https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67443
--- Comment #7 from Dominik Vogt <vogt at linux dot vnet.ibm.com> ---
Almost. Note the strange bit numbering on s390. The highest order bit in any
operation always has the number 0, and the lowest order bit has the highest
number. So the 8-bit-move instruction "mvi" stores a byte into *a (bit 0-7),
then the 32-bit-and instruction "n" reads *a as a 32 bit value. The bits 0-7
are the highest order bits of the result, so the value read is actually
0x03xxxxxx (bits 0-7 have the value 0x03, the rest is random data).
(Actually, I couldn't read this assembly code right without the help of a
debugger.)
# r2 contains the address a at start of function
larl %r5,.L3
mvi 0(%r2),3 # b-byte store of value 3 to the *(a + 0)
# -> memory at a: 03 ** ** ** ** ** ** **
l %r1,.L4-.L3(%r5) # 32-bit-load 0xff000000 to r1
n %r1,0(%r2) # 32-bit "and" of *(a + 0) and $1, result
# stored in r1 -> 0x03000000
oill %r1,5 # 64-bit "or" of r1 with the immediate value
# 0x00000000 00000005
# r1 -> 0x03000005
st %r1,0(%r2) # 32-bit store or r1 to a
# -> memory at a: 03 00 00 05 ** ** ** **
br %r14
.L3:
.L4:
.long -16777216
