriteLine("Circumference = {0}\n", Circumference(Diameter, Radius));
    
    return 0;
}

double Diameter(double Rad)
{
    return Rad * 2;
}

double Circumference(double (*FDiam)(double R), double Rad)
{
    double Circf;
    const double PI = 3.14159;
    Circf = (*FDiam)(Rad);

    return Circf * PI;
}

This would produce:

Cylinder Summary
Radius:   25.52
Circumference = 160.347

Press any key to continue...

To simplify the declaration of a pointer to function, we saw that you can create a programmer-defined type using the typedef keyword. This can also help when passing a function as argument. Here is an example:

typedef double (*FDiam)(double R);
double Circumference(FDiam, double);

When creating such a programmer-defined type, remember that you must give a name to the alias, in this case FDiam. After this creation, FDiam is an alias to a pointer to function of a double-precision type and which takes one double-precision number as argument.

Remember, as we learned when studying functions that return a value, that the item on the right side of the return keyword can be a value or a complete expression. Therefore, you can simplify the implementation of the Circumference() function as follows:

double Circumference(double (*FDiam)(double R), double Rad)
{
    const double PI = 3.14159;
    return (*FDiam)(Rad) * PI;
}

 

 

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