#include <iostream>
using namespace std;
#define squareOf(x) x*x
int main() {
// your code goes here
int x;
cout<<squareOf(x+4);
return 0;
}
I thought the answer would come as 16 but it came out as 4. I am confused how does this works.
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Srishti Mittal
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8`x` is not initialized, this is just *undefined behavior* – UnholySheep Aug 11 '18 at 16:09
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@UnholySheep hehe correct but please explain this behavior of macros – Srishti Mittal Aug 11 '18 at 16:10
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2@SrishtiMittal macros are pure text replacement, there is no magic here: https://en.cppreference.com/w/cpp/preprocessor/replace - and if you replace `squareOf(x+4)` you do get: `x+4*x+4` – UnholySheep Aug 11 '18 at 16:12
1 Answers
8
16 would never be the result here. Let's say you would have initialized x with 0, then you would have your x+4 replaced by x+4*x+4 which would evaluate as 0+4*0+4 = 4.
Preprocessor macros replace source code, they are not functions.
You might now think that maybe
#define squareOf(x) (x)*(x)
would be better, but consider that then
int x = 2;
int y = squareOf(x++);
would result in y = (2)*(3) = 6, not in 4.
If you do not have a really good reason, avoid preprocessor macros. There are good reasons, but if something behaves like a function, better make it a function.
Now take a look at this:
template <class T>
inline T squareOf(const T& number)
{
return number*number;
}
As inline, it does also replace code (at least if the compiler wants so), but here, this one actually behaves like a function (since it is one). Wouldn't expect a bad outcome from that one.
Aziuth
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