pointer to function. why doesn't this compiler

[2boxers]

Can somebody tell me why the following program fails to compile?
I am using gcc-3.4.0.

#include <iostream>
using namespace std;

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

int subtraction (int a, int b)
{ return (a-b); }

int (*minus)(int,int) = subtraction;

int operation (int x, int y, int (*functocall)(int,int))
{
int g;
g = (*functocall)(x,y);
return (g);
}

int main ()
{
int m,n;
m = operation (7, 5, addition);
n = operation (20, m, minus);
cout <<n;
return 0;
}

here is what i get when i try to compile:
g++ pointer4.cpp -o pointer4

pointer4.cpp: In function `int main()':
pointer4.cpp:23: error: `minus' undeclared (first use this function)
pointer4.cpp:23: error: (Each undeclared identifier is reported only
once for each function it appears in.)

It would seem to me that minus is being declared in line 10. i.e.
minus is being declared as a pointer to a function that accepts 2 int
arguments, returns type int, and is being assigned the address of
subtraction.

Thanks in advance for any suggestions,
ChasW

 

[Ron Natalie]

Can somebody tell me why the following program fails to compile?
I am using gcc-3.4.0.

#include <iostream>
using namespace std;

Try changing the above line to
using std::cout;

 

[Juergen Heinzl]

Can somebody tell me why the following program fails to compile?
I am using gcc-3.4.0.

#include <iostream>
using namespace std;

[-]
Bad idea, because ...

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

int subtraction (int a, int b)
{ return (a-b); }

int (*minus)(int,int) = subtraction;

int operation (int x, int y, int (*functocall)(int,int))
{
int g;
g = (*functocall)(x,y);
return (g);
}

int main ()
{
int m,n;
m = operation (7, 5, addition);
n = operation (20, m, minus);
cout <<n;
return 0;
}

here is what i get when i try to compile:
g++ pointer4.cpp -o pointer4

pointer4.cpp: In function `int main()':
pointer4.cpp:23: error: `minus' undeclared (first use this function)
pointer4.cpp:23: error: (Each undeclared identifier is reported only
once for each function it appears in.)

[-]
.... minus is also a STL function object.

Thanks in advance for any suggestions,

[-]
Either don't use using namespace std a./o. put your stuff into a separate
namespace (you know what namespaces are good for now ;-) ) or use a
different naming scheme like minus_( ... ) of minus_fp( ... ).

The namespace solution is the most flexible, though,
Juergen

 

[2boxers]

Try changing the above line to
using std::cout;

Since posting I realized that minus was the name of a template
function in one of the compiler's header files.

gcc-3.2.2 was kind enough to give an error that called my use of
"minus" ambiguous and then pointed out that minus was a function being
used elsewhere as well.

gcc-3.4.0 erroneuosly says that the "minus" is undeclared which is
just wrong.

using std::cout; works as does changing the name minus to Minus as I
realized just before my post.

Thanks,
ChasW