passing the input right through to the function

[bob323]

I may be just experiencing a moment of stupidity, but is there a way to pass input from the keyboard right through to a function without putting it first in a variable?
normally I would go
cin>>var_1;
function(var_1);

what I would rather do though is something like
cin >> function;

any ideas??

[kedaman]

Unless the function is overloaded you could overload the istream operator >> for function pointers, but i'm not enteriely sure if it's possible.

[CornedBee]

you could use the C stdio functions. They have the input as return values.

[Zaei]

Interesting. This works...

typedef VOID f(int x);
operator >> (istream_withassign lhs, f* rhs)
{
      int x;
      lhs >> x;
      f(x);
}

VOID foo(int x)
{
    cout << x * x;
}

int main
{
   cin >> foo;
   return 0;
}

Z.

[CornedBee]

Really???
I would have said that the typedef instruction can't be compiled.
This is a very atypical function pointer declaration. I would have used
typedef void (*pf)(int);
and pf as argument type of the operator overload...
Strange...

[jim mcnamara]

stdio -

getch()
getche()
getc()

using stdin - this work for you?

[CornedBee]

Originally posted by Zaei

Besides, a pointer should tell you it's a pointer. So, had I typedefed the way you said, It would have been a "LPf" =).

Yeah, I wrote pf.
lp is a relict from 16-bit programming when there were near (16-bit) pointers and far (32-bit) pointers. (or something like that).
the l stands for long and means a far pointer. In 32-bit enviroment, all pointers are 32-bit.
I still think it's interesting that you can give a function type a name and not only function pointer types.

[Zaei]

Yeah, I am still running DOS on my laptop, so i get to worry about near and far pointers =). Ive never actually tried to pass a typedef int f(); to a function. I wonder if it would work...

Z.