There is a diff in how C and C++ sees this.
For C add( ) is an function taking undefined argument(s), so you can
send it whatever you want. C++ on other hand will see add( ) as add(
void ), and would complain. To say that add( ) would be equal to add(
int(s) ) is bogus thou.
In real life this does not matter. This is a completely academic
question. Thou *I think* even the C compiler should have given a
warning. (Note -Wall does not turn on all warnings, just almost all).
// Jarmo
--
Unless you're writing a compiler this does not matter. Even if an int
argument in implicitly used it has no meaning to the programmer. Since
void is a well defined type, although an incomplete one, I have
doubts that int is used internally. I simply can't see the rationale
behind that (but I'd be happy to be enlightened). Could you please
try to transport your collegue's argumentation?
Here is what he sent me -
#include <stdio.h>
void add ()
{
printf ("inside function: add. \n");
return;
}
int main (void)
{
/* call function add with some parameters */
add (5, 1);
system ("PAUSE");
return (0);
}
How can this work, if not specifying any argument, is equivalent to
specifying as void?
However, one thing I was able observe was that it accepts any kind of
arguments, and also any number of arguments, as against his theory of
only accepting "int" types.
I even tried compiling with "-Wall" option to see if any warnings are
being thrown by the compiler, but found to my disappointment that there
was none.
Am I fundamentally going wrong in my understanding of functions?
_z33
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