Pass 64-bit int as output to 32-bit inline asm - c++

Question

#include <stdarg.h>
#include <stdint.h>
uint64_t test_func(int n)
{
return 9223372036854775805;
}
int main()
{
uint64_t r = test_func(10);
return 0;
}
converts to:
test_func(int):
push ebp
mov ebp, esp
mov eax, -3
mov edx, 2147483647
pop ebp
ret
main:
push ebp
mov ebp, esp
and esp, -8
sub esp, 24
mov DWORD PTR [esp], 10
call test_func(int)
mov DWORD PTR [esp+16], eax
mov DWORD PTR [esp+20], edx
mov eax, 0
leave
ret
You can see it uses 2 registers to store that 64-bit integer. However, in the C/C++ code, it is only ONE variable.
I tried to replicate this in inline-assembly but I had to do:
#include <stdarg.h>
#include <stdint.h>
int64_t test_func(int n)
{
return 9223372036854775805;
}
int main()
{
int32_t rlow = 0, rhigh = 0;
asm(
"push $10\n"
"\tcall %P2"
: "=a"(rlow), "=d"(rhigh)
: "i"(&test_func) : "memory");
return 0;
}
And the output is:
test_func(int):
push ebp
mov ebp, esp
mov eax, -3
mov edx, 2147483647
pop ebp
ret
main:
push ebp
mov ebp, esp
sub esp, 16
mov DWORD PTR [ebp-8], 0
mov DWORD PTR [ebp-4], 0
push $10
call test_func(int)
mov DWORD PTR [ebp-8], eax
mov DWORD PTR [ebp-4], edx
mov eax, 0
leave
ret
Now you can see I had to manually place the lower and higher order bits into two separate integers. Then I perform shifting to make it into one 64-bit integer.
Is there a way to automatically get it to place into a single 64-bit integer without me having to supply it two 32-bit integers and then shifting the bits?

Answer 1

You want the "A" constraint, which binds a 64-bit value to the eax/edx register pair. Something like:
uint64_t r;
asm("push $10\n"
"\tcall %P1"
: "=A"(r) : "i"(&test_func) : "memory");
should do the trick.