What does this mean: #define x 5+2? And how would C++ colpile it?

Question

I fond this:

#define x 5+2
int main(){
int i= x*x*x;
cout << i;
return 0;
}

// the out put is 27

!!! i don't understand how !!!
Can any one explane, please!

Answer 1

@ KrOW @Ace, thanks for the good explanation 👍😉

Answer 2

#define run in preprocessor context and when it run before compilation... it essentially replace the content of defined identifier with corrispondent value in your case:
#define x 5+2

int i= x*x*x;

here x will be substituited to 5+2 as:

int i= 5+2*5+2*5+2;

that is 5+10+10+2 = 27

Answer 3

@KrOW @Ace ; Thanks a lot for answering me. So, if i understand well; in this example:

#define x 10+10
int i=x/10;
cout<<i;

It will be compiled like
int i=10+10/10;
and the result will be 11

Answer 4

👍