[Bug c/59593] New: [arm big-endian] using "ldrh" access a immediate which stored in a memory by word

[spf_zju at 126 dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=59593

            Bug ID: 59593
           Summary: [arm big-endian] using "ldrh" access a  immediate
                    which stored in a memory by word
           Product: gcc
           Version: 4.7.1
            Status: UNCONFIRMED
          Severity: critical
          Priority: P3
         Component: c
          Assignee: unassigned at gcc dot gnu.org
          Reporter: spf_zju at 126 dot com

the C code like this:

short int a = 1;
int test()
{
  return 922 * a;
}

compile the C code : arm-eabi-gcc -mbig-endian -O2 -S bar.c -o bar.s 

the bar.s like this :

movw r3, #:lower16:.LANCHOR0
movt r3, #:upper16:.LANCHOR0
ldrh r0, [r3,#0]
ldrh r3, .L2
smulbb r0, r0, r3
bx lr

.L3:
   .align 2
.L2:
   .word 922 

The immediate 922 is stored in .L2, in arm big-endian mode  ,the memory of .L2
is like this 0x0000039a, so when executing this "ldrh r0, [r3,#0]", the r0 is
zero,so the result is wrong .

also see the disassembly of bar.o :

   <test>:
0: e3003000  movw r3, #0
4: e3403000  movt r3, #0
8: e1d300b0  ldrh r0,[r3]
c: e1df30b4  ldrh r3,[pc,#4]   ;18
10:e1b00380  smulbb r0,r0,r3
14:e12fff1e  bx lr 
18:0000039a  muleq r0,sl,r3

So the return value of the function test is zero. it is wrong .