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

```asm
<sub>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  test   %edi,%edi
  jg     L55
L57:
  cmp    %ebx,%ebp
  jle    L9
L46:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L55:
  lea    D10(%rip),%r13
  mov    %edi,%esi
  lea    -0x5(%rbp),%r12d
  mov    %r13,%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jg     L47
L19:
  cmp    %ebx,%r12d
  jg     L57
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %ebx,%ebp
  jg     L46
L9:
  pop    %rbx
  mov    %ebp,%esi
  lea    D10(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L47:
  mov    %r12d,%esi
  mov    %r13,%rdi
  lea    -0xa(%rbp),%r14d
  call   <printf@plt>
  test   %r14d,%r14d
  jle    L38
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  lea    -0xf(%rbp),%edi
  mov    %ebx,%esi
  call   <sub>
L38:
  cmp    %r14d,%ebx
  jl     L19
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L19
```

tool_calls: name: retrieve_data
arguments: data_label: D10
data_type: string

tool: D10: %d 

assistant: ```c
int sub(int n, int a) {
  if (n > 0) {
    printf("%d ", n);
    sub(n - 5, a);
  }
  if (n <= a)
    printf("%d ", n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sub>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  test   %edi,%edi
  jg     L36
L24:
  cmp    %ebx,%ebp
  jle    L66
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L36:
  lea    D0(%rip),%r13
  mov    %edi,%esi
  lea    -0x5(%rbp),%r12d
  mov    %r13,%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jg     L19
L57:
  cmp    %ebx,%r12d
  jg     L24
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %ebx,%ebp
  jg     L47
L66:
  pop    %rbx
  mov    %ebp,%esi
  lea    D0(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L19:
  mov    %r12d,%esi
  mov    %r13,%rdi
  lea    -0xa(%rbp),%r14d
  call   <printf@plt>
  test   %r14d,%r14d
  jle    L23
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  lea    -0xf(%rbp),%edi
  mov    %ebx,%esi
  call   <sub>
L23:
  cmp    %r14d,%ebx
  jl     L57
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L57
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type: string

tool: D0: %d 

assistant: ```c
int sub(int n, int a) {
  if (n > 0) {
    printf("%d ", n);
    sub(n - 5, a);
  }
  if (n <= a)
    printf("%d ", n);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <clrscr@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  clrscr();
  exit(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <clrscr@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  clrscr();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clrscr@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  clrscr();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clrscr@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  clrscr();
  exit(0);
}
```
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    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_f@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int main() {
  int a;
  return ptr_f(&a);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <ptr_f@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a;
  return ptr_f(&a);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <ptr_f@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a;
  return ptr_f(&a);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <ptr_f@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a;
  return ptr_f(&a);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xml_text>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L43
  lea    D8(%rip),%rcx
  jmp    L82
  movb   $0x26,(%rsi)
  movb   $0x71,0x1(%rsi)
  movb   $0x75,0x2(%rsi)
  movb   $0x6f,0x3(%rsi)
  movb   $0x74,0x4(%rsi)
  movb   $0x3b,0x5(%rsi)
  lea    0x6(%rsi),%rsi
L29:
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L43
L82:
  lea    -0x22(%rdx),%eax
  cmp    $0x1c,%al
  ja     L77
  movzbl %al,%eax
  movslq (%rcx,%rax,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  movb   $0x26,(%rsi)
  movb   $0x61,0x1(%rsi)
  movb   $0x6d,0x2(%rsi)
  movb   $0x70,0x3(%rsi)
  movb   $0x3b,0x4(%rsi)
  lea    0x5(%rsi),%rsi
  jmp    L29
  movb   $0x26,(%rsi)
  movb   $0x61,0x1(%rsi)
  movb   $0x70,0x2(%rsi)
  movb   $0x6f,0x3(%rsi)
  movb   $0x73,0x4(%rsi)
  movb   $0x3b,0x5(%rsi)
  lea    0x6(%rsi),%rsi
  jmp    L29
  movb   $0x26,(%rsi)
  movb   $0x6c,0x1(%rsi)
  movb   $0x74,0x2(%rsi)
  movb   $0x3b,0x3(%rsi)
  lea    0x4(%rsi),%rsi
  jmp    L29
  movb   $0x26,(%rsi)
  movb   $0x67,0x1(%rsi)
  movb   $0x74,0x2(%rsi)
  movb   $0x3b,0x3(%rsi)
  lea    0x4(%rsi),%rsi
  jmp    L29
L77:
  mov    %dl,(%rsi)
  lea    0x1(%rsi),%rsi
  jmp    L29
L43:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void xml_text(char *in, char *out) {
  char *p, *q;
  q = out;
  for (p = in; *p != '\0'; p++) {
    if (*p == '"') {
      *q++ = '&';
      *q++ = 'q';
      *q++ = 'u';
      *q++ = 'o';
      *q++ = 't';
      *q++ = ';';
    } else if (*p == '&') {
      *q++ = '&';
      *q++ = 'a';
      *q++ = 'm';
      *q++ = 'p';
      *q++ = ';';
    } else if (*p == '\'') {
      *q++ = '&';
      *q++ = 'a';
      *q++ = 'p';
      *q++ = 'o';
      *q++ = 's';
      *q++ = ';';
    } else if (*p == '<') {
      *q++ = '&';
      *q++ = 'l';
      *q++ = 't';
      *q++ = ';';
    } else if (*p == '>') {
      *q++ = '&';
      *q++ = 'g';
      *q++ = 't';
      *q++ = ';';
    } else {
      *q++ = *p;
    }
  }
  *q = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xml_text>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L10
  lea    D6(%rip),%rcx
  nopw   0x0(%rax,%rax,1)
L9:
  lea    -0x22(%rdx),%eax
  cmp    $0x1c,%al
  ja     L64
  movzbl %al,%eax
  movslq (%rcx,%rax,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
L64:
  mov    %dl,(%rsi)
  add    $0x1,%rsi
  cs nopw 0x0(%rax,%rax,1)
L48:
  movzbl 0x1(%rdi),%edx
  add    $0x1,%rdi
  test   %dl,%dl
  jne    L9
L10:
  movb   $0x0,(%rsi)
  ret
  movl   $0x3b746726,(%rsi)
  add    $0x4,%rsi
  jmp    L48
  nopl   0x0(%rax)
  movl   $0x3b746c26,(%rsi)
  add    $0x4,%rsi
  jmp    L48
  nopl   0x0(%rax)
  mov    $0x3b73,%eax
  movl   $0x6f706126,(%rsi)
  add    $0x6,%rsi
  mov    %ax,-0x2(%rsi)
  jmp    L48
  nopl   (%rax)
  movl   $0x706d6126,(%rsi)
  add    $0x5,%rsi
  movb   $0x3b,-0x1(%rsi)
  jmp    L48
  nopl   0x0(%rax,%rax,1)
  mov    $0x3b74,%edx
  movl   $0x6f757126,(%rsi)
  add    $0x6,%rsi
  mov    %dx,-0x2(%rsi)
  jmp    L48
```

assistant: ```c
void xml_text(char *in, char *out) {
  char *p, *q;
  q = out;
  for (p = in; *p != '\0'; p++) {
    if (*p == '"') {
      *q++ = '&';
      *q++ = 'q';
      *q++ = 'u';
      *q++ = 'o';
      *q++ = 't';
      *q++ = ';';
    } else if (*p == '&') {
      *q++ = '&';
      *q++ = 'a';
      *q++ = 'm';
      *q++ = 'p';
      *q++ = ';';
    } else if (*p == '\'') {
      *q++ = '&';
      *q++ = 'a';
      *q++ = 'p';
      *q++ = 'o';
      *q++ = 's';
      *q++ = ';';
    } else if (*p == '<') {
      *q++ = '&';
      *q++ = 'l';
      *q++ = 't';
      *q++ = ';';
    } else if (*p == '>') {
      *q++ = '&';
      *q++ = 'g';
      *q++ = 't';
      *q++ = ';';
    } else {
      *q++ = *p;
    }
  }
  *q = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xml_text>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L70
  lea    D2(%rip),%rcx
  nopw   0x0(%rax,%rax,1)
L4:
  lea    -0x22(%rdx),%eax
  cmp    $0x1c,%al
  ja     L16
  movzbl %al,%eax
  movslq (%rcx,%rax,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
L16:
  mov    %dl,(%rsi)
  add    $0x1,%rsi
  cs nopw 0x0(%rax,%rax,1)
L56:
  movzbl 0x1(%rdi),%edx
  add    $0x1,%rdi
  test   %dl,%dl
  jne    L4
L70:
  movb   $0x0,(%rsi)
  ret
  movl   $0x3b746726,(%rsi)
  add    $0x4,%rsi
  jmp    L56
  nopl   0x0(%rax)
  movl   $0x3b746c26,(%rsi)
  add    $0x4,%rsi
  jmp    L56
  nopl   0x0(%rax)
  mov    $0x3b73,%eax
  movl   $0x6f706126,(%rsi)
  add    $0x6,%rsi
  mov    %ax,-0x2(%rsi)
  jmp    L56
  nopl   (%rax)
  movl   $0x706d6126,(%rsi)
  add    $0x5,%rsi
  movb   $0x3b,-0x1(%rsi)
  jmp    L56
  nopl   0x0(%rax,%rax,1)
  mov    $0x3b74,%edx
  movl   $0x6f757126,(%rsi)
  add    $0x6,%rsi
  mov    %dx,-0x2(%rsi)
  jmp    L56
```

assistant: ```c
void xml_text(char *in, char *out) {
  char *p, *q;
  q = out;
  for (p = in; *p != '\0'; p++) {
    if (*p == '"') {
      *q++ = '&';
      *q++ = 'q';
      *q++ = 'u';
      *q++ = 'o';
      *q++ = 't';
      *q++ = ';';
    } else if (*p == '&') {
      *q++ = '&';
      *q++ = 'a';
      *q++ = 'm';
      *q++ = 'p';
      *q++ = ';';
    } else if (*p == '\'') {
      *q++ = '&';
      *q++ = 'a';
      *q++ = 'p';
      *q++ = 'o';
      *q++ = 's';
      *q++ = ';';
    } else if (*p == '<') {
      *q++ = '&';
      *q++ = 'l';
      *q++ = 't';
      *q++ = ';';
    } else if (*p == '>') {
      *q++ = '&';
      *q++ = 'g';
      *q++ = 't';
      *q++ = ';';
    } else {
      *q++ = *p;
    }
  }
  *q = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<catfrq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <catlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
  jmp    L89
L59:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <catrd@plt>
  test   %eax,%eax
  je     L54
  mov    $0xffffffff,%eax
  jmp    L93
L54:
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0xd,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  js     L56
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L89
L56:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L89:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L59
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L86
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L93
L86:
  mov    -0x8(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jne    L31
  mov    -0x10(%rbp),%eax
  jmp    L93
L31:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <catrd@plt>
  test   %eax,%eax
  je     L70
  mov    $0xffffffff,%eax
  jmp    L93
L70:
  mov    -0x10(%rbp),%eax
L93:
  leave
  ret
```

assistant: ```c
int catfrq(molec, cfreq, buff)
int molec;
char *cfreq, *buff;
{
  int line, nline, l, r;
  nline = catlen(molec);
  l = line = 1;
  r = nline + 1;
  buff[0] = 0;
  while (l < r) {
    line = (l + r) >> 1;
    if (catrd(molec, line, buff))
      return (-1);
    if (strncmp(cfreq, buff, 13) >= 0) {
      l = line + 1;
    } else {
      r = line;
    }
  }
  if (l > nline) {
    buff[0] = 0;
    return (0);
  }
  if (r == line)
    return (line);
  line = r;
  if (catrd(molec, line, buff))
    return (-1);
  return (line);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<catfrq>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r14d
  mov    %rsi,%r15
  mov    %rdx,%r13
  mov    $0x0,%eax
  call   <catlen@plt>
  mov    %eax,0xc(%rsp)
  lea    0x1(%rax),%ebp
  movb   $0x0,0x0(%r13)
  cmp    $0x1,%ebp
  jle    L31
  mov    $0x1,%r12d
  jmp    L77
L36:
  mov    %ebx,%ebp
L41:
  cmp    %ebp,%r12d
  jge    L15
L77:
  lea    (%r12,%rbp,1),%ebx
  sar    %ebx
  mov    %r13,%rdx
  mov    %ebx,%esi
  mov    %r14d,%edi
  mov    $0x0,%eax
  call   <catrd@plt>
  test   %eax,%eax
  jne    L46
  mov    $0xd,%edx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  js     L36
  lea    0x1(%rbx),%r12d
  jmp    L41
L31:
  mov    $0x1,%r12d
  mov    $0x1,%ebx
L15:
  cmp    %r12d,0xc(%rsp)
  jl     L44
  cmp    %ebp,%ebx
  je     L54
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %r14d,%edi
  mov    $0x0,%eax
  call   <catrd@plt>
  test   %eax,%eax
  mov    $0xffffffff,%eax
  cmovne %eax,%ebp
  jmp    L54
L44:
  movb   $0x0,0x0(%r13)
  mov    $0x0,%ebp
  jmp    L54
L46:
  mov    $0xffffffff,%ebp
L54:
  mov    %ebp,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int catfrq(molec, cfreq, buff)
int molec;
char *cfreq, *buff;
{
  int line, nline, l, r;
  nline = catlen(molec);
  l = line = 1;
  r = nline + 1;
  buff[0] = 0;
  while (l < r) {
    line = (l + r) >> 1;
    if (catrd(molec, line, buff))
      return (-1);
    if (strncmp(cfreq, buff, 13) >= 0) {
      l = line + 1;
    } else {
      r = line;
    }
  }
  if (l > nline) {
    buff[0] = 0;
    return (0);
  }
  if (r == line)
    return (line);
  line = r;
  if (catrd(molec, line, buff))
    return (-1);
  return (line);
}
```
user: What is the c source code of the assembly code below:

```asm
<catfrq>:
  endbr64
  push   %r15
  xor    %eax,%eax
  mov    %rdx,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  call   <catlen@plt>
  movb   $0x0,(%r15)
  lea    0x1(%rax),%r12d
  mov    %eax,0xc(%rsp)
  cmp    $0x1,%r12d
  jg     L71
  jmp    L10
  xchg   %ax,%ax
L62:
  lea    0x1(%rbx),%ebp
  cmp    %r12d,%ebp
  jge    L104
L71:
  lea    (%r12,%rbp,1),%ebx
  xor    %eax,%eax
  mov    %r15,%rdx
  mov    %r13d,%edi
  sar    %ebx
  mov    %ebx,%esi
  call   <catrd@plt>
  test   %eax,%eax
  jne    L3
  mov    $0xd,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jns    L62
  mov    %ebx,%r12d
  cmp    %r12d,%ebp
  jl     L71
L104:
  cmp    %ebp,0xc(%rsp)
  jl     L16
  cmp    %ebx,%r12d
  je     L40
  xor    %eax,%eax
  mov    %r15,%rdx
  mov    %r12d,%esi
  mov    %r13d,%edi
  call   <catrd@plt>
  test   %eax,%eax
  jne    L3
L40:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L3:
  mov    $0xffffffff,%r12d
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L16:
  movb   $0x0,(%r15)
  xor    %r12d,%r12d
  jmp    L40
L10:
  mov    $0x1,%ebx
  jmp    L104
```

assistant: ```c
int catfrq(molec, cfreq, buff)
int molec;
char *cfreq, *buff;
{
  int line, nline, l, r;
  nline = catlen(molec);
  l = line = 1;
  r = nline + 1;
  buff[0] = 0;
  while (l < r) {
    line = (l + r) >> 1;
    if (catrd(molec, line, buff))
      return (-1);
    if (strncmp(cfreq, buff, 13) >= 0) {
      l = line + 1;
    } else {
      r = line;
    }
  }
  if (l > nline) {
    buff[0] = 0;
    return (0);
  }
  if (r == line)
    return (line);
  line = r;
  if (catrd(molec, line, buff))
    return (-1);
  return (line);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<catfrq>:
  endbr64
  push   %r15
  xor    %eax,%eax
  mov    %rdx,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  call   <catlen@plt>
  movb   $0x0,(%r15)
  lea    0x1(%rax),%r12d
  mov    %eax,0xc(%rsp)
  cmp    $0x1,%r12d
  jle    L24
  nopl   (%rax)
L92:
  lea    (%r12,%rbp,1),%ebx
  xor    %eax,%eax
  mov    %r15,%rdx
  mov    %r13d,%edi
  sar    %ebx
  mov    %ebx,%esi
  call   <catrd@plt>
  test   %eax,%eax
  jne    L34
  mov    $0xd,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  js     L100
  lea    0x1(%rbx),%ebp
  cmp    %r12d,%ebp
  jl     L92
L54:
  cmp    %ebp,0xc(%rsp)
  jl     L17
  cmp    %ebx,%r12d
  je     L112
  xor    %eax,%eax
  mov    %r15,%rdx
  mov    %r12d,%esi
  mov    %r13d,%edi
  call   <catrd@plt>
  test   %eax,%eax
  jne    L34
L112:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L100:
  cmp    %ebp,%ebx
  jle    L107
  mov    %ebx,%r12d
  jmp    L92
  nopl   0x0(%rax)
L34:
  mov    $0xffffffff,%r12d
  jmp    L112
  nopl   0x0(%rax,%rax,1)
L107:
  cmp    %ebp,0xc(%rsp)
  jl     L17
  mov    %ebx,%r12d
  jmp    L112
  nopl   0x0(%rax,%rax,1)
L17:
  movb   $0x0,(%r15)
  xor    %r12d,%r12d
  jmp    L112
L24:
  mov    $0x1,%ebx
  jmp    L54
```

assistant: ```c
int catfrq(molec, cfreq, buff)
int molec;
char *cfreq, *buff;
{
  int line, nline, l, r;
  nline = catlen(molec);
  l = line = 1;
  r = nline + 1;
  buff[0] = 0;
  while (l < r) {
    line = (l + r) >> 1;
    if (catrd(molec, line, buff))
      return (-1);
    if (strncmp(cfreq, buff, 13) >= 0) {
      l = line + 1;
    } else {
      r = line;
    }
  }
  if (l > nline) {
    buff[0] = 0;
    return (0);
  }
  if (r == line)
    return (line);
  line = r;
  if (catrd(molec, line, buff))
    return (-1);
  return (line);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<E>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <T@plt>
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  je     L4
  mov    $0x0,%eax
  call   <cnext@plt>
  mov    $0x0,%eax
  call   <E>
L4:
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D23
type_infered: string
name: read_data
arguments: data_tag: D21
type_infered: string
name: read_data
arguments: data_tag: D25
type_infered: string

tool: D23: <E>

tool: D21: +-*/

tool: D25: </E>

assistant: ```c
int E() {
  printf("<E>");
  T();
  if (isNext("+-*/")) {
    cnext();
    E();
  }
  printf("</E>");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<E>:
  endbr64
  sub    $0x8,%rsp
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <T@plt>
  lea    D16(%rip),%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L2
L14:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
L2:
  mov    $0x0,%eax
  call   <cnext@plt>
  mov    $0x0,%eax
  call   <E>
  jmp    L14
```

tool_calls: name: read_data
arguments: label: D26
type: string
name: read_data
arguments: label: D16
type: string
name: read_data
arguments: label: D11
type: string

tool: D26: <E>

tool: D16: +-*/

tool: D11: </E>

assistant: ```c
int E() {
  printf("<E>");
  T();
  if (isNext("+-*/")) {
    cnext();
    E();
  }
  printf("</E>");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<E>:
  endbr64
  push   %r13
  lea    D26(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  lea    D9(%rip),%r12
  push   %rbp
  lea    D36(%rip),%rbp
  call   <printf@plt>
  xor    %eax,%eax
  call   <T@plt>
  mov    %r12,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L22
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10a0 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L22:
  xor    %eax,%eax
  call   <cnext@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <T@plt>
  mov    %r12,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L4
L19:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10a0 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <cnext@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <T@plt>
  mov    %r12,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L14
L23:
  lea    D36(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L19
  nopl   0x0(%rax)
L14:
  xor    %eax,%eax
  call   <cnext@plt>
  xor    %eax,%eax
  call   <E>
  jmp    L23
```

tool_calls: name: retrieve_data
arguments: tag: D26
datatype: string
name: retrieve_data
arguments: tag: D9
datatype: string
name: retrieve_data
arguments: tag: D36
datatype: string

tool: D26: <E>

tool: D9: +-*/

tool: D36: </E>

assistant: ```c
int E() {
  printf("<E>");
  T();
  if (isNext("+-*/")) {
    cnext();
    E();
  }
  printf("</E>");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<E>:
  endbr64
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  lea    D15(%rip),%r12
  push   %rbp
  lea    D7(%rip),%rbp
  call   <printf@plt>
  xor    %eax,%eax
  call   <T@plt>
  mov    %r12,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L30
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10a0 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L30:
  xor    %eax,%eax
  call   <cnext@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <T@plt>
  mov    %r12,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L43
L22:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10a0 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L43:
  xor    %eax,%eax
  call   <cnext@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <T@plt>
  mov    %r12,%rdi
  call   <isNext@plt>
  test   %rax,%rax
  jne    L44
L13:
  lea    D7(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L22
  nopl   0x0(%rax)
L44:
  xor    %eax,%eax
  call   <cnext@plt>
  xor    %eax,%eax
  call   <E>
  jmp    L13
```

tool_calls: name: retrieve_data
arguments: label: D6
data_type: string
name: retrieve_data
arguments: label: D15
data_type: string
name: retrieve_data
arguments: label: D7
data_type: string

tool: D6: <E>

tool: D15: +-*/

tool: D7: </E>

assistant: ```c
int E() {
  printf("<E>");
  T();
  if (isNext("+-*/")) {
    cnext();
    E();
  }
  printf("</E>");
}
```