What is the output and why?

Question

#define square(x)(x*x)
Int main() {
Int x, y=1;
x=square(y+1);
printf("%d\n",x);
return 0;
}

Answer 1

#define is preprocessor directive .
It creates a macro.
It'll replace square(x) with (x*x) in code.

Remember macro arguments are NOT evaluated before macro expansion.
Meaning it'll first replace all square(x) with (x*x ) without evaluating values of x .

Here square(y+1) is replaced with
y+1*y+1 then after expansion it evaluates this expression.
y+1*y+1=>1+1*1+1

Due to operator precedence multiplication will be performed firstl
1+(1*1)+1=>1+1+1=>3
Thus output is 3

Read this also.
https://www.sololearn.com/discuss/1874614/?ref=app

refer this for more info
https://www.geeksforgeeks.org/interesting-facts-preprocessors-c/