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General discussion (archive)Need help understanding the calculation...(7 posts)solved(7 posts)solved
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Posted November 30, 2009
I'm having an exam tomorrow... C language... I am having a little problem understanding the logic of this program. Can anyone explain me how the things are calculated? In particular, the j variable. I don't understand how it becomes 20.
#include <stdafx.h>
#include <stdio.h>
#define A 5
#define MACRO(j,i) i=j/i
#define PR printf("i=%d, j=%d, k=%d\n",i,j,k);
void main()
{
int i=8,j=2,k;
int kraj;
k=MACRO(i,j)*A;
PR;
j/=A;
k+=MACRO(++j*A,i);
PR;
scanf_s("%d",&kraj);
}
Post edited November 30, 2009 by KavazovAngel
This question / problem has been solved by Barefoot_Monkeyimage
Posted November 30, 2009
I don't see how it could possibly be 20. I think I don't understand the code fully. Can you give us the printout of the two calls to the PR macro?
Posted November 30, 2009
int i=8,j=2,k;
int kraj;
k=MACRO(i,j)*A;
// expand MACRO to get:
// k=j=i/j*A
// k=j=8/2*5
// k=j=4*5
// k=j=20
// so both k and j are now 20
PR;
// this should print "i=8,j=20,k=20
j/=A;
// j = j / 5, so j is now 4
k+=MACRO(++j*A,i);
// *sigh*, don't they have something more useful to test you on than deliberate obfuscation?
// oh well - you'll probably get something like this in your test then if this is how they think,
// so once you understand what's going on you should practice reading the program
// on your own a few times.
//
// k += i = ++j*5 / i
// this basically says j = j + 1 and i = j*5 / i and k += i
// in other words, j = 5, i = 5*5 / 8 and k += i
// so now j is 5, k is 23, and i is 3
There's plenty of room for me to have made a mistake there, so run the program and see if my results are correct.
I hope this helps you to understand what's going on.
Posted November 30, 2009
avatar
Barefoot_Monkey: k=MACRO(i,j)*A;
// expand MACRO to get:
// k=j=i/j*A
// k=j=8/2*5
// k=j=4*5
// k=j=20
// so both k and j are now 20

Okay, I've never encountered the macro function before. It makes absolutely no sense to me that the multiplication by A outside the macro call should influence the assignment taking place inside it. What an extraordinarily weird bit of functionality. But then I never learned C, my school taught programming using C++.
Posted November 30, 2009
avatar
Barefoot_Monkey: k=MACRO(i,j)*A;
// expand MACRO to get:
// k=j=i/j*A
// k=j=8/2*5
// k=j=4*5
// k=j=20
// so both k and j are now 20
avatar
Wishbone: Okay, I've never encountered the macro function before. It makes absolutely no sense to me that the multiplication by A outside the macro call should influence the assignment taking place inside it. What an extraordinarily weird bit of functionality. But then I never learned C, my school taught programming using C++.

MACRO is not a function (persay) but instead an actual macro/preprocessor directive (I think it counts as one of those).
Basically, when you see
#define A 5
That means that every time the preprocessor sees "A" it replaces it with "5", BEFORE compiling.
So MACRO(i,j) becomes i=j/i , BEFORE compiling.
Macros are very useful (even though I almost always would suggest a typedef or an inline if you are doing anything more than a #ifndef), but they are also very difficult to debug and tend to do little but obfuscate code.
Posted November 30, 2009
avatar
Wishbone: Okay, I've never encountered the macro function before. It makes absolutely no sense to me that the multiplication by A outside the macro call should influence the assignment taking place inside it. What an extraordinarily weird bit of functionality. But then I never learned C, my school taught programming using C++.
avatar
Gundato: MACRO is not a function (persay) but instead an actual macro/preprocessor directive (I think it counts as one of those).

In this context, MACRO is just the name of the macro being defined. I meant that I didn't know about the macro functionality, meaning the ability to define macros in the syntax.
avatar
Gundato: Macros are very useful (even though I almost always would suggest a typedef or an inline if you are doing anything more than a #ifndef), but they are also very difficult to debug and tend to do little but obfuscate code.

Yeah, I'd certainly prefer an actual function. I can imagine the difficulty in debugging code with lots of macros in it. Especially with nested macro calls (I just read about that).
Posted November 30, 2009
Thank you for the help, guys. :) It is much clearer and easier now. :)
Wish me luck tomorrow. :p
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General discussion (archive)Need help understanding the calculation...(7 posts)solved(7 posts)solved
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