CODEWITHSUNDEEP
cmacros
ZULHELMI FAUZI
ZULHELMI FAUZI

Reputation: 23

Confusing on output result

I just curious why the output is 4 instead of 6 . Any of you can help explain their proccess?

#include <stdio.h>
#define AMT1 a+a+a
#define AMT2 AMT1 - AMT1
main()
{
int a=1;
printf (“Amount is %d\n”,AMT2);
}

Thanks

Upvotes: 2

Views: 59

Answers (4)

JoelFan
JoelFan

Reputation: 38724

AMT2 = AMT1 - AMT1
AMT2 = a+a+a - a+a+a
AMT2 = 1+1+1 - 1+1+1
AMT2 = 3 -     1 + 1 + 1
AMT2 = 2         + 1 + 1
AMT2 = 3             + 1
AMT2 = 4

The first rule of macros is: you need parentheses around everything. You want:

#define AMT2 (AMT1) - (AMT1)

Upvotes: 1

mr.loop
mr.loop

Reputation: 1005

a simple

AMT2 (AMT1) - (AMT1)

and/or

AMT1 (a+a+a)

should do your work.

Upvotes: 0

Barmar
Barmar

Reputation: 782509

After you do the macro replacements, the code becomes

printf("Amount is %d\n", 1 + 1 + 1 - 1 + 1 + 1);

Addition and subtraction are left-associative, so the expression is equivalent to

((((1 + 1) + 1) - 1) + 1) + 1)

If you want the macros to act more like variables, you need to wrap the expansions in parentheses:

#define AMT1 (a+a+a)
#define AMT2 (AMT1 - AMT1)

Upvotes: 0

Mustafa Quraish
Mustafa Quraish

Reputation: 694

Macros like this do literal text replacement. When you compile your code, AMT2 is literally being replaced by AMT1, so you're effectively running:

printf (“Amount is %d\n”,a+a+a - a+a+a);

You can verify this yourself by looking at the output of the preprocessor (after the macros have been replaced) by running

gcc file.c -E

To solve your problem, you can either:

  1. Make a function instead
  2. Put parenthesis around all your macro definitions (this is always recommended)

So your code would look like:

#include <stdio.h>

#define AMT1 (a+a+a)
#define AMT2 (AMT1 - AMT1)

int main() {
    int a = 1;
    printf ("Amount is %d\n", AMT2);
}

Upvotes: 1

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