#define print(args) printf args print(("Hello")); I got output
Hello If I call print it works fine. Could you explain how it works?
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1 Answer
This is an example of a macro.
When you compile a program, the first step is the preprocessor.
The preprocessor finds your macro:
#define print(args) printf args This means that if in your program you have something like
print(<some text>) Then the value of <some text> will be processed as args from your macro, i.e. code
print(<some text>) will be replaced with
printf <some text> Now, you have this line of code:
print(("Hello")); If you put <some text> = args = ("Hello"), then preprocessor will replace
print(("Hello")) with
printf ("Hello") and the whole line will be:
printf ("Hello"); which is legal c code to print Hello.
2 Comments
alk
Nice detailed answer. Still, regarding the style, this: "first you run compiler. Before compiler is activated, the program finds" I feel is a bit confusing, or at least quiet unclear. What is "the program"? Only after having read through half of your answer you give a hint ("the preprocessor") what it might have been.
saleph
Imo it's misspelling. The 'program' here is probably a preprocessor. A preprocessor macros are in fact a dummy replacement of text, without any semantic/syntactic analysis (that's why macros are advised to avoid - use constexpr functions instead)