arm-none-eabi-ld是否改写了bl指令?
Does arm-none-eabi-ld rewrite the bl instruction?
我想了解为什么某些 Cortex-M0 代码在链接和未链接时表现不同。在这两种情况下,它都被加载到 0x20000000
。看起来尽管我尽了最大努力通过将 -fPIC
传递给编译器来生成与位置无关的代码,但在代码通过链接器后 bl
指令似乎有所不同。我没看错吗,这只是 ARM Thumb 中链接器工作的一部分吗?是否有更好的方法来生成与位置无关的函数调用?
已链接:
20000000:
20000000: 0003 movs r3, r0
20000002: 4852 ldr r0, [pc, #328]
20000004: 4685 mov sp, r0
20000006: 0018 movs r0, r3
20000008: f000 f802 bl 20000010
2000000c: 46c0 nop ; (mov r8, r8)
2000000e: 46c0 nop ; (mov r8, r8)
未链接:
00000000:
0: 0003 movs r3, r0
2: 4852 ldr r0, [pc, #328]
4: 4685 mov sp, r0
6: 0018 movs r0, r3
8: f7ff fffe bl 10
c: 46c0 nop ; (mov r8, r8)
e: 46c0 nop ; (mov r8, r8)
start.s
.globl _start
_start:
.word 0x20001000
.word reset
.word hang
.word hang
.thumb
.thumb_func
reset:
bl notmain
.thumb_func
hang:
b .
notmain.c
unsigned int x;
unsigned int fun ( unsigned int );
void notmain ( void )
{
x=fun(x+5);
}
fun.c
unsigned int y;
unsigned int fun ( unsigned int z )
{
return(y+z+1);
}
内存映射
MEMORY
{
ram : ORIGIN = 0x20000000, LENGTH = 0x1000
}
SECTIONS
{
.text : { *(.text*) } > ram
.bss : { *(.bss*) } > ram
}
建造
arm-none-eabi-as start.s -o start.o
arm-none-eabi-gcc -fPIC -O2 -c -mthumb fun.c -o fun.o
arm-none-eabi-gcc -fPIC -O2 -c -mthumb notmain.c -o notmain.o
arm-none-eabi-ld -T memmap start.o notmain.o fun.o -o so.elf
产生
20000000 <_start>:
20000000: 20001000 andcs r1, r0, r0
20000004: 20000011 andcs r0, r0, r1, lsl r0
20000008: 20000015 andcs r0, r0, r5, lsl r0
2000000c: 20000015 andcs r0, r0, r5, lsl r0
20000010 <reset>:
20000010: f000 f802 bl 20000018 <notmain>
20000014 <hang>:
20000014: e7fe b.n 20000014 <hang>
...
20000018 <notmain>:
20000018: b510 push {r4, lr}
2000001a: 4b06 ldr r3, [pc, #24] ; (20000034 <notmain+0x1c>)
2000001c: 4a06 ldr r2, [pc, #24] ; (20000038 <notmain+0x20>)
2000001e: 447b add r3, pc
20000020: 589c ldr r4, [r3, r2]
20000022: 6823 ldr r3, [r4, #0]
20000024: 1d58 adds r0, r3, #5
20000026: f000 f809 bl 2000003c <fun>
2000002a: 6020 str r0, [r4, #0]
2000002c: bc10 pop {r4}
2000002e: bc01 pop {r0}
20000030: 4700 bx r0
20000032: 46c0 nop ; (mov r8, r8)
20000034: 00000032 andeq r0, r0, r2, lsr r0
20000038: 00000000 andeq r0, r0, r0
2000003c <fun>:
2000003c: 4b03 ldr r3, [pc, #12] ; (2000004c <fun+0x10>)
2000003e: 4a04 ldr r2, [pc, #16] ; (20000050 <fun+0x14>)
20000040: 447b add r3, pc
20000042: 589b ldr r3, [r3, r2]
20000044: 681b ldr r3, [r3, #0]
20000046: 3301 adds r3, #1
20000048: 1818 adds r0, r3, r0
2000004a: 4770 bx lr
2000004c: 00000010 andeq r0, r0, r0, lsl r0
20000050: 00000004 andeq r0, r0, r4
Disassembly of section .got:
20000054 <.got>:
20000054: 20000068 andcs r0, r0, r8, rrx
20000058: 2000006c andcs r0, r0, ip, rrx
Disassembly of section .got.plt:
2000005c <_GLOBAL_OFFSET_TABLE_>:
...
Disassembly of section .bss:
20000068 <x>:
20000068: 00000000 andeq r0, r0, r0
2000006c <y>:
2000006c: 00000000 andeq r0, r0, r0
当它想要查找全局变量 x 时,它所做的似乎是获取程序计数器和一个 linker supplied/modfied 偏移量 0x32 并使用它来查找全局偏移 table。然后从中获取一个偏移量来找到 X。Y 也是如此。所以看起来当你重新定位时你需要在运行时或加载时修改全局偏移量 table 取决于。
如果我去掉那些全局变量,除了矢量 table 是硬编码的而不是 PIC(无论如何都没有编译),这都是位置无关的。
20000000 <_start>:
20000000: 20001000 andcs r1, r0, r0
20000004: 20000011 andcs r0, r0, r1, lsl r0
20000008: 20000015 andcs r0, r0, r5, lsl r0
2000000c: 20000015 andcs r0, r0, r5, lsl r0
20000010 <reset>:
20000010: f000 f802 bl 20000018 <notmain>
20000014 <hang>:
20000014: e7fe b.n 20000014 <hang>
...
20000018 <notmain>:
20000018: b508 push {r3, lr}
2000001a: 2005 movs r0, #5
2000001c: f000 f804 bl 20000028 <fun>
20000020: 3006 adds r0, #6
20000022: bc08 pop {r3}
20000024: bc02 pop {r1}
20000026: 4708 bx r1
20000028 <fun>:
20000028: 3001 adds r0, #1
2000002a: 4770 bx lr
回到这个版本
unsigned int y;
unsigned int fun ( unsigned int z )
{
return(y+z+1);
}
位置无关
00000000 <fun>:
0: 4b03 ldr r3, [pc, #12] ; (10 <fun+0x10>)
2: 4a04 ldr r2, [pc, #16] ; (14 <fun+0x14>)
4: 447b add r3, pc
6: 589b ldr r3, [r3, r2]
8: 681b ldr r3, [r3, #0]
a: 3301 adds r3, #1
c: 1818 adds r0, r3, r0
e: 4770 bx lr
10: 00000008 andeq r0, r0, r8
14: 00000000 andeq r0, r0, r0
与位置无关
00000000 <fun>:
0: 4b02 ldr r3, [pc, #8] ; (c <fun+0xc>)
2: 681b ldr r3, [r3, #0]
4: 3301 adds r3, #1
6: 1818 adds r0, r3, r0
8: 4770 bx lr
a: 46c0 nop ; (mov r8, r8)
c: 00000000 andeq r0, r0, r0
代码必须做更多的工作才能访问外部变量。位置相关,有些工作因为它是外部的但不是那么多。 link 用户将填写所需的项目以使其正常工作...到 link 它...
elf 文件包含 linker 知道如何执行此操作的信息。
Relocation section '.rel.text' at offset 0x1a4 contains 2 entries:
Offset Info Type Sym.Value Sym. Name
00000010 00000a19 R_ARM_BASE_PREL 00000000 _GLOBAL_OFFSET_TABLE_
00000014 00000b1a R_ARM_GOT_BREL 00000004 y
或
Relocation section '.rel.text' at offset 0x174 contains 1 entries:
Offset Info Type Sym.Value Sym. Name
0000000c 00000a02 R_ARM_ABS32 00000004 y
不是主要有这些 PIC
Relocation section '.rel.text' at offset 0x1cc contains 3 entries:
Offset Info Type Sym.Value Sym. Name
0000000e 00000a0a R_ARM_THM_CALL 00000000 fun
0000001c 00000b19 R_ARM_BASE_PREL 00000000 _GLOBAL_OFFSET_TABLE_
00000020 00000c1a R_ARM_GOT_BREL 00000004 x
无。
Relocation section '.rel.text' at offset 0x198 contains 2 entries:
Offset Info Type Sym.Value Sym. Name
00000008 00000a0a R_ARM_THM_CALL 00000000 fun
00000014 00000b02 R_ARM_ABS32 00000004 x
所以简而言之,工具链正在做它的工作,你不需要重新做它的工作。请注意,这与手臂或拇指无关。任何时候您使用对象和 linker 模型并允许来自对象的外部项目时,linker 都必须修补一些东西以将代码粘合在一起。这就是它的工作原理。
我想了解为什么某些 Cortex-M0 代码在链接和未链接时表现不同。在这两种情况下,它都被加载到 0x20000000
。看起来尽管我尽了最大努力通过将 -fPIC
传递给编译器来生成与位置无关的代码,但在代码通过链接器后 bl
指令似乎有所不同。我没看错吗,这只是 ARM Thumb 中链接器工作的一部分吗?是否有更好的方法来生成与位置无关的函数调用?
已链接:
20000000:
20000000: 0003 movs r3, r0
20000002: 4852 ldr r0, [pc, #328]
20000004: 4685 mov sp, r0
20000006: 0018 movs r0, r3
20000008: f000 f802 bl 20000010
2000000c: 46c0 nop ; (mov r8, r8)
2000000e: 46c0 nop ; (mov r8, r8)
未链接:
00000000:
0: 0003 movs r3, r0
2: 4852 ldr r0, [pc, #328]
4: 4685 mov sp, r0
6: 0018 movs r0, r3
8: f7ff fffe bl 10
c: 46c0 nop ; (mov r8, r8)
e: 46c0 nop ; (mov r8, r8)
start.s
.globl _start
_start:
.word 0x20001000
.word reset
.word hang
.word hang
.thumb
.thumb_func
reset:
bl notmain
.thumb_func
hang:
b .
notmain.c
unsigned int x;
unsigned int fun ( unsigned int );
void notmain ( void )
{
x=fun(x+5);
}
fun.c
unsigned int y;
unsigned int fun ( unsigned int z )
{
return(y+z+1);
}
内存映射
MEMORY
{
ram : ORIGIN = 0x20000000, LENGTH = 0x1000
}
SECTIONS
{
.text : { *(.text*) } > ram
.bss : { *(.bss*) } > ram
}
建造
arm-none-eabi-as start.s -o start.o
arm-none-eabi-gcc -fPIC -O2 -c -mthumb fun.c -o fun.o
arm-none-eabi-gcc -fPIC -O2 -c -mthumb notmain.c -o notmain.o
arm-none-eabi-ld -T memmap start.o notmain.o fun.o -o so.elf
产生
20000000 <_start>:
20000000: 20001000 andcs r1, r0, r0
20000004: 20000011 andcs r0, r0, r1, lsl r0
20000008: 20000015 andcs r0, r0, r5, lsl r0
2000000c: 20000015 andcs r0, r0, r5, lsl r0
20000010 <reset>:
20000010: f000 f802 bl 20000018 <notmain>
20000014 <hang>:
20000014: e7fe b.n 20000014 <hang>
...
20000018 <notmain>:
20000018: b510 push {r4, lr}
2000001a: 4b06 ldr r3, [pc, #24] ; (20000034 <notmain+0x1c>)
2000001c: 4a06 ldr r2, [pc, #24] ; (20000038 <notmain+0x20>)
2000001e: 447b add r3, pc
20000020: 589c ldr r4, [r3, r2]
20000022: 6823 ldr r3, [r4, #0]
20000024: 1d58 adds r0, r3, #5
20000026: f000 f809 bl 2000003c <fun>
2000002a: 6020 str r0, [r4, #0]
2000002c: bc10 pop {r4}
2000002e: bc01 pop {r0}
20000030: 4700 bx r0
20000032: 46c0 nop ; (mov r8, r8)
20000034: 00000032 andeq r0, r0, r2, lsr r0
20000038: 00000000 andeq r0, r0, r0
2000003c <fun>:
2000003c: 4b03 ldr r3, [pc, #12] ; (2000004c <fun+0x10>)
2000003e: 4a04 ldr r2, [pc, #16] ; (20000050 <fun+0x14>)
20000040: 447b add r3, pc
20000042: 589b ldr r3, [r3, r2]
20000044: 681b ldr r3, [r3, #0]
20000046: 3301 adds r3, #1
20000048: 1818 adds r0, r3, r0
2000004a: 4770 bx lr
2000004c: 00000010 andeq r0, r0, r0, lsl r0
20000050: 00000004 andeq r0, r0, r4
Disassembly of section .got:
20000054 <.got>:
20000054: 20000068 andcs r0, r0, r8, rrx
20000058: 2000006c andcs r0, r0, ip, rrx
Disassembly of section .got.plt:
2000005c <_GLOBAL_OFFSET_TABLE_>:
...
Disassembly of section .bss:
20000068 <x>:
20000068: 00000000 andeq r0, r0, r0
2000006c <y>:
2000006c: 00000000 andeq r0, r0, r0
当它想要查找全局变量 x 时,它所做的似乎是获取程序计数器和一个 linker supplied/modfied 偏移量 0x32 并使用它来查找全局偏移 table。然后从中获取一个偏移量来找到 X。Y 也是如此。所以看起来当你重新定位时你需要在运行时或加载时修改全局偏移量 table 取决于。
如果我去掉那些全局变量,除了矢量 table 是硬编码的而不是 PIC(无论如何都没有编译),这都是位置无关的。
20000000 <_start>:
20000000: 20001000 andcs r1, r0, r0
20000004: 20000011 andcs r0, r0, r1, lsl r0
20000008: 20000015 andcs r0, r0, r5, lsl r0
2000000c: 20000015 andcs r0, r0, r5, lsl r0
20000010 <reset>:
20000010: f000 f802 bl 20000018 <notmain>
20000014 <hang>:
20000014: e7fe b.n 20000014 <hang>
...
20000018 <notmain>:
20000018: b508 push {r3, lr}
2000001a: 2005 movs r0, #5
2000001c: f000 f804 bl 20000028 <fun>
20000020: 3006 adds r0, #6
20000022: bc08 pop {r3}
20000024: bc02 pop {r1}
20000026: 4708 bx r1
20000028 <fun>:
20000028: 3001 adds r0, #1
2000002a: 4770 bx lr
回到这个版本
unsigned int y;
unsigned int fun ( unsigned int z )
{
return(y+z+1);
}
位置无关
00000000 <fun>:
0: 4b03 ldr r3, [pc, #12] ; (10 <fun+0x10>)
2: 4a04 ldr r2, [pc, #16] ; (14 <fun+0x14>)
4: 447b add r3, pc
6: 589b ldr r3, [r3, r2]
8: 681b ldr r3, [r3, #0]
a: 3301 adds r3, #1
c: 1818 adds r0, r3, r0
e: 4770 bx lr
10: 00000008 andeq r0, r0, r8
14: 00000000 andeq r0, r0, r0
与位置无关
00000000 <fun>:
0: 4b02 ldr r3, [pc, #8] ; (c <fun+0xc>)
2: 681b ldr r3, [r3, #0]
4: 3301 adds r3, #1
6: 1818 adds r0, r3, r0
8: 4770 bx lr
a: 46c0 nop ; (mov r8, r8)
c: 00000000 andeq r0, r0, r0
代码必须做更多的工作才能访问外部变量。位置相关,有些工作因为它是外部的但不是那么多。 link 用户将填写所需的项目以使其正常工作...到 link 它...
elf 文件包含 linker 知道如何执行此操作的信息。
Relocation section '.rel.text' at offset 0x1a4 contains 2 entries:
Offset Info Type Sym.Value Sym. Name
00000010 00000a19 R_ARM_BASE_PREL 00000000 _GLOBAL_OFFSET_TABLE_
00000014 00000b1a R_ARM_GOT_BREL 00000004 y
或
Relocation section '.rel.text' at offset 0x174 contains 1 entries:
Offset Info Type Sym.Value Sym. Name
0000000c 00000a02 R_ARM_ABS32 00000004 y
不是主要有这些 PIC
Relocation section '.rel.text' at offset 0x1cc contains 3 entries:
Offset Info Type Sym.Value Sym. Name
0000000e 00000a0a R_ARM_THM_CALL 00000000 fun
0000001c 00000b19 R_ARM_BASE_PREL 00000000 _GLOBAL_OFFSET_TABLE_
00000020 00000c1a R_ARM_GOT_BREL 00000004 x
无。
Relocation section '.rel.text' at offset 0x198 contains 2 entries:
Offset Info Type Sym.Value Sym. Name
00000008 00000a0a R_ARM_THM_CALL 00000000 fun
00000014 00000b02 R_ARM_ABS32 00000004 x
所以简而言之,工具链正在做它的工作,你不需要重新做它的工作。请注意,这与手臂或拇指无关。任何时候您使用对象和 linker 模型并允许来自对象的外部项目时,linker 都必须修补一些东西以将代码粘合在一起。这就是它的工作原理。