Could someone explain to me why there are 3 variants of the fused multiply-accumulate instruction: vfmadd132pd, vfmadd231pd and vfmadd213pd, while there is only one C intrinsics _mm256_fmadd_pd?
To make things simple, what is the difference between (in AT&T syntax)
vfmadd132pd %ymm0, %ymm1, %ymm2
vfmadd231pd %ymm0, %ymm1, %ymm2
vfmadd213pd %ymm0, %ymm1, %ymm2
I did not get any idea from Intel's intrinsics guide. I ask because I see all of them in the assembler output of a chunk of C code I wrote. Thanks.
A clean answer (re-formating answers below)
For variant ijk, the meaning of vfmaddijkpd:
op(i) * op(j) + op(k) -> op(1)op(4-i) * op(4-j) + op(4-k) -> op(3)where op(n) denotes the n-th operand after the instruction. So there is a reverse transform between the two:
n <- 4 - n
The fused multiply-add instructions multiply two (packed) values, add a third value, and then overwrite one of the values with the result. Only one of the three values can be a memory operand rather than a register.
The way it works is that all three instructions overwrite ymm0 and allow only ymm2 to be a memory operand. The choice of instruction determines which two operands are multiplied and which is added.
Assuming that ymm0 is the first operand in Intel syntax (or the last in AT&T syntax):
vfmadd132pd: ymm0 = ymm0 * ymm2/mem + ymm1
vfmadd231pd: ymm0 = ymm1 * ymm2/mem + ymm0
vfmadd213pd: ymm0 = ymm1 * ymm0 + ymm2/mem
When using the C intrinsics, this choice isn't necessary: The intrinsic does not overwrite a value but returns its result instead, and it allows all three values to be read from memory. The compiler will add memory reads/writes if needed, and will allocate a temporary register to store the result if it does not want any of the three values to be overwritten. It will choose one of the three instructions as it sees fit.