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armcc problem?

Note: This was originally posted on 20th April 2009 at http://forums.arm.com

Hello,

We're using armcc here version:
ARM/Thumb C/C++ Compiler, RVCT3.1 [Build 739]

And it seems to have an issue. Here's some simple c-code:

main () {
    volatile unsigned int tmp32;

    printf ("GPIO: 1_reset: before\n");

    // touch tb_gpio
    *(unsigned int *)(0x50000004) = 0x12345678;
    tmp32 = *((volatile unsigned int *)(0x50000004));
    printf ("GPIO: tmp32 = 0x%0x\n", tmp32);
    printf ("GPIO: tmp32 = 0x%0x\n", tmp32);
    if (tmp32 != 0x12345678) printf("FAIL: 1st tb_gpio WRV failed!\n");

Here's the resulting output:

# GPIO: 1_reset: before
# GPIO: tmp32 = 0x12345678
# GPIO: tmp32 = 0x0
# FAIL: 1st tb_gpio WRV failed!

Why is the 2nd printf of tmp32 different? Here's the disassembled code, and it seems that the first printf is printing not tmp32 which is in r4, but the literal that was written, which is in r6:

       0x00040082:    4e2b        +N      LDR      r6,[pc,#172] ; [0x40130] = 0x12345678
        0x00040084:    f04f45a0    O..E    MOV      r5,#0x50000000
        0x00040088:    606e        n`      STR      r6,[r5,#4]
        0x0004008a:    686c        lh      LDR      r4,[r5,#4]
        0x0004008c:    4631        1F      MOV      r1,r6
        0x0004008e:    a029        ).      ADR      r0,{pc}+0xa6 ; 0x40134
        0x00040090:    f000fcf6    ....    BL       __0printf ; 0x40a80
        0x00040094:    4621        !F      MOV      r1,r4
        0x00040096:    a027        '.      ADR      r0,{pc}+0x9e ; 0x40134
        0x00040098:    f000fcf2    ....    BL       __0printf ; 0x40a80

This make sense? Is this an armcc bug? If so, what next?

Cheers,
Ric Peregrino
Dust Networks

Parents

Samar Samar over 12 years ago

Note: This was originally posted on 24th April 2009 at http://forums.arm.com

Can you try to make tmp32 global and check the assembly? Also please make sure you can actually write and read back from 0x50000004 location.

The move from r6 to r1 may be because it does not think tmp32 to be volatile. In fact it did not even assign a memory location to tmp32.

Move from r4 to r1 looks correct to me but i am surprised to see 0 in the second line of output. That makes me think that the location 0x50000004 wasnt written to properly or is not read properly.

Another thing you can try is:
volatile unsigned int *loc = (volatile unsigned int *)0x50000004;
main () {
unsigned int tmp32;

printf ("GPIO: 1_reset: before\n");

// touch tb_gpio
*loc = 0x12345678;
tmp32 = *loc;
printf ("GPIO: tmp32 = 0x%0x\n", tmp32);
*loc = 0x87654321;
tmp32 = *loc;
printf ("GPIO: tmp32 = 0x%0x\n", tmp32);
if (tmp32 != 0x12345678) printf("FAIL: 1st tb_gpio WRV failed!\n");

Please share the findings.

Samar
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Samar Samar over 12 years ago

Note: This was originally posted on 24th April 2009 at http://forums.arm.com

Can you try to make tmp32 global and check the assembly? Also please make sure you can actually write and read back from 0x50000004 location.

The move from r6 to r1 may be because it does not think tmp32 to be volatile. In fact it did not even assign a memory location to tmp32.

Move from r4 to r1 looks correct to me but i am surprised to see 0 in the second line of output. That makes me think that the location 0x50000004 wasnt written to properly or is not read properly.

Another thing you can try is:
volatile unsigned int *loc = (volatile unsigned int *)0x50000004;
main () {
unsigned int tmp32;

printf ("GPIO: 1_reset: before\n");

// touch tb_gpio
*loc = 0x12345678;
tmp32 = *loc;
printf ("GPIO: tmp32 = 0x%0x\n", tmp32);
*loc = 0x87654321;
tmp32 = *loc;
printf ("GPIO: tmp32 = 0x%0x\n", tmp32);
if (tmp32 != 0x12345678) printf("FAIL: 1st tb_gpio WRV failed!\n");

Please share the findings.

Samar
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