[Bug target/107339] RISC V -mstrict-align

[yy172179 at 163 dot com via Gcc-bugs]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=107339

--- Comment #2 from yy <yy172179 at 163 dot com> ---
(In reply to Andrew Pinski from comment #1)
> The code is broken.
> 
> uint32_t here is aligned to 32bits and you access it via that type.
> 
> 
> this is the correct fix:
> 
> 
> typedef uint32_t uint32_t_ua __attribute__((aligned(1)));
> 
> void  tttt (misalign_t* t){
> 
>     printf("%x\n",(*(const uint32_t_ua *)&t->rseq[0]));
>     printf("%x\n",(*(const uint32_t_ua *)&t->rseq[1]));
> }
(In reply to Andrew Pinski from comment #1)
> The code is broken.
> 
> uint32_t here is aligned to 32bits and you access it via that type.
> 
> 
> this is the correct fix:
> 
> 
> typedef uint32_t uint32_t_ua __attribute__((aligned(1)));
> 
> void  tttt (misalign_t* t){
> 
>     printf("%x\n",(*(const uint32_t_ua *)&t->rseq[0]));
>     printf("%x\n",(*(const uint32_t_ua *)&t->rseq[1]));
> }
Maybe I didn't make the problem clear. To add, I think the non aligned address
"lw" will be converted to "lbu" to access this address indirectly. Change the
above code parameters to the structure, and the situation I think is correct
will occur.

#include<stdio.h>

typedef  struct  __attribute__((packed)) misalign
{
    uint8_t rseq[16];
    uint8_t type;
    uint8_t cnt;
} misalign_t;
void  tttt (misalign_t t){

    printf("%x\n",(*(const uint32_t*)&t.rseq[0]));
    printf("%x\n",(*(const uint32_t*)&t.rseq[1]));
}
int main()
{

    misalign_t  abc;
    misalign_t *test=&abc;
    tttt(abc);
        while(1); 
   }