ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<read_config_line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%r15
  mov    %rdx,0x8(%rsp)
  movq   $0x0,(%rdx)
  lea    -0x1(%rcx),%r14d
  test   %r14d,%r14d
  jle    L13
  mov    %rdi,%r13
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  jmp    L82
L83:
  mov    %r13,%rdi
  call   <skip_line@plt>
  jmp    L22
L68:
  movslq %ebp,%rax
  movb   $0x0,(%r15,%rax,1)
L138:
  movslq %ebp,%rax
  movb   $0x0,0x1(%r15,%rax,1)
  mov    %ebp,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  mov    $0x0,%ebp
  jmp    L138
L58:
  cmp    $0xa,%ebx
  je     L68
L103:
  mov    D1(%rip),%rax
  cmp    %ebx,(%rax)
  je     L68
  movslq %ebp,%rax
  mov    %bl,(%r15,%rax,1)
L110:
  add    $0x1,%ebp
  mov    $0x1,%r12d
L22:
  cmp    %r14d,%ebp
  jge    L138
L82:
  mov    %r13,%rdi
  call   <next_char@plt>
  mov    %eax,%ebx
  test   %r12d,%r12d
  jne    L109
  cmp    $0x23,%eax
  je     L83
  cmp    $0x3b,%eax
  je     L83
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L22
  cmp    $0x3d,%ebx
  je     L152
  cmp    $0xa,%ebx
  jne    L103
  mov    %r12d,%ebp
  jmp    L82
L109:
  cmp    $0x3d,%eax
  jne    L58
L152:
  mov    0x8(%rsp),%rax
  cmpq   $0x0,(%rax)
  jne    L103
  movslq %ebp,%rax
  movb   $0x0,(%r15,%rax,1)
  lea    0x1(%r15,%rax,1),%rax
  mov    0x8(%rsp),%rdx
  mov    %rax,(%rdx)
  jmp    L110
```

assistant: ```c
int read_config_line(FILE *f, char *line, const char **value, int bufsize) {
  int i = 0, isname = 0;
  *value = NULL;
  while (i < bufsize - 1) {
    int c = next_char(f);
    if (!isname && (c == '#' || c == ';')) {
      skip_line(f);
      continue;
    }
    if (!isname && isspace(c))
      continue;
    if (c == '=' && !*value) {
      line[i] = 0;
      *value = &line[i + 1];
    } else if (c == '\n' && !isname) {
      i = 0;
      continue;
    } else if (c == '\n' || c == EOF) {
      line[i] = 0;
      break;
    } else {
      line[i] = c;
    }
    isname = 1;
    i++;
  }
  line[i + 1] = 0;
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_config_line>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  lea    -0x1(%rcx),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  movq   $0x0,(%rdx)
  mov    %rdx,0x8(%rsp)
  test   %r13d,%r13d
  jle    L157
  mov    %rdi,%r12
  xor    %ebp,%ebp
  xor    %r15d,%r15d
  jmp    L55
  nopl   0x0(%rax,%rax,1)
L33:
  cmp    $0xa,%ebx
  je     L7
L152:
  mov    D3(%rip),%rdx
  movslq %r15d,%rax
  lea    (%r14,%rax,1),%rcx
  cmp    %ebx,(%rdx)
  je     L17
  mov    %bl,(%rcx)
L175:
  add    $0x1,%r15d
  mov    $0x1,%ebp
L95:
  cmp    %r13d,%r15d
  jge    L91
L55:
  mov    %r12,%rdi
  call   <next_char@plt>
  mov    %eax,%ebx
  test   %ebp,%ebp
  je     L57
  cmp    $0x3d,%eax
  jne    L33
L39:
  mov    0x8(%rsp),%rax
  cmpq   $0x0,(%rax)
  je     L108
  mov    $0x3d,%ebx
  jmp    L152
  nopl   0x0(%rax)
L57:
  cmp    $0x23,%ebx
  je     L85
L48:
  cmp    $0x3b,%ebx
  je     L85
  mov    %ebx,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L95
  cmp    $0x3d,%ebx
  je     L39
  cmp    $0xa,%ebx
  jne    L152
  mov    %r12,%rdi
  xor    %r15d,%r15d
  call   <next_char@plt>
  mov    %eax,%ebx
  cmp    $0x23,%ebx
  jne    L48
L85:
  mov    %r12,%rdi
  call   <skip_line@plt>
  cmp    %r13d,%r15d
  jl     L55
  nopw   0x0(%rax,%rax,1)
L91:
  lea    0x1(%r15),%eax
  cltq
L100:
  movb   $0x0,(%r14,%rax,1)
  add    $0x18,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L108:
  movslq %r15d,%rax
  mov    0x8(%rsp),%rsi
  movb   $0x0,(%r14,%rax,1)
  lea    0x1(%r14,%rax,1),%rax
  mov    %rax,(%rsi)
  jmp    L175
  nopw   0x0(%rax,%rax,1)
L7:
  movslq %r15d,%rax
L17:
  movb   $0x0,(%r14,%rax,1)
  lea    0x1(%r15),%eax
  cltq
  jmp    L100
L157:
  mov    $0x1,%eax
  xor    %r15d,%r15d
  jmp    L100
```

assistant: ```c
int read_config_line(FILE *f, char *line, const char **value, int bufsize) {
  int i = 0, isname = 0;
  *value = NULL;
  while (i < bufsize - 1) {
    int c = next_char(f);
    if (!isname && (c == '#' || c == ';')) {
      skip_line(f);
      continue;
    }
    if (!isname && isspace(c))
      continue;
    if (c == '=' && !*value) {
      line[i] = 0;
      *value = &line[i + 1];
    } else if (c == '\n' && !isname) {
      i = 0;
      continue;
    } else if (c == '\n' || c == EOF) {
      line[i] = 0;
      break;
    } else {
      line[i] = c;
    }
    isname = 1;
    i++;
  }
  line[i + 1] = 0;
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<read_config_line>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  lea    -0x1(%rcx),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  movq   $0x0,(%rdx)
  mov    %rdx,0x8(%rsp)
  test   %r13d,%r13d
  jle    L7
  mov    %rdi,%r12
  xor    %ebp,%ebp
  xor    %r15d,%r15d
  jmp    L164
  nopl   0x0(%rax,%rax,1)
L39:
  cmp    $0xa,%ebx
  je     L181
L115:
  mov    D6(%rip),%rdx
  movslq %r15d,%rax
  lea    (%r14,%rax,1),%rcx
  cmp    %ebx,(%rdx)
  je     L51
  mov    %bl,(%rcx)
L15:
  add    $0x1,%r15d
  mov    $0x1,%ebp
L105:
  cmp    %r13d,%r15d
  jge    L74
L164:
  mov    %r12,%rdi
  call   <next_char@plt>
  mov    %eax,%ebx
  test   %ebp,%ebp
  je     L28
  cmp    $0x3d,%eax
  jne    L39
L163:
  mov    0x8(%rsp),%rax
  cmpq   $0x0,(%rax)
  je     L89
  mov    $0x3d,%ebx
  jmp    L115
  nopl   0x0(%rax)
L28:
  cmp    $0x23,%ebx
  je     L12
L5:
  cmp    $0x3b,%ebx
  je     L12
  mov    %ebx,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L105
  cmp    $0x3d,%ebx
  je     L163
  cmp    $0xa,%ebx
  jne    L115
  mov    %r12,%rdi
  xor    %r15d,%r15d
  call   <next_char@plt>
  mov    %eax,%ebx
  cmp    $0x23,%ebx
  jne    L5
L12:
  mov    %r12,%rdi
  call   <skip_line@plt>
  cmp    %r13d,%r15d
  jl     L164
  nopw   0x0(%rax,%rax,1)
L74:
  lea    0x1(%r15),%eax
  cltq
L37:
  movb   $0x0,(%r14,%rax,1)
  add    $0x18,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L89:
  mov    %rax,%rsi
  movslq %r15d,%rax
  movb   $0x0,(%r14,%rax,1)
  lea    0x1(%r14,%rax,1),%rax
  mov    %rax,(%rsi)
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L181:
  movslq %r15d,%rax
L51:
  movb   $0x0,(%r14,%rax,1)
  lea    0x1(%r15),%eax
  cltq
  jmp    L37
L7:
  mov    $0x1,%eax
  xor    %r15d,%r15d
  jmp    L37
```

assistant: ```c
int read_config_line(FILE *f, char *line, const char **value, int bufsize) {
  int i = 0, isname = 0;
  *value = NULL;
  while (i < bufsize - 1) {
    int c = next_char(f);
    if (!isname && (c == '#' || c == ';')) {
      skip_line(f);
      continue;
    }
    if (!isname && isspace(c))
      continue;
    if (c == '=' && !*value) {
      line[i] = 0;
      *value = &line[i + 1];
    } else if (c == '\n' && !isname) {
      i = 0;
      continue;
    } else if (c == '\n' || c == EOF) {
      line[i] = 0;
      break;
    } else {
      line[i] = c;
    }
    isname = 1;
    i++;
  }
  line[i + 1] = 0;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<OneUTF82Unicode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L24
  cmpl   $0x3,-0x4(%rbp)
  jg     L79
  cmpl   $0x1,-0x4(%rbp)
  je     L14
  cmpl   $0x2,-0x4(%rbp)
  je     L37
  jmp    L79
L14:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  jmp    L23
L37:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  and    $0x3f,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x6,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sar    $0x2,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  and    $0xf,%edx
  mov    %dl,(%rax)
  jmp    L23
L24:
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  and    $0x3f,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x6,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sar    $0x2,%eax
  and    $0xf,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  or     %ecx,%edx
  mov    %dl,(%rax)
  jmp    L23
L79:
  nop
L23:
  nop
  leave
  ret
```

assistant: ```c
void OneUTF82Unicode(const char *utf_char, char *unicode_char) {
  int utf_length = strlen(utf_char);
  switch (utf_length) {
  case 1:
    unicode_char[0] = utf_char[0];
    unicode_char[1] = 0;
    break;
  case 2:
    unicode_char[0] = (utf_char[1] & 0x3F) | ((utf_char[0] & 0x3) << 6);
    unicode_char[1] = (utf_char[0] & 0x3C) >> 2;
    break;
  case 3:
    unicode_char[0] = (utf_char[2] & 0x3F) | ((utf_char[1] & 0x3) << 6);
    unicode_char[1] = ((utf_char[1] & 0x3C) >> 2) | ((utf_char[0] & 0xF) << 4);
    break;
  default:
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<OneUTF82Unicode>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  cmp    $0x2,%eax
  je     L16
  cmp    $0x3,%eax
  je     L45
  cmp    $0x1,%eax
  je     L40
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L40:
  movzbl (%rbx),%eax
  mov    %al,0x0(%rbp)
  movb   $0x0,0x1(%rbp)
  jmp    L2
L16:
  movzbl 0x1(%rbx),%eax
  and    $0x3f,%eax
  movsbl (%rbx),%edx
  shl    $0x6,%edx
  or     %edx,%eax
  mov    %al,0x0(%rbp)
  movsbl (%rbx),%eax
  sar    $0x2,%eax
  and    $0xf,%eax
  mov    %al,0x1(%rbp)
  jmp    L2
L45:
  movzbl 0x2(%rbx),%eax
  and    $0x3f,%eax
  movsbl 0x1(%rbx),%edx
  shl    $0x6,%edx
  or     %edx,%eax
  mov    %al,0x0(%rbp)
  movsbl 0x1(%rbx),%eax
  sar    $0x2,%eax
  and    $0xf,%eax
  movsbl (%rbx),%edx
  shl    $0x4,%edx
  or     %edx,%eax
  mov    %al,0x1(%rbp)
  jmp    L2
```

assistant: ```c
void OneUTF82Unicode(const char *utf_char, char *unicode_char) {
  int utf_length = strlen(utf_char);
  switch (utf_length) {
  case 1:
    unicode_char[0] = utf_char[0];
    unicode_char[1] = 0;
    break;
  case 2:
    unicode_char[0] = (utf_char[1] & 0x3F) | ((utf_char[0] & 0x3) << 6);
    unicode_char[1] = (utf_char[0] & 0x3C) >> 2;
    break;
  case 3:
    unicode_char[0] = (utf_char[2] & 0x3F) | ((utf_char[1] & 0x3) << 6);
    unicode_char[1] = ((utf_char[1] & 0x3C) >> 2) | ((utf_char[0] & 0xF) << 4);
    break;
  default:
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<OneUTF82Unicode>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x2,%eax
  je     L23
  cmp    $0x3,%eax
  je     L0
  cmp    $0x1,%eax
  je     L18
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  movzbl 0x1(%rbx),%eax
  movsbl (%rbx),%edx
  and    $0x3f,%eax
  shl    $0x6,%edx
  or     %edx,%eax
  mov    %al,0x0(%rbp)
  movsbl (%rbx),%eax
  sar    $0x2,%eax
  and    $0xf,%eax
  mov    %al,0x1(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L18:
  movzbl (%rbx),%eax
  movb   $0x0,0x1(%rbp)
  mov    %al,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L0:
  movzbl 0x2(%rbx),%eax
  movsbl 0x1(%rbx),%edx
  and    $0x3f,%eax
  shl    $0x6,%edx
  or     %edx,%eax
  mov    %al,0x0(%rbp)
  movsbl 0x1(%rbx),%eax
  movsbl (%rbx),%edx
  sar    $0x2,%eax
  and    $0xf,%eax
  shl    $0x4,%edx
  or     %edx,%eax
  mov    %al,0x1(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void OneUTF82Unicode(const char *utf_char, char *unicode_char) {
  int utf_length = strlen(utf_char);
  switch (utf_length) {
  case 1:
    unicode_char[0] = utf_char[0];
    unicode_char[1] = 0;
    break;
  case 2:
    unicode_char[0] = (utf_char[1] & 0x3F) | ((utf_char[0] & 0x3) << 6);
    unicode_char[1] = (utf_char[0] & 0x3C) >> 2;
    break;
  case 3:
    unicode_char[0] = (utf_char[2] & 0x3F) | ((utf_char[1] & 0x3) << 6);
    unicode_char[1] = ((utf_char[1] & 0x3C) >> 2) | ((utf_char[0] & 0xF) << 4);
    break;
  default:
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<OneUTF82Unicode>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x2,%eax
  je     L14
  cmp    $0x3,%eax
  je     L11
  cmp    $0x1,%eax
  je     L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  movzbl 0x1(%rbx),%eax
  movsbl (%rbx),%edx
  and    $0x3f,%eax
  shl    $0x6,%edx
  or     %edx,%eax
  mov    %al,0x0(%rbp)
  movsbl (%rbx),%eax
  sar    $0x2,%eax
  and    $0xf,%eax
  mov    %al,0x1(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L6:
  movzbl (%rbx),%eax
  movb   $0x0,0x1(%rbp)
  mov    %al,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L11:
  movzbl 0x2(%rbx),%eax
  movsbl 0x1(%rbx),%edx
  and    $0x3f,%eax
  shl    $0x6,%edx
  or     %edx,%eax
  mov    %al,0x0(%rbp)
  movsbl 0x1(%rbx),%eax
  movsbl (%rbx),%edx
  sar    $0x2,%eax
  and    $0xf,%eax
  shl    $0x4,%edx
  or     %edx,%eax
  mov    %al,0x1(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void OneUTF82Unicode(const char *utf_char, char *unicode_char) {
  int utf_length = strlen(utf_char);
  switch (utf_length) {
  case 1:
    unicode_char[0] = utf_char[0];
    unicode_char[1] = 0;
    break;
  case 2:
    unicode_char[0] = (utf_char[1] & 0x3F) | ((utf_char[0] & 0x3) << 6);
    unicode_char[1] = (utf_char[0] & 0x3C) >> 2;
    break;
  case 3:
    unicode_char[0] = (utf_char[2] & 0x3F) | ((utf_char[1] & 0x3) << 6);
    unicode_char[1] = ((utf_char[1] & 0x3C) >> 2) | ((utf_char[0] & 0xF) << 4);
    break;
  default:
    break;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<multiTable_v3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x8(%rbp)
  jmp    L44
L52:
  movl   $0x1,-0x4(%rbp)
  jmp    L34
L43:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L17
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L24
L17:
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L24:
  addl   $0x1,-0x4(%rbp)
L34:
  cmpl   $0x9,-0x4(%rbp)
  jle    L43
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
L44:
  cmpl   $0x9,-0x8(%rbp)
  jle    L52
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D0
type_guessed: string

tool: D5:         

tool: D0: %d*%d=%2d	

assistant: ```c
void multiTable_v3() {
  int i;
  int j;
  for (i = 1; i <= 9; i++) {
    for (j = 1; j <= 9; j++) {
      if (j > i) {
        printf("        ");
      } else {
        printf("%d*%d=%2d\t", i, j, i * j);
      }
    }
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multiTable_v3>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1,%r12d
  lea    D6(%rip),%r15
  lea    D15(%rip),%r14
  jmp    L3
L11:
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L8:
  add    $0x1,%ebx
  add    %r13d,%ebp
  cmp    $0xa,%ebx
  je     L21
L30:
  cmp    %ebx,%r12d
  jge    L11
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L8
L21:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r12d
  cmp    $0xa,%r12d
  je     L18
L3:
  mov    %r12d,%r13d
  mov    %r12d,%ebp
  mov    $0x1,%ebx
  jmp    L30
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
type_guessed: string
name: parse_data
arguments: data_label: D15
type_guessed: string

tool: D6: %d*%d=%2d	

tool: D15:         

assistant: ```c
void multiTable_v3() {
  int i;
  int j;
  for (i = 1; i <= 9; i++) {
    for (j = 1; j <= 9; j++) {
      if (j > i) {
        printf("        ");
      } else {
        printf("%d*%d=%2d\t", i, j, i * j);
      }
    }
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiTable_v3>:
  endbr64
  push   %r14
  lea    D4(%rip),%r14
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  nop
L6:
  mov    %ebp,%r12d
  mov    $0x1,%ebx
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L48:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    %ebp,%r12d
  call   <printf@plt>
  cmp    $0xa,%ebx
  je     L34
L14:
  cmp    %ebx,%ebp
  jl     L48
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    %ebp,%r12d
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L14
L34:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    $0xa,%ebp
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
datatype: string
name: unpack_data
arguments: label: D15
datatype: string

tool: D4: %d*%d=%2d	

tool: D15:         

assistant: ```c
void multiTable_v3() {
  int i;
  int j;
  for (i = 1; i <= 9; i++) {
    for (j = 1; j <= 9; j++) {
      if (j > i) {
        printf("        ");
      } else {
        printf("%d*%d=%2d\t", i, j, i * j);
      }
    }
    printf("\n");
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<multiTable_v3>:
  endbr64
  push   %r14
  lea    D19(%rip),%r14
  push   %r13
  lea    D10(%rip),%r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  nop
L21:
  mov    %ebp,%r12d
  mov    $0x1,%ebx
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    %ebp,%r12d
  call   <printf@plt>
  cmp    $0xa,%ebx
  je     L10
L34:
  cmp    %ebx,%ebp
  jl     L32
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    %ebp,%r12d
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L34
L10:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    $0xa,%ebp
  jne    L21
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D19
datatype: string
name: parse_data
arguments: data_tag: D10
datatype: string

tool: D19: %d*%d=%2d	

tool: D10:         

assistant: ```c
void multiTable_v3() {
  int i;
  int j;
  for (i = 1; i <= 9; i++) {
    for (j = 1; j <= 9; j++) {
      if (j > i) {
        printf("        ");
      } else {
        printf("%d*%d=%2d\t", i, j, i * j);
      }
    }
    printf("\n");
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<isPalindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L33
L31:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L24
  mov    $0x0,%eax
  jmp    L25
L24:
  addl   $0x1,-0x4(%rbp)
L33:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x4(%rbp)
  jl     L31
  mov    $0x1,%eax
L25:
  pop    %rbp
  ret
```

assistant: ```c
int isPalindrome(int seq[], int length) {
  int i = 0;
  for (i = 0; i < length / 2; i++)
    if (seq[i] != seq[length - 1 - i])
      return 0;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<isPalindrome>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L17
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  mov    $0x0,%eax
L22:
  mov    -0x4(%rdx),%esi
  cmp    %esi,(%rdi,%rax,4)
  jne    L20
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jg     L22
  mov    $0x1,%eax
  ret
L17:
  mov    $0x1,%eax
  ret
L20:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int isPalindrome(int seq[], int length) {
  int i = 0;
  for (i = 0; i < length / 2; i++)
    if (seq[i] != seq[length - 1 - i])
      return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<isPalindrome>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L17
  sub    $0x1,%esi
  xor    %eax,%eax
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  jmp    L4
L12:
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jle    L17
L4:
  mov    (%rdx),%esi
  cmp    %esi,(%rdi,%rax,4)
  je     L12
  xor    %eax,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L17:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isPalindrome(int seq[], int length) {
  int i = 0;
  for (i = 0; i < length / 2; i++)
    if (seq[i] != seq[length - 1 - i])
      return 0;
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<isPalindrome>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L14
  sub    $0x1,%esi
  xor    %eax,%eax
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rdx
  jmp    L10
L13:
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jle    L14
L10:
  mov    (%rdx),%esi
  cmp    %esi,(%rdi,%rax,4)
  je     L13
  xor    %eax,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isPalindrome(int seq[], int length) {
  int i = 0;
  for (i = 0; i < length / 2; i++)
    if (seq[i] != seq[length - 1 - i])
      return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<led_on>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    $0xf0000028,%eax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    $0xf0000028,%eax
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void led_on(int led) {
  int current_value;
  current_value = *(unsigned int *)0xf0000028;
  *(unsigned int *)0xf0000028 = current_value | (1 << led);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<led_on>:
  endbr64
  mov    $0xf0000028,%edx
  mov    $0x1,%eax
  mov    %edi,%ecx
  shl    %cl,%eax
  or     %eax,(%rdx)
  ret
```

assistant: ```c
void led_on(int led) {
  int current_value;
  current_value = *(unsigned int *)0xf0000028;
  *(unsigned int *)0xf0000028 = current_value | (1 << led);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<led_on>:
  endbr64
  mov    $0xf0000028,%edx
  mov    $0x1,%eax
  mov    %edi,%ecx
  shl    %cl,%eax
  or     %eax,(%rdx)
  ret
```

assistant: ```c
void led_on(int led) {
  int current_value;
  current_value = *(unsigned int *)0xf0000028;
  *(unsigned int *)0xf0000028 = current_value | (1 << led);
}
```