The below example does not work as I expected (using ">" in the conditional) with GCC on an 8 bit machine, and neither on Linux 64 bit. If I make Timer_Secs and FutureTimeout_1Minute f.ex. 16 bits wide (int in those cases is 16 bits) it works as it should (using ">" in the conditional).
So, to get this to work for 8 bits with sign I seem to have to use "==" in the conditional.
{
// Example with 8-bit only as tested with GCC 3.2 on AVR ATmega 128
unsigned int Cnt1 = 0; // 16 bits (kind of system clock)
unsigned int Cnt2 = 0; // 16 bits (observer, see what happens)
signed char Timer_Secs = 0; // 8 bits with sign
signed char FutureTimeout_1Minute = 60; // 8 bits with sign
while (Cnt1 < 602)
{ // ##
if ((Timer_Secs - FutureTimeout_1Minute) == 0)
{
FutureTimeout_1Minute += 60; // Overflow/underflow allowed,
// so wraps and changes sign
Cnt2++;
} else {} // No code
Timer_Secs++;
Cnt1++;
}
// ##
// Cnt2 is 35 if > 0 above #define AFTER_F(a,b) ((a-b)>0)
// Cnt2 is 35 if >= 0 above #define AFTER_EQ_F(a,b) ((a-b)>=0)
// Cnt2 is 10 if == 0 above #define AFTER_ON_F(a,b) ((a-b)==0) **EXPECTED!**
}
The compare seems to involve some kind of sign conversion. I have not studied the assembly code for the different cases, since I thought this could be solved at code level.
- Why is this so?
- To me this looks plain wrong?
- Still I assume it's compiled like this by design?
- Is there some kind of type transfer or type decoration I could have done?
The context here is a blog note called "Know your timer's type" at: [blog]: http://www.teigfam.net/oyvind/home/technology/109-know-your-timers-type/
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It works as expected in Go, see blog note or http://play.golang.org/p/lH2SzZQEHG
It does not seem to work as expected in C++ either.
