Inline assembly in C - gcc

Question

Pretty self explanatory code. Why doesn't it work!
#include <stdio.h>
int main() {
__asm__("number dw 0"); // declare number?
printf("%d",number);
__asm__("mov %eax,number"
"inc %eax"
"mov number,%eax");
printf("%d",number);
return 0;
}
cc ex1.c -o ex1
ex1.c: In function ‘main’:
ex1.c:22:17: error: ‘number’ undeclared (first use in this function)
ex1.c:22:17: note: each undeclared identifier is reported only once for each function it appears in
make: *** [ex1] Error 1
Thanks.
I have a lot of knowledge gaps to fill... the gcc manual was confusing me with regards to inline assembly as was google results for tutorials...
working on an intel i7 processor

Answer 1

Use this syntax, you can access variables declared in C from the inline assembly
#include <stdio.h>
int main() {
int number = 0;
printf("%d\n",number);
asm(
"mov %[number],%%eax\n"
"inc %%eax\n"
"mov %%eax,%[number]\n"
: [number] "=m" (number) : "m" (number) : "eax", "cc" );
printf("%d\n",number);
return 0;
}
You can let the compiler load number into the eax register for you by specifying the "a" constraint on the input
#include <stdio.h>
int main() {
int number = 0;
printf("%d\n",number);
asm(
"inc %%eax\n"
"mov %%eax,%[number]\n"
: [number] "=m" (number) : "a" (number) : "cc" );
printf("%d\n",number);
return 0;
}
And since x86 inc instruction can operate on memory directly you could reduce it to this
#include <stdio.h>
int main() {
int number = 0;
printf("%d\n",number);
asm(
"incl %[number]\n" /* incl -> "long" (32-bits) */
: [number] "=m" (number) : "m" (number) : "cc" );
printf("%d\n",number);
return 0;
}
For more information see gcc documentation:
6.41 Assembler Instructions with C Expression Operands