Go4Expert - Why this program outputs 225 istead of 1 ?

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Why this program outputs 225 istead of 1 ?

Code:

#include<stdio.h>

#define SQR(x) x * x



int main()

{



    printf("%d", 225/SQR(15));



}

why this program outputs 225 instead of 1 (according to calculation).

Re: Why this program outputs 225 istead of 1 ?

Precedence is not working as you expect with the MACROS.

Try

Code:

printf("%d", 225/(SQR(15)));

Re: Why this program outputs 225 istead of 1 ?

The macro is substituted like the following ,

Code:

#include<stdio.h>

#define SQR(x) x * x

int main()

{



    printf("%d", 225/15*15);



}

Because of the operator precedence 225/15 executed first and have the result 17 .Then the 17 multiplied by 15 .So the result is 225.So use parenthesis .

Re: Why this program outputs 225 istead of 1 ?

Better to use parentheses in the macro so that the macro can be used "normally". You wouldn't normally have to bracket a subexpression so requiring the programmer to remember for SQR that they have to do

Code:

225/(SQR(15))

will lead to bugs because you don't remember stuff like this.

So define the macro as

Code:

#define SQR(x) ( (x) * (x) )

so that it can be used normally, i.e.

Code:

225/SQR(15)

225/SQR(7+8)

etc.
The brackets around the individual x's mean you can put subexpressions into SQR. Without them, the second would be substituted as:

Code:

225/(7+8*7+8)

Of course you always have to be very careful with side effect operators. This won't have the expected result:

Code:

int i=15;

int j=225/SQR(i++);

What's i? 16? Nope. No amount of bracketing will fix this, which is why we make macros upper case. And relying on this is definitely a bad idea, cos if some smart alec rewrites SQR as

Code:

#define SQR(x) (pow((x),2))

then i will be 16 and not 17 as you expected.