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/*----------------------------------------------------------------
// //
// libc_asm.S //
// //
// This file is part of the Amber project //
// http://www.opencores.org/project,amber //
// //
// Description //
// Assembly routines for the mini-libc library. //
// //
// Author(s): //
// - Conor Santifort, csantifort.amber@gmail.com //
// //
//////////////////////////////////////////////////////////////////
// //
// Copyright (C) 2010 Authors and OPENCORES.ORG //
// //
// This source file may be used and distributed without //
// restriction provided that this copyright statement is not //
// removed from the file and that any derivative work contains //
// the original copyright notice and the associated disclaimer. //
// //
// This source file is free software; you can redistribute it //
// and/or modify it under the terms of the GNU Lesser General //
// Public License as published by the Free Software Foundation; //
// either version 2.1 of the License, or (at your option) any //
// later version. //
// //
// This source is distributed in the hope that it will be //
// useful, but WITHOUT ANY WARRANTY; without even the implied //
// warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR //
// PURPOSE. See the GNU Lesser General Public License for more //
// details. //
// //
// You should have received a copy of the GNU Lesser General //
// Public License along with this source; if not, download it //
// from http://www.opencores.org/lgpl.shtml //
// //
----------------------------------------------------------------*/
#include "amber_registers.h"
/* _testfail: Used to terminate execution in Verilog simulations */
/* On the board just puts the processor into an infinite loop */
.section .text
.globl _testfail
_testfail:
ldr r11, AdrTestStatus
str r0, [r11]
b _testfail
/* _testpass: Used to terminate execution in Verilog simulations */
/* On the board just puts the processor into an infinite loop */
.globl _testpass
_testpass:
ldr r11, AdrTestStatus
mov r10, #17
str r10, [r11]
b _testpass
/* _outbyte: Output a single character through UART 0 */
@ if the uart tx fifo is stuck full
@ this routine will cycle forever
.globl _outbyte
_outbyte:
ldr r1, AdrUARTDR
ldr r3, AdrUARTFR
@ Check the tx_full flag
1: ldr r2, [r3]
and r2, r2, #0x20
cmp r2, #0
streqb r0, [r1]
moveqs pc, lr @ return
bne 1b
/* _inbyte: Input a single character from UART 0 */
@ r0 is the timeout in mS
.globl _inbyte
_inbyte:
ldr r2, AdrUARTDR @ data
ldr r3, AdrUARTFR @ flags
@ Multiple delay value by 2560
@ as the delay loop takes about 12 clock cycles running cached
@ so that factor gives 1:1mS @33MHz
mov r1, r0, lsl #11
add r1, r1, r0, lsl #9
@ Check the r2 empty flag
2: ldr r0, [r3]
ands r0, r0, #0x10
ldreqb r0, [r2]
moveq pc, lr
@ decrement timeout
subs r1, r1, #1
bne 2b
mov r0, #-1
movs pc, lr
/* _div: Integer division function */
@ Divide r0 by r1
@ Answer returned in r1
.globl _div
.globl __aeabi_idiv
__aeabi_idiv:
_div:
stmdb sp!, {r4, lr}
@ set r4 to 1 if one of the two inputs is negative
and r2, r0, #0x80000000
and r3, r1, #0x80000000
eor r4, r2, r3
@ Invert negative numbers
tst r0, #0x80000000
mvnne r0, r0
addne r0, r0, #1
tst r1, #0x80000000
mvnne r1, r1
addne r1, r1, #1
@ divide r1 by r2, also use registers r0 and r4
mov r2, r1
mov r1, r0
cmp r2, #0
beq 3f
@ In order to divide r1 by r2, the first thing we need to do is to shift r2
@ left by the necessary number of places. The easiest method of doing this
@ is simply by trial and error - shift until we discover that r2 has become
@ too big, then stop.
mov r0,#0 @ clear r0 to accumulate result
mov r3,#1 @ set bit 0 in r3, which will be
@ shifted left then right
1: cmp r3, #0 @ escape on error
moveq r3, #0x10000000
beq 2f
cmp r2,r1
movls r2,r2,lsl#1
movls r3,r3,lsl#1
bls 1b
@ shift r2 left until it is about to be bigger than r1
@ shift r3 left in parallel in order to flag how far we have to go
@ r0 will be used to hold the result. The role of r3 is more complicated.
@ In effect, we are using r3 to mark where the right-hand end of r2 has got to
@ - if we shift r2 three places left, this will be indicated by a value of %1000
@ in r3. However, we also add it to r0 every time we manage a successful subtraction,
@ since it marks the position of the digit currently being calculated in the answer.
@ so at the time of the first subtraction, r3 would have been %100, at the time
@ of the second (which failed) it would have been %10, and at the time of the
@ third %1. Adding it to r0 after each successful subtraction would have
@ given us, once again, the answer of %101!
@ Now for the loop that actually does the work:
2: cmp r1,r2 @ carry set if r1>r2 (don't ask why)
subcs r1,r1,r2 @ subtract r2 from r1 if this would
@ give a positive answer
addcs r0,r0,r3 @ and add the current bit in r3 to
@ the accumulating answer in r0
@ In subtraction (a cmp instruction simulates a subtraction in
@ order to set the flags), if r1 - r2 gives a positive answer and no 'borrow'
@ is required, the carry flag is set. This is required in order to make SBC
@ (Subtract with Carry) work properly when used to carry out a 64-bit subtraction,
@ but it is confusing!
@ In this case, we are turning it to our advantage. The carry flag is set to
@ indicate that a successful subtraction is possible, i.e. one that doesn't
@ generate a negative result, and the two following instructions are carried
@ out only when the condition Carry Set applies. Note that the 'S' on the end
@ of these instructions is part of the 'CS' condition code and does not mean
@ that they set the flags!
movs r3,r3,lsr #1 @ Shift r3 right into carry flag
movcc r2,r2,lsr #1 @ and if bit 0 of r3 was zero, also
@ shift r2 right
bcc 2b @ If carry not clear, r3 has shifted
@ back to where it started, and we
@ can end
@ if one of the inputs is negetive then return a negative result
tst r4, #0x80000000
mvnne r0, r0
addne r0, r0, #1
3: ldmia sp!, {r4, pc}^
/* strcpy: String copy function
char * strcpy ( char * destination, const char * source );
destination is returned
*/
@ r0 points to destination
@ r1 points to source string which terminates with a 0
.globl strcpy
strcpy:
stmdb sp!, {r4-r6, lr}
@ Use r6 to process the destination pointer.
@ At the end of the function, r0 is returned, so need to preserve it
mov r6, r0
@ only if both strings are zero-aligned use the fast 'aligned' algorithm
orr r2, r6, r1
tst r2, #3
bne strcpy_slow
strcpy_fast:
@ process strings 12 bytes at a time
ldmia r1!, {r2-r5}
@ check for a zero byte
@ only need to examine one of the strings because
@ they are equal up to this point!
tst r2, #0xff
tstne r2, #0xff00
tstne r2, #0xff0000
tstne r2, #0xff000000
strne r2, [r6], #4
subeq r1, r1, #4
tstne r3, #0xff
tstne r3, #0xff00
tstne r3, #0xff0000
tstne r3, #0xff000000
strne r3, [r6], #4
subeq r1, r1, #4
tstne r4, #0xff
tstne r4, #0xff00
tstne r4, #0xff0000
tstne r4, #0xff000000
strne r4, [r6], #4
subeq r1, r1, #4
tstne r5, #0xff
tstne r5, #0xff00
tstne r5, #0xff0000
tstne r5, #0xff000000
strne r5, [r6], #4
subeq r1, r1, #4
@ loop back to look at next 12 bytes
bne strcpy_fast
@ the source string contains a zero character
strcpy_aligned_slow:
@ unroll the loop 4 times
ldr r3, [r1], #4
strb r3, [r6], #1
ands r4, r3, #0xff
ldmeqia sp!, {r4-r6, pc}^
lsr r3, r3, #8
strb r3, [r6], #1
ands r4, r3, #0xff
ldmeqia sp!, {r4-r6, pc}^
lsr r3, r3, #8
strb r3, [r6], #1
ands r4, r3, #0xff
ldmeqia sp!, {r4-r6, pc}^
lsr r3, r3, #8
strb r3, [r6], #1
ands r4, r3, #0xff
ldmeqia sp!, {r4-r6, pc}^
b strcpy_aligned_slow
strcpy_slow:
@ unroll the loop 4 times
ldrb r3, [r1], #1
strb r3, [r6], #1
cmp r3, #0
ldmeqia sp!, {r4-r6, pc}^
ldrb r3, [r1], #1
strb r3, [r6], #1
cmp r3, #0
ldmeqia sp!, {r4-r6, pc}^
ldrb r3, [r1], #1
strb r3, [r6], #1
cmp r3, #0
ldmeqia sp!, {r4-r6, pc}^
ldrb r3, [r1], #1
strb r3, [r6], #1
cmp r3, #0
ldmeqia sp!, {r4-r6, pc}^
b strcpy_slow
/* int strcmp ( const char * str1, const char * str2 );
A value greater than zero indicates that the first character
that does not match has a greater value in str1 than in str2;
And a value less than zero indicates the opposite.
*/
.globl strcmp
strcmp:
stmdb sp!, {r4-r8, lr}
@ only if both strings are zero-aligned use the fast 'aligned' algorithm
orr r2, r0, r1
tst r2, #3
bne strcmp_slow
strcmp_fast:
@ process strings 12 bytes at a time
ldmia r0!, {r2-r4}
ldmia r1!, {r5-r7}
cmp r2, r5
bne 1f
cmpeq r3, r6
bne 2f
cmpeq r4, r7
bne 3f
@ strings are equal - find a zero byte
@ only need to examine one of the strings because
@ they are equal up to this point!
tst r2, #0xff
tstne r2, #0xff00
tstne r2, #0xff0000
tstne r2, #0xff000000
tstne r3, #0xff
tstne r3, #0xff00
tstne r3, #0xff0000
tstne r3, #0xff000000
tstne r4, #0xff
tstne r4, #0xff00
tstne r4, #0xff0000
tstne r4, #0xff000000
@ loop back to look at next 12 bytes
bne strcmp_fast
@ the first string contains a zero character
@ the strings are the same, so both strings end
moveq r0, #0
ldmeqia sp!, {r4-r8, pc}^
@ Roll back the string pointers to before the mismatch
@ then handle the remaining part byte by byte
1: sub r0, r0, #12
sub r1, r1, #12
strcmp_slow:
ldrb r2, [r0], #1
ldrb r3, [r1], #1
eors r4, r2, r3 @ are the bytes equal ?
bne bytes_different
ldrb r5, [r0], #1
ldrb r6, [r1], #1
cmp r2, #0 @ are they equal and zero ?
beq bytes_zero
eors r7, r5, r6 @ are the bytes equal ?
bne bytes_different
ldrb r2, [r0], #1
ldrb r3, [r1], #1
cmp r5, #0 @ are they equal and zero ?
beq bytes_zero
eors r4, r2, r3 @ are the bytes equal ?
bne bytes_different
ldrb r5, [r0], #1
ldrb r6, [r1], #1
cmp r2, #0 @ are they equal and zero ?
beq bytes_zero
eors r7, r5, r6 @ are the bytes equal ?
bne bytes_different
cmp r5, #0 @ are they equal and zero ?
beq bytes_zero
bne strcmp_slow
@ Skipping first 4 bytes so just check they
@ don't contain an end of string 0 character
2: tst r2, #0xff
tstne r2, #0xff00
tstne r2, #0xff0000
tstne r2, #0xff000000
beq bytes_zero
@ start looking at 5th byte
sub r0, r0, #8
sub r1, r1, #8
ldrb r2, [r0], #1
ldrb r3, [r1], #1
eors r4, r2, r3 @ are the bytes equal ?
bne bytes_different
ldrb r5, [r0], #1
ldrb r6, [r1], #1
cmp r2, #0 @ are they equal and zero ?
beq bytes_zero
eors r7, r5, r6 @ are the bytes equal ?
bne bytes_different
ldrb r2, [r0], #1
ldrb r3, [r1], #1
cmp r5, #0 @ are they equal and zero ?
beq bytes_zero
eors r4, r2, r3 @ are the bytes equal ?
bne bytes_different
ldrb r5, [r0], #1
ldrb r6, [r1], #1
cmp r2, #0 @ are they equal and zero ?
beq bytes_zero
eors r7, r5, r6 @ are the bytes equal ?
bne bytes_different
cmp r5, #0 @ are they equal and zero ?
beq bytes_zero
bne strcmp_slow
@ Skipping first 8 bytes so just check they
@ don't contain an end of string 0 character
3: tst r2, #0xff
tstne r2, #0xff00
tstne r2, #0xff0000
tstne r2, #0xff000000
tstne r3, #0xff
tstne r3, #0xff00
tstne r3, #0xff0000
tstne r3, #0xff000000
beq bytes_zero
sub r0, r0, #4
sub r1, r1, #4
ldrb r2, [r0], #1
ldrb r3, [r1], #1
eors r4, r2, r3 @ are the bytes equal ?
bne bytes_different
ldrb r5, [r0], #1
ldrb r6, [r1], #1
cmp r2, #0 @ are they equal and zero ?
beq bytes_zero
eors r7, r5, r6 @ are the bytes equal ?
bne bytes_different
ldrb r2, [r0], #1
ldrb r3, [r1], #1
cmp r5, #0 @ are they equal and zero ?
beq bytes_zero
eors r4, r2, r3 @ are the bytes equal ?
bne bytes_different
ldrb r5, [r0], #1
ldrb r6, [r1], #1
cmp r2, #0 @ are they equal and zero ?
beq bytes_zero
eors r7, r5, r6 @ are the bytes equal ?
bne bytes_different
cmp r5, #0 @ are they equal and zero ?
beq bytes_zero
bne strcmp_slow
bytes_zero:
moveq r0, #0 @ if equal and zero, return zero
ldmeqia sp!, {r4-r8, pc}^
bytes_different:
sub r0, r5, r6
ldmia sp!, {r4-r8, pc}^
@ initialize malloc
.globl init_malloc
init_malloc:
ldr r1, AdrMalloc
str r1, [r1]
mov pc, lr
/* void *malloc(size_t size); */
.globl malloc
malloc:
ldr r1, AdrMalloc
ldr r0, [r1]
add r0, r0, #0x10000
str r0, [r1]
mov pc, lr
/* strncpy: String copy function */
@ r0 points to destination
@ r1 points to source string
@ r2 is the number of bytes to copy
.globl strncpy
strncpy:
stmdb sp!, {r4, lr}
cmp r2, #0
beq 2f
add r4, r0, r2 @ set r4 to the address of the last byte copied
1: ldrb r3, [r1], #1
strb r3, [r0], #1
cmp r0, r4
bne 1b
2: ldmia sp!, {r4, pc}^
/* strncpy: String compare function */
@ return the difference if the strings don't match
.globl strncmp
strncmp:
stmdb sp!, {r4, r5, r6, lr}
@ check for 0 length
cmp r2, #0
moveq r0, #1
beq 2f
mov r3, #0
1: add r3, r3, #1
ldrb r4, [r0], #1
ldrb r5, [r1], #1
subs r6, r4, r5
movne r0, r6
bne 2f
cmp r3, r2
moveq r0, #0
beq 2f
b 1b
2: ldmia sp!, {r4, r5, r6, pc}^
AdrMalloc: .word 0x7000000
AdrTestStatus: .word ADR_AMBER_TEST_STATUS
AdrUARTDR: .word ADR_AMBER_UART0_DR
AdrUARTFR: .word ADR_AMBER_UART0_FR
/* ========================================================================= */
/* ========================================================================= */
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