/* * arch/alpha/lib/ev6-memset.S * * This is an efficient (and relatively small) implementation of the C library * "memset()" function for the 21264 implementation of Alpha. * * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> * * Much of the information about 21264 scheduling/coding comes from: * Compiler Writer's Guide for the Alpha 21264 * abbreviated as 'CWG' in other comments here * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html * Scheduling notation: * E - either cluster * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 * The algorithm for the leading and trailing quadwords remains the same, * however the loop has been unrolled to enable better memory throughput, * and the code has been replicated for each of the entry points: __memset * and __memsetw to permit better scheduling to eliminate the stalling * encountered during the mask replication. * A future enhancement might be to put in a byte store loop for really * small (say < 32 bytes) memset()s. Whether or not that change would be * a win in the kernel would depend upon the contextual usage. * WARNING: Maintaining this is going to be more work than the above version, * as fixes will need to be made in multiple places. The performance gain * is worth it. */ .set noat .set noreorder .text .globl memset .globl __memset .globl ___memset .globl __memsetw .globl __constant_c_memset .ent ___memset .align 5 ___memset: .frame $30,0,$26,0 .prologue 0 /* * Serious stalling happens. The only way to mitigate this is to * undertake a major re-write to interleave the constant materialization * with other parts of the fall-through code. This is important, even * though it makes maintenance tougher. * Do this later. */ and $17,255,$1 # E : 00000000000000ch insbl $17,1,$2 # U : 000000000000ch00 bis $16,$16,$0 # E : return value ble $18,end_b # U : zero length requested? addq $18,$16,$6 # E : max address to write to bis $1,$2,$17 # E : 000000000000chch insbl $1,2,$3 # U : 0000000000ch0000 insbl $1,3,$4 # U : 00000000ch000000 or $3,$4,$3 # E : 00000000chch0000 inswl $17,4,$5 # U : 0000chch00000000 xor $16,$6,$1 # E : will complete write be within one quadword? inswl $17,6,$2 # U : chch000000000000 or $17,$3,$17 # E : 00000000chchchch or $2,$5,$2 # E : chchchch00000000 bic $1,7,$1 # E : fit within a single quadword? and $16,7,$3 # E : Target addr misalignment or $17,$2,$17 # E : chchchchchchchch beq $1,within_quad_b # U : nop # E : beq $3,aligned_b # U : target is 0mod8 /* * Target address is misaligned, and won't fit within a quadword */ ldq_u $4,0($16) # L : Fetch first partial bis $16,$16,$5 # E : Save the address insql $17,$16,$2 # U : Insert new bytes subq $3,8,$3 # E : Invert (for addressing uses) addq $18,$3,$18 # E : $18 is new count ($3 is negative) mskql $4,$16,$4 # U : clear relevant parts of the quad subq $16,$3,$16 # E : $16 is new aligned destination bis $2,$4,$1 # E : Final bytes nop stq_u $1,0($5) # L : Store result nop nop .align 4 aligned_b: /* * We are now guaranteed to be quad aligned, with at least * one partial quad to write. */ sra $18,3,$3 # U : Number of remaining quads to write and $18,7,$18 # E : Number of trailing bytes to write bis $16,$16,$5 # E : Save dest address beq $3,no_quad_b # U : tail stuff only /* * it's worth the effort to unroll this and use wh64 if possible * Lifted a bunch of code from clear_user.S * At this point, entry values are: * $16 Current destination address * $5 A copy of $16 * $6 The max quadword address to write to * $18 Number trailer bytes * $3 Number quads to write */ and $16, 0x3f, $2 # E : Forward work (only useful for unrolled loop) subq $3, 16, $4 # E : Only try to unroll if > 128 bytes subq $2, 0x40, $1 # E : bias counter (aligning stuff 0mod64) blt $4, loop_b # U : /* * We know we've got at least 16 quads, minimum of one trip * through unrolled loop. Do a quad at a time to get us 0mod64 * aligned. */ nop # E : nop # E : nop # E : beq $1, $bigalign_b # U : $alignmod64_b: stq $17, 0($5) # L : subq $3, 1, $3 # E : For consistency later addq $1, 8, $1 # E : Increment towards zero for alignment addq $5, 8, $4 # E : Initial wh64 address (filler instruction) nop nop addq $5, 8, $5 # E : Inc address blt $1, $alignmod64_b # U : $bigalign_b: /* * $3 - number quads left to go * $5 - target address (aligned 0mod64) * $17 - mask of stuff to store * Scratch registers available: $7, $2, $4, $1 * we know that we'll be taking a minimum of one trip through * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle * Assumes the wh64 needs to be for 2 trips through the loop in the future * The wh64 is issued on for the starting destination address for trip +2 * through the loop, and if there are less than two trips left, the target * address will be for the current trip. */ $do_wh64_b: wh64 ($4) # L1 : memory subsystem write hint subq $3, 24, $2 # E : For determining future wh64 addresses stq $17, 0($5) # L : nop # E : addq $5, 128, $4 # E : speculative target of next wh64 stq $17, 8($5) # L : stq $17, 16($5) # L : addq $5, 64, $7 # E : Fallback address for wh64 (== next trip addr) stq $17, 24($5) # L : stq $17, 32($5) # L : cmovlt $2, $7, $4 # E : Latency 2, extra mapping cycle nop stq $17, 40($5) # L : stq $17, 48($5) # L : subq $3, 16, $2 # E : Repeat the loop at least once more? nop stq $17, 56($5) # L : addq $5, 64, $5 # E : subq $3, 8, $3 # E : bge $2, $do_wh64_b # U : nop nop nop beq $3, no_quad_b # U : Might have finished already .align 4 /* * Simple loop for trailing quadwords, or for small amounts * of data (where we can't use an unrolled loop and wh64) */ loop_b: stq $17,0($5) # L : subq $3,1,$3 # E : Decrement number quads left addq $5,8,$5 # E : Inc address bne $3,loop_b # U : more? no_quad_b: /* * Write 0..7 trailing bytes. */ nop # E : beq $18,end_b # U : All done? ldq $7,0($5) # L : mskqh $7,$6,$2 # U : Mask final quad insqh $17,$6,$4 # U : New bits bis $2,$4,$1 # E : Put it all together stq $1,0($5) # L : And back to memory ret $31,($26),1 # L0 : within_quad_b: ldq_u $1,0($16) # L : insql $17,$16,$2 # U : New bits mskql $1,$16,$4 # U : Clear old bis $2,$4,$2 # E : New result mskql $2,$6,$4 # U : mskqh $1,$6,$2 # U : bis $2,$4,$1 # E : stq_u $1,0($16) # L : end_b: nop nop nop ret $31,($26),1 # L0 : .end ___memset /* * This is the original body of code, prior to replication and * rescheduling. Leave it here, as there may be calls to this * entry point. */ .align 4 .ent __constant_c_memset __constant_c_memset: .frame $30,0,$26,0 .prologue 0 addq $18,$16,$6 # E : max address to write to bis $16,$16,$0 # E : return value xor $16,$6,$1 # E : will complete write be within one quadword? ble $18,end # U : zero length requested? bic $1,7,$1 # E : fit within a single quadword beq $1,within_one_quad # U : and $16,7,$3 # E : Target addr misalignment beq $3,aligned # U : target is 0mod8 /* * Target address is misaligned, and won't fit within a quadword */ ldq_u $4,0($16) # L : Fetch first partial bis $16,$16,$5 # E : Save the address insql $17,$16,$2 # U : Insert new bytes subq $3,8,$3 # E : Invert (for addressing uses) addq $18,$3,$18 # E : $18 is new count ($3 is negative) mskql $4,$16,$4 # U : clear relevant parts of the quad subq $16,$3,$16 # E : $16 is new aligned destination bis $2,$4,$1 # E : Final bytes nop stq_u $1,0($5) # L : Store result nop nop .align 4 aligned: /* * We are now guaranteed to be quad aligned, with at least * one partial quad to write. */ sra $18,3,$3 # U : Number of remaining quads to write and $18,7,$18 # E : Number of trailing bytes to write bis $16,$16,$5 # E : Save dest address beq $3,no_quad # U : tail stuff only /* * it's worth the effort to unroll this and use wh64 if possible * Lifted a bunch of code from clear_user.S * At this point, entry values are: * $16 Current destination address * $5 A copy of $16 * $6 The max quadword address to write to * $18 Number trailer bytes * $3 Number quads to write */ and $16, 0x3f, $2 # E : Forward work (only useful for unrolled loop) subq $3, 16, $4 # E : Only try to unroll if > 128 bytes subq $2, 0x40, $1 # E : bias counter (aligning stuff 0mod64) blt $4, loop # U : /* * We know we've got at least 16 quads, minimum of one trip * through unrolled loop. Do a quad at a time to get us 0mod64 * aligned. */ nop # E : nop # E : nop # E : beq $1, $bigalign # U : $alignmod64: stq $17, 0($5) # L : subq $3, 1, $3 # E : For consistency later addq $1, 8, $1 # E : Increment towards zero for alignment addq $5, 8, $4 # E : Initial wh64 address (filler instruction) nop nop addq $5, 8, $5 # E : Inc address blt $1, $alignmod64 # U : $bigalign: /* * $3 - number quads left to go * $5 - target address (aligned 0mod64) * $17 - mask of stuff to store * Scratch registers available: $7, $2, $4, $1 * we know that we'll be taking a minimum of one trip through * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle * Assumes the wh64 needs to be for 2 trips through the loop in the future * The wh64 is issued on for the starting destination address for trip +2 * through the loop, and if there are less than two trips left, the target * address will be for the current trip. */ $do_wh64: wh64 ($4) # L1 : memory subsystem write hint subq $3, 24, $2 # E : For determining future wh64 addresses stq $17, 0($5) # L : nop # E : addq $5, 128, $4 # E : speculative target of next wh64 stq $17, 8($5) # L : stq $17, 16($5) # L : addq $5, 64, $7 # E : Fallback address for wh64 (== next trip addr) stq $17, 24($5) # L : stq $17, 32($5) # L : cmovlt $2, $7, $4 # E : Latency 2, extra mapping cycle nop stq $17, 40($5) # L : stq $17, 48($5) # L : subq $3, 16, $2 # E : Repeat the loop at least once more? nop stq $17, 56($5) # L : addq $5, 64, $5 # E : subq $3, 8, $3 # E : bge $2, $do_wh64 # U : nop nop nop beq $3, no_quad # U : Might have finished already .align 4 /* * Simple loop for trailing quadwords, or for small amounts * of data (where we can't use an unrolled loop and wh64) */ loop: stq $17,0($5) # L : subq $3,1,$3 # E : Decrement number quads left addq $5,8,$5 # E : Inc address bne $3,loop # U : more? no_quad: /* * Write 0..7 trailing bytes. */ nop # E : beq $18,end # U : All done? ldq $7,0($5) # L : mskqh $7,$6,$2 # U : Mask final quad insqh $17,$6,$4 # U : New bits bis $2,$4,$1 # E : Put it all together stq $1,0($5) # L : And back to memory ret $31,($26),1 # L0 : within_one_quad: ldq_u $1,0($16) # L : insql $17,$16,$2 # U : New bits mskql $1,$16,$4 # U : Clear old bis $2,$4,$2 # E : New result mskql $2,$6,$4 # U : mskqh $1,$6,$2 # U : bis $2,$4,$1 # E : stq_u $1,0($16) # L : end: nop nop nop ret $31,($26),1 # L0 : .end __constant_c_memset /* * This is a replicant of the __constant_c_memset code, rescheduled * to mask stalls. Note that entry point names also had to change */ .align 5 .ent __memsetw __memsetw: .frame $30,0,$26,0 .prologue 0 inswl $17,0,$5 # U : 000000000000c1c2 inswl $17,2,$2 # U : 00000000c1c20000 bis $16,$16,$0 # E : return value addq $18,$16,$6 # E : max address to write to ble $18, end_w # U : zero length requested? inswl $17,4,$3 # U : 0000c1c200000000 inswl $17,6,$4 # U : c1c2000000000000 xor $16,$6,$1 # E : will complete write be within one quadword? or $2,$5,$2 # E : 00000000c1c2c1c2 or $3,$4,$17 # E : c1c2c1c200000000 bic $1,7,$1 # E : fit within a single quadword and $16,7,$3 # E : Target addr misalignment or $17,$2,$17 # E : c1c2c1c2c1c2c1c2 beq $1,within_quad_w # U : nop beq $3,aligned_w # U : target is 0mod8 /* * Target address is misaligned, and won't fit within a quadword */ ldq_u $4,0($16) # L : Fetch first partial bis $16,$16,$5 # E : Save the address insql $17,$16,$2 # U : Insert new bytes subq $3,8,$3 # E : Invert (for addressing uses) addq $18,$3,$18 # E : $18 is new count ($3 is negative) mskql $4,$16,$4 # U : clear relevant parts of the quad subq $16,$3,$16 # E : $16 is new aligned destination bis $2,$4,$1 # E : Final bytes nop stq_u $1,0($5) # L : Store result nop nop .align 4 aligned_w: /* * We are now guaranteed to be quad aligned, with at least * one partial quad to write. */ sra $18,3,$3 # U : Number of remaining quads to write and $18,7,$18 # E : Number of trailing bytes to write bis $16,$16,$5 # E : Save dest address beq $3,no_quad_w # U : tail stuff only /* * it's worth the effort to unroll this and use wh64 if possible * Lifted a bunch of code from clear_user.S * At this point, entry values are: * $16 Current destination address * $5 A copy of $16 * $6 The max quadword address to write to * $18 Number trailer bytes * $3 Number quads to write */ and $16, 0x3f, $2 # E : Forward work (only useful for unrolled loop) subq $3, 16, $4 # E : Only try to unroll if > 128 bytes subq $2, 0x40, $1 # E : bias counter (aligning stuff 0mod64) blt $4, loop_w # U : /* * We know we've got at least 16 quads, minimum of one trip * through unrolled loop. Do a quad at a time to get us 0mod64 * aligned. */ nop # E : nop # E : nop # E : beq $1, $bigalign_w # U : $alignmod64_w: stq $17, 0($5) # L : subq $3, 1, $3 # E : For consistency later addq $1, 8, $1 # E : Increment towards zero for alignment addq $5, 8, $4 # E : Initial wh64 address (filler instruction) nop nop addq $5, 8, $5 # E : Inc address blt $1, $alignmod64_w # U : $bigalign_w: /* * $3 - number quads left to go * $5 - target address (aligned 0mod64) * $17 - mask of stuff to store * Scratch registers available: $7, $2, $4, $1 * we know that we'll be taking a minimum of one trip through * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle * Assumes the wh64 needs to be for 2 trips through the loop in the future * The wh64 is issued on for the starting destination address for trip +2 * through the loop, and if there are less than two trips left, the target * address will be for the current trip. */ $do_wh64_w: wh64 ($4) # L1 : memory subsystem write hint subq $3, 24, $2 # E : For determining future wh64 addresses stq $17, 0($5) # L : nop # E : addq $5, 128, $4 # E : speculative target of next wh64 stq $17, 8($5) # L : stq $17, 16($5) # L : addq $5, 64, $7 # E : Fallback address for wh64 (== next trip addr) stq $17, 24($5) # L : stq $17, 32($5) # L : cmovlt $2, $7, $4 # E : Latency 2, extra mapping cycle nop stq $17, 40($5) # L : stq $17, 48($5) # L : subq $3, 16, $2 # E : Repeat the loop at least once more? nop stq $17, 56($5) # L : addq $5, 64, $5 # E : subq $3, 8, $3 # E : bge $2, $do_wh64_w # U : nop nop nop beq $3, no_quad_w # U : Might have finished already .align 4 /* * Simple loop for trailing quadwords, or for small amounts * of data (where we can't use an unrolled loop and wh64) */ loop_w: stq $17,0($5) # L : subq $3,1,$3 # E : Decrement number quads left addq $5,8,$5 # E : Inc address bne $3,loop_w # U : more? no_quad_w: /* * Write 0..7 trailing bytes. */ nop # E : beq $18,end_w # U : All done? ldq $7,0($5) # L : mskqh $7,$6,$2 # U : Mask final quad insqh $17,$6,$4 # U : New bits bis $2,$4,$1 # E : Put it all together stq $1,0($5) # L : And back to memory ret $31,($26),1 # L0 : within_quad_w: ldq_u $1,0($16) # L : insql $17,$16,$2 # U : New bits mskql $1,$16,$4 # U : Clear old bis $2,$4,$2 # E : New result mskql $2,$6,$4 # U : mskqh $1,$6,$2 # U : bis $2,$4,$1 # E : stq_u $1,0($16) # L : end_w: nop nop nop ret $31,($26),1 # L0 : .end __memsetw memset = ___memset __memset = ___memset