What does #x inside a C macro mean?

问题:

For example I have a macro:

#define PRINT(int) printf(#int "%d\n",int)

I kinda know what is the result. But how come #int repersent the whole thing?

I kinda forget this detail. Can anybody kindely give me a hint?

Thanks!

回答1:

In this context (applied to a parameter reference in a macro definition), the pound sign means to expand this parameter to the literal text of the argument that was passed to the macro.

In this case, if you call PRINT(5) the macro expansion will be printf("5" "%d\n", 5); which will print 5 5; not very useful; however if you call PRINT(5+5) the macro expansion will be printf("5+5" "%d\n", 5+5); which will print 5+5 10, a little less trivial.

This very example is explained in this tutorial on the C preprocessor (which, incidentally, is the first Google hit for c macro pound sign).

回答2:

"#" can show the name of a variable, it's better to define the macro as this:

#define PRINT(i) printf(#i "= %d\n", i)

and use it like this:

int i = 5;
PRINT(i);

Result shown:

i = 5

回答3:

That is a bad choice of name for the macro parameter, but harmless (thanks dreamlax).

Basically if i write like so

PRINT(5);

It will be replaced as

printf("5" "%d\n",5);

or

printf("5 %d\n",5);

It is a process called Stringification, #int is replaced with a string consisting of its content, 5 -> "5"