OV - attention-gcc-skl256 - Loop 31

0x2dc0 VMOVSS	(%R14),%XMM0    [2]

0x2dc5 ADD	$0x4,%R14

0x2dc9 VSUBSS	-0x38(%RBP),%XMM0,%XMM0    [1]

0x2dce CALL	1110 <expf@plt>

0x2dd3 VADDSS	-0x34(%RBP),%XMM0,%XMM5    [1]

0x2dd8 VMOVSS	%XMM5,-0x34(%RBP)    [1]

0x2ddd CMP	%RBX,%R14

0x2de0 JNE	2dc0

/home/eoseret/Applications/llm-attention/attention_v2.cpp: 52 - 53

--------------------------------------------------------------------------------

52:       for (int idx = 0; idx <= row; ++idx) // vectorised

53:          sum += expf(S_row[idx] - max_val);

Coverage (%)	Name	Source Location	Module
►100.00+	main	attention_v2.cpp:283	attention-gcc-skl256
○	__libc_init_first		libc.so.6
○	__libc_start_main		libc.so.6
○	main	new_allocator.h:183	attention-gcc-skl256

min	med	avg	max

Percentile Index	10	20	30	40	50	60	70	80	90	100
Value

min	med	avg	max

Percentile Index	10	20	30	40	50	60	70	80	90	100
Value

Path /

Metric	Value
CQA speedup if no scalar integer	1.33
CQA speedup if FP arith vectorized	1.33
CQA speedup if fully vectorized	16.00
CQA speedup if no inter-iteration dependency	NA
CQA speedup if next bottleneck killed	1.09
Bottlenecks	micro-operation queue, P4,
Function	softmax(float const, float, float*, int)
Source	attention_v2.cpp:52-53
Source loop unroll info	not unrolled or unrolled with no peel/tail loop
Source loop unroll confidence level	max
Unroll/vectorization loop type	NA
Unroll factor	NA
CQA cycles	2.00
CQA cycles if no scalar integer	1.50
CQA cycles if FP arith vectorized	1.50
CQA cycles if fully vectorized	0.13
Front-end cycles	2.00
P0 cycles	1.50
P1 cycles	1.00
P2 cycles	1.83
P3 cycles	1.50
P4 cycles	2.00
P5 cycles	1.00
P6 cycles	1.50
P7 cycles	1.67
DIV/SQRT cycles	0.00
Inter-iter dependencies cycles	1
FE+BE cycles (UFS)	NA
Stall cycles (UFS)	NA
Nb insns	8.00
Nb uops	8.00
Nb loads	3.00
Nb stores	1.00
Nb stack references	2.00
FLOP/cycle	1.00
Nb FLOP add-sub	2.00
Nb FLOP mul	0.00
Nb FLOP fma	0.00
Nb FLOP div	0.00
Nb FLOP rcp	0.00
Nb FLOP sqrt	0.00
Nb FLOP rsqrt	0.00
Bytes/cycle	8.00
Bytes prefetched	0.00
Bytes loaded	12.00
Bytes stored	4.00
Stride 0	1.00
Stride 1	1.00
Stride n	0.00
Stride unknown	0.00
Stride indirect	0.00
Vectorization ratio all	0.00
Vectorization ratio load	0.00
Vectorization ratio store	0.00
Vectorization ratio mul	NA
Vectorization ratio add_sub	0.00
Vectorization ratio fma	NA
Vectorization ratio div_sqrt	NA
Vectorization ratio other	NA
Vector-efficiency ratio all	6.25
Vector-efficiency ratio load	6.25
Vector-efficiency ratio store	6.25
Vector-efficiency ratio mul	NA
Vector-efficiency ratio add_sub	6.25
Vector-efficiency ratio fma	NA
Vector-efficiency ratio div_sqrt	NA
Vector-efficiency ratio other	NA

Metric	Value
CQA speedup if no scalar integer	1.33
CQA speedup if FP arith vectorized	1.33
CQA speedup if fully vectorized	16.00
CQA speedup if no inter-iteration dependency	NA
CQA speedup if next bottleneck killed	1.09
Bottlenecks	micro-operation queue, P4,
Function	softmax(float const, float, float*, int)
Source	attention_v2.cpp:52-53
Source loop unroll info	not unrolled or unrolled with no peel/tail loop
Source loop unroll confidence level	max
Unroll/vectorization loop type	NA
Unroll factor	NA
CQA cycles	2.00
CQA cycles if no scalar integer	1.50
CQA cycles if FP arith vectorized	1.50
CQA cycles if fully vectorized	0.13
Front-end cycles	2.00
P0 cycles	1.50
P1 cycles	1.00
P2 cycles	1.83
P3 cycles	1.50
P4 cycles	2.00
P5 cycles	1.00
P6 cycles	1.50
P7 cycles	1.67
DIV/SQRT cycles	0.00
Inter-iter dependencies cycles	1
FE+BE cycles (UFS)	NA
Stall cycles (UFS)	NA
Nb insns	8.00
Nb uops	8.00
Nb loads	3.00
Nb stores	1.00
Nb stack references	2.00
FLOP/cycle	1.00
Nb FLOP add-sub	2.00
Nb FLOP mul	0.00
Nb FLOP fma	0.00
Nb FLOP div	0.00
Nb FLOP rcp	0.00
Nb FLOP sqrt	0.00
Nb FLOP rsqrt	0.00
Bytes/cycle	8.00
Bytes prefetched	0.00
Bytes loaded	12.00
Bytes stored	4.00
Stride 0	1.00
Stride 1	1.00
Stride n	0.00
Stride unknown	0.00
Stride indirect	0.00
Vectorization ratio all	0.00
Vectorization ratio load	0.00
Vectorization ratio store	0.00
Vectorization ratio mul	NA
Vectorization ratio add_sub	0.00
Vectorization ratio fma	NA
Vectorization ratio div_sqrt	NA
Vectorization ratio other	NA
Vector-efficiency ratio all	6.25
Vector-efficiency ratio load	6.25
Vector-efficiency ratio store	6.25
Vector-efficiency ratio mul	NA
Vector-efficiency ratio add_sub	6.25
Vector-efficiency ratio fma	NA
Vector-efficiency ratio div_sqrt	NA
Vector-efficiency ratio other	NA

Path /

Average path: Display a virtual path defined by average values of all real paths

Function	softmax(float const, float, float*, int)
Source file and lines	attention_v2.cpp:52-53
Module	attention-gcc-skl256

The loop is defined in /home/eoseret/Applications/llm-attention/attention_v2.cpp:52-53.

The related source loop is not unrolled or unrolled with no peel/tail loop.

gain
potential
hint
expert

Code clean check

Detected a slowdown caused by scalar integer instructions (typically used for address computation). By removing them, you can lower the cost of an iteration from 2.00 to 1.50 cycles (1.33x speedup).

Workaround

Try to reorganize arrays of structures to structures of arrays
Consider to permute loops (see vectorization gain report)

Vectorization

Your loop is not vectorized. 16 data elements could be processed at once in vector registers.

By vectorizing your loop, you can lower the cost of an iteration from 2.00 to 0.12 cycles (16.00x speedup).

Details

All SSE/AVX instructions are used in scalar version (process only one data element in vector registers). Since your execution units are vector units, only a vectorized loop can use their full power.

Workaround

Try another compiler or update/tune your current one
Remove inter-iterations dependences from your loop and make it unit-stride:
- If your arrays have 2 or more dimensions, check whether elements are accessed contiguously and, otherwise, try to permute loops accordingly: C storage order is row-major: for(i) for(j) a[j][i] = b[j][i]; (slow, non stride 1) => for(i) for(j) a[i][j] = b[i][j]; (fast, stride 1)
- If your loop streams arrays of structures (AoS), try to use structures of arrays instead (SoA): for(i) a[i].x = b[i].x; (slow, non stride 1) => for(i) a.x[i] = b.x[i]; (fast, stride 1)

Execution units bottlenecks

Performance is limited by writing data to caches/RAM (the store unit is a bottleneck). By removing all these bottlenecks, you can lower the cost of an iteration from 2.00 to 1.83 cycles (1.09x speedup).

Workaround

Write less array elements
Provide more information to your compiler:
- hardcode the bounds of the corresponding 'for' loop

No data for this section

CALL instructions

Detected function call instructions.

Details

Calling (and then returning from) a function prevents many compiler optimizations (like vectorization), breaks control flow (which reduces pipeline performance) and executes extra instructions to save/restore the registers used inside it, which is very expensive (dozens of cycles). Consider to inline small functions.

expf@plt: 1 occurrences►
- /home/eoseret/Applications/llm-attention/attention_v2.cpp:53

Type of elements and instruction set

2 SSE or AVX instructions are processing arithmetic or math operations on single precision FP elements in scalar mode (one at a time).

Matching between your loop (in the source code) and the binary loop

The binary loop is composed of 2 FP arithmetical operations:

2: addition or subtraction

The binary loop is loading 12 bytes (3 single precision FP elements). The binary loop is storing 4 bytes (1 single precision FP elements).

Arithmetic intensity

Arithmetic intensity is 0.12 FP operations per loaded or stored byte.

Unroll opportunity

Loop is data access bound.

Workaround

Unroll your loop if trip count is significantly higher than target unroll factor and if some data references are common to consecutive iterations. This can be done manually. Or by recompiling with -funroll-loops and/or -floop-unroll-and-jam. Or with the unroll (resp. unroll_and_jam) directive on top of the inner (resp. surrounding) loop. You can enforce an unroll factor: #pragma GCC unroll N

General properties

nb instructions	8
nb uops	8
loop length	34
used x86 registers	3
used mmx registers	0
used xmm registers	2
used ymm registers	0
used zmm registers	0
nb stack references	2

Front-end

ASSUMED MACRO FUSION FIT IN UOP CACHE

micro-operation queue	2.00 cycles
front end	2.00 cycles

Back-end

	P0	P1	P2	P3	P4	P5	P6	P7
uops	1.50	1.00	1.83	1.50	2.00	1.00	1.50	1.67
cycles	1.50	1.00	1.83	1.50	2.00	1.00	1.50	1.67

Execution ports to units layout:

P0 (256 bits): VPU, BRU, ALU, DIV/SQRT
P1 (256 bits): ALU, VPU
P2 (512 bits): store address, load
P3 (512 bits): store address, load
P4 (512 bits): store data
P5 (512 bits): ALU, VPU
P6: ALU, BRU
P7: store address

Cycles executing div or sqrt instructions	NA
Longest recurrence chain latency (RecMII)	1.00

Cycles summary

Front-end	2.00
Dispatch	2.00
Data deps.	1.00
Overall L1	2.00

Vectorization ratios

all	0%
load	0%
store	0%
mul	NA (no mul vectorizable/vectorized instructions)
add-sub	0%
fma	NA (no fma vectorizable/vectorized instructions)
div/sqrt	NA (no div/sqrt vectorizable/vectorized instructions)
other	NA (no other vectorizable/vectorized instructions)

Vector efficiency ratios

all	6%
load	6%
store	6%
mul	NA (no mul vectorizable/vectorized instructions)
add-sub	6%
fma	NA (no fma vectorizable/vectorized instructions)
div/sqrt	NA (no div/sqrt vectorizable/vectorized instructions)
other	NA (no other vectorizable/vectorized instructions)

Cycles and memory resources usage

Assuming all data fit into the L1 cache, each iteration of the binary loop takes 2.00 cycles. At this rate:

4% of peak load performance is reached (6.00 out of 128.00 bytes loaded per cycle (GB/s @ 1GHz))
3% of peak store performance is reached (2.00 out of 64.00 bytes stored per cycle (GB/s @ 1GHz))

Front-end bottlenecks

Performance is limited by instruction throughput (loading/decoding program instructions to execution core) (front-end is a bottleneck). By removing all these bottlenecks, you can lower the cost of an iteration from 2.00 to 1.83 cycles (1.09x speedup).

ASM code

In the binary file, the address of the loop is: 2dc0

Instruction	Nb FU	P0	P1	P2	P3	P4	P5	P6	P7	Latency	Recip. throughput	Vectorization
VMOVSS (%R14),%XMM0	1	0	0	0.50	0.50	0	0	0	0	4-5	0.50	scal (6.3%)
ADD $0x4,%R14	1	0.25	0.25	0	0	0	0.25	0.25	0	1	0.25	N/A
VSUBSS -0x38(%RBP),%XMM0,%XMM0	1	0.50	0.50	0.50	0.50	0	0	0	0	4	0.50	scal (6.3%)
CALL 1110 <expf@plt>	2	0	0	0.33	0.33	1	0	1	0.33	0	1	N/A
VADDSS -0x34(%RBP),%XMM0,%XMM5	1	0.50	0.50	0.50	0.50	0	0	0	0	4	0.50	scal (6.3%)
VMOVSS %XMM5,-0x34(%RBP)	1	0	0	0.33	0.33	1	0	0	0.33	3	1	scal (6.3%)
CMP %RBX,%R14	1	0.25	0.25	0	0	0	0.25	0.25	0	1	0.25	N/A
JNE 2dc0 <_Z7softmaxPKfPfS1_i+0x150>	1	0.50	0	0	0	0	0	0.50	0	0	0.50-1	N/A

Function	softmax(float const, float, float*, int)
Source file and lines	attention_v2.cpp:52-53
Module	attention-gcc-skl256

The loop is defined in /home/eoseret/Applications/llm-attention/attention_v2.cpp:52-53.

The related source loop is not unrolled or unrolled with no peel/tail loop.

gain
potential
hint
expert

Code clean check

Detected a slowdown caused by scalar integer instructions (typically used for address computation). By removing them, you can lower the cost of an iteration from 2.00 to 1.50 cycles (1.33x speedup).

Workaround

Try to reorganize arrays of structures to structures of arrays
Consider to permute loops (see vectorization gain report)

Vectorization

Your loop is not vectorized. 16 data elements could be processed at once in vector registers.

By vectorizing your loop, you can lower the cost of an iteration from 2.00 to 0.12 cycles (16.00x speedup).

Details

All SSE/AVX instructions are used in scalar version (process only one data element in vector registers). Since your execution units are vector units, only a vectorized loop can use their full power.

Workaround

Try another compiler or update/tune your current one
Remove inter-iterations dependences from your loop and make it unit-stride:
- If your arrays have 2 or more dimensions, check whether elements are accessed contiguously and, otherwise, try to permute loops accordingly: C storage order is row-major: for(i) for(j) a[j][i] = b[j][i]; (slow, non stride 1) => for(i) for(j) a[i][j] = b[i][j]; (fast, stride 1)
- If your loop streams arrays of structures (AoS), try to use structures of arrays instead (SoA): for(i) a[i].x = b[i].x; (slow, non stride 1) => for(i) a.x[i] = b.x[i]; (fast, stride 1)

Execution units bottlenecks

Performance is limited by writing data to caches/RAM (the store unit is a bottleneck). By removing all these bottlenecks, you can lower the cost of an iteration from 2.00 to 1.83 cycles (1.09x speedup).

Workaround

Write less array elements
Provide more information to your compiler:
- hardcode the bounds of the corresponding 'for' loop

No data for this section

CALL instructions

Detected function call instructions.

Details

Calling (and then returning from) a function prevents many compiler optimizations (like vectorization), breaks control flow (which reduces pipeline performance) and executes extra instructions to save/restore the registers used inside it, which is very expensive (dozens of cycles). Consider to inline small functions.

expf@plt: 1 occurrences►
- /home/eoseret/Applications/llm-attention/attention_v2.cpp:53

Type of elements and instruction set

2 SSE or AVX instructions are processing arithmetic or math operations on single precision FP elements in scalar mode (one at a time).

Matching between your loop (in the source code) and the binary loop

The binary loop is composed of 2 FP arithmetical operations:

2: addition or subtraction

The binary loop is loading 12 bytes (3 single precision FP elements). The binary loop is storing 4 bytes (1 single precision FP elements).

Arithmetic intensity

Arithmetic intensity is 0.12 FP operations per loaded or stored byte.

Unroll opportunity

Loop is data access bound.

Workaround

Unroll your loop if trip count is significantly higher than target unroll factor and if some data references are common to consecutive iterations. This can be done manually. Or by recompiling with -funroll-loops and/or -floop-unroll-and-jam. Or with the unroll (resp. unroll_and_jam) directive on top of the inner (resp. surrounding) loop. You can enforce an unroll factor: #pragma GCC unroll N

General properties

nb instructions	8
nb uops	8
loop length	34
used x86 registers	3
used mmx registers	0
used xmm registers	2
used ymm registers	0
used zmm registers	0
nb stack references	2

Front-end

ASSUMED MACRO FUSION FIT IN UOP CACHE

micro-operation queue	2.00 cycles
front end	2.00 cycles

Back-end

	P0	P1	P2	P3	P4	P5	P6	P7
uops	1.50	1.00	1.83	1.50	2.00	1.00	1.50	1.67
cycles	1.50	1.00	1.83	1.50	2.00	1.00	1.50	1.67

Execution ports to units layout:

P0 (256 bits): VPU, BRU, ALU, DIV/SQRT
P1 (256 bits): ALU, VPU
P2 (512 bits): store address, load
P3 (512 bits): store address, load
P4 (512 bits): store data
P5 (512 bits): ALU, VPU
P6: ALU, BRU
P7: store address

Cycles executing div or sqrt instructions	NA
Longest recurrence chain latency (RecMII)	1.00

Cycles summary

Front-end	2.00
Dispatch	2.00
Data deps.	1.00
Overall L1	2.00

Vectorization ratios

all	0%
load	0%
store	0%
mul	NA (no mul vectorizable/vectorized instructions)
add-sub	0%
fma	NA (no fma vectorizable/vectorized instructions)
div/sqrt	NA (no div/sqrt vectorizable/vectorized instructions)
other	NA (no other vectorizable/vectorized instructions)

Vector efficiency ratios

all	6%
load	6%
store	6%
mul	NA (no mul vectorizable/vectorized instructions)
add-sub	6%
fma	NA (no fma vectorizable/vectorized instructions)
div/sqrt	NA (no div/sqrt vectorizable/vectorized instructions)
other	NA (no other vectorizable/vectorized instructions)

Cycles and memory resources usage

Assuming all data fit into the L1 cache, each iteration of the binary loop takes 2.00 cycles. At this rate:

4% of peak load performance is reached (6.00 out of 128.00 bytes loaded per cycle (GB/s @ 1GHz))
3% of peak store performance is reached (2.00 out of 64.00 bytes stored per cycle (GB/s @ 1GHz))

Front-end bottlenecks

Performance is limited by instruction throughput (loading/decoding program instructions to execution core) (front-end is a bottleneck). By removing all these bottlenecks, you can lower the cost of an iteration from 2.00 to 1.83 cycles (1.09x speedup).

ASM code

In the binary file, the address of the loop is: 2dc0

Instruction	Nb FU	P0	P1	P2	P3	P4	P5	P6	P7	Latency	Recip. throughput	Vectorization
VMOVSS (%R14),%XMM0	1	0	0	0.50	0.50	0	0	0	0	4-5	0.50	scal (6.3%)
ADD $0x4,%R14	1	0.25	0.25	0	0	0	0.25	0.25	0	1	0.25	N/A
VSUBSS -0x38(%RBP),%XMM0,%XMM0	1	0.50	0.50	0.50	0.50	0	0	0	0	4	0.50	scal (6.3%)
CALL 1110 <expf@plt>	2	0	0	0.33	0.33	1	0	1	0.33	0	1	N/A
VADDSS -0x34(%RBP),%XMM0,%XMM5	1	0.50	0.50	0.50	0.50	0	0	0	0	4	0.50	scal (6.3%)
VMOVSS %XMM5,-0x34(%RBP)	1	0	0	0.33	0.33	1	0	0	0.33	3	1	scal (6.3%)
CMP %RBX,%R14	1	0.25	0.25	0	0	0	0.25	0.25	0	1	0.25	N/A
JNE 2dc0 <_Z7softmaxPKfPfS1_i+0x150>	1	0.50	0	0	0	0	0	0.50	0	0	0.50-1	N/A

Report Configuration

Code clean check

Workaround

Vectorization

Details

Workaround

Execution units bottlenecks

Workaround

CALL instructions

Details

Type of elements and instruction set

Matching between your loop (in the source code) and the binary loop

Arithmetic intensity

Unroll opportunity

Workaround

General properties

Front-end

Back-end

Cycles summary

Vectorization ratios

Vector efficiency ratios

Cycles and memory resources usage

Front-end bottlenecks

ASM code

Code clean check

Workaround

Vectorization

Details

Workaround

Execution units bottlenecks

Workaround

CALL instructions

Details

Type of elements and instruction set

Matching between your loop (in the source code) and the binary loop

Arithmetic intensity

Unroll opportunity

Workaround

General properties

Front-end

Back-end

Cycles summary

Vectorization ratios

Vector efficiency ratios

Cycles and memory resources usage

Front-end bottlenecks

ASM code