What is the instruction number per cycle in fma with minus?

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If I use fma(a, b, c) in cuda, it means that the formula ab+c is calculated in a single ternary operation. But if I want to calculate -ab+c, does the invoking fma(-a, b, c) take one more multiply operation ?

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Unfortunately shader assembly language is undocumented at that level.

However we can try it out:

#!/bin/bash
cat <<EOF > fmatest.cu
__global__ void fma_plus(float *res, float a, float b, float c)
{
    *res = fma(a, b, c);
}

__global__ void fma_minus(float *res, float a, float b, float c)
{
    *res = fma(-a, b, c);
}
EOF
nvcc -arch sm_60 -c fmatest.cu
cuobjdump -sass fmatest.o

gives

code for sm_60
    Function : _Z9fma_minusPffff
.headerflags    @"EF_CUDA_SM60 EF_CUDA_PTX_SM(EF_CUDA_SM60)"
                                                                 /* 0x001fc400fe2007f6 */
    /*0008*/                   MOV R1, c[0x0][0x20];             /* 0x4c98078000870001 */
    /*0010*/                   MOV R0, c[0x0][0x148];            /* 0x4c98078005270000 */
    /*0018*/                   MOV R5, c[0x0][0x14c];            /* 0x4c98078005370005 */
                                                                 /* 0x001fc800fe8007f1 */
    /*0028*/                   MOV R2, c[0x0][0x140];            /* 0x4c98078005070002 */
    /*0030*/                   MOV R3, c[0x0][0x144];            /* 0x4c98078005170003 */
    /*0038*/                   FFMA R0, R0, -R5, c[0x0][0x150];  /* 0x5181028005470000 */
                                                                 /* 0x001ffc00ffe000f1 */
    /*0048*/                   STG.E [R2], R0;                   /* 0xeedc200000070200 */
    /*0050*/                   EXIT;                             /* 0xe30000000007000f */
    /*0058*/                   BRA 0x58;                         /* 0xe2400fffff87000f */
                                                                 /* 0x001f8000fc0007e0 */
    /*0068*/                   NOP;                              /* 0x50b0000000070f00 */
    /*0070*/                   NOP;                              /* 0x50b0000000070f00 */
    /*0078*/                   NOP;                              /* 0x50b0000000070f00 */
    ..................................


    Function : _Z8fma_plusPffff
.headerflags    @"EF_CUDA_SM60 EF_CUDA_PTX_SM(EF_CUDA_SM60)"
                                                                /* 0x001fc400fe2007f6 */
    /*0008*/                   MOV R1, c[0x0][0x20];            /* 0x4c98078000870001 */
    /*0010*/                   MOV R0, c[0x0][0x148];           /* 0x4c98078005270000 */
    /*0018*/                   MOV R5, c[0x0][0x14c];           /* 0x4c98078005370005 */
                                                                /* 0x001fc800fe8007f1 */
    /*0028*/                   MOV R2, c[0x0][0x140];           /* 0x4c98078005070002 */
    /*0030*/                   MOV R3, c[0x0][0x144];           /* 0x4c98078005170003 */
    /*0038*/                   FFMA R0, R0, R5, c[0x0][0x150];  /* 0x5180028005470000 */
                                                                /* 0x001ffc00ffe000f1 */
    /*0048*/                   STG.E [R2], R0;                  /* 0xeedc200000070200 */
    /*0050*/                   EXIT;                            /* 0xe30000000007000f */
    /*0058*/                   BRA 0x58;                        /* 0xe2400fffff87000f */
                                                                /* 0x001f8000fc0007e0 */
    /*0068*/                   NOP;                             /* 0x50b0000000070f00 */
    /*0070*/                   NOP;                             /* 0x50b0000000070f00 */
    /*0078*/                   NOP;                             /* 0x50b0000000070f00 */
    .................................

So the FFMA instruction can indeed take an additional sign to apply to the product (note that it is applied to b in the shader assembly instruction, however this gives the same result). You can try the same with double precision operands and other compute capabilities instead of sm_60 as well, which will give you similar results.