ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<look_up_flag>:
  endbr64
  push   %r12
  mov    %edi,%ecx
  lea    D9(%rip),%rdx
  mov    $0x32,%esi
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rbp
  mov    %rsp,%r12
  mov    %rbp,%rdi
  call   <snprintf@plt>
  mov    %rbp,%rsi
  mov    $0x5,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  lea    0x20(%rsp),%rbp
  lea    0x45(%rsp),%rsi
  mov    $0x3,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    %r12,%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    %ebx,%eax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type: string

tool: D9: %d

assistant: ```c
int look_up_flag(int number) {
  int index = 0;
  char temp[50];
  char part1[25];
  char part2[25];
  snprintf(temp, 50, "%d", number);
  strncpy(part1, temp, 5);
  strncpy(part2, &temp[5], 3);
  index = atoi(part1) + atoi(part2);
  return index;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<look_up_flag>:
  endbr64
  push   %r12
  mov    %edi,%ecx
  lea    D0(%rip),%rdx
  mov    $0x32,%esi
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rbp
  mov    %rsp,%r12
  mov    %rbp,%rdi
  call   <snprintf@plt>
  mov    %rbp,%rsi
  mov    $0x5,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  lea    0x20(%rsp),%rbp
  lea    0x45(%rsp),%rsi
  mov    $0x3,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    %r12,%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    %ebx,%eax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D0
type_infered: string

tool: D0: %d

assistant: ```c
int look_up_flag(int number) {
  int index = 0;
  char temp[50];
  char part1[25];
  char part2[25];
  snprintf(temp, 50, "%d", number);
  strncpy(part1, temp, 5);
  strncpy(part2, &temp[5], 3);
  index = atoi(part1) + atoi(part2);
  return index;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4,-0x30(%rbp)
  movl   $0x41,-0x2c(%rbp)
  movl   $0x2,-0x28(%rbp)
  movl   $0xffffffe1,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x63,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  movl   $0x53,-0x14(%rbp)
  movl   $0x30e,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  movl   $0xa,-0x34(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L0
L3:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L0:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L3
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <selectionsort@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L34
L5:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L34:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L5
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type: string

tool: D16: %d 

assistant: ```c
int main() {
  int a[] = {4, 65, 2, -31, 0, 99, 2, 83, 782, 1};
  int n = sizeof a / sizeof a[0];
  int i;
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  selectionsort(a, n);
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x4,(%rsp)
  movl   $0x41,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0xffffffe1,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x63,0x14(%rsp)
  movl   $0x2,0x18(%rsp)
  movl   $0x53,0x1c(%rsp)
  movl   $0x30e,0x20(%rsp)
  movl   $0x1,0x24(%rsp)
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  mov    %rbx,%rbp
  lea    D18(%rip),%r13
L15:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L15
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rsp,%rdi
  mov    $0xa,%esi
  call   <selectionsort@plt>
  lea    D18(%rip),%rbp
L21:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L21
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D18
type_infered: string

tool: D18: %d 

assistant: ```c
int main() {
  int a[] = {4, 65, 2, -31, 0, 99, 2, 83, 782, 1};
  int n = sizeof a / sizeof a[0];
  int i;
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  selectionsort(a, n);
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x4,%esi
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x4100000004,%rax
  mov    %rsp,%r14
  lea    0x4(%rsp),%rbp
  lea    0x28(%rsp),%r12
  mov    %r14,%rbx
  mov    %rax,(%rsp)
  movabs $0xffffffe100000002,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x6300000000,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x5300000002,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x10000030e,%rax
  mov    %rax,0x20(%rsp)
  jmp    L21
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    0x0(%rbp),%esi
  add    $0x4,%rbp
L21:
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L27
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%esi
  mov    %r14,%rdi
  call   <selectionsort@plt>
  nopl   (%rax)
L29:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L29
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
type: string

tool: D9: %d 

assistant: ```c
int main() {
  int a[] = {4, 65, 2, -31, 0, 99, 2, 83, 782, 1};
  int n = sizeof a / sizeof a[0];
  int i;
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  selectionsort(a, n);
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D36(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D14(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D27(%rip),%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D20(%rip),%xmm0
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  nopw   0x0(%rax,%rax,1)
L13:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%esi
  mov    %r14,%rdi
  call   <selectionsort@plt>
  nopl   (%rax)
L11:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D36
type: string
name: retrieve_data
arguments: data_label: D14
type: byte[16]
name: retrieve_data
arguments: data_label: D27
type: byte[8]
name: retrieve_data
arguments: data_label: D20
type: byte[16]

tool: D36: %d 

tool: {"D14": [0x04,0x00,0x00,0x00,0x41,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0xe1,0xff,0xff,0xff]}

tool: {"D27": [0x0e,0x03,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D20": [0x00,0x00,0x00,0x00,0x63,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x53,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int a[] = {4, 65, 2, -31, 0, 99, 2, 83, 782, 1};
  int n = sizeof a / sizeof a[0];
  int i;
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  selectionsort(a, n);
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<a_minus>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  call   <Equalize@plt>
  mov    0x8(%rsp),%rdi
  call   <my_strdup@plt>
  mov    %rax,%rbp
  mov    0x8(%rsp),%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  je     L67
  movslq %eax,%rcx
  mov    $0x0,%edx
  jmp    L35
L69:
  movsbl %al,%eax
  sub    $0x30,%eax
  jmp    L49
L82:
  movsbl %sil,%esi
  sub    $0x30,%esi
L10:
  sub    %esi,%eax
  add    %edx,%eax
  lea    0x1f(%rax),%edx
  add    $0x10,%eax
  cmovns %eax,%edx
  sar    $0x4,%edx
  sub    $0x1,%edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  shr    $0x1c,%esi
  add    %esi,%eax
  and    $0xf,%eax
  sub    %esi,%eax
  lea    0x37(%rax),%edi
  lea    0x30(%rax),%esi
  cmp    $0xa,%eax
  mov    %edi,%eax
  cmovl  %esi,%eax
  mov    %al,0x0(%rbp,%rcx,1)
  sub    $0x1,%rcx
  test   %ecx,%ecx
  je     L67
L35:
  cmpb   $0x78,(%rbx,%rcx,1)
  je     L67
  mov    0x8(%rsp),%rax
  movzbl (%rax,%rcx,1),%eax
  cmp    $0x39,%al
  jle    L69
  movsbl %al,%eax
  sub    $0x37,%eax
L49:
  mov    0x10(%rsp),%rsi
  movzbl (%rsi,%rcx,1),%esi
  cmp    $0x39,%sil
  jle    L82
  movsbl %sil,%esi
  sub    $0x37,%esi
  jmp    L10
L67:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    %rbp,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *a_minus(char *pcL, char *pcR) {
  char *pcResult;
  char *pcNewLeft, *pcNewRight;
  int iDigitDiff;
  int i, iCarry;
  Equalize(pcL, pcR, &pcNewLeft, &pcNewRight);
  pcResult = my_strdup(pcNewLeft);
  i = strlen(pcNewLeft);
  i--;
  iCarry = 0;
  while ((i != 0) && (pcL[i] != 'x')) {
    iDigitDiff = (((pcNewLeft[i]) > '9') ? ((pcNewLeft[i]) - 'A' + 10) : ((pcNewLeft[i]) - '0')) - (((pcNewRight[i]) > '9') ? ((pcNewRight[i]) - 'A' + 10) : ((pcNewRight[i]) - '0'));
    iDigitDiff = iDigitDiff + iCarry;
    iDigitDiff += 0x10;
    iCarry = iDigitDiff / 0x10;
    iCarry--;
    iDigitDiff = iDigitDiff % 0x10;
    pcResult[i] = (((iDigitDiff) > 9) ? ((iDigitDiff)-10 + 'A') : ((iDigitDiff) + '0'));
    i--;
  }
  free(pcNewLeft);
  free(pcNewRight);
  return pcResult;
}
```
user: Translate the below assembly code into C programming language:

```asm
<a_minus>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  call   <Equalize@plt>
  mov    0x8(%rsp),%rdi
  call   <my_strdup@plt>
  mov    0x8(%rsp),%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  sub    $0x1,%eax
  je     L47
  movslq %eax,%rcx
  xor    %edx,%edx
  jmp    L59
  nopl   0x0(%rax)
L38:
  mov    0x10(%rsp),%rsi
  sub    $0x37,%eax
  movsbl %al,%eax
  movsbl (%rsi,%rcx,1),%esi
  cmp    $0x39,%sil
  jle    L5
L0:
  sub    $0x37,%esi
  movsbl %sil,%esi
L14:
  sub    %esi,%eax
  add    %edx,%eax
  lea    0x1f(%rax),%edx
  add    $0x10,%eax
  mov    %eax,%esi
  cmovns %eax,%edx
  sar    $0x1f,%esi
  shr    $0x1c,%esi
  sar    $0x4,%edx
  add    %esi,%eax
  sub    $0x1,%edx
  and    $0xf,%eax
  sub    %esi,%eax
  lea    0x37(%rax),%edi
  cmp    $0xa,%eax
  lea    0x30(%rax),%esi
  mov    %edi,%eax
  cmovl  %esi,%eax
  mov    %al,(%r12,%rcx,1)
  sub    $0x1,%rcx
  test   %ecx,%ecx
  je     L47
L59:
  cmpb   $0x78,(%rbx,%rcx,1)
  mov    0x8(%rsp),%rdi
  je     L57
  movsbl (%rdi,%rcx,1),%eax
  cmp    $0x39,%al
  jg     L38
  mov    0x10(%rsp),%rsi
  sub    $0x30,%eax
  movsbl (%rsi,%rcx,1),%esi
  cmp    $0x39,%sil
  jg     L0
L5:
  sub    $0x30,%esi
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L47:
  mov    0x8(%rsp),%rdi
L57:
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x28,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *a_minus(char *pcL, char *pcR) {
  char *pcResult;
  char *pcNewLeft, *pcNewRight;
  int iDigitDiff;
  int i, iCarry;
  Equalize(pcL, pcR, &pcNewLeft, &pcNewRight);
  pcResult = my_strdup(pcNewLeft);
  i = strlen(pcNewLeft);
  i--;
  iCarry = 0;
  while ((i != 0) && (pcL[i] != 'x')) {
    iDigitDiff = (((pcNewLeft[i]) > '9') ? ((pcNewLeft[i]) - 'A' + 10) : ((pcNewLeft[i]) - '0')) - (((pcNewRight[i]) > '9') ? ((pcNewRight[i]) - 'A' + 10) : ((pcNewRight[i]) - '0'));
    iDigitDiff = iDigitDiff + iCarry;
    iDigitDiff += 0x10;
    iCarry = iDigitDiff / 0x10;
    iCarry--;
    iDigitDiff = iDigitDiff % 0x10;
    pcResult[i] = (((iDigitDiff) > 9) ? ((iDigitDiff)-10 + 'A') : ((iDigitDiff) + '0'));
    i--;
  }
  free(pcNewLeft);
  free(pcNewRight);
  return pcResult;
}
```
user: What is the c source code of the assembly code below?

```asm
<a_minus>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  call   <Equalize@plt>
  mov    0x8(%rsp),%rdi
  call   <my_strdup@plt>
  mov    0x8(%rsp),%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  sub    $0x1,%eax
  je     L72
  movslq %eax,%rdx
  xor    %ecx,%ecx
  xchg   %ax,%ax
L18:
  cmpb   $0x78,(%rbx,%rdx,1)
  mov    0x8(%rsp),%rdi
  je     L77
  movsbl (%rdi,%rdx,1),%eax
  cmp    $0x39,%al
  jle    L94
  mov    0x10(%rsp),%rsi
  sub    $0x37,%eax
  movsbl %al,%eax
  movsbl (%rsi,%rdx,1),%esi
  cmp    $0x39,%sil
  jle    L76
L57:
  sub    $0x37,%esi
  movsbl %sil,%esi
L43:
  sub    %esi,%eax
  add    %ecx,%eax
  lea    0x1f(%rax),%ecx
  add    $0x10,%eax
  mov    %eax,%esi
  cmovns %eax,%ecx
  sar    $0x1f,%esi
  shr    $0x1c,%esi
  sar    $0x4,%ecx
  add    %esi,%eax
  sub    $0x1,%ecx
  and    $0xf,%eax
  sub    %esi,%eax
  cmp    $0x9,%eax
  jle    L88
  add    $0x37,%eax
  mov    %al,(%r12,%rdx,1)
  sub    $0x1,%rdx
  test   %edx,%edx
  jne    L18
L72:
  mov    0x8(%rsp),%rdi
L77:
  call   <free@plt>
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x28,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L88:
  add    $0x30,%eax
  mov    %al,(%r12,%rdx,1)
  sub    $0x1,%rdx
  test   %edx,%edx
  jne    L18
  jmp    L72
  nopl   (%rax)
L94:
  mov    0x10(%rsp),%rsi
  sub    $0x30,%eax
  movsbl (%rsi,%rdx,1),%esi
  cmp    $0x39,%sil
  jg     L57
L76:
  sub    $0x30,%esi
  jmp    L43
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *a_minus(char *pcL, char *pcR) {
  char *pcResult;
  char *pcNewLeft, *pcNewRight;
  int iDigitDiff;
  int i, iCarry;
  Equalize(pcL, pcR, &pcNewLeft, &pcNewRight);
  pcResult = my_strdup(pcNewLeft);
  i = strlen(pcNewLeft);
  i--;
  iCarry = 0;
  while ((i != 0) && (pcL[i] != 'x')) {
    iDigitDiff = (((pcNewLeft[i]) > '9') ? ((pcNewLeft[i]) - 'A' + 10) : ((pcNewLeft[i]) - '0')) - (((pcNewRight[i]) > '9') ? ((pcNewRight[i]) - 'A' + 10) : ((pcNewRight[i]) - '0'));
    iDigitDiff = iDigitDiff + iCarry;
    iDigitDiff += 0x10;
    iCarry = iDigitDiff / 0x10;
    iCarry--;
    iDigitDiff = iDigitDiff % 0x10;
    pcResult[i] = (((iDigitDiff) > 9) ? ((iDigitDiff)-10 + 'A') : ((iDigitDiff) + '0'));
    i--;
  }
  free(pcNewLeft);
  free(pcNewRight);
  return pcResult;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<c2_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <gpio_init@plt>
  mov    $0xd,%esi
  mov    $0x4,%edi
  call   <gpio_input@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_output@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_low@plt>
  mov    $0x14,%edi
  call   <usleep@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_high@plt>
  mov    $0x2,%edi
  call   <usleep@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void c2_init(void) {
  gpio_init();
  gpio_input(4, 13);
  gpio_output(4, 11);
  gpio_low(4, 11);
  usleep(20);
  gpio_high(4, 11);
  usleep(2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<c2_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <gpio_init@plt>
  mov    $0xd,%esi
  mov    $0x4,%edi
  call   <gpio_input@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_output@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_low@plt>
  mov    $0x14,%edi
  call   <usleep@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_high@plt>
  mov    $0x2,%edi
  call   <usleep@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void c2_init(void) {
  gpio_init();
  gpio_input(4, 13);
  gpio_output(4, 11);
  gpio_low(4, 11);
  usleep(20);
  gpio_high(4, 11);
  usleep(2);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<c2_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <gpio_init@plt>
  mov    $0xd,%esi
  mov    $0x4,%edi
  call   <gpio_input@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_output@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_low@plt>
  mov    $0x14,%edi
  call   <usleep@plt>
  mov    $0x4,%edi
  mov    $0xb,%esi
  call   <gpio_high@plt>
  mov    $0x2,%edi
  add    $0x8,%rsp
  jmp    10f0 <usleep@plt>
```

assistant: ```c
void c2_init(void) {
  gpio_init();
  gpio_input(4, 13);
  gpio_output(4, 11);
  gpio_low(4, 11);
  usleep(20);
  gpio_high(4, 11);
  usleep(2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<c2_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <gpio_init@plt>
  mov    $0xd,%esi
  mov    $0x4,%edi
  call   <gpio_input@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_output@plt>
  mov    $0xb,%esi
  mov    $0x4,%edi
  call   <gpio_low@plt>
  mov    $0x14,%edi
  call   <usleep@plt>
  mov    $0x4,%edi
  mov    $0xb,%esi
  call   <gpio_high@plt>
  mov    $0x2,%edi
  add    $0x8,%rsp
  jmp    10f0 <usleep@plt>
```

assistant: ```c
void c2_init(void) {
  gpio_init();
  gpio_input(4, 13);
  gpio_output(4, 11);
  gpio_low(4, 11);
  usleep(20);
  gpio_high(4, 11);
  usleep(2);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<gpio_iic_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0xe02000c0,%eax
  mov    (%rax),%eax
  and    $0xfffffff0,%eax
  or     $0x1,%eax
  mov    %eax,%edx
  mov    $0xe02000c0,%eax
  mov    %edx,(%rax)
  mov    $0xe02000c4,%eax
  mov    (%rax),%edx
  mov    $0xe02000c4,%eax
  or     $0x1,%edx
  mov    %edx,(%rax)
  mov    $0xe02000c0,%eax
  mov    (%rax),%edx
  mov    $0xe02000c0,%eax
  and    $0xfffffff0,%edx
  mov    %edx,(%rax)
  movb   $0x0,-0x1(%rbp)
  jmp    L0
L10:
  mov    $0xe02000c4,%eax
  mov    (%rax),%edx
  mov    $0xe02000c4,%eax
  or     $0x2,%edx
  mov    %edx,(%rax)
  mov    $0xa,%edi
  call   <delay@plt>
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    $0xe02000c4,%eax
  mov    (%rax),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L25
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
L25:
  mov    $0xe02000c4,%eax
  mov    (%rax),%edx
  mov    $0xe02000c4,%eax
  and    $0xfffffffd,%edx
  mov    %edx,(%rax)
  mov    $0xa,%edi
  call   <delay@plt>
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L0:
  cmpb   $0x7,-0x1(%rbp)
  jbe    L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void gpio_iic_read(unsigned char *rbyte) {
  unsigned char i;
  (*((volatile uint32_t *)0xE02000C0)) = ((*((volatile uint32_t *)0xE02000C0)) & 0xfffffff0) | 0x00000001;
  (*((volatile uint32_t *)0xE02000C4)) |= 1 << 0x00;
  (*((volatile uint32_t *)0xE02000C0)) = ((*((volatile uint32_t *)0xE02000C0)) & 0xfffffff0);
  for (i = 0; i < 8; i++) {
    (*((volatile uint32_t *)0xE02000C4)) |= 1 << 0x01;
    delay((10));
    (*rbyte) <<= 1;
    if (((*((volatile uint32_t *)0xE02000C4)) & 0x01))
      (*rbyte)++;
    (*((volatile uint32_t *)0xE02000C4)) &= ~(1 << 0x01);
    delay((10));
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<gpio_iic_read>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0xe02000c0,%edx
  mov    (%rdx),%eax
  and    $0xfffffff0,%eax
  or     $0x1,%eax
  mov    %eax,(%rdx)
  mov    $0xe02000c4,%ecx
  mov    (%rcx),%eax
  or     $0x1,%eax
  mov    %eax,(%rcx)
  mov    (%rdx),%eax
  and    $0xfffffff0,%eax
  mov    %eax,(%rdx)
  mov    $0x8,%r12d
  mov    %rcx,%rbx
L4:
  mov    (%rbx),%eax
  or     $0x2,%eax
  mov    %eax,(%rbx)
  mov    $0xa,%edi
  call   <delay@plt>
  movzbl 0x0(%rbp),%eax
  add    %eax,%eax
  mov    %al,0x0(%rbp)
  mov    (%rbx),%ecx
  lea    0x1(%rax),%edx
  test   $0x1,%cl
  cmovne %edx,%eax
  mov    %al,0x0(%rbp)
  mov    (%rbx),%eax
  and    $0xfffffffd,%eax
  mov    %eax,(%rbx)
  mov    $0xa,%edi
  call   <delay@plt>
  sub    $0x1,%r12b
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void gpio_iic_read(unsigned char *rbyte) {
  unsigned char i;
  (*((volatile uint32_t *)0xE02000C0)) = ((*((volatile uint32_t *)0xE02000C0)) & 0xfffffff0) | 0x00000001;
  (*((volatile uint32_t *)0xE02000C4)) |= 1 << 0x00;
  (*((volatile uint32_t *)0xE02000C0)) = ((*((volatile uint32_t *)0xE02000C0)) & 0xfffffff0);
  for (i = 0; i < 8; i++) {
    (*((volatile uint32_t *)0xE02000C4)) |= 1 << 0x01;
    delay((10));
    (*rbyte) <<= 1;
    if (((*((volatile uint32_t *)0xE02000C4)) & 0x01))
      (*rbyte)++;
    (*((volatile uint32_t *)0xE02000C4)) &= ~(1 << 0x01);
    delay((10));
  }
}
```