[CaHbKo]

I've been wondering if it's possible to make this

pawn Код:

if(var == 3 ... 5)

be read by compiler like this

pawn Код:

if(3 <= var <= 5)

I tried to make a macro for that but I'm not very familiar with macros. Any ideas?

[CaHbKo]

Quote:

Originally Posted by ****** Yes, but the only way to do it as you want is to name the macro "if", which isn't that nice (as you're redefining a keyword): pawn Код: #define if%0(%1==%2...%3) if(%2<=%1<=%3) Note that, although it isn't used, "%0" is important to account for different spacing styles (for example you wrote "if(", whereas I write "if (" - this macro will cover both cases).

Thanks. But would it work if I do something more complex like this

pawn Код:

if(Function() == true && (var1 > 10 || var2 == 1...3) || var3 != 100...999)