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Freddie
Joined: 06 Sep 2003
Posts: 49
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| How do you fprintf a #define ? |
 Posted: Wed Sep 21, 2005 7:59 pm |
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I wrote some code that has many #defines, which will be changed by other programmers as the code is recompiled in the future.
For example:
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#define _parameter1 200
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Later in the code I have some fprintf statements:
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fprintf(serialPC,"Parameter1= %U", _parameter1);
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Now, if later the #define is changed to the following, I get an error with the fprinf statement because the "type" assinged to _parameter1 is not longer printable with %U.
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#define _parameter1 200
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I guess this is because the compiler automatically types the #define? I thought I could get around this by copying all the #defines to strings and them printing them but that takes up allot of ROM/RAM.
Anyone have and thoughts on how to do this a better way?
Thanks. |
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Mark
Joined: 07 Sep 2003
Posts: 2838
Location: Atlanta, GA
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| Posted: Wed Sep 21, 2005 8:58 pm |
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Try this:
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int8 parameter;
parameter = _parameter1;
fprintf(serialPC,"Parameter1= %U", parameter); |
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Ttelmah
Guest
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| Posted: Thu Sep 22, 2005 2:15 am |
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You can 'force type' a define.
So:
#define _parameter1 (int8)(200)
Will force the define to always be treated as an int8. Or:
#define _parameter1 (200L)
Will force the define to be a 'long' (you could also use an int16 cast the same way).
You should also get into the habit of bracketting defines. If a define contains arithmetic, and this is not done, the evaluation order may not be what is expected, and hence always bracketting is a good way of avoiding problems in the future. -)
Best Wishes |
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