writing an expression with #define

[Guest_imported]

I actually have two questions.
#define x(wal) cout<<(wal) //why do I also
have
to declare x as
variable?

Next question:Author of a book I'm reading says I shouldn't use macros that generate expressions and than gives an example:
#define max(x,y) x>y ? x:y
result=max(myval++,99);
Here It says myval is incremented twice and if I was using a function it would not be incremented twice.Well,I did also used this expresion in a function and it also was incremented twice.Can someone explain to me why is this?


thank you

 

[SeekerOfKnowledge]

First, I do not think you need to declare x as a variable.
#define x(wal) cout<<wal

and later
int h;
x(h);
will output h to cout, no need to re-declare x.

Second, the reason the variable x is inc'd twice is because
the define uses text replacement and plants &quot;x++&quot; whenever
&quot;X&quot; is declared in the #define statement, each such time causes x to be inc'd

 

[Guest_imported]

Thank you for your replie.I guess you're right on the first one,but I still don't understand why variable x is also inc'd twice when I used that expresion in a function?

 

[NiceButDim]

As seeker says ‘the define uses text replacement’, therefore
result=max(myval++,99);
becomes
result=myval++>99 ? myval++:99
in this example myval will in fact only be incremented once if it’s value is less than (or equal to) 99, otherwise it will be incremented twice.

 

[Hellburn]

Check that you are not using the define in the function. Where ever you use the define, whether in the main code or a function, the anomalous effect will be the same.

What the book is saying is to rather write a function to do the same thing, ie:
int max(int a, int b)
{
return( a>b ? a : b );
}

If this was to be called as result = max(myval++,99) the initial value of myval would be passed to the function. After myval has been passed it would then be incremented.

 

[Guest_imported]

Thank you for helping me out