«

void Blah(int k) { k; }


int main()
{
int a;
int b;

Blah( ++a, ++b );
}


Is there any rules to say in what way a comma is interpreted within function
arguments?

If I were to guess, I'd say that it's always interpreted as an argument
seperator, except when enclosed in its own brackets, as in:


void Blah(int k) { k; }


int main()
{
int a;
int b;

Blah( (++a, ++b) );
}


But then again I'm just guessing!


-JKop

"JKop" <> schrieb im Newsbeitrag
news:ZgIid.40966$...
>
> void Blah(int k) { k; }
>
>
> int main()
> {
> int a;
> int b;
>
> Blah( ++a, ++b );
> }
>
>
> Is there any rules to say in what way a comma is interpreted within
> function
> arguments?
>
> If I were to guess, I'd say that it's always interpreted as an
> argument
> seperator, except when enclosed in its own brackets, as in:
>
>
> void Blah(int k) { k; }
>
>
> int main()
> {
> int a;
> int b;
>
> Blah( (++a, ++b) );
> }
>
>
> But then again I'm just guessing!

The , is an operator with highest prescendence (will be evaluated at
very last, even after braces). It simply divides operations, but does
not change the lvalue nor the rvalue.

Common example:
for (int i=0, j=5; i<5; i++, j++) {}

Was that your question?
-Gernot

Gernot Frisch posted:

>
> "JKop" <> schrieb im Newsbeitrag
> news:ZgIid.40966$...
>>
>> void Blah(int k) { k; }
>>
>>
>> int main()
>> {
>> int a;
>> int b;
>>
>> Blah( ++a, ++b ); }
>>
>>
>> Is there any rules to say in what way a comma is interpreted within
>> function arguments?
>>
>> If I were to guess, I'd say that it's always interpreted as an
>> argument seperator, except when enclosed in its own brackets, as in:
>>
>>
>> void Blah(int k) { k; }
>>
>>
>> int main()
>> {
>> int a;
>> int b;
>>
>> Blah( (++a, ++b) ); }
>>
>>
>> But then again I'm just guessing!
>
> The , is an operator with highest prescendence (will be evaluated at
> very last, even after braces). It simply divides operations, but does
> not change the lvalue nor the rvalue.
>
> Common example:
> for (int i=0, j=5; i<5; i++, j++) {}
>
> Was that your question?
> -Gernot

Consider a function that takes ONE argument. You call it as so:


Func( a, b );


What are the rules to dictate whether the function has been supplied:

A) Two arguments

- First argument: a
- Second argument: b


B) One argument

- First argument: a, b


-JKop

"JKop" <> wrote in message
news:ZgIid.40966$...
>
> void Blah(int k) { k; }
>
>
> int main()
> {
> int a;
> int b;
>
> Blah( ++a, ++b );
> }
>
>
> Is there any rules to say in what way a comma is interpreted within
function
> arguments?
>
> If I were to guess, I'd say that it's always interpreted as an argument
> seperator, except when enclosed in its own brackets, as in:
>
>
> void Blah(int k) { k; }
>
>
> int main()
> {
> int a;
> int b;
>
> Blah( (++a, ++b) );
> }
>
>
> But then again I'm just guessing!
>

I think you guess right, but then again I am only guessing.

The rule you are talking about definitely applies to macro invocations I
don't see why it shouldn't apply to function calls as well.

john

JKop wrote:
rnot
>
>
> Consider a function that takes ONE argument. You call it as so:
>
>
> Func( a, b );
>
>
> What are the rules to dictate whether the function has been supplied:
>
> A) Two arguments
>
> - First argument: a
> - Second argument: b
>
A. The language grammar says that the comma in the above is always
the punctuation separating the function call. It doesn't matter
how Func is declared (it can't really consider that anyhow, the
whole overload process depends on knowing the function signature
first).

',' in an argument list is not considered the same as ',' in general usage. So
if you use it in an argument list, it separates the arguments, unless you do as
you suggest and put it in parenthesis (there are a number of such
context-dependent usages for symbols in the language)

David

JKop wrote:

> void Blah(int k) { k; }
>
> int main()
> {
> int a;
> int b;
>
> Blah( ++a, ++b );
> }
>
> Is there any rules to say in what way a comma is interpreted within function
> arguments?
>
> If I were to guess, I'd say that it's always interpreted as an argument
> seperator, except when enclosed in its own brackets, as in:
>
> void Blah(int k) { k; }
>
> int main()
> {
> int a;
> int b;
>
> Blah( (++a, ++b) );
> }
>
> But then again I'm just guessing!
>
> -JKop