How to make a function return a pointer to a function? (C++)
C++Return ValueFunction PointersC++ Problem Overview
I'm trying to make a function that takes a character, then returns a pointer to a function depending on what the character was. I just am not sure how to make a function return a pointer to a function.
C++ Solutions
Solution 1 - C++
int f(char) {
return 0;
}
int (*return_f())(char) {
return f;
}
No, seriously, use a typedef :)
Solution 2 - C++
#include <iostream>
using namespace std;
int f1() {
return 1;
}
int f2() {
return 2;
}
typedef int (*fptr)();
fptr f( char c ) {
if ( c == '1' ) {
return f1;
}
else {
return f2;
}
}
int main() {
char c = '1';
fptr fp = f( c );
cout << fp() << endl;
}
Solution 3 - C++
Create a typedef for the function signature:
typedef void (* FuncSig)(int param);
Then declare your function as returning FuncSig:
FuncSig GetFunction();
Solution 4 - C++
Assuming int f(char)
and ret_f
which returns &f
.
C++98/C++03 compatible ways:
-
Ugly way:
int (*ret_f()) (char) { return &f; }
-
With typedef:
typedef int (sig)(char); sig* ret_f() { return &f; }
or:
typedef int (*sig_ptr)(char); sig_ptr ret_f() { return &f; }
Since C++11, in addition we have:
-
with
decltype
:decltype(&f) ret_f() { return &f; }
-
trailing return type:
auto ret_f() -> int(*)(char) { return &f; }
or:
auto ret_f() -> decltype(&f) { return &f; }
-
typedef
withusing
:using sig = int(char); sig* ret_f() { return &f; }
or:
using sig_ptr = int (*)(char); sig_ptr ret_f() { return &f; }
C++14 adds:
-
auto
deduction:auto ret_f() { return &f; }
Solution 5 - C++
In C++11 you can use trailing return types to simplify the syntax, e.g. assuming a function:
int c(int d) { return d * 2; }
This can be returned from a function (that takes a double to show that):
int (*foo(double e))(int)
{
e;
return c;
}
Using a trailing return type, this becomes a bit easier to read:
auto foo2(double e) -> int(*)(int)
{
e;
return c;
}
Solution 6 - C++
Here is how to do it without using a typedef:
int c(){ return 0; }
int (* foo (void))(){ //compiles
return c;
}
Solution 7 - C++
Syntax for returning the function:
return_type_of_returning_function (*function_name_which_returns_function)(actual_function_parameters) (returning_function_parameters)
Eg: Consider the function that need to be returned as follows,
void* (iNeedToBeReturend)(double iNeedToBeReturend_par)
{
}
Now the iNeedToBeReturend function can be returned as
void* (*iAmGoingToReturn(int iAmGoingToReturn_par))(double)
{
return iNeedToBeReturend;
}
I Felt very bad to learn this concept after 3 years of professional programming life.
Bonus for you waiting down for dereferencing function pointer.
Example for function which returns the function pointer is dlopen in dynamic library in c++
Solution 8 - C++
typedef void (*voidFn)();
void foo()
{
}
voidFn goo(char c)
{
if (c == 'f') {
return foo;
}
else {
//..
}
// ..
}
Solution 9 - C++
This is the code to show return of a function pointer. You need to define the "function signature" to return first:
int returnsOne() {
return 1;
}
typedef int(*fp)();
fp returnsFPtoReturnsOne() {
&returnsOne;
}
In your specific case:
fp getFunctionFor(char code) {
switch (code) {
case 'a': return &functionA;
case 'b': return &functionB;
}
return NULL;
}
Solution 10 - C++
Check out this site - http://cdecl.org
Helps you convert english to C declarations and back!
Cool Stuff!
This link decodes the example in erikallen's answer. int (*return_f())(char)
Solution 11 - C++
Easiest way is to typedef the pointer-to-function type you want, and then use that
typedef void (*fnptr_t)(int, int);
fptr_t myfunc(char *) { ....
Solution 12 - C++
I prefer returning objects and call the operator(). This way your function can return an interface and all classes can inherit from this. That is, if you're using C++ and not C.
Then you can use the parametrized factor method to return the objects based on your input.
Solution 13 - C++
I'm assuming C here (no objects) :) :
// Type of function which takes a char and returns an int:
typedef int (*Func)(char a);
// An example of the function you're trying to return and which does something
// with char:
int exampleFunc(char a)
{
return (int)(a + 42);
}
// The function returning the pointer to a function:
Func *returnAfunc(void)
{
return exampleFunc;
}
Solution 14 - C++
Something like this
#include <iostream>
typedef char (*fn_ptr_t)(char);
char a_fn(char c)
{
return c + 1;
}
char b_fn(char c)
{
return c + 2;
}
fn_ptr_t
return_function(char c)
{
fn_ptr_t result = 0;
switch (c)
{
case 'a':
result = a_fn;
break;
case 'b':
result = b_fn;
break;
}
return result;
}
int
main()
{
fn_ptr_t fn = return_function('a');
std::cout << "a(l) = " << (fn)('l') << std::endl;
return 0;
}