simple #define question
Author
Can you use a #define in an if statement? Here''s an example of what I''m trying:
int a = 0;
//manipulate a
if(a == 1)
{
#define example 4
}
else if(a == 2)
{
#define example 2
}
//etc...
#define are compile-time statements, they are not evaluated at run-time.
You can use #if and compare with compile-time constants
#if defined(__BLAH)
#define EXAMPLE 5
#else
#define EXAMPLE 4
#endif
You can use #if and compare with compile-time constants
#if defined(__BLAH)
#define EXAMPLE 5
#else
#define EXAMPLE 4
#endif
Author
so what would be the best alternative to use if i needed a Run-Time #define-like variable?
[edited by - Armedies on August 7, 2003 11:59:31 AM]
[edited by - Armedies on August 7, 2003 11:59:31 AM]
Um, if it''s a value that can change at run-time, then use a plain-old variable!
int example = 0;
if(a == 1)
{
example=4;
}
else if(a == 2)
{
example=2;
}
int example = 0;
if(a == 1)
{
example=4;
}
else if(a == 2)
{
example=2;
}
A normal variable would do nicely for run-time data 
I don''t understand what you mean by a run-time #define-like variable?
#defines are NOT variables, they''re replacements. If you do:
the compiler will see this code (after the pre-processing, which is where it does the #include and #define work (and anything else starting with ''#'') as:
It''s as simple as that. #define is like a find-and-replace
Anything that needs to change its value during runtime has to be a variable of some form.
int example;if(a == 1){ example = 4;}else{ example = 2;} I don''t understand what you mean by a run-time #define-like variable?
#defines are NOT variables, they''re replacements. If you do:
#define WABADOO (x+5)int something(int x){ return WABADOO;} the compiler will see this code (after the pre-processing, which is where it does the #include and #define work (and anything else starting with ''#'') as:
int something(int x){ return (x+5);} It''s as simple as that. #define is like a find-and-replace
Anything that needs to change its value during runtime has to be a variable of some form.
Author
well, I don''t want to change it''s value. I just want to run a switch statement at the beginning of my program, and depending on it''s value, use on or another #define. nothing will be changed after that.
just use:
const int EXAMPLE_1 = 1;
const int EXAMPLE_2 = 2;
const int* EXAMPLE = NULL;
if(a == 1)
EXAMPLE = &EXAMPLE_1;
else
EXAMPLE = &EXAMPLE_2;
const int EXAMPLE_1 = 1;
const int EXAMPLE_2 = 2;
const int* EXAMPLE = NULL;
if(a == 1)
EXAMPLE = &EXAMPLE_1;
else
EXAMPLE = &EXAMPLE_2;
quote:Original post by Armedies
well, I don''t want to change it''s value. I just want to run a switch statement at the beginning of my program, and depending on it''s value, use on or another #define. nothing will be changed after that.
What''s the difference between using define and using a variable for you?
If you''re concerned about someone accidentally changing the value, then I would say make a class that can wrap up your settings and only provide read-only access:
class Settings { friend void SetSettings(Settings& settings); const int m_example;public: Settings() : m_example(0); Settings(_example) : m_example(_example) {} int example() const { return m_example; }};Settings g_Settings;void SetSettings(Settings& settings) { if(a == 1) { settings.m_example = 4; } else { settings.m_example = 5; }}Call SetSettings(g_Settings) at the beginning of your app.
Regards,
Jeff
Author
"What's the difference between using define and using a variable for you?"
-well, i started the program with a define, and changing it to an actual variable means a lot of chasing. however, the class idea, isn't a bad one, perhaps I'll try that, and see how well it works
Thank you for the help.
[edited by - Armedies on August 7, 2003 12:31:53 PM]
-well, i started the program with a define, and changing it to an actual variable means a lot of chasing. however, the class idea, isn't a bad one, perhaps I'll try that, and see how well it works
Thank you for the help.
[edited by - Armedies on August 7, 2003 12:31:53 PM]
What kind of chasing do you need to do? You''ve changed a literal integer value (the #define) into an integer variable but how does that screw up your code?
This topic is closed to new replies.
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