Swapping the value of two variables

[haminin]

Hi there

Swapping the value of two variables (a and b) without a temporary variable, can be done as: a = a + b; b = a - b; a = a - b.

But could someone let me know if there is a direct command forthis purpose in FORTRAN.

Many thanks

Hamid

[TimP]

Quoting - haminin

Swapping the value of two variables (a and b) without a temporary variable, can be done as: a = a + b; b = a - b; a = a - b.

But could someone let me know if there is a direct command forthis purpose in FORTRAN.

No, any optimizing compiler in any language can do a good job with it.

[ZlamalJakub]

Why to do swaping in this manner. I suppose compiler creates code which change variables in processor registers (and then puts values to the memory) and does not create intermediate variable in memory, so swap using intermediate variable will be faster than mathematic operatins as You suggested.

I think there is no FORTRAN routine to do swap of variables, You must program it yourself.

Jakub

[jimdempseyatthecove]

try using ieor (xor) intrinsic function

[cpp]a = ieor(a,b)
b = ieor(b,a)
a = ieor(a,b)
[/cpp]

This may eliminate a neg instruction or two
Also experiment with swapping the order of the a,b or b,a in the function arguments.

Look at the dissassembly code.

Jim Dempsey

[jimdempseyatthecove]

Quoting - jimdempseyatthecove

try using ieor (xor) intrinsic function

[cpp]a = ieor(a,b)
b = ieor(b,a)
a = ieor(a,b)
[/cpp]

This may eliminate a neg instruction or two
Also experiment with swapping the order of the a,b or b,a in the function arguments.

Look at the dissassembly code.

Jim Dempsey

Also, an inline subroutine that declares a temp may optimize the temp out of the code.
You can declare the subroutine as PURE as long as a and b are declared with INTENT(INOUT).

Jim Dempsey

[jparsly1]

Go ahead and create a temporary variable. It's a lot clearer what you are doing, and a good compiler is likely
to optimize the temp variable out of existence anyhow and just do the swap in registers.

Also, if 'a' and 'b' are floating point variables, the math method can alter the value of the variables through truncation error.

Consider this carefully constructed case:

read*4 a,b
a = 100000000.
b = 1.
write(*,*) a,b
a = a + b
b = a - b
a = a - b
write(*,*) a,b
stop
end

Depending on your compiler, or just using different options on the same compiler, you can get
different results.The first two compilers I just tried both give a = 0, b = 1.e8 as the final result.

Quoting - haminin

Hi there

Swapping the value of two variables (a and b) without a temporary variable, can be done as: a = a + b; b = a - b; a = a - b.

But could someone let me know if there is a direct command forthis purpose in FORTRAN.

Many thanks

Hamid

[TimP]

Quoting - Quba

Why to do swaping in this manner. I suppose compiler creates code which change variables in processor registers (and then puts values to the memory) and does not create intermediate variable in memory, so swap using intermediate variable will be faster than mathematic operatins as You suggested.

I think there is no FORTRAN routine to do swap of variables, You must program it yourself.

Yes, on the face of it, inserting the arithmetic or xor operations involves the latency of 3 integer operations rather than 2 moves, plus another move which can be done in parallel on any pipeline CPU, saving at best 1 register use. It's been over 20 years since I had access to a machine with only 2 integer registers.

[TimP]

Quoting - jparsly@charter.net

read*4 a,b
a = 100000000.
b = 1.

"carefully" not using integer operations?
As Jim pointed out, integer operations have to be used, not floating point. If you were using x87 registers, the move method undoubtedly would be superior. Even floating point moves might raise a question on some machines about whether subnormals are preserved.

[jimdempseyatthecove]

4 movs 2 integer operations

[cpp]; a = ieor(a,b)
  mov rax, qword ptr   ; assuming a not in register already
  mov rbx, qword ptr   ; assuming b not in register already
  xor rbx, rax
; b = ieor(b,a)
  mov qword ptr , rbx
; a = ieor(a,b)
  xor rax,rbx
  mov qword ptr , rax
[/cpp]

Too bad there isn't an intrinsic for swap

[cpp]; swap(a,b)
  mov rax, qword ptr   ; assuming a not in register already
  xchg qword ptr , rax
  mov qword ptr , rax

[/cpp]

Jim Dempsey