This discussion has been locked.
You can no longer post new replies to this discussion. If you have a question you can start a new discussion

Timer, t1-t0, integral promotion

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.

Parents
  • 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.

Reply
  • 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.

Children
No data