7.8 — Function Pointers

In lesson 6.7 -- Introduction to pointers, you learned that a pointer is a variable that holds the address of another variable. Function pointers are similar, except that instead of pointing to variables, they point to functions!

Consider the following function:

Identifier foo is the function’s name. But what type is the function? Functions have their own l-value function type -- in this case, a function type that returns an integer and takes no parameters. Much like variables, functions live at an assigned address in memory.

When a function is called (via the () operator), execution jumps to the address of the function being called:

At some point in your programming career (if you haven’t already), you’ll probably make a simple mistake:

Instead of calling function foo() and printing the return value, we’ve unintentionally sent function foo directly to std::cout. What happens in this case?

On the author’s machine, this printed:


…but it may print some other value (e.g. 1) on your machine, depending on how your compiler decides to convert the function pointer to another type for printing. If your machine doesn’t print the function’s address, you may be able to force it to do so by converting the function to a void pointer and printing that:

Just like it is possible to declare a non-constant pointer to a normal variable, it’s also possible to declare a non-constant pointer to a function. In the rest of this lesson, we’ll examine these function pointers and their uses. Function pointers are a fairly advanced topic, and the rest of this lesson can be safely skipped or skimmed by those only looking for C++ basics.

Pointers to functions

The syntax for creating a non-const function pointer is one of the ugliest things you will ever see in C++:

In the above snippet, fcnPtr is a pointer to a function that has no parameters and returns an integer. fcnPtr can point to any function that matches this type.

The parenthesis around *fcnPtr are necessary for precedence reasons, as int *fcnPtr() would be interpreted as a forward declaration for a function named fcnPtr that takes no parameters and returns a pointer to an integer.

To make a const function pointer, the const goes after the asterisk:

If you put the const before the int, then that would indicate the function being pointed to would return a const int.

Assigning a function to a function pointer

Function pointers can be initialized with a function (and non-const function pointers can be assigned a function):

One common mistake is to do this:

This would actually assign the return value from a call to function goo() to fcnPtr, which isn’t what we want. We want fcnPtr to be assigned the address of function goo, not the return value from function goo(). So no parenthesis are needed.

Note that the type (parameters and return type) of the function pointer must match the type of the function. Here are some examples of this:

Unlike fundamental types, C++ will implicitly convert a function into a function pointer if needed (so you don’t need to use the address-of operator (&) to get the function’s address). However, it will not implicitly convert function pointers to void pointers, or vice-versa.

Calling a function using a function pointer

The other primary thing you can do with a function pointer is use it to actually call the function. There are two ways to do this. The first is via explicit dereference:

The second way is via implicit dereference:

As you can see, the implicit dereference method looks just like a normal function call -- which is what you’d expect, since normal function names are pointers to functions anyway! However, some older compilers do not support the implicit dereference method, but all modern compilers should.

One interesting note: Default parameters won’t work for functions called through function pointers. Default parameters are resolved at compile-time (that is, if you don’t supply an argument for a defaulted parameter, the compiler substitutes one in for you when the code is compiled). However, function pointers are resolved at run-time. Consequently, default parameters can not be resolved when making a function call with a function pointer. You’ll explicitly have to pass in values for any defaulted parameters in this case.

Passing functions as arguments to other functions

One of the most useful things to do with function pointers is pass a function as an argument to another function. Functions used as arguments to another function are sometimes called callback functions.

Consider a case where you are writing a function to perform a task (such as sorting an array), but you want the user to be able to define how a particular part of that task will be performed (such as whether the array is sorted in ascending or descending order). Let’s take a closer look at this problem as applied specifically to sorting, as an example that can be generalized to other similar problems.

All sorting algorithms work on a similar concept: the sorting algorithm iterates through a list of numbers, does comparisons on pairs of numbers, and reorders the numbers based on the results of those comparisons. Consequently, by varying the comparison, we can change the way the function sorts without affecting the rest of the sorting code.

Here is our selection sort routine from a previous lesson:

Let’s replace that comparison with a function to do the comparison. Because our comparison function is going to compare two integers and return a boolean value to indicate whether the elements should be swapped, it will look something like this:

And here’s our selection sort routine using the ascending() function to do the comparison:

Now, in order to let the caller decide how the sorting will be done, instead of using our own hard-coded comparison function, we’ll allow the caller to provide their own sorting function! This is done via a function pointer.

Because the caller’s comparison function is going to compare two integers and return a boolean value, a pointer to such a function would look something like this:

So, we’ll allow the caller to pass our sort routine a pointer to their desired comparison function as the third parameter, and then we’ll use the caller’s function to do the comparison.

Here’s a full example of a selection sort that uses a function pointer parameter to do a user-defined comparison, along with an example of how to call it:

This program produces the result:

9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 9

Is that cool or what? We’ve given the caller the ability to control how our selection sort does its job.

The caller can even define their own “strange” comparison functions:

The above snippet produces the following result:

2 4 6 8 1 3 5 7 9

As you can see, using a function pointer in this context provides a nice way to allow a caller to “hook” their own functionality into something you’ve previously written and tested, which helps facilitate code reuse! Previously, if you wanted to sort one array in descending order and another in ascending order, you’d need multiple version of the sort routine. Now you can have one version that can sort any way the caller desires!

Providing default functions

If you’re going to allow the caller to pass in a function as a parameter, it can often be useful to provide some standard functions for the caller to use for their convenience. For example, in the selection sort example above, providing the ascending() and descending() function along with the selectionSort() function would make the callers life easier, as they wouldn’t have to rewrite ascending() or descending() every time they want to use them.

You can even set one of these as a default parameter:

In this case, as long as the user calls selectionSort normally (not through a function pointer), the comparisonFcn parameter will default to ascending.

Making function pointers prettier with typedef or type aliases

Let’s face it -- the syntax for pointers to functions is ugly. However, typedefs can be used to make pointers to functions look more like regular variables:

This defines a typedef called “validateFcn” that is a pointer to a function that takes two ints and returns a bool.

Now instead of doing this:

You can do this:

Which reads a lot nicer! However, the syntax to define the typedef itself can be difficult to remember.

In C++11, you can instead use type aliases to create aliases for function pointers types:

This reads more naturally than the equivalent typedef, since the name of the alias and the alias definition are placed on opposite sides of the equals sign.

Using a type alias is identical to using a typedef:

Using std::function in C++11

Introduced in C++11, an alternate method of defining and storing function pointers is to use std::function, which is part of the standard library <functional> header. To define a function pointer using this method, declare a std::function object like so:

As you see, both the return type and parameters go inside angled brackets, with the parameters inside parenthesis. If there are no parameters, the parentheses can be left empty. Although this reads a little more verbosely, it’s also more explicit, as it makes it clear what the return type and parameters expected are (whereas the typedef method obscures them).

Updating our earlier example with std::function:


Function pointers are useful primarily when you want to store functions in an array (or other structure), or when you need to pass a function to another function. Because the native syntax to declare function pointers is ugly and error prone, we recommend you use typedefs (or in C++11, std::function).

Quiz time!

1) In this quiz, we’re going to write a version of our basic calculator using function pointers.

1a) Create a short program asking the user for two integer inputs and a mathematical operation (‘+’, ‘-‘, ‘*’, ‘/’). Ensure the user enters a valid operation.

Show Solution

1b) Write functions named add(), subtract(), multiply(), and divide(). These should take two integer parameters and return an integer.

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1c) Create a typedef named arithmeticFcn for a pointer to a function that takes two integer parameters and returns an integer.

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1d) Write a function named getArithmeticFunction() that takes an operator character and returns the appropriate function as a function pointer.

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1e) Modify your main() function to call getArithmeticFunction(). Call the return value from that function with your inputs and print the result.

Show Solution

Here’s the full program:

Show Solution

2) Now let’s modify the program we wrote in quiz 1 to move the logic out of the getArithmeticFcn and into an array.

2a) Create a struct named arithmeticStruct that has two members: a mathematical operator char, and an arithmeticFcn function pointer.

Show Solution

2b) Create a static global array of arithmeticStruct named arithmeticArray, initialized with each of the four arithmetic functions.

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2c) Modify getArithmeticFcn to loop through the array and return the appropriate function pointer.

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Here’s the full program:

Show Solution

7.9 -- The stack and the heap
7.7 -- Default parameters

225 comments to 7.8 — Function Pointers

  • Sivasankar

    Hi Alex,

    By reading this lesson, I finally got an understanding on how register callbacks mechanisms might be implemented in the Teamcenter PLM tool that is built in C++. Thank you for that.

    I got confusion at below line in this lesson.
    "One interesting note: Default parameters won’t work with function pointers."

    But in "providing default functions" section, a code snippet is put as below which I hope ascending is the default argument for comparisonFcn function parameter. And user may call with only the 1st 2 arguments if I'm correct.

    Could you please explain ?

    • Alex

      Yep. Two separate concepts.

      When defining a function, you can have function pointers as parameters, and those function pointers can have default values (see the selectionSort() example above). This works just like any other defaulted parameter.

      If you call any given function normally (not through a function pointer), any default parameters will resolve as you'd expect.
      If you call any given function through a function pointer, any default parameters will not resolve, so you'll always have to pass in values for all parameters in this case.

      I've update the lesson text slightly to try to make this clearer.

  • Matt

    Under "Assigning a function to a function pointer", you wrote:
    "Note that the signature (parameters and return value) of the function pointer must match the signature of the function."

    Should "return value" actually be return "type"?

  • Yash Verma

    I was understand function pointers from C++ Primer. So I decided to read it from the tutorial and now all my doubts are clear. Thank you for this tutorial.

  • Mauricio Mirabetti

    Alex, I'm following your course and I'm actually learning. Thanks again (and again).
    Regarding typedef to simplify functions declarations that return function pointers, I totally get it. But I haven't seem an example where typedef is not used, with the same functionality. Is it possible?
    I include a code where line comments show originals (working fine) and what I understand would it be without using typedefs:

    Got several errors, first being C2091 "function returns function" (function returning function is not allowed), on this line below:
    int(*getArithmeticFunction)(int,int) (char oper) {

    Thanks a lot!



    I enjoyed this chapter. Thank you.

    I have one question to ask.
    If doesn't "typedef" or "std::function" , How could we declare a function which return function pointer??

    • Alex

      The syntax to do this is super ugly. Here's how you'd write a function that takes a bool and returns a pointer to a function that takes two integers and returns a bool:

  • Nephilimac

    A word "unintentially" below the 3rd code sample may be aimed at "unintentionally" or "inadvertently".

  • Chris


    i have question about number 2c, why we need return add at bottom of getArithmeticFcn function? i think its not necessarry because if the "op" variable is + so arith.fcn must be "add" and it is sure the input is just beetween that 4 operation because we have a do-while loop to make sure it, isnt it?

    Thank you.

    • Chris

      sorry double post, there is more question i want to ask.
      if i make function pointer argument just like normal function (argument "int x") it seem the compiler not complaint or warning me. there is use for this? int (*ptrFcn)(int x)

    • Alex

      What happens if none of the arith.op match the op parameter?

      It would probably be better to assert out or something, but in this case, I've just assumed the user meant to add (for simplicity).

      • Chris

        but i think op parameter always will match with arith.op because in getOperation() there is do-while loop and it make sure the op parameter just can have + or - or * or / and arith.op just have them too.

        • Alex

          Yes, in this particular case, getOperation() will ensure that op has a valid value. But when you're writing functions, you should never assume anything about the inputs.

          For example, what if I updated getOperation() to also support modulus (%)? In that case, getOperation() would allow % as a valid input, but if I forgot to update arithmeticArray then nothing would match in getArithmeticFcn().

  • Vlad

    Hello Alex

    I can't compile this part:

    I get this:

    There is a caret pointing to the closing bracket around (foo) in the static_cast. I am using archlinux x64 (codeblocks or console "g++ -o ...", same thing). Is this a g++ whim?

    • Alex

      Hmmm, I"m not sure. Does the following work?

      • Vlad

        Yes, that works. Also std::cout << foo, and both return 1. Just now I remembered about the other type casts, so, looking at the list you gave, the "reinterpret" one seemed more appropiate, so I tried it, and, to my surprise, it works. At least, it returns an address: 0x400796. Maybe it's some new change in the compiler version?

        • Alex

          It seems like reinterpret_cast to void would be a valid way to print a normal function's address. This doesn't work for member functions though (which is fine for these examples since these aren't member functions).

  • Arslan

    foo() in the first case is an lvalue function type. Not a pointer to a function.

    lvalue functions get implicitly converted to pointer-to-functions when passed to a non-reference parameter. So in the sorting code in your post, 'ascending' and 'descending' are lvalue functions types, but are implicitly converted to pointer-to-functions when passed to the sort function.

  • Rob G.

    I finished 7.8_e and now have a working primitive calculator!!

    I used auto for determining my function type above as well as the caller (below):

    So its up and working fine:

    But what did my type(s) end up being that were predicted by the auto fx?

    • Alex

      Your auto is inferring type int(*)(int, int) (what you typedef'd with arithmeticFcn).

      It's best practice not to auto your function return types (it makes it harder for callers to discover what type your function is returning).

      • Rob G.

        It’s best practice not to auto your function return types

        I agree but couldn’t get the result any other way. Would you mind modifying and posting my code with changes you mentioned so it works without auto keyword? I think its only a few.This is key to my understanding. I would rather master this aspect than rely on auto.
        Thx in advance

        • Rob G.

          "Would you mind modifying and posting my code" ... that sounded terrible sorry. Just modification to the points of interest. 🙂

          • Rob G.

            Well I didn't think I could do it, but auto keyword removed successfully. arithmeticFcn type replacing auto keyword. A typedef is a type.

  • Rob G.

    Alex I get an error with the return pointer addresses: I've tried them without & and with, but I still get errors. Assume I have included all of the other functions necessary to run the program to save space here.

    ..srcmain.cpp: In function 'int* getArithmeticFunction(char)':
    ..srcmain.cpp:28:21: error: cannot convert 'int (*)(int, int)' to 'int*' in return
    case '+' : return &add;
    ... the same error for -,/,*

    • Rob G.

      To overcome the challenges above I changed the int * get...   to "auto".
      I also had to change compiler settings to ISO C++1y. I don't line just auto-ing hit; I'd like to know what the proper type is.

      • Rob G.

        Still having a hard time with this:

        Where are they returned to and how do we know what are we retrunig.

        • Alex

          The return values are returned to the caller just like any other return value. The caller can then do what it wants with those values (including ignore them).

          You know what you are returning because you have return statements indicating what is being returned in each case.

    • Alex

      Yes, your getArithmeticFunction function is declared as returning a pointer to an integer. However, add, subtract, divide, and multiply have type int (*)(int, int).

      You'll need to update the return type of your getArithmeticFunction to match the type of the function you're wanting to return.

      • J3ANP3T3R

        bool validate(int x, int y, std::function<bool(int, int)>); // std::function method that returns a bool and takes two int parameters

        i don't see any temporary variable for the function variable

        std::function<bool(int, int)>

        how do i use it from within the function ?

  • Rob G.

    Alex you mentioned that I was using pointers to functions, not nested loops. Could you give a brief example of a nested loop (maybe from my code where I pass the pointer)? Also, any recommendations on the legibility of the code? I've got indentations going -- other suggestions? Thx for your time.


    • Alex

      > Alex you mentioned that I was using pointers to functions, not nested loops

      No, I said you were using pointers to functions, not nested _functions_.

      Using a nested loop wouldn't make sense here.

      • Rob G.

        I meant nested functions --
        Not nested loops.

        • Rob G.

          Hi Alex, I meant nested functions not loops -- sorry. If you have anything to say abt how this exercise could relate to nested functions, I’d lke to hear it. If not -- onward. This is a great website and thank you for your time. I have thus far learned so much.

          • Alex

            Nested functions are illegal in C++ (though there are some workaround via classes/lambdas/closures), so there isn't much to say here, actually. 🙂

  • Rob G.

    I finally get it! The key to understanding for me was that we are creating functions within functions. I tried doing this other ways, but found that this was what function pointers were suited for. I was either asleep at the wheel when you mentioned this above, or it needs to be made more clear in the start of the exercise.

    cool !

    do stuff
    do other stuff

    • Alex

      It's misleading to say we're creating functions within functions (that sounds like nested functions).

      What we're doing is using a pointer to hold the address of a function. Then we can dereference that pointer to call the function the pointer is pointing to!

      • Rob G.

        I stand corrected, but I basically get the exercise. I'm still scratching my head a little on when to use a pointer to function, but its immediate utility to me is to dereference the pointer to another function while in a function already. Yes, very cool.

  • Sachin Singh

    Hey, I'm just wondering why didn't you mention the functional header. It makes defining function pointers a lot cleaner.

    Here pf can point to a function that returns a double and has one parameter which is a double as well.

  • Rob G.

    Ok, strategy: when faced with multiple functions, how does one know whether to use the function pointers or more standard setup of referencing the functions as needed?

    • Alex

      I always use regular functions unless I need to store functions in an array or pass a function as a parameter to another function.

  • Rob G.

    Hi Alex,

    I was getting this to work with ints which I no longer need. Trying to use void since I have no return value.
    What is wrong with my code? The fx count_up is never called. Thx in advance.

    • Alex

      Your function is running just fine, but your loop condition is failing immediately since 10 < 3 is false. If you switch your 10 and 3 parameters, you'll see it's working as you expect.

  • Satish

    Hi Alex,
    I am having a hard time understanding the below line of code, and would appreciate your help:

    arithmeticFcn fcn = getArithmeticFcn(op);

    Here arithmeticFcn takes two integer values as parameters and return type is int. However, function getArithmeticFcn takes char as an input. Could you please explain how this works?

    • Alex

      Sure. In this case, getArithmeticFcn(op) is just a normal function call. So when we call getArithmeticFcn('+'), the function executes, and the switch statement inside returns a pointer to function add. This return value (function add) is then assigned to variable fcn.

      There's no mismatch here since we're not trying to assign function getArithmeticFcn() (which takes a char parameter) to variable fcn (which expects two int parameters) -- we're assigning the return value from getArithmeticFcn() to variable fcn.

      Make sense?

      • matt senko

        i have the same issue of wrapping my head around this.

        arithmeticFcn fcn = getArithmeticFcn(op);

        i get that we are callnig getArithmeticFcn(op) to get add, subtract, multiply, divide.  what i dont get is how do you know you are getting a pointer to those functions?  can you please try and explain that more to me, im getting stuck on it.

  • Rob G.

    Hi Alex, when I create a function ptr, I have to specify the type and arguments:

    However when I call the function using it with type int and including the arguments
    an error occurs and I have to remove the type to make it work. Why is this so?

    • Alex

      Same as with normal variables. You need the type when you declare the variable so the compiler knows what kind of variable it is (in this case, to tell the compiler it's a pointer to a function taking an integer parameter and returning and integer).

      However, whenever you use the variable, you don't need to include the type again.


  • Rob G.




    worked fine. Thx!

  • Rob G.

    Alex I am trying to create a function ptr to a void type with no arguments. I am getting no errors but the void function characters aren't printing out: its not going to the function.

  • Anonymous

    Hi Alex, a couple of questions:

    1) Why do we need the <static> keyword when declaring <arithmeticArray>? From what I recall, since we are declaring the array outside of any functions, so globally, this restricts the array to internal linkage. Is that the only reason it is used here?

    2) In the for loop, should we not be using a reference for each element in the array since they are structs? Such as

    for (auto &arith: arithmeticArray) {...}



  • Connor

    Hi Alex, WRT Q2a.

    I created my struct as:

    but when I try to initialise the array in Q 2b. as

    it gives me the error "no instance of constructor "arythmeticStruct::arythmeticStruct" matches the argument list ". When I don't give

    a default value then it compiles.
    I don't understand why this is.

    • Alex

      I'm actually a little surprised that this doesn't work. It must be a side effect of how the non-static member initialization was implemented.

      For now, just leave your struct uninitialized. Defaulting it to '+' doesn't really make sense anyway.

  • I have been studying c++ and in particular this tutorial on and off for around 18 months now and find it really helpful.  The first time I read this page I didn't get it at all, but after moving further through the tutorial and returning to various pages as and when I thought it necessary to brush up or advance my skills, I returned to this page and with my added experience and knowledge I suddenly got it.  The problem that I had though was that I was trying to apply this knowledge to member functions and simply could not get it to work.  Unless I have missed it, I cannot find anything about pointers to member functions and would like to see an article on the subject.  I have taken a guess at what the code should be and it works when I compile it, but I would be interested to know if there are any particular downsides.

    Here is my header

    my .cpp file

    and my main

    It appears to me that pointer to member functions have to be grafted on to classes rather than being properly built into their functionality.  Would I be right in assuming this?

    • Alex

      I've taken a note to add a section about pointers to member functions once we cover that member functions are. Pointer to member functions are a little different than pointers to functions because of the implicit "this" pointer parameter. Consequently, pointers to member functions need to contain the class scope qualifier, so the compiler knows what type *this should be. And because a pointer to a member function can only point to functions within the specified class, they aren't used all that often.

      • I am writing something that involved an over the top switch statement to handle user input and it just seemed a little easier to read and maintain using function pointers instead.  I am very much still learning but I am also learning on the fly because I am writing a sizeable program that interfaces with a database.  Somehow my code just seems a bit messier with switch statements than it does with function pointers although perhaps that is just because I have not developed a proper style yet and function pointers discipline me better.

    • Rob G.

      Your history concerning your c++ road traveled was very inspiring. Thanks for sharing that.

  • Daniel P.

    I have a problem with the first part of the quiz. I declared a pointer to an int function:

    then to initialize it  I used a function like this:

    and in main(), I call the function like so:

    The problem is that the pointer operation isn’t innitialized. Anyone knows what I’m doing wrong? Thanks

    • Alex

      The problem here is that you're passing your function pointer by address (which is essentially a pass by value of a pointer). Because of this, when your function changes the value of parameter operation, that isn't passed back through the argument.

      There are two ways to fix this:
      1) Pass your pointer operation by reference
      2) Return your operator as a return value rather than a function parameter

      • Daniel P.

        Thank you very much :)! I got confused because of the way we use arrays in functions but I see now that you have to dereference the pointer to change the value of the parameter. I'd like to take the opportunity to thank you for these very well-made tutorials and all the effort you put into them, it shows.

  • Soul

    Thanks for your great lessons, this website has been awesome.

    Could you also please tell me which lesson to go back to, to understand the answer to 2b? I don't understand how to tie two things together for one choice from the array (for instance: {'+', add}, I don't understand how the array understands the two things together as one element).


    • Alex

      You'll want to review lesson 4.7 -- Structs.

      Each array element of arithmeticArray is a single struct. Each struct contains two values (the op and the fcn).

      So we use the array index to pick the struct, and then use the member selection operator to determine whether we want op or fcn.

  • Bede

    Hi Alex, I found a small typo: "as a argument" --> "as an argument"

  • BlueTooth4269

    Just wanted to point out that your solution for 1a will not work assuming the user enters more than one character for "operator".
    The loop will go around several times with cin feeding off the additional characters, resulting in "Enter operator" being printed several times without a cin happening.

    • Alex

      Yes, typically the quiz answers given here don't have robust error handling, because doing so makes the solutions significantly more complicated and obscures the point of the lessons. It's a tradeoff for simplicity.

  • Shmid

    Oh my god this solves so many problems.

  • Pisces

    This returns 1 with codeblocks. In fact, any function name when passed to cout without () returns 1 on my machine. Why?

    • Alex

      It looks like GCC is converting the function pointer to a boolean before printing it (hence the value of 1). Visual studio prints it as a hex value. My guess is that GCC is interpreting the C++ standard strictly, whereas Visual Studio is doing something non-standard to better reflect the programmer's likely intent.

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