What's the efficient way to swap two register variables in CUDA?

Question

I'm starting to write some CUDA code, and I want to do the equivalent of std::swap() for two variables within a kernel; they're in the register file (no spillage, not in some buffer, etc.). Suppose I have the following device code:

__device__ foo(/* some args here */) {

    /* etc. */

    int x = /* value v1 */;
    int y = /* value v2 */;

    /* etc. */

    swap(x,y);

    /* etc. */
}        

Now, I could just write

template <typename T> void swap ( T& a, T& b )
{
  T c(a); a=b; b=c;
}

but I wonder - isn't there some CUDA built-in for this functionality?

Notes:

• Yes, I want this to run for all threads.
• Never mind about whether I have enough registers or not. Assume that I have them.

Answer 1

I have considered the following test program

template <typename T> __device__ void inline swap_test_device1(T& a, T& b)
{
    T c(a); a=b; b=c;
}

template <typename T> __device__ void inline swap_test_device2(T a, T b)
{
    T c(a); a=b; b=c;
}

__global__ void swap_test_global(const int* __restrict__ input1, const int* __restrict__ input2, int* output1, int* output2) {

    int tx = threadIdx.x + blockIdx.x * blockDim.x;

    int x = input1[tx]*input1[tx];
    int y = input2[tx]*input2[tx];

    //swap_test_device2(x,y);
    swap_test_device1(x,y);

    output1[tx] = x;
    output2[tx] = y;

} 

and I have disassembled it. The result when using swap_test_device1 and swap_test_device2 is the same. The common disassembled code is the following

MOV R1, c[0x1][0x100];
S2R R0, SR_CTAID.X;
S2R R2, SR_TID.X;
MOV32I R9, 0x4;  
IMAD R3, R0, c[0x0][0x8], R2;
IMAD R6.CC, R3, R9, c[0x0][0x28];
IMAD.HI.X R7, R3, R9, c[0x0][0x2c];
IMAD R10.CC, R3, R9, c[0x0][0x20];
LD.E R2, [R6];                         loads input1[tx] and stores it in R2
IMAD.HI.X R11, R3, R9, c[0x0][0x24];
IMAD R4.CC, R3, R9, c[0x0][0x30];
LD.E R0, [R10];                        loads input2[tx] and stores it in R0
IMAD.HI.X R5, R3, R9, c[0x0][0x34];
IMAD R8.CC, R3, R9, c[0x0][0x38];
IMAD.HI.X R9, R3, R9, c[0x0][0x3c];
IMUL R2, R2, R2;                       R2 = R2 * R2
ST.E [R4], R2;                         stores input1[tx]*input1[tx] in global memory
IMUL R0, R0, R0;                       R0 = R0 * R0
ST.E [R8], R0;                         stores input2[tx]*input2[tx] in global memory
EXIT ;

It seems that the there is not an explicit swap in the disassembled code. In other words, the compiler, for this simple example, is capable to optimize the code directly writing x and y in the proper global memory locations.

EDIT

I have now considered the following more involved test case

__global__ void swap_test_global(const char* __restrict__ input1, const char* __restrict__ input2, char* output1, char* output2) {

    int tx = threadIdx.x + blockIdx.x * blockDim.x;

    char x = input1[tx];
    char y = input2[tx];

    //swap_test_device2(x,y);
    swap_test_device1(x,y);

    output1[tx] = (x >> 3) & y;
    output2[tx] = (y >> 5) & x;

 }

with the same above __device__ functions. The disassembled code is

MOV R1, c[0x1][0x100];              
S2R R0, SR_CTAID.X;                 
S2R R2, SR_TID.X;           
IMAD R0, R0, c[0x0][0x8], R2;       R0 = threadIdx.x + blockIdx.x * blockDim.x
BFE R7, R0, 0x11f;
IADD R8.CC, R0, c[0x0][0x28];
IADD.X R9, R7, c[0x0][0x2c];
IADD R10.CC, R0, c[0x0][0x20];
LD.E.S8 R4, [R8];                   R4 = x = input1[tx]
IADD.X R11, R7, c[0x0][0x24];
IADD R2.CC, R0, c[0x0][0x30];
LD.E.S8 R5, [R10];                  R5 = y = input2[tx]
IADD.X R3, R7, c[0x0][0x34];
IADD R12.CC, R0, c[0x0][0x38];
IADD.X R13, R7, c[0x0][0x3c];
SHR.U32 R0, R4, 0x3;                R0 = x >> 3
SHR.U32 R6, R5, 0x5;                R6 = y >> 5
LOP.AND R5, R0, R5;                 R5 = (x >> 3) & y
LOP.AND R0, R6, R4;                 R0 = (y >> 5) & x
ST.E.U8 [R2], R5;                   global memory store
ST.E.U8 [R12], R0;                  global memory store
EXIT ;

As it can be seen, there is still no apparent register swap.