when assigning a function pointer, what is the difference between these two notations?

Question

Consider following code which declares a function pointer, and points it to a function.

void MyIntFunction( int x ){
    std::cout << x << '\n';
}

void ( *funcPtr )( int ) = MyIntFunction;
void ( *funcRef )( int ) = &MyIntFunction;


(*funcPtr)( 2 );
(*funcRef)( 2 );

This code runs fine in my Xcode, and the question is, when assigning the function pointer, what is the difference between MyIntFunction and &MyIntFunction

"assigning reference"? What reference? There are no references in the code you posted. — AnT stands with Russia, Mar 23 '16 at 19:51
Your terminology is a little confused. `T* a = &b` is a pointer to `b`, not a reference. `T& a = b` is a reference to `b`. — Jonathan Potter, Mar 23 '16 at 19:57
@JonathanPotter yes, you're right, I edited the question , thank you : ) — Curtis2, Mar 23 '16 at 20:01

SergeyA · Answer 1 · 2016-03-23T20:02:40.123

3

Formally MyIntFunction is a function type, and &MyIntFunction is a pointer-to-function type. However, function type decays to the pointer-to-function type in almost all contexts, so there is no real difference between using MyIntFunction and &MyIntFunction in this context.

Please note, there is no reference anywhere in the posted code. Both funcPtr and funcRef are pointers-to-function.

edited Mar 23 '16 at 20:02

answered Mar 23 '16 at 19:49

SergeyA

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Indeed, the `&` just takes the address of the function – kmdreko Mar 23 '16 at 19:51
I see. Thank you so much. I will keep the variable name funcRef unchanged, so that other people cloud notice this :) – Curtis2 Mar 23 '16 at 19:54
2

Also, because of the peculiarities of function pointer decay, `(*funcPtr)(2)` is the same as `funcPtr(2)` which is also the same as `(*******funcPtr)(2)` – kmdreko Mar 23 '16 at 19:56
@vu1p3n0x, good and funny point. – SergeyA Mar 23 '16 at 20:05

when assigning a function pointer, what is the difference between these two notations?

1 Answers1