Consider the following code:
#define P1(x) x+x
#define P2(x) 2*P1(x)
int main()
{
int a = P1(1) ? 1 : 0;
int b = P2(a)&a;
return 0;
}
Now, I thought that the compiler first replacing the macros with their values and so int b = 2*a+a&a; (and since a=1 then b=3). Why isn't it so?
This is because & has lower precedence than that of addition + operator. The grouping of operands will take place as
int b = ( (2*a + a) & a );
Therefore, (2*a + a) = 3 and 3 & 1 = 1 (011 & 001 = 001).
There's no precedence in your operation (it is just a textual substitution) thus, as you've noted,
#define P1(x) x+x
#define P2(x) 2*P1(x)
int a = P1(1) ? 1 : 0; // 1
and since & has lower precedence than +, it is equivalent to
int b = ((2 * a) + a) & a;
i.e. only the rightmost bit is set on b.
((2 * a) + a) 011 &
a 001 =
--------------------
b 001
