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Problem with uint32_t accesses using LDRD

To illustrate my problem, I just implemented demonstration code:

#include <stdint.h>

uint64_t read(const uint32_t *wordStream) {
    uint32_t low = *wordStream++;
    uint64_t high = *wordStream++;
    high <<= 32;
    return high | low;
}

This code reads (according to C) two unsigned 32 bit words.
It assumes a little endian notation.
Then it assembles these two words into a 64 bit value and returns it.

Unfortunately, the compiler assumes, that the first word is 64 bit aligned
(by using the LDRD instruction which is only allowed on aligned 64 bit values):

; generated by Component: ARM Compiler 5.06 update 1 (build 61) Tool: armcc [4d35ad]
; commandline armcc [--thumb -c --asm -otestCode.o --cpu=SC300 testCode.c]
        THUMB
        REQUIRE8
        PRESERVE8

        AREA ||.text||, CODE, READONLY, ALIGN=1

read PROC
        LDRD     r2,r1,[r0,#0]
        MOV      r0,r2
        BX       lr
        ENDP

...

Where did I tell the compiler that the first word is aligned?

Parents
  • The LDRD has bitten once in recent times, where the compiler was loading a double, and the byte packed records would cause it to fault. It was an easily identifiable issue for anyone familiar with the old ARM7/9 parts, and remedied with an memcpy(), but it might frustrate the average PC coder porting "working" code.

    double dx = *((double *)&p[8]); // where p is a byte pointer, alignment hazard
    

Reply
  • The LDRD has bitten once in recent times, where the compiler was loading a double, and the byte packed records would cause it to fault. It was an easily identifiable issue for anyone familiar with the old ARM7/9 parts, and remedied with an memcpy(), but it might frustrate the average PC coder porting "working" code.

    double dx = *((double *)&p[8]); // where p is a byte pointer, alignment hazard
    

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