Simple question

Its possible to declare something like...

#define n1to9 (i == (1 to 9))

or something?
Like if i is 1 to 9, but in a declaration..
I dont want answers like tell me to do:
if(i <= 9 && i > 0) or something...

i just wanna know if is possible and how to declare a variable like that..
1 to 9 or something....

You mean an array?

int i [] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Changed it to C++ only had to move 1 thing xD

#include <vector>

vector oneToNine;
for(int i = 1; i <= 9; i++)
oneToNine.push_back(i);

use vectors

Originally Posted by Virtual Void

#include <vector>

vector oneToNine;
for(int i = 1; i <= 9; i++)
oneToNine.push_back(i);

use vectors

Oh thanks! I learned something even though i didn't ask the question xD Thanks!

Nope none of that....
If i make a array im gonna need to make like shit[9] = {0,1,2,3,4,5,6,7,9,+,9,859,89,8,96............}

There is not a way like..
shit[]= {1-9}
?

You can define almost anything you want, but you need to remember that the compiler won't give you a numerical result from #define.

#define is a preprocessor(Hence the '#'). The compiler will go through anything with a '#' before compiling the program, and with #define (in reference to your original post) it will replace every 'n1to9' it finds with '(i == (1 to 9))'. Then compile the program.

As long as whatever you are using it with can take an expression instead of just a variable you shouldn't have a problem.

But i don't know if it will work with any spaces. You will have to figure that out yourself.
@-Away

Originally Posted by Virtual Void

#include <vector>

vector oneToNine;
for(int i = 1; i <= 9; i++)
oneToNine.push_back(i);

use vectors

I always end up mindfucking myself over vectors.

Originally Posted by -Away

Well, I guess you could...

Since C++ isn't VB, you can't use VB syntax <<

Replace

Code:

#define n1to9  (i == (1 to 9))

with

Code:

#define n1to9  (i >= 1 && i <= 9)

Originally Posted by Disturbed

You can define almost anything you want, but you need to remember that the compiler won't give you a numerical result from #define.

#define is a preprocessor(Hence the '#'). The compiler will go through anything with a '#' before compiling the program, and with #define (in reference to your original post) it will replace every 'n1to9' it finds with '(i == (1 to 9))'. Then compile the program.

As long as whatever you are using it with can take an expression instead of just a variable you shouldn't have a problem.

But i don't know if it will work with any spaces. You will have to figure that out yourself.
@-Away

I always end up mindfucking myself over vectors. :confused:

Vectors aren't thread safe :(

Well ty, I see what i want is not possible...Thank you all anyway

Simple question

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