I'm reading some C code embedded with a few assembly code. I understand that __asm__ is a statement to run assembly code, but what does __asm__ do in the following code? According to the output (i.e., r = 16), it seems that __asm__ does not effect the variable r. Isn't it?
#include <stdio.h>
static void foo()
{
static volatile unsigned int r __asm__ ("0x0019");
r |= 1 << 4;
printf("foo: %u\n", r);
}
Platform: Apple LLVM version 6.0 (clang-600.0.56) (based on LLVM 3.5svn) on OSX Yosemite
Strictly speaking, your "asm" snippet simply loads a constant (0x0019).
Here's a 32-bit example:
#include <stdio.h>
static void foo()
{
static volatile unsigned int r __asm__ ("0x0019");
static volatile unsigned int s __asm__ ("0x1122");
static volatile unsigned int t = 0x3344;
printf("foo: %u %u %u\n", r, s, t);
}
gcc -O0 -S x.c
cat x.c
.file "x.c"
.data
.align 4
.type t.1781, @object
.size t.1781, 4
t.1781:
.long 13124 # Note: 13124 decimal == 0x3344 hex
.local 0x1122
.comm 0x1122,4,4
.local 0x0019
.comm 0x0019,4,4
.section .rodata
.LC0:
.string "foo: %u %u %u\n"
.text
.type foo, @function
foo:
pushl %ebp
movl %esp, %ebp
subl $24, %esp
movl t.1781, %eax
movl 0x1122, %edx
movl 0x0019, %ecx
movl %eax, 12(%esp)
movl %edx, 8(%esp)
movl %ecx, 4(%esp)
movl $.LC0, (%esp)
call printf
leave
ret
PS: The "asm" syntax is applicable to all gcc-based compilers.
PPS: I absolutely encourage you to experiment with assembly anywhere you please: embedded systems, Ubuntu, Mac OSX - whatever pleases you.
Here is an excellent book. It's about Linux, but it's also very largely applicable to your OSX:
Programming from the Ground Up, Jonathan Bartlett
Also:
https://www.hackerschool.com/blog/7-understanding-c-by-learning-assembly
http://fabiensanglard.net/macosxassembly/
PPS: x86 assembly syntax comes in two variants: "Intel" and "ATT" syntax. Gcc uses ATT. The ATT syntax is also applicable for any other architecture supported by GCC (MIPS, PPC, etc etc). I encourage you to start off with ATT syntax ("gcc/gas"), rather than Intel ("nasm").
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