ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<string_testEqualN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L67
  cmpq   $0x0,-0x30(%rbp)
  jne    L3
L67:
  mov    $0x0,%eax
  jmp    L41
L3:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L46
L25:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  cmp    %al,%cl
  je     L51
  mov    $0x0,%eax
  jmp    L41
L51:
  addl   $0x1,-0x14(%rbp)
L46:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jae    L58
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L25
L58:
  mov    $0x1,%eax
L41:
  pop    %rbp
  ret
```

assistant: ```c
int string_testEqualN(const char *one, const char *two, const unsigned int n) {
  if (!one || !two)
    return 0;
  char *m1 = (char *)one;
  char *m2 = (char *)two;
  unsigned int i;
  for (i = 0; i < n && *m1; i++)
    if (*m1++ != *m2++)
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<string_testEqualN>:
  endbr64
  test   %rdi,%rdi
  je     L35
  test   %rsi,%rsi
  je     L35
  test   %edx,%edx
  je     L50
  mov    %edx,%edx
  mov    $0x0,%eax
L45:
  movzbl (%rdi,%rax,1),%ecx
  test   %cl,%cl
  je     L31
  cmp    %cl,(%rsi,%rax,1)
  jne    L16
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L45
  mov    $0x1,%eax
  ret
L31:
  mov    $0x1,%eax
  ret
L35:
  mov    $0x0,%eax
  ret
L50:
  mov    $0x1,%eax
  ret
L16:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int string_testEqualN(const char *one, const char *two, const unsigned int n) {
  if (!one || !two)
    return 0;
  char *m1 = (char *)one;
  char *m2 = (char *)two;
  unsigned int i;
  for (i = 0; i < n && *m1; i++)
    if (*m1++ != *m2++)
      return 0;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<string_testEqualN>:
  endbr64
  test   %rdi,%rdi
  je     L69
  test   %rsi,%rsi
  je     L69
  test   %edx,%edx
  je     L55
  mov    %edx,%edx
  xor    %eax,%eax
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L66:
  cmp    %cl,(%rsi,%rax,1)
  jne    L69
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L55
L26:
  movzbl (%rdi,%rax,1),%ecx
  test   %cl,%cl
  jne    L66
L55:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L69:
  xor    %eax,%eax
  ret
```

assistant: ```c
int string_testEqualN(const char *one, const char *two, const unsigned int n) {
  if (!one || !two)
    return 0;
  char *m1 = (char *)one;
  char *m2 = (char *)two;
  unsigned int i;
  for (i = 0; i < n && *m1; i++)
    if (*m1++ != *m2++)
      return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<string_testEqualN>:
  endbr64
  test   %rdi,%rdi
  je     L54
  test   %rsi,%rsi
  je     L54
  test   %edx,%edx
  je     L64
  mov    %edx,%edx
  xor    %eax,%eax
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L2:
  cmp    %cl,(%rsi,%rax,1)
  jne    L54
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L64
L23:
  movzbl (%rdi,%rax,1),%ecx
  test   %cl,%cl
  jne    L2
L64:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L54:
  xor    %eax,%eax
  ret
```

assistant: ```c
int string_testEqualN(const char *one, const char *two, const unsigned int n) {
  if (!one || !two)
    return 0;
  char *m1 = (char *)one;
  char *m2 = (char *)two;
  unsigned int i;
  for (i = 0; i < n && *m1; i++)
    if (*m1++ != *m2++)
      return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<xrop3_113_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  xor    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  and    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x10(%rbp),%eax
  not    %eax
  or     -0x14(%rbp),%eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  not    %eax
  and    -0x10(%rbp),%eax
  and    -0x18(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_113_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk2 = S ^ D;
  stk3 = T ^ D;
  stk2 = stk2 & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xrop3_113_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%eax
  xor    $0xff,%al
  mov    %esi,%edx
  xor    $0xff,%dl
  and    %eax,%edx
  xor    %edi,%edx
  mov    %esi,%eax
  not    %eax
  or     %edi,%eax
  not    %edx
  and    %edx,%eax
  not    %edi
  and    %esi,%edi
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_113_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk2 = S ^ D;
  stk3 = T ^ D;
  stk2 = stk2 & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: Translate the below assembly code into C programming language:

```asm
<xrop3_113_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%eax
  mov    %esi,%edx
  xor    $0xff,%al
  xor    $0xff,%dl
  and    %eax,%edx
  mov    %esi,%eax
  not    %eax
  xor    %edi,%edx
  or     %edi,%eax
  not    %edi
  not    %edx
  and    %esi,%edi
  and    %edx,%eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_113_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk2 = S ^ D;
  stk3 = T ^ D;
  stk2 = stk2 & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<xrop3_113_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%eax
  mov    %esi,%edx
  xor    $0xff,%al
  xor    $0xff,%dl
  and    %eax,%edx
  mov    %esi,%eax
  not    %eax
  xor    %edi,%edx
  or     %edi,%eax
  not    %edi
  not    %edx
  and    %esi,%edi
  and    %edx,%eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_113_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk2 = S ^ D;
  stk3 = T ^ D;
  stk2 = stk2 & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<keypress>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,%eax
  mov    %al,-0x8(%rbp)
  movsbl -0x8(%rbp),%eax
  sub    $0x20,%eax
  cmp    $0x5a,%eax
  ja     L184
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D15(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D15(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x4,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x4,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
  mov    $0x2,%eax
  jmp    L75
  mov    $0x3,%eax
  jmp    L75
  mov    $0x4,%eax
  jmp    L75
  mov    $0x1,%eax
  jmp    L75
L184:
  mov    $0x0,%eax
L75:
  pop    %rbp
  ret
```

assistant: ```c
int keypress(int sum, char s) {
  switch (s) {
  case 'a':
    return 1;
    break;
  case 'b':
    return 2;
    break;
  case 'c':
    return 3;
    break;
  case 'd':
    return 1;
    break;
  case 'e':
    return 2;
    break;
  case 'f':
    return 3;
    break;
  case 'g':
    return 1;
    break;
  case 'h':
    return 2;
    break;
  case 'i':
    return 3;
    break;
  case 'j':
    return 1;
    break;
  case 'k':
    return 2;
    break;
  case 'l':
    return 3;
    break;
  case 'm':
    return 1;
    break;
  case 'n':
    return 2;
    break;
  case 'o':
    return 3;
    break;
  case 'p':
    return 1;
    break;
  case 'q':
    return 2;
    break;
  case 'r':
    return 3;
    break;
  case 's':
    return 4;
    break;
  case 't':
    return 1;
    break;
  case 'u':
    return 2;
    break;
  case 'v':
    return 4;
    break;
  case 'w':
    return 1;
    break;
  case 'x':
    return 2;
    break;
  case 'y':
    return 3;
    break;
  case 'z':
    return 4;
    break;
  case ' ':
    return 1;
    break;
  default:
    return 0;
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<keypress>:
  endbr64
  sub    $0x20,%esi
  cmp    $0x5a,%sil
  ja     L6
  movzbl %sil,%esi
  lea    D1(%rip),%rdx
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x2,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x4,%eax
  ret
  mov    $0x4,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x4,%eax
  ret
L6:
  mov    $0x0,%eax
  ret
  mov    $0x1,%eax
  ret
```

assistant: ```c
int keypress(int sum, char s) {
  switch (s) {
  case 'a':
    return 1;
    break;
  case 'b':
    return 2;
    break;
  case 'c':
    return 3;
    break;
  case 'd':
    return 1;
    break;
  case 'e':
    return 2;
    break;
  case 'f':
    return 3;
    break;
  case 'g':
    return 1;
    break;
  case 'h':
    return 2;
    break;
  case 'i':
    return 3;
    break;
  case 'j':
    return 1;
    break;
  case 'k':
    return 2;
    break;
  case 'l':
    return 3;
    break;
  case 'm':
    return 1;
    break;
  case 'n':
    return 2;
    break;
  case 'o':
    return 3;
    break;
  case 'p':
    return 1;
    break;
  case 'q':
    return 2;
    break;
  case 'r':
    return 3;
    break;
  case 's':
    return 4;
    break;
  case 't':
    return 1;
    break;
  case 'u':
    return 2;
    break;
  case 'v':
    return 4;
    break;
  case 'w':
    return 1;
    break;
  case 'x':
    return 2;
    break;
  case 'y':
    return 3;
    break;
  case 'z':
    return 4;
    break;
  case ' ':
    return 1;
    break;
  default:
    return 0;
    break;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<keypress>:
  endbr64
  sub    $0x20,%esi
  xor    %eax,%eax
  cmp    $0x5a,%sil
  ja     L4
  movzbl %sil,%esi
  lea    D7(%rip),%rax
  movsbl (%rax,%rsi,1),%eax
L4:
  ret
```

tool_calls: name: read_data
arguments: tag: D7
data_type: string

tool: D7: 

assistant: ```c
int keypress(int sum, char s) {
  switch (s) {
  case 'a':
    return 1;
    break;
  case 'b':
    return 2;
    break;
  case 'c':
    return 3;
    break;
  case 'd':
    return 1;
    break;
  case 'e':
    return 2;
    break;
  case 'f':
    return 3;
    break;
  case 'g':
    return 1;
    break;
  case 'h':
    return 2;
    break;
  case 'i':
    return 3;
    break;
  case 'j':
    return 1;
    break;
  case 'k':
    return 2;
    break;
  case 'l':
    return 3;
    break;
  case 'm':
    return 1;
    break;
  case 'n':
    return 2;
    break;
  case 'o':
    return 3;
    break;
  case 'p':
    return 1;
    break;
  case 'q':
    return 2;
    break;
  case 'r':
    return 3;
    break;
  case 's':
    return 4;
    break;
  case 't':
    return 1;
    break;
  case 'u':
    return 2;
    break;
  case 'v':
    return 4;
    break;
  case 'w':
    return 1;
    break;
  case 'x':
    return 2;
    break;
  case 'y':
    return 3;
    break;
  case 'z':
    return 4;
    break;
  case ' ':
    return 1;
    break;
  default:
    return 0;
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<keypress>:
  endbr64
  sub    $0x20,%esi
  xor    %eax,%eax
  cmp    $0x5a,%sil
  ja     L2
  movzbl %sil,%esi
  lea    D5(%rip),%rax
  movsbl (%rax,%rsi,1),%eax
L2:
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
type_guessed: string

tool: D5: 

assistant: ```c
int keypress(int sum, char s) {
  switch (s) {
  case 'a':
    return 1;
    break;
  case 'b':
    return 2;
    break;
  case 'c':
    return 3;
    break;
  case 'd':
    return 1;
    break;
  case 'e':
    return 2;
    break;
  case 'f':
    return 3;
    break;
  case 'g':
    return 1;
    break;
  case 'h':
    return 2;
    break;
  case 'i':
    return 3;
    break;
  case 'j':
    return 1;
    break;
  case 'k':
    return 2;
    break;
  case 'l':
    return 3;
    break;
  case 'm':
    return 1;
    break;
  case 'n':
    return 2;
    break;
  case 'o':
    return 3;
    break;
  case 'p':
    return 1;
    break;
  case 'q':
    return 2;
    break;
  case 'r':
    return 3;
    break;
  case 's':
    return 4;
    break;
  case 't':
    return 1;
    break;
  case 'u':
    return 2;
    break;
  case 'v':
    return 4;
    break;
  case 'w':
    return 1;
    break;
  case 'x':
    return 2;
    break;
  case 'y':
    return 3;
    break;
  case 'z':
    return 4;
    break;
  case ' ':
    return 1;
    break;
  default:
    return 0;
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<destroy_tokens>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L12
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L29:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L29
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L12:
  nop
  leave
  ret
```

assistant: ```c
void destroy_tokens(char **tokens) {
  if (tokens) {
    for (int i = 0; tokens[i]; i++)
      free(tokens[i]);
    free(tokens);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<destroy_tokens>:
  endbr64
  test   %rdi,%rdi
  je     L25
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L10
  lea    0x8(%rbp),%rbx
L7:
  call   <free@plt>
  add    $0x8,%rbx
  mov    -0x8(%rbx),%rdi
  test   %rdi,%rdi
  jne    L7
L10:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L25:
  ret
```

assistant: ```c
void destroy_tokens(char **tokens) {
  if (tokens) {
    for (int i = 0; tokens[i]; i++)
      free(tokens[i]);
    free(tokens);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<destroy_tokens>:
  endbr64
  test   %rdi,%rdi
  je     L16
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L15
  lea    0x8(%rbp),%rbx
  xchg   %ax,%ax
L23:
  call   <free@plt>
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  jne    L23
L15:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <free@plt>
  nop
L16:
  ret
```

assistant: ```c
void destroy_tokens(char **tokens) {
  if (tokens) {
    for (int i = 0; tokens[i]; i++)
      free(tokens[i]);
    free(tokens);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<destroy_tokens>:
  endbr64
  test   %rdi,%rdi
  je     L5
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L15
  lea    0x8(%rbp),%rbx
  xchg   %ax,%ax
L24:
  call   <free@plt>
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  jne    L24
L15:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <free@plt>
  nop
L5:
  ret
```

assistant: ```c
void destroy_tokens(char **tokens) {
  if (tokens) {
    for (int i = 0; tokens[i]; i++)
      free(tokens[i]);
    free(tokens);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<parse_image_format>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x18(%rbp),%rdx
  lea    0x6(%rdx),%rbx
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x18(%rbp),%rdx
  lea    0x7(%rdx),%rbx
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  mov    -0x28(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x18(%rbp),%rdx
  lea    0x8(%rdx),%rbx
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  mov    -0x30(%rbp),%rax
  movl   $0x3,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type_guessed: string

tool: D7: image/xxx

assistant: ```c
char *parse_image_format(char *data, int *len) {
  char *mime_type = strdup("image/xxx");
  mime_type[6] = tolower(data[0]);
  mime_type[7] = tolower(data[1]);
  mime_type[8] = tolower(data[2]);
  *len = 3;
  return mime_type;
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_image_format>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  lea    D3(%rip),%rdi
  call   <strdup@plt>
  mov    %rax,%rbx
  movsbl 0x0(%rbp),%edi
  call   <tolower@plt>
  mov    %al,0x6(%rbx)
  movsbl 0x1(%rbp),%edi
  call   <tolower@plt>
  mov    %al,0x7(%rbx)
  movsbl 0x2(%rbp),%edi
  call   <tolower@plt>
  mov    %al,0x8(%rbx)
  movl   $0x3,(%r12)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D3: image/xxx

assistant: ```c
char *parse_image_format(char *data, int *len) {
  char *mime_type = strdup("image/xxx");
  mime_type[6] = tolower(data[0]);
  mime_type[7] = tolower(data[1]);
  mime_type[8] = tolower(data[2]);
  *len = 3;
  return mime_type;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_image_format>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    D8(%rip),%rdi
  call   <strdup@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r12
  call   <tolower@plt>
  movsbl 0x1(%rbx),%edi
  mov    %al,0x6(%r12)
  call   <tolower@plt>
  movsbl 0x2(%rbx),%edi
  mov    %al,0x7(%r12)
  call   <tolower@plt>
  pop    %rbx
  movl   $0x3,0x0(%rbp)
  mov    %al,0x8(%r12)
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

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

tool: D8: image/xxx

assistant: ```c
char *parse_image_format(char *data, int *len) {
  char *mime_type = strdup("image/xxx");
  mime_type[6] = tolower(data[0]);
  mime_type[7] = tolower(data[1]);
  mime_type[8] = tolower(data[2]);
  *len = 3;
  return mime_type;
}
```
user: Translate the below assembly code into C programming language:

```asm
<parse_image_format>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    D8(%rip),%rdi
  call   <strdup@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r12
  call   <tolower@plt>
  movsbl 0x1(%rbx),%edi
  mov    %al,0x6(%r12)
  call   <tolower@plt>
  movsbl 0x2(%rbx),%edi
  mov    %al,0x7(%r12)
  call   <tolower@plt>
  pop    %rbx
  movl   $0x3,0x0(%rbp)
  mov    %al,0x8(%r12)
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
type_guessed: string

tool: D8: image/xxx

assistant: ```c
char *parse_image_format(char *data, int *len) {
  char *mime_type = strdup("image/xxx");
  mime_type[6] = tolower(data[0]);
  mime_type[7] = tolower(data[1]);
  mime_type[8] = tolower(data[2]);
  *len = 3;
  return mime_type;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ucase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  lea    -0x5000(%rsp),%r11
L0:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L0
  sub    $0x20,%rsp
  mov    %rdi,-0x5028(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    -0x5028(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%r12
  lea    0x1(%r12),%rdx
  lea    -0x5020(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    $0x0,%ebx
  jmp    L28
L19:
  mov    -0x5028(%rbp),%rax
  add    %rbx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <uch@plt>
  mov    %al,-0x5020(%rbp,%rbx,1)
  add    $0x1,%rbx
L28:
  cmp    %r12,%rbx
  jb     L19
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  add    $0x5020,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
char *ucase(char *str) {
  register unsigned long int i, l;
  char uc[20480];
  l = strlen(str);
  memset(uc, '\0', l + 1);
  for (i = 0; i < l; i++)
    uc[i] = uch(str[i]);
  return uc;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ucase>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x5000(%rsp),%r11
L20:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L20
  sub    $0x10,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x5008(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rax,%rbp
  lea    0x1(%rax),%rdx
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  test   %rbp,%rbp
  je     L5
  mov    $0x0,%ebx
L13:
  movsbl (%r12,%rbx,1),%edi
  call   <uch@plt>
  mov    %al,(%rsp,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L13
L5:
  mov    0x5008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x5010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ucase(char *str) {
  register unsigned long int i, l;
  char uc[20480];
  l = strlen(str);
  memset(uc, '\0', l + 1);
  for (i = 0; i < l; i++)
    uc[i] = uch(str[i]);
  return uc;
}
```