C #define macros

问题

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("%d %d\n", FTIMES(j+5), FTIMES((j+5)));
}

It produces two integers: 26 and 30.

How does it do that?

回答1:

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

回答2:

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.

回答3:

define is just a string substitution.

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

回答4:

The compiler pre-process simply does a substitution of FTIMES wherever it sees it, and then compiles the code. So in reality, the code that the compiler sees is this:

#define NUM 5
#define FTIMES(x)(x*5)

int main(void)
{

    int j = 1;

    printf("%d %d\n", j+5*5,(j+5)*5);
}

Then, taking operator preference into account, you can see why you get 26 and 30.

回答5:

And if you want to fix it:

#define FTIMES(x) ((x) * 5)

回答6:

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.

回答7:

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

来源：https://stackoverflow.com/questions/480939/c-define-macros