MACRO question

**baverlyhills2** · 02-15-2012

Hey guys, I am studying some C from K&R book and I just play around with these. Anyway, I have a question.

Lets say this is a macro I have gotten.
#define ADD(X,Y)(X+Y)

(I know it will be wiser to code it: #define ADD(X,Y)((X)+(Y)) )

so when I get:
int x = 100, y = 3;
int z = ADD(x,++y);
printf("x %d y %d z %d\n",x,y,z);
x = 100; y = 3;
z = ADD(x+,+y);
printf("x %d y %d z %d\n",x,y,z);

why do I get a different results? I mean, the compiler during the pre-processing should replace ADD(x,++y) and ADD(x+,+y) with x+++y. Thus, they should give the same result.

Thanks in adv.

**TheBigH** · 02-15-2012

I believe (though I may be mistaken) that a sequence of three consecutive plusses is undefined because it is ambiguous. The compiler could legitimately interpret x+++y as (x++) + y or x + (++y), or even (x+)+(+y) if it wants to. And it can interpret them differently even within the same program.

**laserlight** · 02-15-2012

Originally Posted by TheBigH

I believe (though I may be mistaken) that a sequence of three consecutive plusses is undefined because it is ambiguous. The compiler could legitimately interpret x+++y as (x++) + y or x + (++y), or even (x+)+(+y) if it wants to. And it can interpret them differently even within the same program.

You're mistaken: greedy parsing is required, so x+++y must be interpreted as x++ + y.

**TheBigH** · 02-15-2012

OK, I stand corrected. But then why is x not incremented in the second example?

If I do z=x+++y directly without going through the macro I get x=101, y=3, z=103 as I would expect.

**laserlight** · 02-15-2012

Originally Posted by TheBigH

But then why is x not incremented in the second example?

Testing with gcc 4.4.4, I get:

Code:

x 100 y 4 z 104
x 100 y 3 z 103

I removed the include and printf statements, and got the compiler to perform preprocessing only. The result:

Code:

int main(int argc, char *argv[])
{
    int x = 100, y = 3;
    int z = (x+ ++y);
    x = 100; y = 3;
    z = (x+ + +y);
    return 0;
}

I believe that this line:

Code:

z = (x+ + +y);

is equivalent to:

Code:

z = x + (+(+y));

Hence neither x nor y is changed.

**baverlyhills2** · 02-15-2012

Thanks a lot for the help.
So from what you're saying, during the pre-processing, the compiler separate the two operands?
I mean: ADD(x++,y) will be x++ + y and not x+++y, even though I haven't mentioned any space between those in the macro?

In addition, can you explain or give me a link on how to run pre-process (and seeing the results) without compiling, it sounds interesting.

Thanks in adv.

**manasij7479** · 02-15-2012

Originally Posted by baverlyhills2

In addition, can you explain or give me a link on how to run pre-process (and seeing the results) without compiling, it sounds interesting.

I believe the cpp command/program does it.

**laserlight** · 02-15-2012

Originally Posted by baverlyhills2

So from what you're saying, during the pre-processing, the compiler separate the two operands?

Yes.

Originally Posted by baverlyhills2

In addition, can you explain or give me a link on how to run pre-process (and seeing the results) without compiling, it sounds interesting.

Refer to your compiler documentation. For example, with gcc, the -E option should be used.

**baverlyhills2** · 02-15-2012

Thanks for everything! (YES, I am using GCC on Linux).

I have another question if you don't mind. It is related to bitwise, though.
Since char is 8 bits, then it can store only 0xFF to 0x00 numbers.
Then, why the output is FFFFFFFE when I am compiling and running the code:
char x = 0x01;
printf("%x\n",~x);
(~x) has to be FE, hasn't it?

Thanks in adv.

**laserlight** · 02-15-2012

Integer promotion happens when you do the ~x.

**baverlyhills2** · 02-15-2012

Alright, that's one thing I haven't known (Doing some research on the net). Thanks for that.

Well, that's all for now. THANKS you all for your time, help and support. I took a lot from your replies, guys!

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