Question

I was reading an article where #define macro was made like this:

#define TEST(level) if (level > 2) ; else std::cout

which then could be used in the code like this:

Test(1) << 3;
Test(3) << 4;

I was under the impression that if you wrote

TEST(1)

it would replace it with:

std::cout

and

TEST(3)

would be replaced with an empty string (in the code file). However, if it worked this way, then it should throw an error, since

<< 3;

is invalid.

How does this macro actually work? and how does the pre-processor change

TEST(3) << 3;

so that it doesn’t output anything (that is, the code doesn’t run)

Answer 1

I believe you got it wrong:

For TEST(3) << 3, the replaced form is:

if (3 > 2) ; else std::cout << 3

It is still a valid expression