Post by thread:function pointer syntax in C/C++

----- Original Message ----- From: "Ashley Roll" <.....ashKILLspam@spam@DIGITALNEMESIS.COM> To: <PICLISTKILLspamMITVMA.MIT.EDU> Sent: Wednesday, December 11, 2002 1:18 AM Subject: Re: function pointer syntax in C/C++ {Quote hidden}> Hi Brendan > > The "nicest" way is to use typedefs: > > int foo(int bar); // function to point to0 > > typedef int (* MyFuncPtrType)(int /* parameters here */); > > Then you can say: > > MyFuncPtrType pFunc = &foo; I'm not quite sure this is right. As far as I know, you don't need the "&" before the function name. It may still work, I haven't tried it. I'm basically adapting the following from my C reference manual: /* This is the function prototype */ int foo(int); /* This is will be the pointer to our function above, called p */ int (*p) (int); /* This is a variable to hold the return value of the function */ int answer; /* This will assign the address of the function foo to p */ /* The function name itself is actually a pointer to the function */ p = foo; /* Now using the function pointer we can call the function foo */ answer = (*p) (4); Look at the above function pointer declaration, the first word is "int". This is the return type of your function. Next is (*p). This is the name of the variable, you can call it whatever you want. We could have said int (*foo_p) (int) if we wanted, its just the variable name. The final part are the parameters of the function. Say we had another function we wanted to make a pointer to called foo2, and say it looked like: int foo2(int, int, float); Then the declaration of the pointer variable would look like this: int (*foo2_p) (int, int, float); We would then assign the pointer like this: foo2_p = foo2; And we could call the function like this: answer = (*foo2_p) (1, 2, 3.0); Again, foo2_p is just the name of the variable, but you can call it whatever you want. This is the way defined by the ANSI C standard, and is guaranteed to work. Ash's answer may still be correct, I didn't check. I really hope this helps. Good luck in your coding. Ricky Hussmann > > {Original Message removed}

Hi Ricky, Interesting, I just tried both your way and mine and they seem to work. I've always just done it the way I said.. #include <stdio.h> typedef int (*FooPtrType)(int); int foo(int bar) { printf("Foo: %d\r\n", bar); return 0; } main() { FooPtrType pFoo1 = foo; FooPtrType pFoo2 = &foo; (*pFoo1)(1); pFoo2(2); } I like to use mine because it makes it explicit that you are taking the address of the function. Also there is no need to dereference the pointer when calling the function. This just makes reading the code that much more difficult. Assuming you name you pointers in a distinct way.. Guess its just preferences. Cheers, Ash. --- Ashley Roll Digital Nemesis Pty Ltd http://www.digitalnemesis.com Mobile: +61 (0)417 705 718 > {Original Message removed}

> > Hi Brendan > > > > The "nicest" way is to use typedefs: > > > > int foo(int bar); // function to point to0 > > > > typedef int (* MyFuncPtrType)(int /* parameters here */); > > > > Then you can say: > > > > MyFuncPtrType pFunc = &foo; > >I'm not quite sure this is right. As far as I know, you don't need the "&" >before the function name. It may still work, I haven't tried it. Thanks for that, guys, it works well. As to the '&', I'm not sure if it's accepted, but it works fine without it. --Brendan -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email .....listservKILLspam.....mitvma.mit.edu with SET PICList DIGEST in the body

The "C" standard says that a standalone function name (without trailing parens) 'degenerates' to the address of the function. So, myFooPtr = foo; and myFooPtr = &foo; do the same thing in ANSI C. Bob Ammerman RAm Systems ----- Original Message ----- From: "Ricky Hussmann" <EraseMEgargoylespam_OUTTakeThisOuTIOLINC.NET> To: <PICLISTspam_OUTMITVMA.MIT.EDU> Sent: Wednesday, December 11, 2002 1:41 AM Subject: Re: function pointer syntax in C/C++ {Quote hidden}> ----- Original Message ----- > From: "Ashley Roll" <@spam@ashKILLspamDIGITALNEMESIS.COM> > To: <KILLspamPICLISTKILLspamMITVMA.MIT.EDU> > Sent: Wednesday, December 11, 2002 1:18 AM > Subject: Re: function pointer syntax in C/C++ > > > > Hi Brendan > > > > The "nicest" way is to use typedefs: > > > > int foo(int bar); // function to point to0 > > > > typedef int (* MyFuncPtrType)(int /* parameters here */); > > > > Then you can say: > > > > MyFuncPtrType pFunc = &foo; > > I'm not quite sure this is right. As far as I know, you don't need the "&" > before the function name. It may still work, I haven't tried it. > > I'm basically adapting the following from my C reference manual: > > /* This is the function prototype */ > int foo(int); > > /* This is will be the pointer to our function above, called p */ > int (*p) (int); > > /* This is a variable to hold the return value of the function */ > int answer; > > /* This will assign the address of the function foo to p */ > /* The function name itself is actually a pointer to the function */ > p = foo; > > /* Now using the function pointer we can call the function foo */ > answer = (*p) (4); > > Look at the above function pointer declaration, the first word is "int". > This is the return type of your function. Next is (*p). This is the name of > the variable, you can call it whatever you want. We could have said int > (*foo_p) (int) if we wanted, its just the variable name. > > The final part are the parameters of the function. Say we had another > function we wanted to make a pointer to called foo2, and say it looked like: {Quote hidden}> > int foo2(int, int, float); > > Then the declaration of the pointer variable would look like this: > > int (*foo2_p) (int, int, float); > > We would then assign the pointer like this: > > foo2_p = foo2; > > And we could call the function like this: > > answer = (*foo2_p) (1, 2, 3.0); > > Again, foo2_p is just the name of the variable, but you can call it whatever > you want. This is the way defined by the ANSI C standard, and is guaranteed > to work. Ash's answer may still be correct, I didn't check. > > I really hope this helps. Good luck in your coding. > > Ricky Hussmann > > > > {Original Message removed}

On Tue, Dec 10, 2002 at 09:40:36PM -0800, Brendan Moran wrote: > On most C syntax I do just fine, but function pointers are always a > head-scratcher for me. > > If I wanted to declare a pointer to this function: > > int foo(int bar); > > What would be the syntax for the declaration? > > The best I came up with was this: > > (int)(*)(int) pfoo = &foo; > > but that doesn't compile. It's actually very simple once you realize that the function () have the highest precedence. Couple that with the fact that you want to define a function pointer, and you're in business. Start be defining all of the elements: int *pfoo(int) = foo; // Note that the & is redundant in front of foo This won't compile but is the correct format. The reason is precedence (i.e. 5+3*2 is 11 not 16). pfoo binds to the function () first so pfoo is a function that returns an int pointer. So to change the precedence add parenthesis. pfoo is first an foremost a pointer, so put parens around the * and pfoo so that it binds first. int (*pfoo)(int) = foo; // Works every time! It's just that simple. pfoo must now be a pointer above all else and since it has function parens following it's now a pointer to a function. BAJ -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email RemoveMElistservTakeThisOuTmitvma.mit.edu with SET PICList DIGEST in the body

Ricky Hussmann wrote: {Quote hidden}> ----- Original Message ----- > From: "Ashley Roll" <spamBeGoneashspamBeGoneDIGITALNEMESIS.COM> > To: <TakeThisOuTPICLISTEraseMEspam_OUTMITVMA.MIT.EDU> > Sent: Wednesday, December 11, 2002 1:18 AM > Subject: Re: function pointer syntax in C/C++ > > > Hi Brendan > > > > The "nicest" way is to use typedefs: > > > > int foo(int bar); // function to point to0 > > > > typedef int (* MyFuncPtrType)(int /* parameters here */); > > > > Then you can say: > > > > MyFuncPtrType pFunc = &foo; > > I'm not quite sure this is right. As far as I know, you don't need the "&" > before the function name. It may still work, I haven't tried it. > You are correct b'cos the compiler is able to determine that this is only a address for the function and not a value example int i = 5; int *x = &i; or int i = 5; int *x; x = &5; the variable i is the value "5" while the &i represents the address. {Quote hidden}> > I'm basically adapting the following from my C reference manual: > > /* This is the function prototype */ > int foo(int); > > /* This is will be the pointer to our function above, called p */ > int (*p) (int); > > /* This is a variable to hold the return value of the function */ > int answer; > > /* This will assign the address of the function foo to p */ > /* The function name itself is actually a pointer to the function */ > p = foo; > > /* Now using the function pointer we can call the function foo */ > answer = (*p) (4); > Suprisingly GCC allows answer = (p)(4); // not sure that this is apart of the ANSI C standard {Quote hidden}> > Look at the above function pointer declaration, the first word is "int". > This is the return type of your function. Next is (*p). This is the name of > the variable, you can call it whatever you want. We could have said int > (*foo_p) (int) if we wanted, its just the variable name. > > The final part are the parameters of the function. Say we had another > function we wanted to make a pointer to called foo2, and say it looked like: > > int foo2(int, int, float); > > Then the declaration of the pointer variable would look like this: > > int (*foo2_p) (int, int, float); > > We would then assign the pointer like this: > > foo2_p = foo2; > > And we could call the function like this: > > answer = (*foo2_p) (1, 2, 3.0); > > Again, foo2_p is just the name of the variable, but you can call it whatever > you want. This is the way defined by the ANSI C standard, and is guaranteed > to work. Ash's answer may still be correct, I didn't check. > > I really hope this helps. Good luck in your coding. > > Ricky Hussmann > > > > {Original Message removed}