ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<__ultoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rsi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    -0x18(%rbp),%rbx
  cmpl   $0x10,-0x1c(%rbp)
  je     L116
  cmpl   $0x10,-0x1c(%rbp)
  jg     L124
  cmpl   $0x8,-0x1c(%rbp)
  je     L49
  cmpl   $0xa,-0x1c(%rbp)
  jne    L124
  cmp    $0x9,%rdi
  ja     L129
  mov    %edi,%eax
  add    $0x30,%eax
  sub    $0x1,%rbx
  mov    %al,(%rbx)
  mov    %rbx,%rax
  jmp    L98
L129:
  mov    D6(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdi
  jbe    L90
  movabs $0xcccccccccccccccd,%rdx
  mov    %rdi,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rdi,%rdx
  sub    %rax,%rdx
  mov    %edx,%eax
  add    $0x30,%eax
  sub    $0x1,%rbx
  mov    %al,(%rbx)
  movabs $0xcccccccccccccccd,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,%r12
  jmp    L32
L90:
  mov    %rdi,%r12
L32:
  movabs $0x6666666666666667,%rdx
  mov    %r12,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %r12,%rcx
  sar    $0x3f,%rcx
  sub    %rcx,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %r12,%rdx
  sub    %rax,%rdx
  mov    %edx,%eax
  add    $0x30,%eax
  sub    $0x1,%rbx
  mov    %al,(%rbx)
  movabs $0x6666666666666667,%rdx
  mov    %r12,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %r12,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,%r12
  test   %r12,%r12
  jne    L32
  jmp    L124
L93:
  nop
L49:
  mov    %edi,%eax
  and    $0x7,%eax
  add    $0x30,%eax
  sub    $0x1,%rbx
  mov    %al,(%rbx)
  shr    $0x3,%rdi
  test   %rdi,%rdi
  jne    L93
  cmpl   $0x0,-0x20(%rbp)
  je     L97
  movzbl (%rbx),%eax
  cmp    $0x30,%al
  je     L97
  sub    $0x1,%rbx
  movb   $0x30,(%rbx)
  jmp    L97
L62:
  nop
L116:
  mov    %rdi,%rdx
  and    $0xf,%edx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  sub    $0x1,%rbx
  movzbl (%rax),%eax
  mov    %al,(%rbx)
  shr    $0x4,%rdi
  test   %rdi,%rdi
  jne    L62
  jmp    L124
L97:
  nop
L124:
  mov    %rbx,%rax
L98:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
char *__ultoa(register unsigned long val, char *endp, int base, int octzero, const char *xdigs) {
  register char *cp = endp;
  register long sval;
  switch (base) {
  case 10:
    if (val < 10) {
      *--cp = ((val) + '0');
      return (cp);
    }
    if (val > LONG_MAX) {
      *--cp = ((val % 10) + '0');
      sval = val / 10;
    } else
      sval = val;
    do {
      *--cp = ((sval % 10) + '0');
      sval /= 10;
    } while (sval != 0);
    break;
  case 8:
    do {
      *--cp = ((val & 7) + '0');
      val >>= 3;
    } while (val);
    if (octzero && *cp != '0')
      *--cp = '0';
    break;
  case 16:
    do {
      *--cp = xdigs[val & 15];
      val >>= 4;
    } while (val);
    break;
  }
  return (cp);
}
```
user: What is the c source code of the assembly code below?

```asm
<__ultoa>:
  endbr64
  cmp    $0xa,%edx
  je     L87
  cmp    $0x10,%edx
  je     L137
  cmp    $0x8,%edx
  je     L116
L95:
  mov    %rsi,%rax
  ret
L87:
  cmp    $0x9,%rdi
  jbe    L35
  mov    %rdi,%rcx
  mov    D5(%rip),%rax
  cmp    %rdi,(%rax)
  jae    L112
  movabs $0xcccccccccccccccd,%rcx
  mov    %rdi,%rax
  mul    %rcx
  mov    %rdx,%rcx
  shr    $0x3,%rcx
  lea    (%rcx,%rcx,4),%rax
  add    %rax,%rax
  sub    %rax,%rdi
  add    $0x30,%edi
  mov    %dil,-0x1(%rsi)
  lea    -0x1(%rsi),%rsi
L112:
  movabs $0x6666666666666667,%rdi
L68:
  sub    $0x1,%rsi
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  add    $0x30,%ecx
  mov    %cl,(%rsi)
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L68
  jmp    L95
L35:
  add    $0x30,%edi
  mov    %dil,-0x1(%rsi)
  lea    -0x1(%rsi),%rsi
  jmp    L95
L116:
  mov    %rsi,%rdx
  sub    $0x1,%rsi
  mov    %edi,%eax
  and    $0x7,%eax
  add    $0x30,%eax
  mov    %al,(%rsi)
  shr    $0x3,%rdi
  jne    L116
  cmp    $0x30,%al
  je     L95
  test   %ecx,%ecx
  je     L95
  movb   $0x30,-0x1(%rsi)
  lea    -0x2(%rdx),%rsi
  jmp    L95
L137:
  sub    $0x1,%rsi
  mov    %rdi,%rax
  and    $0xf,%eax
  movzbl (%r8,%rax,1),%eax
  mov    %al,(%rsi)
  shr    $0x4,%rdi
  jne    L137
  jmp    L95
```

assistant: ```c
char *__ultoa(register unsigned long val, char *endp, int base, int octzero, const char *xdigs) {
  register char *cp = endp;
  register long sval;
  switch (base) {
  case 10:
    if (val < 10) {
      *--cp = ((val) + '0');
      return (cp);
    }
    if (val > LONG_MAX) {
      *--cp = ((val % 10) + '0');
      sval = val / 10;
    } else
      sval = val;
    do {
      *--cp = ((sval % 10) + '0');
      sval /= 10;
    } while (sval != 0);
    break;
  case 8:
    do {
      *--cp = ((val & 7) + '0');
      val >>= 3;
    } while (val);
    if (octzero && *cp != '0')
      *--cp = '0';
    break;
  case 16:
    do {
      *--cp = xdigs[val & 15];
      val >>= 4;
    } while (val);
    break;
  }
  return (cp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<__ultoa>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%r9
  cmp    $0xa,%edx
  je     L3
  cmp    $0x10,%edx
  je     L78
  cmp    $0x8,%edx
  je     L57
L137:
  mov    %r9,%rax
  ret
  nopl   (%rax)
L78:
  mov    %rax,%rdx
  sub    $0x1,%r9
  and    $0xf,%edx
  shr    $0x4,%rax
  movzbl (%r8,%rdx,1),%edx
  mov    %dl,(%r9)
  jne    L78
  jmp    L137
  nopw   0x0(%rax,%rax,1)
L57:
  mov    %eax,%edx
  mov    %r9,%rsi
  sub    $0x1,%r9
  and    $0x7,%edx
  add    $0x30,%edx
  shr    $0x3,%rax
  mov    %dl,(%r9)
  jne    L57
  cmp    $0x30,%dl
  je     L137
  test   %ecx,%ecx
  je     L137
  movb   $0x30,-0x1(%r9)
  lea    -0x2(%rsi),%r9
  jmp    L137
  nopl   0x0(%rax)
L3:
  lea    -0x1(%rsi),%rsi
  cmp    $0x9,%rdi
  jbe    L119
  mov    D8(%rip),%rax
  mov    %rdi,%rcx
  cmp    %rdi,(%rax)
  jae    L38
  movabs $0xcccccccccccccccd,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdi,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  mov    %rdx,%rcx
  add    %rsi,%rsi
  sub    %rsi,%rax
  lea    -0x2(%r9),%rsi
  add    $0x30,%eax
  mov    %al,-0x1(%r9)
L38:
  movabs $0x6666666666666667,%r8
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L33:
  sub    $0x1,%rsi
L14:
  mov    %rcx,%rax
  mov    %rsi,%r9
  imul   %r8
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  mov    %rcx,%rax
  lea    (%rdx,%rdx,4),%rdi
  mov    %rdx,%rcx
  add    %rdi,%rdi
  sub    %rdi,%rax
  add    $0x30,%eax
  mov    %al,(%rsi)
  test   %rdx,%rdx
  jne    L33
  mov    %r9,%rax
  ret
  nopl   0x0(%rax)
L119:
  add    $0x30,%eax
  mov    %al,-0x1(%r9)
  mov    %rsi,%r9
  jmp    L137
```

assistant: ```c
char *__ultoa(register unsigned long val, char *endp, int base, int octzero, const char *xdigs) {
  register char *cp = endp;
  register long sval;
  switch (base) {
  case 10:
    if (val < 10) {
      *--cp = ((val) + '0');
      return (cp);
    }
    if (val > LONG_MAX) {
      *--cp = ((val % 10) + '0');
      sval = val / 10;
    } else
      sval = val;
    do {
      *--cp = ((sval % 10) + '0');
      sval /= 10;
    } while (sval != 0);
    break;
  case 8:
    do {
      *--cp = ((val & 7) + '0');
      val >>= 3;
    } while (val);
    if (octzero && *cp != '0')
      *--cp = '0';
    break;
  case 16:
    do {
      *--cp = xdigs[val & 15];
      val >>= 4;
    } while (val);
    break;
  }
  return (cp);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__ultoa>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%r9
  cmp    $0xa,%edx
  je     L62
  cmp    $0x10,%edx
  je     L21
  cmp    $0x8,%edx
  je     L32
L65:
  mov    %r9,%rax
  ret
  nopl   (%rax)
L21:
  mov    %rax,%rdx
  sub    $0x1,%r9
  and    $0xf,%edx
  shr    $0x4,%rax
  movzbl (%r8,%rdx,1),%edx
  mov    %dl,(%r9)
  jne    L21
  jmp    L65
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %eax,%edx
  mov    %r9,%rsi
  sub    $0x1,%r9
  and    $0x7,%edx
  add    $0x30,%edx
  shr    $0x3,%rax
  mov    %dl,(%r9)
  jne    L32
  cmp    $0x30,%dl
  je     L65
  test   %ecx,%ecx
  je     L65
  movb   $0x30,-0x1(%r9)
  lea    -0x2(%rsi),%r9
  jmp    L65
  nopl   0x0(%rax)
L62:
  lea    -0x1(%rsi),%rsi
  cmp    $0x9,%rdi
  jbe    L11
  mov    D6(%rip),%rax
  mov    %rdi,%rcx
  cmp    %rdi,(%rax)
  jae    L19
  movabs $0xcccccccccccccccd,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdi,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  mov    %rdx,%rcx
  add    %rsi,%rsi
  sub    %rsi,%rax
  lea    -0x2(%r9),%rsi
  add    $0x30,%eax
  mov    %al,-0x1(%r9)
L19:
  movabs $0x6666666666666667,%r8
  jmp    L108
  nopl   0x0(%rax,%rax,1)
L113:
  sub    $0x1,%rsi
L108:
  mov    %rcx,%rax
  mov    %rsi,%r9
  imul   %r8
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  mov    %rcx,%rax
  lea    (%rdx,%rdx,4),%rdi
  mov    %rdx,%rcx
  add    %rdi,%rdi
  sub    %rdi,%rax
  add    $0x30,%eax
  mov    %al,(%rsi)
  test   %rdx,%rdx
  jne    L113
  mov    %r9,%rax
  ret
  nopl   0x0(%rax)
L11:
  add    $0x30,%eax
  mov    %al,-0x1(%r9)
  mov    %rsi,%r9
  jmp    L65
```

assistant: ```c
char *__ultoa(register unsigned long val, char *endp, int base, int octzero, const char *xdigs) {
  register char *cp = endp;
  register long sval;
  switch (base) {
  case 10:
    if (val < 10) {
      *--cp = ((val) + '0');
      return (cp);
    }
    if (val > LONG_MAX) {
      *--cp = ((val % 10) + '0');
      sval = val / 10;
    } else
      sval = val;
    do {
      *--cp = ((sval % 10) + '0');
      sval /= 10;
    } while (sval != 0);
    break;
  case 8:
    do {
      *--cp = ((val & 7) + '0');
      val >>= 3;
    } while (val);
    if (octzero && *cp != '0')
      *--cp = '0';
    break;
  case 16:
    do {
      *--cp = xdigs[val & 15];
      val >>= 4;
    } while (val);
    break;
  }
  return (cp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prime_geq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
L28:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  je     L26
  mov    -0x4(%rbp),%eax
  jmp    L12
L26:
  addl   $0x1,-0x4(%rbp)
  jmp    L28
L12:
  leave
  ret
```

assistant: ```c
unsigned int prime_geq(int n) {
  for (int m = n + 1;; m++) {
    if (is_prime(m)) {
      return m;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<prime_geq>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%ebx
  jmp    L12
L17:
  add    $0x1,%ebx
L12:
  mov    %ebx,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  je     L17
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned int prime_geq(int n) {
  for (int m = n + 1;; m++) {
    if (is_prime(m)) {
      return m;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<prime_geq>:
  endbr64
  push   %r12
  lea    0x1(%rdi),%r12d
  jmp    L8
  nopl   0x0(%rax)
L19:
  add    $0x1,%r12d
L8:
  mov    %r12d,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  je     L19
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
unsigned int prime_geq(int n) {
  for (int m = n + 1;; m++) {
    if (is_prime(m)) {
      return m;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prime_geq>:
  endbr64
  push   %r12
  lea    0x1(%rdi),%r12d
  jmp    L13
  nopl   0x0(%rax)
L19:
  add    $0x1,%r12d
L13:
  mov    %r12d,%edi
  call   <is_prime@plt>
  test   %rax,%rax
  je     L19
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
unsigned int prime_geq(int n) {
  for (int m = n + 1;; m++) {
    if (is_prime(m)) {
      return m;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strupr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L0:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <islower@plt>
  test   %rax,%rax
  je     L8
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
L8:
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strupr(char *s) {
  char *p = s;
  while (*p != '\0') {
    if (islower(*p)) {
      *p = toupper(*p);
    }
    p++;
  }
  return s;
}
```
user: What is the c source code of the assembly code below?

```asm
<strupr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L24
  mov    %rbp,%rbx
  jmp    L10
L27:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L24
L10:
  movsbl %dil,%edi
  call   <islower@plt>
  test   %rax,%rax
  je     L27
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  jmp    L27
L24:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strupr(char *s) {
  char *p = s;
  while (*p != '\0') {
    if (islower(*p)) {
      *p = toupper(*p);
    }
    p++;
  }
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strupr>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L1
  mov    %r12,%rbx
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L9:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L1
L31:
  call   <islower@plt>
  test   %rax,%rax
  je     L9
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L31
L1:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *strupr(char *s) {
  char *p = s;
  while (*p != '\0') {
    if (islower(*p)) {
      *p = toupper(*p);
    }
    p++;
  }
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strupr>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L13
  mov    %r12,%rbx
  nopl   0x0(%rax)
L38:
  call   <islower@plt>
  test   %rax,%rax
  jne    L26
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L38
L13:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L38
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *strupr(char *s) {
  char *p = s;
  while (*p != '\0') {
    if (islower(*p)) {
      *p = toupper(*p);
    }
    p++;
  }
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<baz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <setjmp@plt>
  endbr64
  test   %eax,%eax
  jne    L14
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <qux@plt>
  jmp    L10
L14:
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L10:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D18
data_type: string
name: unpack_data
arguments: data_tag: D0
data_type: string

tool: D18: Inside baz


tool: D0: Error: Shouldn't be here in baz


assistant: ```c
void baz(jmp_buf buf) {
  jmp_buf local_buf;
  printf("Inside baz\n");
  if (!setjmp(local_buf))
    qux(buf);
  else
    printf("Error: Shouldn't be here in baz\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<baz>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <setjmp@plt>
  endbr64
  test   %eax,%eax
  jne    L3
  mov    0xc(%rsp),%edi
  call   <qux@plt>
L16:
  add    $0x18,%rsp
  ret
L3:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L16
```

tool_calls: name: unpack_data
arguments: data_tag: D15
type_guessed: string
name: unpack_data
arguments: data_tag: D12
type_guessed: string

tool: D15: Inside baz


tool: D12: Error: Shouldn't be here in baz


assistant: ```c
void baz(jmp_buf buf) {
  jmp_buf local_buf;
  printf("Inside baz\n");
  if (!setjmp(local_buf))
    qux(buf);
  else
    printf("Error: Shouldn't be here in baz\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<baz>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    D11(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <setjmp@plt>
  endbr64
  test   %eax,%eax
  jne    L3
  mov    0xc(%rsp),%edi
  call   <qux@plt>
  add    $0x18,%rsp
  ret
  nopl   (%rax)
L3:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D11
datatype: string
name: unpack_data
arguments: data_tag: D9
datatype: string

tool: D11: Inside baz


tool: D9: Error: Shouldn't be here in baz


assistant: ```c
void baz(jmp_buf buf) {
  jmp_buf local_buf;
  printf("Inside baz\n");
  if (!setjmp(local_buf))
    qux(buf);
  else
    printf("Error: Shouldn't be here in baz\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<baz>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <setjmp@plt>
  endbr64
  test   %eax,%eax
  jne    L6
  mov    0xc(%rsp),%edi
  call   <qux@plt>
  add    $0x18,%rsp
  ret
  nopl   (%rax)
L6:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string

tool: D7: Inside baz


tool: D8: Error: Shouldn't be here in baz


assistant: ```c
void baz(jmp_buf buf) {
  jmp_buf local_buf;
  printf("Inside baz\n");
  if (!setjmp(local_buf))
    qux(buf);
  else
    printf("Error: Shouldn't be here in baz\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x38(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x4,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  movl   $0x9,0x1c(%rsp)
  movl   $0xd,0x20(%rsp)
  movl   $0x10,0x24(%rsp)
  movl   $0x13,0x28(%rsp)
  movl   $0x1c,0x2c(%rsp)
  movl   $0x28,0x30(%rsp)
  movl   $0x64,0x34(%rsp)
  lea    D24(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rbx
  lea    0x38(%rsp),%r13
  mov    %rbx,%rbp
  lea    D9(%rip),%r12
L69:
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r13,%rbp
  jne    L69
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D27(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  cmp    $0x64,%ecx
  jle    L51
  mov    %ecx,0x38(%rsp)
L25:
  lea    D47(%rip),%rdi
  call   <puts@plt>
  lea    0x2c(%rbx),%r12
  lea    D9(%rip),%rbp
L76:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%r12
  jne    L76
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L51:
  mov    $0x0,%eax
L59:
  mov    0x10(%rsp,%rax,4),%edx
  cmp    %edx,%ecx
  jl     L22
  add    $0x1,%rax
  cmp    $0xa,%rax
  jne    L59
  jmp    L25
L22:
  movslq %eax,%rdi
  mov    %ecx,0x10(%rsp,%rdi,4)
  cmp    $0x9,%eax
  jg     L25
  mov    %rdi,%rsi
  lea    0x10(%rsp,%rdi,4),%rcx
  mov    $0x9,%edi
  sub    %eax,%edi
  lea    (%rdi,%rsi,1),%rax
  lea    0x4(%rbx,%rax,4),%rsi
L33:
  mov    %edx,%eax
  mov    0x4(%rcx),%edx
  mov    %eax,0x4(%rcx)
  add    $0x4,%rcx
  cmp    %rsi,%rcx
  jne    L33
  jmp    L25
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D24
data_type: string
name: unpack_data
arguments: data_tag: D9
data_type: string
name: unpack_data
arguments: data_tag: D19
data_type: string
name: unpack_data
arguments: data_tag: D27
data_type: string
name: unpack_data
arguments: data_tag: D47
data_type: string

tool: D24: array a:

tool: D9: %5d

tool: D19: insert data:

tool: D27: %d

tool: D47: Now array a:

assistant: ```c
int main() {
  int a[11] = {1, 4, 6, 9, 13, 16, 19, 28, 40, 100};
  int temp1, temp2, number, end, i, j;
  printf("array a:\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  printf("insert data:");
  scanf("%d", &number);
  end = a[9];
  if (number > end)
    a[10] = number;
  else {
    for (i = 0; i < 10; i++) {
      if (a[i] > number) {
        temp1 = a[i];
        a[i] = number;
        for (j = i + 1; j < 11; j++) {
          temp2 = a[j];
          a[j] = temp1;
          temp1 = temp2;
        }
        break;
      }
    }
  }
  printf("Now array a:\n");
  for (i = 0; i < 11; i++)
    printf("%5d", a[i]);
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc4(%rbp)
  lea    -0xcc(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0xc0(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0xc8(%rbp)
  jmp    L39
L33:
  mov    -0xc8(%rbp),%eax
  cltq
  movzbl -0x60(%rbp,%rax,1),%edx
  mov    -0xc8(%rbp),%eax
  cltq
  movzbl -0xc0(%rbp,%rax,1),%eax
  cmp    %al,%dl
  jne    L37
  addl   $0x1,-0xc4(%rbp)
L37:
  addl   $0x1,-0xc8(%rbp)
L39:
  mov    -0xcc(%rbp),%eax
  cmp    %eax,-0xc8(%rbp)
  jl     L33
  mov    -0xc4(%rbp),%eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D26
datatype: string
name: parse_data
arguments: tag: D31
datatype: string
name: parse_data
arguments: tag: D32
datatype: string

tool: D26:  %d

tool: D31:  %s

tool: D32: 
%d

assistant: ```c
int main() {
  int i, acert = 0, n;
  char resp[85], gaba[85];
  scanf(" %d", &n);
  scanf(" %s", resp);
  scanf(" %s", gaba);
  for (i = 0; i < n; i++) {
    if (gaba[i] == resp[i])
      acert++;
  }
  printf("\n%d", acert);
  return 0;
}
```