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gangti
Total Points:
35
Registered User
July 1, 2009 9:18 PM PDT
x64/ia64 Assemly Instructions in Code Porting
refer to:
http://software.intel.com/en-us/articles/use-explicit-prototyping/#comment-26974

May someone give an answer???

Contents:
file.c
==================================================
#undef NULL
#define NULL 0
void main() {
func(NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL);
}
void func(char *arg1, char *arg2, char *arg3,
char *arg4, char *arg5, char *arg6,
char *arg7, char *arg8, char *arg9)
{
//
}
==================================================

When I port above code from x86 to x64/ia64, I found:
A.
In windows x64 platform, when compiled with cl
(vs2005's x64 cross compiler)the first four param
will be transferred to func with universal registers,
using mov(same as movzx) assembly instructions, so
the upper 32 bits will be zeroed.
So in the above code, arg1~arg4 will be transferred
correctly, although arg5~arg9 may have problem.

But in linux x64 platform, when compiled with cc(gcc's
component), I found the first sixth parm will be
transferred with universal registers.
So in the above code, arg1~arg6 will be transferred
correctly, although arg7~arg9 may have problem.

B.
In windows ia64 platform, when compiled with cl
(vs2005's ia64 cross compiler)the first eight param
will be transferred to func with universal registers,
using mov(upper bits will be signed extended)assembly
instruction. In this case, that is 0 extended for the
upper bits. The other params will be put in the stack
memory, but they will be first put to registers, and
then stack memory.
So in the above code, all params will be transferred
correctly.

So the questions is, whether I am right or not?
May I rely on the knowledge of intermediate assembly
code, and just modify the code as follows for windows
x64??????
==================================================
#undef NULL
#define NULL 0
void main() {
func(NULL,NULL,NULL,NULL,(char*)NULL,
(char*)NULL,(char*)NULL,(char*)NULL,(char*)NULL);
}
void func(char *arg1, char *arg2, char *arg3,
char *arg4, char *arg5, char *arg6,
char *arg7, char *arg8, char *arg9)
{
//
}
==================================================

Another questions is that, why linux x64 put the first
six params to registers while windows x64 put the first
four params to registers????
The hardware(cpu especially) is the same, so why shouldn't
the calling conventions is the same????

--the end--
srimks
Total Points:
5,884
Status Points:
5,384
Brown Belt
July 2, 2009 12:03 PM PDT
Rate
 
#1
Quoting - gangti
refer to:
http://software.intel.com/en-us/articles/use-explicit-prototyping/#comment-26974

May someone give an answer???

Contents:
file.c
==================================================
#undef NULL
#define NULL 0
void main() {
func(NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL);
}
void func(char *arg1, char *arg2, char *arg3,
char *arg4, char *arg5, char *arg6,
char *arg7, char *arg8, char *arg9)
{
//
}
==================================================

When I port above code from x86 to x64/ia64, I found:
A.
In windows x64 platform, when compiled with cl
(vs2005's x64 cross compiler)the first four param
will be transferred to func with universal registers,
using mov(same as movzx) assembly instructions, so
the upper 32 bits will be zeroed.
So in the above code, arg1~arg4 will be transferred
correctly, although arg5~arg9 may have problem.

But in linux x64 platform, when compiled with cc(gcc's
component), I found the first sixth parm will be
transferred with universal registers.
So in the above code, arg1~arg6 will be transferred
correctly, although arg7~arg9 may have problem.

B.
In windows ia64 platform, when compiled with cl
(vs2005's ia64 cross compiler)the first eight param
will be transferred to func with universal registers,
using mov(upper bits will be signed extended)assembly
instruction. In this case, that is 0 extended for the
upper bits. The other params will be put in the stack
memory, but they will be first put to registers, and
then stack memory.
So in the above code, all params will be transferred
correctly.

So the questions is, whether I am right or not?
May I rely on the knowledge of intermediate assembly
code, and just modify the code as follows for windows
x64??????
==================================================
#undef NULL
#define NULL 0
void main() {
func(NULL,NULL,NULL,NULL,(char*)NULL,
(char*)NULL,(char*)NULL,(char*)NULL,(char*)NULL);
}
void func(char *arg1, char *arg2, char *arg3,
char *arg4, char *arg5, char *arg6,
char *arg7, char *arg8, char *arg9)
{
//
}
==================================================

Another questions is that, why linux x64 put the first
six params to registers while windows x64 put the first
four params to registers????
The hardware(cpu especially) is the same, so why shouldn't
the calling conventions is the same????

--the end--

Ans.[A]: The calling convention for 64-bit Windows is - the first parameter is transferred in RCX if it is an integer or in XMM0 if it is a float or double, second in RDX or XMM1, third in RB or XMM2 and fourth in R9 or in XMM9. No more than 4 parameters can be transferred in registers, ragardless of types. Any further parameters are transferred on the stack with the first parameter at the lowest address & aligned by 8.
                                                                       While in Linux, first six parameters are transferred in RDI, RSI, RDX, RCX, R8 & R9 resp.. The first 8 floating-point parameters are transferred in XMM0 - XMM7. All these registers can be used, so in maximum fourteen parameters can be transferred in registers. Any further parameters are transferred in stack with first parameter at the lowest address and aligned by 8.

Note that 64-bit Linux does not use the same registers for parameters transfer as done by 64-bit Windows. Also, for 64-bit, the calling save status registers for Windows(SI, RDI, XMM6 - XMM15) are different form Linux which means that calling conventions are different for Windows and Linux for 64-bit.

All compilers for 64-bit Windows use the COFF/PE32+ format, while compilers for Linux use ELF64 format.

HIH.

~BR
Mukkaysh Srivastav

gangti
Total Points:
35
Registered User
July 2, 2009 9:52 PM PDT
Rate
 
#2 Reply to #1
Quoting - srimks

Ans.[A]: The calling convention for 64-bit Windows is - the first parameter is transferred in RCX if it is an integer or in XMM0 if it is a float or double, second in RDX or XMM1, third in RB or XMM2 and fourth in R9 or in XMM9. No more than 4 parameters can be transferred in registers, ragardless of types. Any further parameters are transferred on the stack with the first parameter at the lowest address & aligned by 8.
                                                                       While in Linux, first six parameters are transferred in RDI, RSI, RDX, RCX, R8 & R9 resp.. The first 8 floating-point parameters are transferred in XMM0 - XMM7. All these registers can be used, so in maximum fourteen parameters can be transferred in registers. Any further parameters are transferred in stack with first parameter at the lowest address and aligned by 8.

Note that 64-bit Linux does not use the same registers for parameters transfer as done by 64-bit Windows. Also, for 64-bit, the calling save status registers for Windows(SI, RDI, XMM6 - XMM15) are different form Linux which means that calling conventions are different for Windows and Linux for 64-bit.

All compilers for 64-bit Windows use the COFF/PE32+ format, while compilers for Linux use ELF64 format.

HIH.

~BR
Mukkaysh Srivastav

I appreciate you answer of my question, and the whole knowledge of calling conventions.

My problem is that I just want to modify the arguments from 5 to 9 as following
code shows, for windows x64. Am I right or not?
I want to get an official answer from intel to persuade my boss. So I am looking
forward to you reply...
==================================================
#undef NULL
#define NULL 0
void main() {
func(NULL,NULL,NULL,NULL,(char*)NULL,
(char*)NULL,(char*)NULL,(char*)NULL,(char*)NULL);
}
void func(char *arg1, char *arg2, char *arg3,
char *arg4, char *arg5, char *arg6,
char *arg7, char *arg8, char *arg9)
{
//
}
==================================================

Though another way to approach this is to give an explictit prototype, I give the
above approach first. It's a pity that  I cannot change the approach now.

jimdempseyatthecove
Total Points:
34,847
Status Points:
34,847
Black Belt
July 6, 2009 9:42 AM PDT
Rate
 
#3 Reply to #2

Use function prototype

#undef NULL
#define NULL 0
// declare forward reference to function func
void func(char *arg1, char *arg2, char *arg3, 
char *arg4, char *arg5, char *arg6, 
char *arg7, char *arg8, char *arg9);

void main() {
  func(NULL,NULL,NULL,NULL,NULL,
  NULL,NULL,NULL,NULL);
}

void func(char *arg1, char *arg2, char *arg3, 
char *arg4, char *arg5, char *arg6, 
char *arg7, char *arg8, char *arg9)
{
//
}




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