ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<of_read>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  lea    D0(%rip),%rax
  mov    %edi,0x10(%rsp)
  mov    %rsp,%rdi
  mov    %rsi,0x18(%rsp)
  mov    %edx,0x20(%rsp)
  mov    %rax,(%rsp)
  mov    D4(%rip),%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <of@plt>
  mov    0x24(%rsp),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x38,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_infered: string
name: unpack_data
arguments: data_label: D4
type_infered: byte[8]

tool: D0: read

tool: {"D4": [0x03,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int of_read(int handle, void *buf, int buf_len) {
  struct {
    const char *name;
    int num_args;
    int num_returns;
    int handle;
    void *buf;
    int buf_len;
    int size;
  } args;
  args.name = "read";
  args.num_args = 3;
  args.num_returns = 1;
  args.handle = handle;
  args.buf = buf;
  args.buf_len = buf_len;
  of(&args);
  return args.size;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_fnmatch_class>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rcx
  mov    %edx,%r11d
  mov    $0x0,%ebp
  cmpb   $0x21,(%rdi)
  je     L59
L55:
  movzbl (%rcx),%edx
  mov    $0x0,%ebx
  cmp    $0x5d,%dl
  je     L198
L4:
  movzbl 0x1(%rcx),%edi
  mov    $0x0,%eax
  mov    $0x1,%r10d
  jmp    L214
L59:
  add    $0x1,%rcx
  mov    $0x1,%ebp
  jmp    L55
L198:
  cmpb   $0x2d,0x1(%rcx)
  je     L201
L14:
  cmpb   $0x5d,(%rsi)
  sete   %bl
  movzbl %bl,%ebx
  mov    %rcx,%rax
L110:
  lea    0x1(%rax),%rcx
  movzbl 0x1(%rax),%edx
  mov    $0x0,%r8d
  cmp    $0x5d,%dl
  jne    L4
  jmp    L44
L201:
  movzbl 0x2(%rcx),%eax
  cmp    $0x5d,%al
  je     L14
  movzbl (%rsi),%edx
  cmp    %dl,%al
  setge  %al
  cmp    $0x5c,%dl
  setg   %bl
  movzbl %bl,%ebx
  and    %eax,%ebx
  lea    0x1(%rcx),%rax
  jmp    L110
L183:
  mov    %r8d,%eax
  mov    $0x1,%ebx
  jmp    L40
L38:
  cmp    %dl,(%rsi)
  cmove  %r10d,%ebx
L40:
  add    $0x1,%eax
  mov    %eax,%edx
  movzbl (%rcx,%rdx,1),%edx
  cmp    $0x5d,%dl
  je     L186
L214:
  cmp    $0x2d,%dil
  jne    L38
  lea    0x1(%rax),%r8d
  mov    %r8d,%r9d
  movzbl (%rcx,%r9,1),%r9d
  cmp    $0x5d,%r9b
  je     L160
  movzbl (%rsi),%edx
  lea    -0x1(%rax),%eax
  cmp    (%rcx,%rax,1),%dl
  jl     L122
  cmp    %dl,%r9b
  jge    L183
L122:
  mov    %r8d,%eax
  jmp    L40
L186:
  mov    %eax,%r8d
  jmp    L44
L57:
  mov    $0x1,%ebx
  jmp    L44
L192:
  add    $0x1,%rsi
  lea    0x1(%r8),%edi
  add    %rcx,%rdi
  mov    %r11d,%edx
  call   <fnmatch@plt>
  test   %eax,%eax
  je     L26
  jmp    L78
L160:
  cmp    (%rsi),%dl
  je     L57
  lea    0x1(%rax),%r8d
L44:
  cmpb   $0x0,(%rsi)
  jne    L192
L78:
  cmp    %ebx,%ebp
  setne  %al
  movzbl %al,%eax
L26:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int _fnmatch_class(char const *pattern, char const *string, int flags) {
  int ret = 0;
  int neg = 0;
  unsigned int i;
  if (pattern[0] == '!') {
    neg = 1;
    pattern++;
  }
  if (*pattern == ']') {
    if (pattern[1] == '-' && pattern[2] != ']') {
      if (*string >= pattern[0] && *string <= pattern[2])
        ret = 1;
      pattern++;
    } else if (string[0] == ']')
      ret = 1;
    pattern++;
  }
  for (i = 0; pattern[i] != ']'; i++) {
    if (pattern[1] == '-' && pattern[i + 1] != ']') {
      if (*string >= pattern[i - 1] && string[0] <= pattern[i + 1])
        ret = 1;
      i++;
    } else if (pattern[i] == *string)
      ret = 1;
  }
  return ((string[0] != '\0' ? fnmatch(&pattern[i + 1], &string[1], flags) : 1) && neg ^ ret) ? 1 : 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_fnmatch_class>:
  endbr64
  push   %r13
  mov    %rdi,%r8
  mov    %rsi,%r10
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%edx
  cmp    $0x21,%al
  jne    L29
  add    $0x1,%r8
  mov    %edx,%eax
  mov    $0x1,%ebp
L29:
  movzbl 0x1(%r8),%ecx
  movzbl (%r10),%edi
  xor    %ebx,%ebx
  cmp    $0x5d,%al
  je     L90
L19:
  movzbl 0x1(%r8),%r9d
  xor    %ecx,%ecx
  mov    $0x1,%r13d
  jmp    L129
  nopl   0x0(%rax)
L149:
  cmp    %dil,%al
  mov    %edx,%ecx
  mov    %esi,%eax
  jne    L104
  mov    $0x1,%ebx
L104:
  cmp    $0x5d,%al
  je     L116
L129:
  lea    0x1(%rcx),%esi
  mov    %rsi,%rdx
  movzbl (%r8,%rsi,1),%esi
  cmp    $0x2d,%r9b
  jne    L149
  cmp    $0x5d,%sil
  je     L13
  lea    -0x1(%rcx),%eax
  cmp    %dil,(%r8,%rax,1)
  lea    0x2(%rcx),%eax
  setle  %r11b
  cmp    %dil,%sil
  mov    %rax,%rcx
  movzbl (%r8,%rax,1),%eax
  setge  %dl
  test   %dl,%r11b
  cmovne %r13d,%ebx
  jmp    L104
  nopw   0x0(%rax,%rax,1)
L13:
  cmp    %dil,%al
  mov    $0x1,%eax
  cmove  %eax,%ebx
L9:
  test   %dil,%dil
  jne    L132
L30:
  add    $0x8,%rsp
  mov    %ebp,%eax
  xor    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L90:
  cmp    $0x2d,%cl
  je     L82
  xor    %ebx,%ebx
  cmp    $0x5d,%dil
  sete   %bl
  add    $0x1,%r8
  xor    %edx,%edx
  cmp    $0x5d,%cl
  je     L9
  mov    %ecx,%eax
  jmp    L19
L132:
  lea    0x1(%rdx),%edi
  lea    0x1(%r10),%rsi
  mov    %r12d,%edx
  add    %r8,%rdi
  call   <fnmatch@plt>
  test   %eax,%eax
  jne    L30
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L82:
  movzbl 0x2(%r8),%eax
  cmp    $0x5d,%al
  je     L79
  cmp    %dil,%al
  setge  %dl
  xor    %ebx,%ebx
  cmp    $0x5c,%dil
  setg   %bl
  add    $0x2,%r8
  and    %edx,%ebx
  jmp    L19
L79:
  xor    %ebx,%ebx
  cmp    $0x5d,%dil
  mov    %ecx,%eax
  sete   %bl
  add    $0x1,%r8
  jmp    L19
L116:
  mov    %ecx,%edx
  jmp    L9
```

assistant: ```c
int _fnmatch_class(char const *pattern, char const *string, int flags) {
  int ret = 0;
  int neg = 0;
  unsigned int i;
  if (pattern[0] == '!') {
    neg = 1;
    pattern++;
  }
  if (*pattern == ']') {
    if (pattern[1] == '-' && pattern[2] != ']') {
      if (*string >= pattern[0] && *string <= pattern[2])
        ret = 1;
      pattern++;
    } else if (string[0] == ']')
      ret = 1;
    pattern++;
  }
  for (i = 0; pattern[i] != ']'; i++) {
    if (pattern[1] == '-' && pattern[i + 1] != ']') {
      if (*string >= pattern[i - 1] && string[0] <= pattern[i + 1])
        ret = 1;
      i++;
    } else if (pattern[i] == *string)
      ret = 1;
  }
  return ((string[0] != '\0' ? fnmatch(&pattern[i + 1], &string[1], flags) : 1) && neg ^ ret) ? 1 : 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_fnmatch_class>:
  endbr64
  push   %r12
  mov    %edx,%r10d
  mov    %rdi,%rcx
  mov    %rsi,%r9
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  movzbl (%rdi),%edx
  movzbl 0x1(%rdi),%eax
  cmp    $0x21,%dl
  jne    L87
  add    $0x1,%rcx
  mov    %eax,%edx
  mov    $0x1,%ebp
L87:
  movzbl 0x1(%rcx),%esi
  movzbl (%r9),%edi
  xor    %ebx,%ebx
  cmp    $0x5d,%dl
  je     L5
L152:
  cmpb   $0x2d,0x1(%rcx)
  je     L20
L84:
  xor    %eax,%eax
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L54:
  add    $0x1,%eax
  mov    %edx,%esi
  mov    %eax,%edx
  cmp    %dil,%sil
  movzbl (%rcx,%rdx,1),%edx
  cmove  %r8d,%ebx
  cmp    $0x5d,%dl
  jne    L54
L67:
  test   %dil,%dil
  jne    L18
L93:
  mov    %ebp,%eax
  xor    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  cmp    $0x2d,%sil
  je     L27
  xor    %ebx,%ebx
  cmp    $0x5d,%dil
  sete   %bl
  add    $0x1,%rcx
  xor    %eax,%eax
  cmp    $0x5d,%sil
  je     L67
  mov    %esi,%edx
L116:
  cmpb   $0x2d,0x1(%rcx)
  jne    L84
  nopw   0x0(%rax,%rax,1)
L20:
  xor    %r8d,%r8d
  mov    $0x1,%r12d
L58:
  lea    0x1(%r8),%esi
  mov    %rsi,%rax
  movzbl (%rcx,%rsi,1),%esi
  cmp    $0x5d,%sil
  je     L106
  lea    -0x1(%r8),%eax
  lea    0x2(%r8),%edx
  cmp    %dil,(%rcx,%rax,1)
  mov    %rdx,%r8
  movzbl (%rcx,%rdx,1),%edx
  setle  %r11b
  cmp    %dil,%sil
  setge  %al
  test   %al,%r11b
  cmovne %r12d,%ebx
  cmp    $0x5d,%dl
  jne    L58
  mov    %r8d,%eax
  jmp    L67
  nopl   0x0(%rax,%rax,1)
L18:
  lea    0x1(%rax),%edi
  lea    0x1(%r9),%rsi
  mov    %r10d,%edx
  add    %rcx,%rdi
  call   <fnmatch@plt>
  test   %eax,%eax
  jne    L93
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L106:
  cmp    %dil,%dl
  mov    $0x1,%edx
  cmove  %edx,%ebx
  jmp    L67
L27:
  movzbl 0x2(%rcx),%edx
  cmp    $0x5d,%dl
  je     L61
  cmp    %dil,%dl
  setge  %al
  xor    %ebx,%ebx
  cmp    $0x5c,%dil
  setg   %bl
  add    $0x2,%rcx
  and    %eax,%ebx
  jmp    L152
L61:
  xor    %ebx,%ebx
  cmp    $0x5d,%dil
  mov    %esi,%edx
  sete   %bl
  add    $0x1,%rcx
  jmp    L116
```

assistant: ```c
int _fnmatch_class(char const *pattern, char const *string, int flags) {
  int ret = 0;
  int neg = 0;
  unsigned int i;
  if (pattern[0] == '!') {
    neg = 1;
    pattern++;
  }
  if (*pattern == ']') {
    if (pattern[1] == '-' && pattern[2] != ']') {
      if (*string >= pattern[0] && *string <= pattern[2])
        ret = 1;
      pattern++;
    } else if (string[0] == ']')
      ret = 1;
    pattern++;
  }
  for (i = 0; pattern[i] != ']'; i++) {
    if (pattern[1] == '-' && pattern[i + 1] != ']') {
      if (*string >= pattern[i - 1] && string[0] <= pattern[i + 1])
        ret = 1;
      i++;
    } else if (pattern[i] == *string)
      ret = 1;
  }
  return ((string[0] != '\0' ? fnmatch(&pattern[i + 1], &string[1], flags) : 1) && neg ^ ret) ? 1 : 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L18:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %eax,%esi
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  lea    (%rsi,%rax,1),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L18
  nop
  nop
  leave
  ret
```

assistant: ```c
void encrypt(char arr[], int key) {
  int i;
  for (i = 0; i < strlen(arr); i++) {
    arr[i] = arr[i] + key % 26;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<encrypt>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r12d
  movslq %esi,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  sar    $0x23,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x1a,%eax,%eax
  sub    %eax,%r12d
  mov    $0x0,%ebx
  jmp    L2
L0:
  add    %r12b,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
L2:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void encrypt(char arr[], int key) {
  int i;
  for (i = 0; i < strlen(arr); i++) {
    arr[i] = arr[i] + key % 26;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<encrypt>:
  endbr64
  movslq %esi,%rax
  push   %r12
  mov    %rax,%r12
  imul   $0x4ec4ec4f,%rax,%rax
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edx
  push   %rbx
  xor    %ebx,%ebx
  sar    $0x1f,%edx
  sar    $0x23,%rax
  sub    %edx,%eax
  imul   $0x1a,%eax,%eax
  sub    %eax,%r12d
  jmp    L16
  xchg   %ax,%ax
L0:
  add    %r12b,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
L16:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void encrypt(char arr[], int key) {
  int i;
  for (i = 0; i < strlen(arr); i++) {
    arr[i] = arr[i] + key % 26;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<encrypt>:
  endbr64
  movslq %esi,%rax
  push   %r12
  mov    %rax,%r12
  imul   $0x4ec4ec4f,%rax,%rax
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edx
  push   %rbx
  xor    %ebx,%ebx
  sar    $0x1f,%edx
  sar    $0x23,%rax
  sub    %edx,%eax
  imul   $0x1a,%eax,%eax
  sub    %eax,%r12d
  jmp    L4
  xchg   %ax,%ax
L1:
  add    %r12b,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void encrypt(char arr[], int key) {
  int i;
  for (i = 0; i < strlen(arr); i++) {
    arr[i] = arr[i] + key % 26;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<solve_by_standard_Laurent_homotopy_continuation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  lea    -0x24(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x306,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
int solve_by_standard_Laurent_homotopy_continuation(int number_of_tasks) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(774, &number_of_tasks, b, c, 0);
  return fail;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<solve_by_standard_Laurent_homotopy_continuation>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x306,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solve_by_standard_Laurent_homotopy_continuation(int number_of_tasks) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(774, &number_of_tasks, b, c, 0);
  return fail;
}
```
user: Translate the below assembly code into C programming language:

```asm
<solve_by_standard_Laurent_homotopy_continuation>:
  endbr64
  sub    $0x18,%rsp
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x306,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solve_by_standard_Laurent_homotopy_continuation(int number_of_tasks) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(774, &number_of_tasks, b, c, 0);
  return fail;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<solve_by_standard_Laurent_homotopy_continuation>:
  endbr64
  sub    $0x18,%rsp
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x306,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int solve_by_standard_Laurent_homotopy_continuation(int number_of_tasks) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(774, &number_of_tasks, b, c, 0);
  return fail;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <set_signals@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <getargs@plt>
  mov    $0x0,%eax
  call   <open_files@plt>
  mov    $0x0,%eax
  call   <reader@plt>
  mov    $0x0,%eax
  call   <lr0@plt>
  mov    $0x0,%eax
  call   <lalr@plt>
  mov    $0x0,%eax
  call   <make_parser@plt>
  mov    $0x0,%eax
  call   <verbose@plt>
  mov    $0x0,%eax
  call   <output@plt>
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <done@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(argc, argv)
int argc;
char *argv[];
{
  set_signals();
  getargs(argc, argv);
  open_files();
  reader();
  lr0();
  lalr();
  make_parser();
  verbose();
  output();
  done(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    $0x0,%eax
  call   <set_signals@plt>
  mov    %rbp,%rsi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <getargs@plt>
  mov    $0x0,%eax
  call   <open_files@plt>
  mov    $0x0,%eax
  call   <reader@plt>
  mov    $0x0,%eax
  call   <lr0@plt>
  mov    $0x0,%eax
  call   <lalr@plt>
  mov    $0x0,%eax
  call   <make_parser@plt>
  mov    $0x0,%eax
  call   <verbose@plt>
  mov    $0x0,%eax
  call   <output@plt>
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <done@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(argc, argv)
int argc;
char *argv[];
{
  set_signals();
  getargs(argc, argv);
  open_files();
  reader();
  lr0();
  lalr();
  make_parser();
  verbose();
  output();
  done(0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <set_signals@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <getargs@plt>
  xor    %eax,%eax
  call   <open_files@plt>
  xor    %eax,%eax
  call   <reader@plt>
  xor    %eax,%eax
  call   <lr0@plt>
  xor    %eax,%eax
  call   <lalr@plt>
  xor    %eax,%eax
  call   <make_parser@plt>
  xor    %eax,%eax
  call   <verbose@plt>
  xor    %eax,%eax
  call   <output@plt>
  xor    %edi,%edi
  xor    %eax,%eax
  call   <done@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(argc, argv)
int argc;
char *argv[];
{
  set_signals();
  getargs(argc, argv);
  open_files();
  reader();
  lr0();
  lalr();
  make_parser();
  verbose();
  output();
  done(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <set_signals@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <getargs@plt>
  xor    %eax,%eax
  call   <open_files@plt>
  xor    %eax,%eax
  call   <reader@plt>
  xor    %eax,%eax
  call   <lr0@plt>
  xor    %eax,%eax
  call   <lalr@plt>
  xor    %eax,%eax
  call   <make_parser@plt>
  xor    %eax,%eax
  call   <verbose@plt>
  xor    %eax,%eax
  call   <output@plt>
  xor    %edi,%edi
  xor    %eax,%eax
  call   <done@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(argc, argv)
int argc;
char *argv[];
{
  set_signals();
  getargs(argc, argv);
  open_files();
  reader();
  lr0();
  lalr();
  make_parser();
  verbose();
  output();
  done(0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_bu@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_td@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
data_type: string
name: parse_data
arguments: data_label: D9
data_type: string

tool: D1: Fibonacci de 10: 

tool: D9: %d


assistant: ```c
int main(void) {
  printf("Fibonacci de 10: ");
  int resultado = fib_bu(10);
  printf("%d\n", resultado);
  resultado = fib_td(10);
  printf("%d\n", resultado);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_bu@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_td@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
type: string
name: parse_data
arguments: tag: D6
type: string

tool: D3: Fibonacci de 10: 

tool: D6: %d


assistant: ```c
int main(void) {
  printf("Fibonacci de 10: ");
  int resultado = fib_bu(10);
  printf("%d\n", resultado);
  resultado = fib_td(10);
  printf("%d\n", resultado);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  lea    D16(%rip),%rbp
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_bu@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_td@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
data_type: string
name: extract_symbolic_data
arguments: data_label: D16
data_type: string

tool: D7: Fibonacci de 10: 

tool: D16: %d


assistant: ```c
int main(void) {
  printf("Fibonacci de 10: ");
  int resultado = fib_bu(10);
  printf("%d\n", resultado);
  resultado = fib_td(10);
  printf("%d\n", resultado);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  lea    D6(%rip),%rbp
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_bu@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <fib_td@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D14
type: string
name: parse_data
arguments: data_label: D6
type: string

tool: D14: Fibonacci de 10: 

tool: D6: %d


assistant: ```c
int main(void) {
  printf("Fibonacci de 10: ");
  int resultado = fib_bu(10);
  printf("%d\n", resultado);
  resultado = fib_td(10);
  printf("%d\n", resultado);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mainpa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <ini_sem@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0xa,%edi
  call   <semget@plt>
  mov    %eax,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <P@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <V@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x3,%esi
  mov    %eax,%edi
  call   <P@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <V@plt>
  jmp    L1
```

tool_calls: name: read_data
arguments: data_label: D10
type_guessed: string

tool: D10: A


assistant: ```c
void mainpa() {
  ini_sem();
  int semid = semget(0XA, 0, 0);
  while (1) {
    P(semid, 2);
    printf("A\n");
    sleep(1);
    V(semid, 0);
    P(semid, 3);
    printf("A\n");
    sleep(1);
    V(semid, 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mainpa>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ini_sem@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0xa,%edi
  call   <semget@plt>
  mov    %eax,%ebx
  lea    D9(%rip),%rbp
L6:
  mov    $0x2,%esi
  mov    %ebx,%edi
  call   <P@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <V@plt>
  mov    $0x3,%esi
  mov    %ebx,%edi
  call   <P@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x1,%esi
  mov    %ebx,%edi
  call   <V@plt>
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
datatype: string

tool: D9: A


assistant: ```c
void mainpa() {
  ini_sem();
  int semid = semget(0XA, 0, 0);
  while (1) {
    P(semid, 2);
    printf("A\n");
    sleep(1);
    V(semid, 0);
    P(semid, 3);
    printf("A\n");
    sleep(1);
    V(semid, 1);
  }
}
```