ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  lea    D13(%rip),%rbp
  call   <puts@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D18
type: string
name: parse_data
arguments: label: D13
type: string

tool: D18: foooo

tool: D13: %d


assistant: ```c
int main() {
  printf("foooo\n");
  int y = {1, 2};
  printf("%d\n", y);
  assert(y == 1);
  printf("%d\n", y);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mymain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8d,-0x20(%rbp)
  mov    %r9,-0x30(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L46
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L46:
  cmpl   $0x0,-0x18(%rbp)
  jle    L56
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L56:
  cmpl   $0x0,0x10(%rbp)
  jle    L33
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L33:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L67:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L22:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L67
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L29:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    0x20(%rbp),%eax
  jl     L29
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string
name: parse_data
arguments: data_label: D48
type_infered: string
name: parse_data
arguments: data_label: D56
type_infered: string
name: parse_data
arguments: data_label: D24
type_infered: string
name: parse_data
arguments: data_label: D29
type_infered: string
name: parse_data
arguments: data_label: D67
type_infered: string
name: parse_data
arguments: data_label: D4
type_infered: string

tool: D2: list files (-l)

tool: D48: recurse through directories (-R)

tool: D56: verbose is enabled (-v)

tool: D24: scalar k=%d


tool: D29: output is "%s"


tool: D67: user defined macro "%s"


tool: D4: infile[%d]="%s"


assistant: ```c
int mymain(int l, int R, int k, const char **defines, int ndefines, const char *outfile, int v, const char **infiles, int ninfiles) {
  int i;
  if (l > 0)
    printf("list files (-l)\n");
  if (R > 0)
    printf("recurse through directories (-R)\n");
  if (v > 0)
    printf("verbose is enabled (-v)\n");
  printf("scalar k=%d\n", k);
  printf("output is \"%s\"\n", outfile);
  for (i = 0; i < ndefines; i++)
    printf("user defined macro \"%s\"\n", defines[i]);
  for (i = 0; i < ninfiles; i++)
    printf("infile[%d]=\"%s\"\n", i, infiles[i]);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<mymain>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r15d
  mov    %edx,%r14d
  mov    %rcx,%r12
  mov    %r8d,%ebp
  mov    %r9,%r13
  mov    0x48(%rsp),%rbx
  test   %edi,%edi
  jg     L49
L78:
  test   %r15d,%r15d
  jg     L52
L95:
  cmpl   $0x0,0x40(%rsp)
  jg     L7
L53:
  mov    %r14d,%esi
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  lea    D56(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L2
  mov    %r12,%r13
  lea    -0x1(%rbp),%eax
  lea    0x8(%r12,%rax,8),%r12
  lea    D31(%rip),%rbp
L96:
  mov    0x0(%r13),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%r13
  cmp    %r12,%r13
  jne    L96
L2:
  cmpl   $0x0,0x50(%rsp)
  jle    L65
  mov    0x50(%rsp),%r12d
  mov    $0x0,%ebp
  lea    D2(%rip),%r13
L58:
  mov    (%rbx,%rbp,8),%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbp
  cmp    %r12,%rbp
  jne    L58
L65:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L49:
  lea    D43(%rip),%rdi
  call   <puts@plt>
  jmp    L78
L52:
  lea    D65(%rip),%rdi
  call   <puts@plt>
  jmp    L95
L7:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  jmp    L53
```

tool_calls: name: read_data
arguments: label: D39
datatype: string
name: read_data
arguments: label: D56
datatype: string
name: read_data
arguments: label: D31
datatype: string
name: read_data
arguments: label: D2
datatype: string
name: read_data
arguments: label: D43
datatype: string
name: read_data
arguments: label: D65
datatype: string
name: read_data
arguments: label: D26
datatype: string

tool: D39: scalar k=%d


tool: D56: output is "%s"


tool: D31: user defined macro "%s"


tool: D2: infile[%d]="%s"


tool: D43: list files (-l)

tool: D65: recurse through directories (-R)

tool: D26: verbose is enabled (-v)

assistant: ```c
int mymain(int l, int R, int k, const char **defines, int ndefines, const char *outfile, int v, const char **infiles, int ninfiles) {
  int i;
  if (l > 0)
    printf("list files (-l)\n");
  if (R > 0)
    printf("recurse through directories (-R)\n");
  if (v > 0)
    printf("verbose is enabled (-v)\n");
  printf("scalar k=%d\n", k);
  printf("output is \"%s\"\n", outfile);
  for (i = 0; i < ndefines; i++)
    printf("user defined macro \"%s\"\n", defines[i]);
  for (i = 0; i < ninfiles; i++)
    printf("infile[%d]=\"%s\"\n", i, infiles[i]);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mymain>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %r9,%r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
  mov    0x58(%rsp),%rbp
  movslq 0x60(%rsp),%r12
  test   %edi,%edi
  jle    L6
  lea    D27(%rip),%rdi
  mov    %esi,0xc(%rsp)
  call   <puts@plt>
  mov    0xc(%rsp),%esi
L6:
  test   %esi,%esi
  jle    L40
  lea    D3(%rip),%rdi
  call   <puts@plt>
L40:
  mov    0x50(%rsp),%eax
  test   %eax,%eax
  jle    L10
  lea    D24(%rip),%rdi
  call   <puts@plt>
L10:
  mov    %r15d,%esi
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r14,%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  jle    L68
  lea    -0x1(%r13),%eax
  lea    D63(%rip),%r13
  lea    0x8(%rbx,%rax,8),%r14
  nopl   0x0(%rax,%rax,1)
L22:
  mov    (%rbx),%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r14,%rbx
  jne    L22
L68:
  test   %r12d,%r12d
  jle    L14
  xor    %ebx,%ebx
  lea    D57(%rip),%r13
  nopl   0x0(%rax)
L29:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L29
L14:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D27
type_guessed: string
name: retrieve_data
arguments: data_label: D3
type_guessed: string
name: retrieve_data
arguments: data_label: D24
type_guessed: string
name: retrieve_data
arguments: data_label: D47
type_guessed: string
name: retrieve_data
arguments: data_label: D16
type_guessed: string
name: retrieve_data
arguments: data_label: D63
type_guessed: string
name: retrieve_data
arguments: data_label: D57
type_guessed: string

tool: D27: list files (-l)

tool: D3: recurse through directories (-R)

tool: D24: verbose is enabled (-v)

tool: D47: scalar k=%d


tool: D16: output is "%s"


tool: D63: user defined macro "%s"


tool: D57: infile[%d]="%s"


assistant: ```c
int mymain(int l, int R, int k, const char **defines, int ndefines, const char *outfile, int v, const char **infiles, int ninfiles) {
  int i;
  if (l > 0)
    printf("list files (-l)\n");
  if (R > 0)
    printf("recurse through directories (-R)\n");
  if (v > 0)
    printf("verbose is enabled (-v)\n");
  printf("scalar k=%d\n", k);
  printf("output is \"%s\"\n", outfile);
  for (i = 0; i < ndefines; i++)
    printf("user defined macro \"%s\"\n", defines[i]);
  for (i = 0; i < ninfiles; i++)
    printf("infile[%d]=\"%s\"\n", i, infiles[i]);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mymain>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %r9,%r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
  mov    0x58(%rsp),%rbp
  movslq 0x60(%rsp),%r12
  test   %edi,%edi
  jle    L33
  lea    D45(%rip),%rdi
  mov    %esi,0xc(%rsp)
  call   <puts@plt>
  mov    0xc(%rsp),%esi
L33:
  test   %esi,%esi
  jle    L59
  lea    D32(%rip),%rdi
  call   <puts@plt>
L59:
  mov    0x50(%rsp),%eax
  test   %eax,%eax
  jle    L65
  lea    D22(%rip),%rdi
  call   <puts@plt>
L65:
  mov    %r15d,%esi
  lea    D59(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r14,%rsi
  lea    D55(%rip),%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  jle    L39
  lea    -0x1(%r13),%eax
  lea    D2(%rip),%r13
  lea    0x8(%rbx,%rax,8),%r14
  nopl   0x0(%rax,%rax,1)
L51:
  mov    (%rbx),%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r14,%rbx
  jne    L51
L39:
  test   %r12d,%r12d
  jle    L50
  xor    %ebx,%ebx
  lea    D16(%rip),%r13
  nopl   0x0(%rax)
L15:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L15
L50:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: label: D45
data_type: string
name: retrieve_data
arguments: label: D32
data_type: string
name: retrieve_data
arguments: label: D22
data_type: string
name: retrieve_data
arguments: label: D59
data_type: string
name: retrieve_data
arguments: label: D55
data_type: string
name: retrieve_data
arguments: label: D2
data_type: string
name: retrieve_data
arguments: label: D16
data_type: string

tool: D45: list files (-l)

tool: D32: recurse through directories (-R)

tool: D22: verbose is enabled (-v)

tool: D59: scalar k=%d


tool: D55: output is "%s"


tool: D2: user defined macro "%s"


tool: D16: infile[%d]="%s"


assistant: ```c
int mymain(int l, int R, int k, const char **defines, int ndefines, const char *outfile, int v, const char **infiles, int ninfiles) {
  int i;
  if (l > 0)
    printf("list files (-l)\n");
  if (R > 0)
    printf("recurse through directories (-R)\n");
  if (v > 0)
    printf("verbose is enabled (-v)\n");
  printf("scalar k=%d\n", k);
  printf("output is \"%s\"\n", outfile);
  for (i = 0; i < ndefines; i++)
    printf("user defined macro \"%s\"\n", defines[i]);
  for (i = 0; i < ninfiles; i++)
    printf("infile[%d]=\"%s\"\n", i, infiles[i]);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<asm_issyslinux>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  movq   $0x3000,-0x28(%rbp)
  mov    $0xffffffff,%eax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    -0x10(%rbp),%rsi
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    %rsi,%rbx
  int    $0x21
  mov    %rbx,%rsi
  mov    %rax,-0x28(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  cmpq   $0x59530000,-0x28(%rbp)
  jne    L7
  cmpq   $0x4c530000,-0x10(%rbp)
  jne    L7
  cmpq   $0x4e490000,-0x18(%rbp)
  jne    L7
  cmpq   $0x58550000,-0x20(%rbp)
  jne    L7
  mov    $0x1,%eax
  jmp    L18
L7:
  mov    $0x0,%eax
L18:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int asm_issyslinux(void) {
  unsigned long eax, ebx, ecx, edx;
  eax = 0x00003000;
  ebx = ecx = edx = 0xFFFFFFFF;
  asm("int $0x21" : "+a"(eax), "+b"(ebx), "+c"(ecx), "+d"(edx));
  return (eax == 0x59530000) && (ebx == 0x4c530000) && (ecx == 0x4e490000) && (edx == 0x58550000);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<asm_issyslinux>:
  endbr64
  push   %rbx
  mov    $0x3000,%eax
  mov    $0xffffffff,%ebx
  mov    %rbx,%rcx
  mov    %rbx,%rdx
  int    $0x21
  cmp    $0x59530000,%rax
  jne    L28
  cmp    $0x4c530000,%rbx
  jne    L28
  cmp    $0x4e490000,%rcx
  sete   %cl
  cmp    $0x58550000,%rdx
  sete   %al
  movzbl %al,%eax
  and    %ecx,%eax
L22:
  pop    %rbx
  ret
L28:
  mov    $0x0,%eax
  jmp    L22
```

assistant: ```c
int asm_issyslinux(void) {
  unsigned long eax, ebx, ecx, edx;
  eax = 0x00003000;
  ebx = ecx = edx = 0xFFFFFFFF;
  asm("int $0x21" : "+a"(eax), "+b"(ebx), "+c"(ecx), "+d"(edx));
  return (eax == 0x59530000) && (ebx == 0x4c530000) && (ecx == 0x4e490000) && (edx == 0x58550000);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<asm_issyslinux>:
  endbr64
  push   %rbx
  mov    $0xffffffff,%ebx
  mov    $0x3000,%eax
  mov    %rbx,%rcx
  mov    %rbx,%rdx
  int    $0x21
  cmp    $0x59530000,%rax
  jne    L12
  cmp    $0x4c530000,%rbx
  jne    L12
  cmp    $0x4e490000,%rcx
  pop    %rbx
  sete   %cl
  xor    %eax,%eax
  cmp    $0x58550000,%rdx
  sete   %al
  and    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int asm_issyslinux(void) {
  unsigned long eax, ebx, ecx, edx;
  eax = 0x00003000;
  ebx = ecx = edx = 0xFFFFFFFF;
  asm("int $0x21" : "+a"(eax), "+b"(ebx), "+c"(ecx), "+d"(edx));
  return (eax == 0x59530000) && (ebx == 0x4c530000) && (ecx == 0x4e490000) && (edx == 0x58550000);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<asm_issyslinux>:
  endbr64
  push   %rbx
  mov    $0xffffffff,%ebx
  mov    $0x3000,%eax
  mov    %rbx,%rcx
  mov    %rbx,%rdx
  int    $0x21
  cmp    $0x59530000,%rax
  jne    L12
  cmp    $0x4c530000,%rbx
  jne    L12
  cmp    $0x4e490000,%rcx
  pop    %rbx
  sete   %cl
  xor    %eax,%eax
  cmp    $0x58550000,%rdx
  sete   %al
  and    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int asm_issyslinux(void) {
  unsigned long eax, ebx, ecx, edx;
  eax = 0x00003000;
  ebx = ecx = edx = 0xFFFFFFFF;
  asm("int $0x21" : "+a"(eax), "+b"(ebx), "+c"(ecx), "+d"(edx));
  return (eax == 0x59530000) && (ebx == 0x4c530000) && (ecx == 0x4e490000) && (edx == 0x58550000);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_in_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  jmp    L21
L10:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  cmp    %al,-0x4(%rbp)
  jne    L21
  mov    $0x1,%eax
  jmp    L30
L21:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    $0x0,%eax
L30:
  pop    %rbp
  ret
```

assistant: ```c
int is_in_str(char c, const char *str) {
  while (*str)
    if (c == *(str++))
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_in_str>:
  endbr64
L3:
  movzbl (%rsi),%eax
  test   %al,%al
  je     L13
  add    $0x1,%rsi
  cmp    %dil,%al
  jne    L3
  mov    $0x1,%eax
  ret
L13:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int is_in_str(char c, const char *str) {
  while (*str)
    if (c == *(str++))
      return 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_in_str>:
  endbr64
  jmp    L8
  cs nopw 0x0(%rax,%rax,1)
L14:
  add    $0x1,%rsi
  cmp    %dil,%al
  je     L3
L8:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L14
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int is_in_str(char c, const char *str) {
  while (*str)
    if (c == *(str++))
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_in_str>:
  endbr64
  jmp    L10
  cs nopw 0x0(%rax,%rax,1)
L12:
  add    $0x1,%rsi
  cmp    %dil,%al
  je     L11
L10:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L12
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int is_in_str(char c, const char *str) {
  while (*str)
    if (c == *(str++))
      return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L51
  mov    $0x0,%eax
  jmp    L52
L51:
  mov    $0x0,%eax
  call   <lp_pr_identifier@plt>
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L58
L49:
  addq   $0x9,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L63
  mov    $0x0,%eax
  jmp    L52
L63:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L4
L44:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  test   %eax,%eax
  je     L14
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  jmp    L52
L14:
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
L4:
  mov    $0x0,%eax
  call   <lp_pr_identifier@plt>
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L44
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L58:
  mov    -0x18(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L49
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L52:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string

tool: D9: 
Subject:

assistant: ```c
int getid(char *mbuf) {
  char *p1, *p2, *p3;
  int lpr;
  mbuf = strdup(mbuf);
  if (!mbuf)
    return 0;
  lpr = strlen(lp_pr_identifier());
  p1 = mbuf;
  while ((p1 = strstr(p1, "\nSubject:"))) {
    p1 += 9;
    p2 = strchr(p1, '\n');
    if (!p2)
      return 0;
    *p2++ = 0;
    p3 = p1;
    while ((p3 = strstr(p3, lp_pr_identifier()))) {
      if (atoi(p3 + lpr))
        return atoi(p3 + lpr);
      p3 += lpr;
    }
    p1 = p2;
  }
  free(mbuf);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getid>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strdup@plt>
  test   %rax,%rax
  je     L32
  mov    %rax,%r12
  mov    $0x0,%eax
  call   <lp_pr_identifier@plt>
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %r12,%r13
  lea    D3(%rip),%r14
  movslq %eax,%rbp
L16:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L37
  lea    0x9(%rax),%rbx
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L56
  lea    0x1(%rax),%r13
  movb   $0x0,(%rax)
L64:
  mov    $0x0,%eax
  call   <lp_pr_identifier@plt>
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L16
  lea    (%rax,%rbp,1),%rbx
  mov    %rbx,%rdi
  call   <atoi@plt>
  test   %eax,%eax
  je     L64
  mov    %rbx,%rdi
  call   <atoi@plt>
L45:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L45
L32:
  mov    $0x0,%eax
  jmp    L45
L56:
  mov    $0x0,%eax
  jmp    L45
```

tool_calls: name: parse_data
arguments: data_label: D3
type: string

tool: D3: 
Subject:

assistant: ```c
int getid(char *mbuf) {
  char *p1, *p2, *p3;
  int lpr;
  mbuf = strdup(mbuf);
  if (!mbuf)
    return 0;
  lpr = strlen(lp_pr_identifier());
  p1 = mbuf;
  while ((p1 = strstr(p1, "\nSubject:"))) {
    p1 += 9;
    p2 = strchr(p1, '\n');
    if (!p2)
      return 0;
    *p2++ = 0;
    p3 = p1;
    while ((p3 = strstr(p3, lp_pr_identifier()))) {
      if (atoi(p3 + lpr))
        return atoi(p3 + lpr);
      p3 += lpr;
    }
    p1 = p2;
  }
  free(mbuf);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getid>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strdup@plt>
  test   %rax,%rax
  je     L14
  mov    %rax,%r12
  xor    %eax,%eax
  lea    D1(%rip),%r14
  call   <lp_pr_identifier@plt>
  mov    %r12,%r13
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
L3:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L8
  lea    0x9(%rax),%rbp
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L14
  movb   $0x0,(%rax)
  lea    0x1(%rax),%r13
  cs nopw 0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  call   <lp_pr_identifier@plt>
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  je     L3
  lea    (%rax,%rbx,1),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  test   %eax,%eax
  je     L7
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10f0 <atoi@plt>
L8:
  mov    %r12,%rdi
  call   <free@plt>
L14:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
datatype: string

tool: D1: 
Subject:

assistant: ```c
int getid(char *mbuf) {
  char *p1, *p2, *p3;
  int lpr;
  mbuf = strdup(mbuf);
  if (!mbuf)
    return 0;
  lpr = strlen(lp_pr_identifier());
  p1 = mbuf;
  while ((p1 = strstr(p1, "\nSubject:"))) {
    p1 += 9;
    p2 = strchr(p1, '\n');
    if (!p2)
      return 0;
    *p2++ = 0;
    p3 = p1;
    while ((p3 = strstr(p3, lp_pr_identifier()))) {
      if (atoi(p3 + lpr))
        return atoi(p3 + lpr);
      p3 += lpr;
    }
    p1 = p2;
  }
  free(mbuf);
  return 0;
}
```