#define process in C++ [duplicate] - c++

Question

This question already has answers here:
Macro Expansion
(7 answers)
How #define works in Programming when define has value with operator?
(4 answers)
How does #define work in C++ [duplicate]
(1 answer)
Closed 1 year ago.
I am trying C++, There is a point about can not understanding about #define addition. The example code below.
#include <iostream>
using namespace std;
#define A 0
#define B A+1
#define C 3-B
int main(){
cout << A << endl;
cout << B << endl;
cout << C;
return 0;
}
The result gives A -> 0, B -> 1, C-> 4. How C equal 4 ?

Answer 1

#define performs simple textual substitution. When you expand B out, you get 0+1 in source code, which is not necessarily identical to "an integer with the value 1".
So, in your example code, if we substitute the values in:
int main(){
cout << 0 << endl;
cout << 0+1 << endl;
cout << 3-0+1;
return 0;
}
3 - 0 + 1 is 4.

Answer 2

Do the text substitution. B expands to 0+1. C expands to 3-0+1 rather than 3-1.
cout << C;
Becomes:
cout << 3-0+1;
Because of order of operations this displays 4 rather than 2.
An easy way to see this would be to try something like:
cout << C * 50;
If we operate under the faulty assumption that C is actually 4, we'd expect to see 200. But because the above is equivalent to:
cout << 3-0+1 * 50;
Then we correctly see 53.