预编译器会评估移位和算术运算吗？

Question

If I define the following in a C++98 program:

#define BITS_PER_FOO 2
#define BITS_PER_BAR 3
#define FOOBARS (1<<(BITS_PER_FOO*BITS_PER_BAR))

then will FOOBARS get evaluated as 64 by the precompiler? Or will a multiplication and bit-shift operation take place at each location that I use FOOBARS in the code?

Answer 1

No, since it's not the preprocessor's business. It does the usual replace thing, not constant folding.

However, any reasonable compiler will do constant folding, so you should not expect this to correspond to instructions being executed at runtime.

Answer 2

The precompiler just does text substitution - it's essentially copy & paste on steroids. This means that the expression you wrote for FOOBAR will be expanded in full at each replacement location.

Now, any decent compiler will evaluate that whole subexpression at compile time anyway. However, you can save it some work (and have some extra advantages, like having a clear type of your expression, clearer diagnostic, less surprises deriving from substitutions in the wrong places, and having an actual lvalue for your constants instead of expressions) by defining these values as actual constants, like:

const int bits_per_foo = 2;
const int bits_per_bar = 3;
const int foobars = 1<<(bits_per_foo*bits_per_bar);

Answer 3

A multiplication and bit-shift operation will take place at each location that you use FOOBARS .

See: #define Directive (C/C++)

#define forms a macro, meaning the identifier is replaced with everything in the token-string. In your case, the preprocessor replaces instances of FOOBARS with the expression (1<<(BITS_PER_FOO*BITS_PER_BAR))