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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

0 John Linq over 13 years ago in reply to Mike Kleshov

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        uint32_t t0 = 4294967290, t1 = 19;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        if ( (t1 - t0) == 25 )
                printf("OK\n");
        else
                printf("NG\n");

        system("PAUSE");
}

t0 u = 4294967290
t1 u = 19
t0 d = -6
t1 d = 19
result u = 25
result d = 25
OK

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        int32_t t0 = 2147483640, t1 = -2147483631;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        if ( (t1 - t0) == 25 )
                printf("OK\n");
        else
                printf("NG\n");

        system("PAUSE");
}

t0 u = 2147483640
t1 u = 2147483665
t0 d = 2147483640
t1 d = -2147483631
result u = 25
result d = 25
OK

Reply

0 John Linq over 13 years ago in reply to Mike Kleshov

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        uint32_t t0 = 4294967290, t1 = 19;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        if ( (t1 - t0) == 25 )
                printf("OK\n");
        else
                printf("NG\n");

        system("PAUSE");
}

t0 u = 4294967290
t1 u = 19
t0 d = -6
t1 d = 19
result u = 25
result d = 25
OK

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        int32_t t0 = 2147483640, t1 = -2147483631;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        if ( (t1 - t0) == 25 )
                printf("OK\n");
        else
                printf("NG\n");

        system("PAUSE");
}

t0 u = 2147483640
t1 u = 2147483665
t0 d = 2147483640
t1 d = -2147483631
result u = 25
result d = 25
OK

Children

0 John Linq over 13 years ago in reply to John Linq

For int32_t/uint32_t, the result is what I want.
Due to the integral promotion, the result of int16_t/uint16_t is not I want.
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0 John Linq over 13 years ago in reply to John Linq

My mistake:

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        int32_t t0 = 2147483640, t1 = 123;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        system("PAUSE");
}

t0 u = 2147483640
t1 u = 123
t0 d = 2147483640
t1 d = 123
result u = 2147483779
result d = -2147483517

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        uint32_t t0 = 2147483640, t1 = 123;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        system("PAUSE");
}

t0 u = 2147483640
t1 u = 123
t0 d = 2147483640
t1 d = 123
result u = 2147483779
result d = -2147483517

0 John Linq over 13 years ago in reply to John Linq

My mistake again:

#include <stdio.h>
#include <stdint.h>


void main(void)
{
        int32_t t0 = 2147483640, t1 = 123;    // t1 < t0

        printf("t0 u = %u\n", t0);
        printf("t1 u = %u\n", t1);
        printf("t0 d = %d\n", t0);
        printf("t1 d = %d\n", t1);

        printf("result u = %u\n", (t1 - t0));
        printf("result d = %d\n", (t1 - t0));

        if ( (t1 - t0) == 2147483779 )
            printf("OK\n");
        else
            printf("NG\n");

        system("PAUSE");
}

t0 u = 2147483640
t1 u = 123
t0 d = 2147483640
t1 d = 123
result u = 2147483779
result d = -2147483517
OK

It is still OK!!!

0 John Linq over 13 years ago in reply to John Linq
I think this topic is too difficult to understand for me.

Many thanks to everyone involved.

My favorite:

/* both t0 and timer_val are unsigned */ t0 = timer_val; while ( (cast to timer-size unsigned)( timer_val - t0 ) < delay_val );
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