[英]Manipulating an array in memory via the arm c inline assembler
int smplSize = 48;
int Smpl[48];
for(int i = 0; i < smplSize; i++) Smpl[i] = 0x0;
Smpl[smplSize-1] = 0x1;
int *ptrToSmpl = &Smpl[0];
printf("Sample @%p of Size %i :\n",(void*)ptrToSmpl,smplSize);
asm volatile(
"@ ------------------------------------------------- \n"
"@ Invert the sample \n"
"@ ------------------------------------------------- \n"
//"0: \n"
"ldr r2,[r3] \n"
//"cmp r2,#0x1 \n"
//"bne 1f \n"
"add r2,#0x1 \n"
//"add r2,#0x1 \n"
"str r2,[r3] \n"
//"ldr r1, .0 \n"
//"bx r1 \n"
//"1: \n"
:
: "r" (ptrToSmpl)
: "r3", "memory"
);
printf("Sample[0] = %i" , Smpl[0]);
Edit: 编辑:
As you can see I want to manipulate a variable of an array via the inline assembler on arm, but I always get a segfault. 如您所见,我想通过arm上的内联汇编器来操作数组的变量,但是我总是遇到段错误。 How can I access the memory without segfault?
如何在没有段错误的情况下访问内存?
printf("Hello inline asmTest start!\n\n");
int smplSize = 48;
int Smpl[48];
for(int i = 0; i < smplSize; i++) Smpl[i] = 0x0;
Smpl[smplSize-1] = 0x1;
int *ptrToSmpl = &Smpl[0];
printf("Sample @%p of Size %i :\n",(void*)ptrToSmpl,smplSize);
asm volatile(
"@ ------------------------------------------------- \n"
"@ Invert the sample \n"
"@ ------------------------------------------------- \n"
//"0: \n"
"ldr r2,%[ptrToSmpl] \n"
//"cmp r2,#0x1 \n"
//"bne 1f \n"
"add r2,#0x1 \n"
//"add r2,#0x1 \n"
"str r2,%[ptrToSmpl] \n"
//"ldr r1, .0 \n"
//"bx r1 \n"
//"1: \n"
:
: [ptrToSmpl]"r" (ptrToSmpl)
:
);
printf("Sample[0] = %i" , Smpl[0]);
Ok, I did so, but now I get a "/tmp/cczQDyiw.s|72|Error: internal_relocation (type: OFFSET_IMM) not fixed up|" 好的,我这样做了,但是现在我得到一个“ /tmp/cczQDyiw.s | 72 |错误:内部定位(类型:OFFSET_IMM)未修复|” when compiling it.
编译时。
BTW: Can I "org [Adress of the programm]" the inline asm code ? 顺便说一句:我可以“组织[程序的地址]”内联的asm代码吗?
printf("Hello inline asmTest start!\n\n");
int smplSize = 48;
int Smpl[48];
for(int i = 0; i < smplSize; i++) Smpl[i] = 0x0;
Smpl[smplSize-1] = 0x1;
int *ptrToSmpl = &Smpl[0];
printf("Sample @%p of Size %i :\n",(void*)ptrToSmpl,smplSize);
asm volatile(
"@ ------------------------------------------------- \n"
"@ Invert the sample \n"
"@ ------------------------------------------------- \n"
"init: \n"
"ldr r0,%[ptrToSmpl] \n"
"loop : \n"
"ldr r4,[r0] \n"
"cmp r4,#0x0 \n"
"bne end \n"
"add r4,#0x1 \n"
"str r4,[r0] \n"
"add r0,#0x1 \n"
"b loop \n"
"end: \n"
:
: [ptrToSmpl]"r" (ptrToSmpl)
: "r0" , "r4", "memory"
);
printf("Sample[0] = %i" , Smpl[0]);
Edit 2: 编辑2:
As you can see above theres still a bug in the adressing I think. 如您在上方看到的,我认为地址定位中仍然存在错误。 The error message now is: "/tmp/ccE69oZd.s|75|Error: undefined symbol r6 used as an immediate value|"
现在的错误消息是:“ / tmp / ccE69oZd.s | 75 |错误:未定义的符号r6用作立即值|” but there's no r6 anywhere.
但哪里都没有r6。
Don't hard code the registers that the assembly is supposed to use. 不要对程序集应该使用的寄存器进行硬编码。 Use the
%0
, %1
etc notation to name the registers that correspond to the parameters that you are passing. 使用
%0
, %1
等符号来命名与您要传递的参数相对应的寄存器。
If you do so correctly you shouldn't need to use constraints such as "memory"
in your case. 如果正确执行此操作,则无需使用
"memory"
约束。
You should put %[ptrToSmpl]
inside square brackets - you want to load/store from/to the address that is in the register. 您应该将
%[ptrToSmpl]
放在方括号内-您要从寄存器中的地址加载/存储。
This ldr r2,%[ptrToSmpl]
gets translated into ldr r2,rX
- nonsense 这个
ldr r2,%[ptrToSmpl]
被翻译成ldr r2,rX
ldr r2,%[ptrToSmpl]
废话
This ldr r2,[%[ptrToSmpl]]
ldr r2,[rX]
- correct 这个
ldr r2,[%[ptrToSmpl]]
ldr r2,[rX]
-正确
Moreover - you should put r2 into the list of clobbered registers, as you overwrite what compiler puts there. 此外,在覆盖编译器放置的内容时,应将r2放入损坏的寄存器列表中。 Add "r2" (with quotes) after the last (third) colon.
在最后一个(第三个)冒号后面添加“ r2”(带引号)。
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