leo1502

I need to write a macro which changes the max value of 2 given integers using this syntax:
SmartMax(x,y)=100

I would expect that this code won't compile or run right.
the ? : operator is suppoesd to make a copy of *w or *z so there is no reason for x or y to change.

#include <stdio.h>
#define SmartMax(x,y) int *z,*w; z=&x; w=&y; (y>x) ? *w : *z

int main()
{
int x=2,y=6;
SmartMax(x,y)=100;
printf("%d\n",y);
return 0;
}


but, I found out that the code works fine when I compile using gcc
in g++ though, it doesn't do what I wanted and returns y 6

what's goin on here?

DaWei

You want free advice, yet you don't feel you should read the forum's rules/FAQ to get a feel for the community you want the advice from? Shame on you. If you decide to read it, keep a sharp eye out for the part about 'code tags'.

Since a macro is a preprocessed thangy that is substituted into your source before the compiler sees it, you might hand-expand your macro into your code in the same way and see for yourself why it doesn't perform up to expectations.

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InfoGeek

It's equivalent to writing:
Can cause problems if you write the macro in the middle of the code as most of the C compilers like all the declarations to be at the start of a function.

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leo1502

Thanks, but you haven't really answered my question.

I know what a macro is, that's not the problem here.
why does the code work right on C but not on C++?

looks like C returns the real value X (or Y) and not a copy of it.

About the variables in the middle of the code, is there a way to solve this problem without using new variables, yet preserving the syntax SmartMax(x,y)=100 ?

nnxion

Since you are using gcc, you can choose to only do the preprocessing and then look at the result to see if that's what you want:
I'm not sure what you mean by that. What do you need to do? Find the highest value, x or y?

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leo1502

How does that answer my question?
I already know it works right with gcc

by SmartMax(x,y)=10

I mean if y>x then y will be 10, else x will be 10
the greater number will change to 10

Narue

>How does that answer my question?
By showing you how to find the problem, genius. Did you expect us to fix all of your problems and hand you a shiny new macro that perfectly does everything you want?

>I already know it works right with gcc
I hope you didn't expect it to work right elsewhere. The result of the conditional operator can't be used as an lvalue. It's not guaranteed to work on other compilers, or even different versions of gcc.

By the way, that macro is pretty much guaranteed to be error prone and frustrating no matter how you write it. Are you trying to duplicate the effect of a C++ function returning a reference?

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leo1502

no, I didn't want you to hand me a shiny new macro.
just tried to figure out what's goin' on there.
someone told me C returns the right lvalue there when I use a dynamic var like with w and z.
is that true?

Narue, what are you so pissed about anyway

OopSheMishtamesh

this is the most crappy piece of code i seen in the past month , it shouldn't compile at all ...
this makes a copy of some value and puts there 100 ..

this doesn't make any sense ..

or maybe the ? : operator doesn't make a copy , and returns the X or the Y (don't ask me how)

if the last assumption is correct, this should work too (y>x) ? y : x, if not, we have some real issues here...


cheers m8s

grumpy

Well, yeah, it will work if you do other things right.

But it is also easily broken.

Try this;
and watch your compiler complain bitterly.

Formally, the result of ?: can be an lvalue if both possible results are lvalues. You would be better off avoiding introducing the variables w and z within your macro; they actually serve no purpose and, as in my example here, open up all sorts of opportunities to write broken program.