Here is what i have and I wonder how this works and what it actually does.
#define NUM 5
#define FTIMES(x)(x*5)
int main(void) {
int j = 1;
printf(\
the preprocessor substitutes all NUM ocurrences in the code with 5, and all the FTIMES(x) with x * 5. The compiler then compiles the code.
Its just text substitution.
And if you want to fix it:
#define FTIMES(x) ((x) * 5)
The answer to your question after that is order of operations:
FTIMES(j+5) = 1+5*5 = 26
FTIMES((j+5)) = (1+5)*5 = 30
Since it hasn't been mentioned yet, the way to fix this problem is to do the following:
#define FTIMES(x) ((x)*5)
The parentheses around x
in the macro expansion prevent the operator associativity problem.
The reason this happens is because your macro expands the print to:
printf("%d %d\n", j+5*5, (j+5)*5);
Meaning:
1+5*5 and (1+5)*5
Order of operations.
FTIMES(j+5) where j=1 evaluates to:
1+5*5
Which is:
25+1
=26
By making FTIMES((j+5)) you've changed it to:
(1+5)*5
6*5
30