ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<new_str_clean>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L79
  nopl   0x0(%rax,%rax,1)
L50:
  movzbl (%rsi),%eax
  test   %al,%al
  je     L17
  lea    0x1(%rsi),%rdx
  jmp    L49
  nopl   (%rax)
L28:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L17
L49:
  cmp    %al,%cl
  jne    L28
  mov    %rdi,%rax
  nopw   0x0(%rax,%rax,1)
L12:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  test   %dl,%dl
  jne    L12
  movzbl (%rdi),%ecx
  test   %cl,%cl
  jne    L50
L31:
  ret
  nop
L17:
  movzbl 0x1(%rdi),%ecx
  add    $0x1,%rdi
  test   %cl,%cl
  jne    L50
  jmp    L31
L79:
  ret
```

assistant: ```c
void new_str_clean(char *base, const char *set) {
  int i;
  while (*base) {
    i = 0;
    while (set[i] && *base != set[i]) {
      i++;
    }
    if (set[i]) {
      i = 0;
      do {
        base[i] = base[i + 1];
      } while (base[++i]);
    } else {
      base++;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<new_str_clean>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L77
L20:
  movzbl (%rsi),%r8d
  mov    %r8d,%eax
  test   %r8b,%r8b
  je     L32
L22:
  lea    0x1(%rsi),%rdx
  jmp    L55
  nopl   (%rax)
L74:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L25
L55:
  cmp    %al,%cl
  jne    L74
  mov    %rdi,%rax
  nopw   0x0(%rax,%rax,1)
L100:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  test   %dl,%dl
  jne    L100
  movzbl (%rdi),%ecx
  test   %cl,%cl
  jne    L20
L77:
  ret
  nop
L25:
  movzbl 0x1(%rdi),%ecx
  lea    0x1(%rdi),%rax
  test   %cl,%cl
  je     L77
  mov    %rax,%rdi
  mov    %r8d,%eax
  test   %r8b,%r8b
  jne    L22
L32:
  cmpb   $0x0,0x1(%rdi)
  lea    0x1(%rdi),%rax
  je     L54
  nopl   0x0(%rax)
L57:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L57
  ret
L54:
  ret
```

assistant: ```c
void new_str_clean(char *base, const char *set) {
  int i;
  while (*base) {
    i = 0;
    while (set[i] && *base != set[i]) {
      i++;
    }
    if (set[i]) {
      i = 0;
      do {
        base[i] = base[i + 1];
      } while (base[++i]);
    } else {
      base++;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dump_sys_getpgrp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
datatype: string

tool: D6: void

assistant: ```c
char *dump_sys_getpgrp(int depth, struct target *target) {
  char *param_str = malloc(5);
  snprintf(param_str, 5, "void");
  return param_str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dump_sys_getpgrp>:
  endbr64
  push   %rbx
  mov    $0x5,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  lea    D3(%rip),%rdx
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
data_type: string

tool: D3: void

assistant: ```c
char *dump_sys_getpgrp(int depth, struct target *target) {
  char *param_str = malloc(5);
  snprintf(param_str, 5, "void");
  return param_str;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dump_sys_getpgrp>:
  endbr64
  push   %r12
  mov    $0x5,%edi
  call   <malloc@plt>
  lea    D8(%rip),%rdx
  mov    $0x5,%esi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type_infered: string

tool: D8: void

assistant: ```c
char *dump_sys_getpgrp(int depth, struct target *target) {
  char *param_str = malloc(5);
  snprintf(param_str, 5, "void");
  return param_str;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dump_sys_getpgrp>:
  endbr64
  push   %r12
  mov    $0x5,%edi
  call   <malloc@plt>
  lea    D0(%rip),%rdx
  mov    $0x5,%esi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
datatype: string

tool: D0: void

assistant: ```c
char *dump_sys_getpgrp(int depth, struct target *target) {
  char *param_str = malloc(5);
  snprintf(param_str, 5, "void");
  return param_str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subtraction>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int subtraction(int num1, int num2) {
  int aux;
  aux = num1 - num2;
  return aux;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<subtraction>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subtraction(int num1, int num2) {
  int aux;
  aux = num1 - num2;
  return aux;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<subtraction>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subtraction(int num1, int num2) {
  int aux;
  aux = num1 - num2;
  return aux;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<subtraction>:
  endbr64
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int subtraction(int num1, int num2) {
  int aux;
  aux = num1 - num2;
  return aux;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rdtsc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  rdtsc
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long int rdtsc(void) {
  unsigned long long int x;
  __asm__ volatile(".byte 0x0f, 0x31" : "=A"(x));
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rdtsc>:
  endbr64
  rdtsc
  ret
```

assistant: ```c
unsigned long long int rdtsc(void) {
  unsigned long long int x;
  __asm__ volatile(".byte 0x0f, 0x31" : "=A"(x));
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rdtsc>:
  endbr64
  rdtsc
  ret
```

assistant: ```c
unsigned long long int rdtsc(void) {
  unsigned long long int x;
  __asm__ volatile(".byte 0x0f, 0x31" : "=A"(x));
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rdtsc>:
  endbr64
  rdtsc
  ret
```

assistant: ```c
unsigned long long int rdtsc(void) {
  unsigned long long int x;
  __asm__ volatile(".byte 0x0f, 0x31" : "=A"(x));
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<format_cpu_method>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0xc,-0x14(%rbp)
  ja     L100
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D4(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D4(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  movb   $0x55,-0x1(%rbp)
  jmp    L46
  movb   $0x49,-0x1(%rbp)
  jmp    L46
  movb   $0x50,-0x1(%rbp)
  jmp    L46
  movb   $0x4b,-0x1(%rbp)
  jmp    L46
  movb   $0x4d,-0x1(%rbp)
  jmp    L46
  movb   $0x45,-0x1(%rbp)
  jmp    L46
  movb   $0x54,-0x1(%rbp)
  jmp    L46
  movb   $0x52,-0x1(%rbp)
  jmp    L46
  movb   $0x4c,-0x1(%rbp)
  jmp    L46
  movb   $0x4e,-0x1(%rbp)
  jmp    L46
  movb   $0x53,-0x1(%rbp)
  jmp    L46
  movb   $0x43,-0x1(%rbp)
  jmp    L46
  movb   $0x4f,-0x1(%rbp)
  jmp    L46
L100:
  movb   $0x3f,-0x1(%rbp)
L46:
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
char format_cpu_method(int method) {
  char method_char;
  switch (method) {
  case 0:
    method_char = 'U';
    break;
  case 1:
    method_char = 'I';
    break;
  case 2:
    method_char = 'P';
    break;
  case 7:
    method_char = 'K';
    break;
  case 11:
    method_char = 'M';
    break;
  case 10:
    method_char = 'E';
    break;
  case 3:
    method_char = 'T';
    break;
  case 5:
    method_char = 'R';
    break;
  case 4:
    method_char = 'L';
    break;
  case 6:
    method_char = 'N';
    break;
  case 8:
    method_char = 'S';
    break;
  case 9:
    method_char = 'C';
    break;
  case 12:
    method_char = 'O';
    break;
  default:
    method_char = '?';
  }
  return method_char;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<format_cpu_method>:
  endbr64
  cmp    $0xc,%edi
  ja     L3
  mov    %edi,%edi
  lea    D0(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x49,%eax
  ret
  mov    $0x50,%eax
  ret
  mov    $0x4b,%eax
  ret
  mov    $0x4d,%eax
  ret
  mov    $0x45,%eax
  ret
  mov    $0x54,%eax
  ret
  mov    $0x52,%eax
  ret
  mov    $0x4c,%eax
  ret
  mov    $0x4e,%eax
  ret
  mov    $0x53,%eax
  ret
  mov    $0x43,%eax
  ret
  mov    $0x4f,%eax
  ret
L3:
  mov    $0x3f,%eax
  ret
  mov    $0x55,%eax
  ret
```

assistant: ```c
char format_cpu_method(int method) {
  char method_char;
  switch (method) {
  case 0:
    method_char = 'U';
    break;
  case 1:
    method_char = 'I';
    break;
  case 2:
    method_char = 'P';
    break;
  case 7:
    method_char = 'K';
    break;
  case 11:
    method_char = 'M';
    break;
  case 10:
    method_char = 'E';
    break;
  case 3:
    method_char = 'T';
    break;
  case 5:
    method_char = 'R';
    break;
  case 4:
    method_char = 'L';
    break;
  case 6:
    method_char = 'N';
    break;
  case 8:
    method_char = 'S';
    break;
  case 9:
    method_char = 'C';
    break;
  case 12:
    method_char = 'O';
    break;
  default:
    method_char = '?';
  }
  return method_char;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<format_cpu_method>:
  endbr64
  mov    $0x3f,%eax
  cmp    $0xc,%edi
  ja     L5
  mov    %edi,%edi
  lea    D8(%rip),%rax
  movzbl (%rax,%rdi,1),%eax
L5:
  ret
```

assistant: ```c
char format_cpu_method(int method) {
  char method_char;
  switch (method) {
  case 0:
    method_char = 'U';
    break;
  case 1:
    method_char = 'I';
    break;
  case 2:
    method_char = 'P';
    break;
  case 7:
    method_char = 'K';
    break;
  case 11:
    method_char = 'M';
    break;
  case 10:
    method_char = 'E';
    break;
  case 3:
    method_char = 'T';
    break;
  case 5:
    method_char = 'R';
    break;
  case 4:
    method_char = 'L';
    break;
  case 6:
    method_char = 'N';
    break;
  case 8:
    method_char = 'S';
    break;
  case 9:
    method_char = 'C';
    break;
  case 12:
    method_char = 'O';
    break;
  default:
    method_char = '?';
  }
  return method_char;
}
```
user: Translate the below assembly code into C programming language:

```asm
<format_cpu_method>:
  endbr64
  mov    $0x3f,%eax
  cmp    $0xc,%edi
  ja     L0
  mov    %edi,%edi
  lea    D3(%rip),%rax
  movzbl (%rax,%rdi,1),%eax
L0:
  ret
```

assistant: ```c
char format_cpu_method(int method) {
  char method_char;
  switch (method) {
  case 0:
    method_char = 'U';
    break;
  case 1:
    method_char = 'I';
    break;
  case 2:
    method_char = 'P';
    break;
  case 7:
    method_char = 'K';
    break;
  case 11:
    method_char = 'M';
    break;
  case 10:
    method_char = 'E';
    break;
  case 3:
    method_char = 'T';
    break;
  case 5:
    method_char = 'R';
    break;
  case 4:
    method_char = 'L';
    break;
  case 6:
    method_char = 'N';
    break;
  case 8:
    method_char = 'S';
    break;
  case 9:
    method_char = 'C';
    break;
  case 12:
    method_char = 'O';
    break;
  default:
    method_char = '?';
  }
  return method_char;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<kfork>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x3,%edi
  call   <syscall@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D19
type_infered: string
name: retrieve_data
arguments: label: D9
type_infered: string

tool: D19: Proc %d entering kernal to fork a child...


tool: D9: Proc %d forked a child (%d)!


assistant: ```c
int kfork() {
  int child, pid;
  pid = getpid();
  printf("Proc %d entering kernal to fork a child...\n", pid);
  child = syscall(3, 0, 0);
  printf("Proc %d forked a child (%d)!\n", pid, child);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<kfork>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x3,%edi
  call   <syscall@plt>
  mov    %eax,%edx
  mov    %ebx,%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pop    %rbx
  ret
```

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

tool: D6: Proc %d entering kernal to fork a child...


tool: D3: Proc %d forked a child (%d)!


assistant: ```c
int kfork() {
  int child, pid;
  pid = getpid();
  printf("Proc %d entering kernal to fork a child...\n", pid);
  child = syscall(3, 0, 0);
  printf("Proc %d forked a child (%d)!\n", pid, child);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<kfork>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x3,%edi
  call   <syscall@plt>
  mov    %ebp,%esi
  lea    D12(%rip),%rdi
  pop    %rbp
  mov    %eax,%edx
  xor    %eax,%eax
  jmp    1080 <printf@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
type_guessed: string
name: unpack_data
arguments: label: D12
type_guessed: string

tool: D9: Proc %d entering kernal to fork a child...


tool: D12: Proc %d forked a child (%d)!


assistant: ```c
int kfork() {
  int child, pid;
  pid = getpid();
  printf("Proc %d entering kernal to fork a child...\n", pid);
  child = syscall(3, 0, 0);
  printf("Proc %d forked a child (%d)!\n", pid, child);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<kfork>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x3,%edi
  call   <syscall@plt>
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  pop    %rbp
  mov    %eax,%edx
  xor    %eax,%eax
  jmp    1080 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D2
type_guessed: string

tool: D9: Proc %d entering kernal to fork a child...


tool: D2: Proc %d forked a child (%d)!


assistant: ```c
int kfork() {
  int child, pid;
  pid = getpid();
  printf("Proc %d entering kernal to fork a child...\n", pid);
  child = syscall(3, 0, 0);
  printf("Proc %d forked a child (%d)!\n", pid, child);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gen_instance_logname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%r8d
  mov    %rax,%rcx
  lea    D21(%rip),%rax
  mov    %rax,%rdx
  mov    $0x50,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    D7(%rip),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D21
type: string

tool: D21: %s-%d.log

assistant: ```c
char *gen_instance_logname(char *template, int instance_num) {
  static char filename[80];
  snprintf(filename, 80, "%s-%d.log", template, instance_num);
  return filename;
}
```
user: What is the c source code of the assembly code below?

```asm
<gen_instance_logname>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %esi,%r8d
  lea    D15(%rip),%rdx
  mov    $0x50,%esi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D15
type_guessed: string

tool: D15: %s-%d.log

assistant: ```c
char *gen_instance_logname(char *template, int instance_num) {
  static char filename[80];
  snprintf(filename, 80, "%s-%d.log", template, instance_num);
  return filename;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gen_instance_logname>:
  endbr64
  push   %r12
  lea    D18(%rip),%r12
  mov    %rdi,%rcx
  mov    %esi,%r8d
  mov    %r12,%rdi
  lea    D1(%rip),%rdx
  mov    $0x50,%esi
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
type_guessed: string

tool: D1: %s-%d.log

assistant: ```c
char *gen_instance_logname(char *template, int instance_num) {
  static char filename[80];
  snprintf(filename, 80, "%s-%d.log", template, instance_num);
  return filename;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<gen_instance_logname>:
  endbr64
  push   %r12
  lea    D13(%rip),%r12
  mov    %rdi,%rcx
  mov    %esi,%r8d
  mov    %r12,%rdi
  lea    D4(%rip),%rdx
  mov    $0x50,%esi
  call   <snprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type_guessed: string

tool: D4: %s-%d.log

assistant: ```c
char *gen_instance_logname(char *template, int instance_num) {
  static char filename[80];
  snprintf(filename, 80, "%s-%d.log", template, instance_num);
  return filename;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x14(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L19
  movss  -0x14(%rbp),%xmm0
  movss  -0xc(%rbp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L19
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L64
L19:
  movss  -0x10(%rbp),%xmm0
  movss  -0x14(%rbp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L68
  movss  -0x10(%rbp),%xmm0
  movss  -0xc(%rbp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L68
  mov    -0x10(%rbp),%eax
  movd   %eax,%xmm0
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L64
L68:
  movss  -0xc(%rbp),%xmm0
  movss  -0x14(%rbp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L64
  movss  -0xc(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L64
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L64:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D26
type_infered: string
name: disassemble_data
arguments: label: D59
type_infered: string

tool: D26: %f

tool: D59: %.1f

assistant: ```c
int main() {
  float k, o, c;
  scanf("%f", &k);
  scanf("%f", &o);
  scanf("%f", &c);
  if (k > o && k > c) {
    printf("%.1f", k);
  } else if (o > k && o > c) {
    printf("%.1f", o);
  } else if (c > k && c > o) {
    printf("%.1f", c);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D34(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm0
  movss  0x10(%rsp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L7
  comiss 0x14(%rsp),%xmm0
  ja     L40
L7:
  comiss %xmm0,%xmm1
  jbe    L88
  comiss 0x14(%rsp),%xmm1
  ja     L21
L88:
  movss  0x14(%rsp),%xmm2
  comiss %xmm0,%xmm2
  jbe    L20
  comiss %xmm1,%xmm2
  ja     L86
L20:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L40:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L21:
  movaps %xmm1,%xmm0
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L86:
  movaps %xmm2,%xmm0
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D34
data_type: string
name: retrieve_data
arguments: data_label: D6
data_type: string

tool: D34: %f

tool: D6: %.1f

assistant: ```c
int main() {
  float k, o, c;
  scanf("%f", &k);
  scanf("%f", &o);
  scanf("%f", &c);
  if (k > o && k > c) {
    printf("%.1f", k);
  } else if (o > k && o > c) {
    printf("%.1f", o);
  } else if (c > k && c > o) {
    printf("%.1f", c);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm0
  movss  0x10(%rsp),%xmm1
  comiss %xmm1,%xmm0
  jbe    L79
  comiss 0x14(%rsp),%xmm0
  ja     L88
L79:
  comiss %xmm0,%xmm1
  movss  0x14(%rsp),%xmm2
  jbe    L42
  comiss %xmm2,%xmm1
  ja     L6
L42:
  comiss %xmm0,%xmm2
  ja     L20
L12:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L20:
  comiss %xmm1,%xmm2
  jbe    L12
  movaps %xmm2,%xmm0
L88:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L12
L6:
  movaps %xmm1,%xmm0
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L12
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D8
type_guessed: string
name: parse_data
arguments: label: D6
type_guessed: string

tool: D8: %f

tool: D6: %.1f

assistant: ```c
int main() {
  float k, o, c;
  scanf("%f", &k);
  scanf("%f", &o);
  scanf("%f", &c);
  if (k > o && k > c) {
    printf("%.1f", k);
  } else if (o > k && o > c) {
    printf("%.1f", o);
  } else if (c > k && c > o) {
    printf("%.1f", c);
  }
  return 0;
}
```