For ARM7, A hardware timer is 32-bits, and unsigned int is 32-bits too; For a 16-bits MCU, A hardware timer is 16-bits, and unsigned int is 16-bits too;
So,
t0 = timer_val; while ( ( timer_val - t0 ) < delay_val );
provide a simple and good delay;
Not so sure about, if a cast is needed.
t0 = timer_val; while ( (unsigned int)( timer_val - t0 ) < delay_val );
Just noticed that, due to the integral promotion,
unsigned short t0, t1; ( t1 - t0 ) is a signed int.
It seems that,
unsigned int t0, t1; ( t1 - t0 ) is still an unsigned int.
I noticed this, because I use unsigned short to present a 16-bits timer_val on my X86-PC for testing/simulating purpose, and the result is not what I expected.
#include <stdio.h> #include <stdint.h> void main(void) { int16_t now = -32768, end; end = now - 1; // while (now - end < 0); printf("now = %d\n", now); printf("end = %d\n", end); if ( (int16_t)(now - end) < 0 ) printf("Less than 0\n"); else printf("It is >= Zero\n"); system("PAUSE"); }
now = -32768 end = 32767 It is >= Zero
#include <stdio.h> #include <stdint.h> void main(void) { int16_t now = -32768, end; end = now - 1; // while (now - end < 0); printf("now = %d\n", now); printf("end = %d\n", end); if ( (now - end) < 0 ) printf("Less than 0\n"); else printf("It is >= Zero\n"); system("PAUSE"); }
now = -32768 end = 32767 Less than 0
It is different.
No wonder. When you are doing ( (now - end) < 0 ), the 16-bit signed integers are promoted to 32-bit ones prior to subtraction, so an overflow is no longer possible. Hence, the result is negative. When you are doing ( (int16_t)(now - end) < 0 ), you are converting the result of subtruction (which is -65535 or 0xFFFF0001) to int16_t. The conversion simply throws away the upper 16 bits, so you are left with 0x0001 which is positive.
@John Linq,
You are _not_ stupid and you are certainly _not_ an idiot !