I want to write unified assembly to get rid of pesky #
in front of my literals as mentioned at: Is the hash required for immediate values in ARM assembly?
This is a minimal non-unified code with #
:
#include <assert.h>
#include <inttypes.h>
int main(void) {
uint32_t io = 0;
__asm__ (
"add %0, %0, #1;"
: "+r" (io)
:
:
);
assert(io == 1);
}
which compiles and later runs fine under QEMU:
arm-linux-gnueabihf-gcc -c -ggdb3 -march=armv7-a -pedantic -std=c99 -Wall -Wextra \
-fno-pie -no-pie -marm -o 'tmp.o' 'tmp.c'
If I try to remove the #
, then the code fails with:
/tmp/user/20321/ccoBzpSK.s: Assembler messages:
/tmp/user/20321/ccoBzpSK.s:51: Error: shift expression expected -- `add r3,r3,1'
as expected, since non-unified seems to be the default.
How to make that work?
I found the promising option:
gcc -masm-syntax-unified
but adding it did not help.
If I write instead:
".syntax unified; add %0, %0, #1;"
then it works, but I would have to do that for every __asm__
which is not practical.
UI also found that without -marm
, then it does use unified assembly, but it generates thumb code, which I don't want.
Maybe this bug is the root cause of the problem: https://gcc.gnu.org/bugzilla/show_bug.cgi?id=88648
Tested in arm-linux-gnueabi-gcc 5.4.0, Ubuntu 18.04.