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
<product>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <unpack@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <unpack@plt>
  movzbl 0x4(%rsp),%esi
  movzbl (%rsp),%edi
  call   <bmul@plt>
  mov    %eax,%ebx
  movzbl 0x5(%rsp),%esi
  movzbl 0x1(%rsp),%edi
  call   <bmul@plt>
  xor    %eax,%ebx
  movzbl 0x6(%rsp),%esi
  movzbl 0x2(%rsp),%edi
  call   <bmul@plt>
  xor    %eax,%ebx
  movzbl 0x7(%rsp),%esi
  movzbl 0x3(%rsp),%edi
  call   <bmul@plt>
  xor    %ebx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char product(unsigned long x, unsigned long y) {
  unsigned char xb[4], yb[4];
  unpack(x, xb);
  unpack(y, yb);
  return bmul(xb[0], yb[0]) ^ bmul(xb[1], yb[1]) ^ bmul(xb[2], yb[2]) ^ bmul(xb[3], yb[3]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<product>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <unpack@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <unpack@plt>
  movzbl 0x4(%rsp),%esi
  movzbl (%rsp),%edi
  call   <bmul@plt>
  movzbl 0x5(%rsp),%esi
  movzbl 0x1(%rsp),%edi
  mov    %eax,%ebx
  call   <bmul@plt>
  movzbl 0x6(%rsp),%esi
  movzbl 0x2(%rsp),%edi
  xor    %eax,%ebx
  call   <bmul@plt>
  movzbl 0x7(%rsp),%esi
  movzbl 0x3(%rsp),%edi
  xor    %eax,%ebx
  call   <bmul@plt>
  xor    %ebx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char product(unsigned long x, unsigned long y) {
  unsigned char xb[4], yb[4];
  unpack(x, xb);
  unpack(y, yb);
  return bmul(xb[0], yb[0]) ^ bmul(xb[1], yb[1]) ^ bmul(xb[2], yb[2]) ^ bmul(xb[3], yb[3]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<product>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <unpack@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <unpack@plt>
  movzbl 0x4(%rsp),%esi
  movzbl (%rsp),%edi
  call   <bmul@plt>
  movzbl 0x5(%rsp),%esi
  movzbl 0x1(%rsp),%edi
  mov    %eax,%ebx
  call   <bmul@plt>
  movzbl 0x6(%rsp),%esi
  movzbl 0x2(%rsp),%edi
  xor    %eax,%ebx
  call   <bmul@plt>
  movzbl 0x7(%rsp),%esi
  movzbl 0x3(%rsp),%edi
  xor    %eax,%ebx
  call   <bmul@plt>
  xor    %ebx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char product(unsigned long x, unsigned long y) {
  unsigned char xb[4], yb[4];
  unpack(x, xb);
  unpack(y, yb);
  return bmul(xb[0], yb[0]) ^ bmul(xb[1], yb[1]) ^ bmul(xb[2], yb[2]) ^ bmul(xb[3], yb[3]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movss  D21(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0xa,-0xc(%rbp)
  jmp    L15
L19:
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0xc(%rbp),%xmm0
  movss  -0x8(%rbp),%xmm1
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0x4(%rbp),%xmm2
  movq   %xmm2,%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0xa,-0xc(%rbp)
L15:
  cmpl   $0x63,-0xc(%rbp)
  jle    L19
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D21
type_infered: f32
name: read_data
arguments: tag: D1
type_infered: string
name: read_data
arguments: tag: D0
type_infered: string
name: read_data
arguments: tag: D11
type_infered: string

tool: D21: 13.36

tool: D1: Dolares    Pesos

tool: D0: ****************

tool: D11: %d             %f


assistant: ```c
int main() {
  float x = 13.36;
  printf("Dolares    Pesos\n");
  printf("****************\n");
  for (int i = 10; i < 100;) {
    float pesos = x * i;
    printf("%d             %f\n", i, pesos);
    i += 10;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D33(%rip),%rdi
  call   <puts@plt>
  lea    D27(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%ebx
  lea    D31(%rip),%rbp
L7:
  pxor   %xmm0,%xmm0
  cvtsi2ss %ebx,%xmm0
  mulss  D18(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0xa,%ebx
  cmp    $0x64,%ebx
  jne    L7
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D33
datatype: string
name: parse_data
arguments: data_label: D27
datatype: string
name: parse_data
arguments: data_label: D31
datatype: string
name: parse_data
arguments: data_label: D18
datatype: f32

tool: D33: Dolares    Pesos

tool: D27: ****************

tool: D31: %d             %f


tool: D18: 13.36

assistant: ```c
int main() {
  float x = 13.36;
  printf("Dolares    Pesos\n");
  printf("****************\n");
  for (int i = 10; i < 100;) {
    float pesos = x * i;
    printf("%d             %f\n", i, pesos);
    i += 10;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D30(%rip),%rdi
  lea    D32(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  xchg   %ax,%ax
L4:
  pxor   %xmm0,%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  cvtsi2ss %ebx,%xmm0
  mulss  D14(%rip),%xmm0
  add    $0xa,%ebx
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    $0x64,%ebx
  jne    L4
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D30
type_guessed: string
name: disassemble_data
arguments: data_tag: D32
type_guessed: string
name: disassemble_data
arguments: data_tag: D1
type_guessed: string
name: disassemble_data
arguments: data_tag: D14
type_guessed: f32

tool: D30: Dolares    Pesos

tool: D32: %d             %f


tool: D1: ****************

tool: D14: 13.36

assistant: ```c
int main() {
  float x = 13.36;
  printf("Dolares    Pesos\n");
  printf("****************\n");
  for (int i = 10; i < 100;) {
    float pesos = x * i;
    printf("%d             %f\n", i, pesos);
    i += 10;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D39(%rip),%rdi
  lea    D3(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  xchg   %ax,%ax
L9:
  pxor   %xmm0,%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  cvtsi2ss %ebx,%xmm0
  mulss  D11(%rip),%xmm0
  add    $0xa,%ebx
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    $0x64,%ebx
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D39
type: string
name: unpack_data
arguments: data_tag: D3
type: string
name: unpack_data
arguments: data_tag: D13
type: string
name: unpack_data
arguments: data_tag: D11
type: f32

tool: D39: Dolares    Pesos

tool: D3: %d             %f


tool: D13: ****************

tool: D11: 13.36

assistant: ```c
int main() {
  float x = 13.36;
  printf("Dolares    Pesos\n");
  printf("****************\n");
  for (int i = 10; i < 100;) {
    float pesos = x * i;
    printf("%d             %f\n", i, pesos);
    i += 10;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strncasecmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  je     L98
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <__tokencap_dump@plt>
L98:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  je     L33
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <__tokencap_dump@plt>
  jmp    L33
L83:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,-0x2(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,-0x1(%rbp)
  cmpb   $0x0,-0x2(%rbp)
  jne    L50
  mov    $0x0,%eax
  jmp    L21
L50:
  movzbl -0x2(%rbp),%eax
  cmp    -0x1(%rbp),%al
  je     L47
  movzbl -0x2(%rbp),%eax
  cmp    -0x1(%rbp),%al
  jbe    L69
  mov    $0x1,%eax
  jmp    L21
L69:
  mov    $0xffffffff,%eax
  jmp    L21
L47:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L33:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L83
  mov    $0x0,%eax
L21:
  leave
  ret
```

assistant: ```c
int strncasecmp(const char *str1, const char *str2, size_t len) {
  if (__tokencap_is_ro(str1))
    __tokencap_dump(str1, len, 1);
  if (__tokencap_is_ro(str2))
    __tokencap_dump(str2, len, 1);
  while (len--) {
    unsigned char c1 = tolower(*str1), c2 = tolower(*str2);
    if (!c1)
      return 0;
    if (c1 != c2)
      return (c1 > c2) ? 1 : -1;
    str1++;
    str2++;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strncasecmp>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L82
L6:
  mov    %r12,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L12
L99:
  test   %r13,%r13
  je     L11
  add    %rbp,%r13
L1:
  movsbl 0x0(%rbp),%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  movsbl (%r12),%edi
  call   <tolower@plt>
  test   %bl,%bl
  je     L76
  cmp    %al,%bl
  jne    L64
  add    $0x1,%rbp
  add    $0x1,%r12
  cmp    %rbp,%r13
  jne    L1
  mov    $0x0,%eax
  jmp    L41
L82:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <__tokencap_dump@plt>
  jmp    L6
L12:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <__tokencap_dump@plt>
  jmp    L99
L64:
  cmp    %bl,%al
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  jmp    L41
L11:
  mov    $0x0,%eax
  jmp    L41
L76:
  mov    $0x0,%eax
L41:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int strncasecmp(const char *str1, const char *str2, size_t len) {
  if (__tokencap_is_ro(str1))
    __tokencap_dump(str1, len, 1);
  if (__tokencap_is_ro(str2))
    __tokencap_dump(str2, len, 1);
  while (len--) {
    unsigned char c1 = tolower(*str1), c2 = tolower(*str2);
    if (!c1)
      return 0;
    if (c1 != c2)
      return (c1 > c2) ? 1 : -1;
    str1++;
    str2++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strncasecmp>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L41
  mov    %rbp,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L70
L61:
  test   %r12,%r12
  je     L14
  add    %rbp,%r12
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L24:
  cmp    %al,%bl
  jne    L75
  add    $0x1,%rbp
  add    $0x1,%r13
  cmp    %r12,%rbp
  je     L14
L28:
  movsbl 0x0(%r13),%edi
  call   <tolower@plt>
  movsbl 0x0(%rbp),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  test   %bl,%bl
  jne    L24
L14:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %r13,%rdi
  mov    $0x1,%edx
  mov    %r12,%rsi
  call   <__tokencap_dump@plt>
  mov    %rbp,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  je     L61
L70:
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <__tokencap_dump@plt>
  jmp    L61
  nopw   0x0(%rax,%rax,1)
L75:
  cmp    %bl,%al
  sbb    %eax,%eax
  add    $0x8,%rsp
  and    $0x2,%eax
  pop    %rbx
  pop    %rbp
  sub    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int strncasecmp(const char *str1, const char *str2, size_t len) {
  if (__tokencap_is_ro(str1))
    __tokencap_dump(str1, len, 1);
  if (__tokencap_is_ro(str2))
    __tokencap_dump(str2, len, 1);
  while (len--) {
    unsigned char c1 = tolower(*str1), c2 = tolower(*str2);
    if (!c1)
      return 0;
    if (c1 != c2)
      return (c1 > c2) ? 1 : -1;
    str1++;
    str2++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strncasecmp>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L51
  mov    %rbp,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L13
L6:
  test   %r12,%r12
  je     L67
  add    %rbp,%r12
  jmp    L83
  nopl   0x0(%rax,%rax,1)
L1:
  cmp    %al,%bl
  jne    L47
  add    $0x1,%rbp
  add    $0x1,%r13
  cmp    %r12,%rbp
  je     L67
L83:
  movsbl 0x0(%r13),%edi
  call   <tolower@plt>
  movsbl 0x0(%rbp),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  test   %bl,%bl
  jne    L1
L67:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L51:
  mov    %r13,%rdi
  mov    $0x1,%edx
  mov    %r12,%rsi
  call   <__tokencap_dump@plt>
  mov    %rbp,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  je     L6
L13:
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <__tokencap_dump@plt>
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L47:
  cmp    %bl,%al
  sbb    %eax,%eax
  add    $0x8,%rsp
  and    $0x2,%eax
  pop    %rbx
  pop    %rbp
  sub    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int strncasecmp(const char *str1, const char *str2, size_t len) {
  if (__tokencap_is_ro(str1))
    __tokencap_dump(str1, len, 1);
  if (__tokencap_is_ro(str2))
    __tokencap_dump(str2, len, 1);
  while (len--) {
    unsigned char c1 = tolower(*str1), c2 = tolower(*str2);
    if (!c1)
      return 0;
    if (c1 != c2)
      return (c1 > c2) ? 1 : -1;
    str1++;
    str2++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reverseWord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rax,%r9
  mov    %eax,%r8d
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L59:
  cmp    %rdx,%rsp
  je     L84
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L59
L84:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L21
  orq    $0x0,-0x8(%rsp,%rax,1)
L21:
  mov    %rsp,%rsi
  mov    %r9d,%edx
  sub    $0x1,%edx
  js     L53
  movslq %edx,%rdx
  mov    $0x0,%edi
  jmp    L176
L58:
  lea    0x1(%rdx),%eax
  cltq
  movzbl (%rbx,%rax,1),%ecx
  test   $0xdf,%cl
  je     L130
  lea    0x1(%rdi),%eax
  cltq
  movslq %edi,%r10
  mov    %rdx,%rdi
  sub    %r10,%rdi
  add    %rbx,%rdi
L57:
  mov    %cl,-0x1(%rsi,%rax,1)
  mov    %eax,%r10d
  movzbl 0x1(%rdi,%rax,1),%ecx
  add    $0x1,%rax
  test   $0xdf,%cl
  jne    L57
L174:
  lea    0x1(%r10),%edi
  movslq %r10d,%r10
  movb   $0x20,(%rsi,%r10,1)
L38:
  sub    $0x1,%rdx
  test   %edx,%edx
  js     L53
L176:
  cmpb   $0x20,(%rbx,%rdx,1)
  je     L58
  test   %edx,%edx
  jne    L38
  movzbl (%rbx),%edx
  cmp    $0x20,%dl
  je     L109
  lea    0x1(%rdi),%eax
  cltq
  movslq %edi,%rcx
  mov    %rbx,%r10
  sub    %rcx,%r10
L3:
  mov    %dl,-0x1(%rsi,%rax,1)
  mov    %eax,%ecx
  add    $0x1,%rax
  movzbl -0x1(%r10,%rax,1),%edx
  cmp    $0x20,%dl
  jne    L3
L100:
  movslq %ecx,%rcx
  movb   $0x0,(%rsi,%rcx,1)
L53:
  test   %r9d,%r9d
  jle    L25
  lea    -0x1(%r9),%ecx
  mov    $0x0,%eax
L87:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rbx,%rax,1)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdx,%rcx
  jne    L87
L125:
  movslq %r8d,%r8
  movb   $0x0,(%rbx,%r8,1)
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  mov    -0x8(%rbp),%rbx
  leave
  ret
L130:
  mov    %edi,%r10d
  jmp    L174
L109:
  mov    %edi,%ecx
  jmp    L100
L25:
  mov    $0x0,%r8d
  jmp    L125
L62:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverseWord(char *str) {
  char *p = str;
  int length = (int)strlen(str);
  int i, j, k = 0;
  char q[length];
  for (i = length - 1; i >= 0; i--) {
    if (' ' == p[i]) {
      j = i + 1;
      while (p[j] != ' ' && p[j] != '\0') {
        q[k] = p[j];
        j++;
        k++;
      }
      q[k++] = ' ';
    } else if (i == 0) {
      j = i;
      while (p[j] != ' ') {
        q[k] = p[j];
        j++;
        k++;
      }
      q[k] = '\0';
    }
  }
  for (i = 0; i < length; i++) {
    str[i] = q[i];
  }
  str[i] = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<reverseWord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  movslq %eax,%rbx
  mov    %rax,%r9
  lea    0xf(%rbx),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L135
L74:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L74
L135:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L16
L143:
  lea    -0x1(%r9),%edx
  mov    %rsp,%r8
  xor    %edi,%edi
  movslq %edx,%rcx
  test   %edx,%edx
  jns    L91
  jmp    L166
  nopl   0x0(%rax)
L102:
  test   %ecx,%ecx
  je     L21
L93:
  sub    $0x1,%rcx
  test   %ecx,%ecx
  js     L166
L91:
  cmpb   $0x20,(%r12,%rcx,1)
  jne    L102
  lea    0x1(%rcx),%eax
  cltq
  movzbl (%r12,%rax,1),%esi
  test   $0xdf,%sil
  je     L79
  lea    0x1(%rdi),%eax
  mov    %rcx,%r10
  movslq %edi,%rdi
  sub    %rdi,%r10
  cltq
  lea    (%r12,%r10,1),%rdi
  nopl   (%rax)
L178:
  mov    %sil,-0x1(%r8,%rax,1)
  movzbl 0x1(%rdi,%rax,1),%esi
  movslq %eax,%r10
  add    $0x1,%rax
  test   $0xdf,%sil
  jne    L178
  movb   $0x20,(%r8,%r10,1)
  lea    0x1(%r10),%edi
  jmp    L93
  nopw   0x0(%rax,%rax,1)
L21:
  movzbl (%r12),%ecx
  cmp    $0x20,%cl
  je     L63
  lea    0x1(%rdi),%eax
  mov    %r12,%r10
  movslq %edi,%rdi
  cltq
  sub    %rdi,%r10
  nopl   0x0(%rax)
L15:
  mov    %cl,-0x1(%r8,%rax,1)
  movslq %eax,%rsi
  add    $0x1,%rax
  movzbl -0x1(%r10,%rax,1),%ecx
  cmp    $0x20,%cl
  jne    L15
L62:
  movb   $0x0,(%r8,%rsi,1)
L166:
  test   %r9d,%r9d
  jle    L92
  mov    %edx,%edx
  mov    %r12,%rdi
  mov    %r8,%rsi
  add    %rbx,%r12
  add    $0x1,%rdx
  call   <memcpy@plt>
L92:
  movb   $0x0,(%r12)
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L143
  nopl   0x0(%rax,%rax,1)
L79:
  movslq %edi,%r10
  movb   $0x20,(%r8,%r10,1)
  lea    0x1(%r10),%edi
  jmp    L93
  nopl   0x0(%rax)
L63:
  movslq %edi,%rsi
  jmp    L62
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverseWord(char *str) {
  char *p = str;
  int length = (int)strlen(str);
  int i, j, k = 0;
  char q[length];
  for (i = length - 1; i >= 0; i--) {
    if (' ' == p[i]) {
      j = i + 1;
      while (p[j] != ' ' && p[j] != '\0') {
        q[k] = p[j];
        j++;
        k++;
      }
      q[k++] = ' ';
    } else if (i == 0) {
      j = i;
      while (p[j] != ' ') {
        q[k] = p[j];
        j++;
        k++;
      }
      q[k] = '\0';
    }
  }
  for (i = 0; i < length; i++) {
    str[i] = q[i];
  }
  str[i] = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<reverseWord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  movslq %eax,%rbx
  mov    %rax,%r9
  lea    0xf(%rbx),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L34
L66:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L66
L34:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L60
L173:
  lea    -0x1(%r9),%edx
  mov    %rsp,%r8
  xor    %edi,%edi
  movslq %edx,%rcx
  test   %edx,%edx
  jns    L49
  jmp    L123
  nopl   0x0(%rax)
L46:
  test   %ecx,%ecx
  je     L149
L3:
  sub    $0x1,%rcx
  test   %ecx,%ecx
  js     L123
L49:
  cmpb   $0x20,(%r12,%rcx,1)
  jne    L46
  lea    0x1(%rcx),%eax
  cltq
  movzbl (%r12,%rax,1),%esi
  test   $0xdf,%sil
  je     L61
  lea    0x1(%rdi),%eax
  mov    %rcx,%r10
  movslq %edi,%rdi
  sub    %rdi,%r10
  cltq
  lea    (%r12,%r10,1),%rdi
  nopl   (%rax)
L169:
  mov    %sil,-0x1(%r8,%rax,1)
  movzbl 0x1(%rdi,%rax,1),%esi
  movslq %eax,%r10
  add    $0x1,%rax
  test   $0xdf,%sil
  jne    L169
  movb   $0x20,(%r8,%r10,1)
  lea    0x1(%r10),%edi
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L149:
  movzbl (%r12),%ecx
  cmp    $0x20,%cl
  je     L99
  lea    0x1(%rdi),%eax
  mov    %r12,%r10
  movslq %edi,%rdi
  cltq
  sub    %rdi,%r10
  nopl   0x0(%rax)
L128:
  mov    %cl,-0x1(%r8,%rax,1)
  movslq %eax,%rsi
  add    $0x1,%rax
  movzbl -0x1(%r10,%rax,1),%ecx
  cmp    $0x20,%cl
  jne    L128
L171:
  movb   $0x0,(%r8,%rsi,1)
L123:
  test   %r9d,%r9d
  jle    L110
  mov    %edx,%edx
  mov    %r12,%rdi
  mov    %r8,%rsi
  add    %rbx,%r12
  add    $0x1,%rdx
  call   <memcpy@plt>
L110:
  movb   $0x0,(%r12)
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L156
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L60:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L173
  nopl   0x0(%rax,%rax,1)
L61:
  movslq %edi,%r10
  movb   $0x20,(%r8,%r10,1)
  lea    0x1(%r10),%edi
  jmp    L3
  nopl   0x0(%rax)
L99:
  movslq %edi,%rsi
  jmp    L171
L156:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverseWord(char *str) {
  char *p = str;
  int length = (int)strlen(str);
  int i, j, k = 0;
  char q[length];
  for (i = length - 1; i >= 0; i--) {
    if (' ' == p[i]) {
      j = i + 1;
      while (p[j] != ' ' && p[j] != '\0') {
        q[k] = p[j];
        j++;
        k++;
      }
      q[k++] = ' ';
    } else if (i == 0) {
      j = i;
      while (p[j] != ' ') {
        q[k] = p[j];
        j++;
        k++;
      }
      q[k] = '\0';
    }
  }
  for (i = 0; i < length; i++) {
    str[i] = q[i];
  }
  str[i] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rop3_58_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  xor    -0x1c(%rbp),%al
  mov    %al,-0x3(%rbp)
  movzbl -0x20(%rbp),%eax
  or     -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x1c(%rbp),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_58_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = *D ^ S;
  stk2 = T | stk3;
  stk1 = S ^ stk2;
  *D = stk1;
}
```