Macroprocessor in c++

Question

Hello everyone I am having doubt in this code

#include<iostream>
#define SQR(x)(x*x) 

int main() {
    int a, b=3;
    a = SQR(b+1);

    std::cout << a;
}

The result is 7 instead of 16 .I am not able to understand it .

The preprocessor is a *text-replacement* engine. It does no semantic parsing. — Cameron, Sep 16 '14 at 16:30

score 5 · Accepted Answer · answered Sep 16 '14 at 16:30

5

The macro expands to a literal

(b + 1 * b + 1)

So your result is:

3 + (1 * 3) + 1.

Change your macro to:

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

And it should work.

answered Sep 16 '14 at 16:30

Serge

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I have a doubt when we write SQR(++x) and pass value of x as 3 then why would the result come out to be 25 in my case. – Freedom911 Sep 16 '14 at 16:51
2

What you are seeing with SQR(++x) is [Undefined Behavior](http://stackoverflow.com/q/4176328/416627). – Dave Rager Sep 16 '14 at 16:55
oh ok thanks but is there any way to do it correctly – Freedom911 Sep 16 '14 at 16:58
2

Yes, do `++x; a = SQR(x);` Or, as james McLaughlin points out, don't use a macro. – Dave Rager Sep 16 '14 at 17:15
Yeah, James McLaughlin has some really good alternatives. – Serge Sep 16 '14 at 17:42
In fact, change your MACRO to a function. – Jarod42 Sep 16 '14 at 18:09

score 5 · Answer 2 · answered Sep 16 '14 at 16:30

After substituting the macro, the code looks like this:

int main() {
    int a, b=3;
    a = b+1*b+1;

    std::cout << a;
}

Which is the same as b+(1*b)+1 with operator precedence. You can fix it by either using a function instead:

inline int SQR(int x) {
    return x*x;
}

Or more generically:

template<class T> inline T SQR(T x) {
    return x*x;
}

Or by surrounding the macro parameters in parentheses:

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

Which would expand to ((b+1)*(b+1)). In general I wouldn't recommend using a macro for this, however, because there are other potential problems, such as SQR(x++) expanding to ((x++)*(x++)).

score 1 · Answer 3 · answered Sep 16 '14 at 16:30

1

This is what your code expands out to:

a = (b + 1 * b + 1)

Since * has more precedence than +, 1 * b is evaluated first.

answered Sep 16 '14 at 16:30

David G

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Macroprocessor in c++

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