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
<ft_atoi>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L56
L12:
  add    $0x1,%rbx
L56:
  movsbl (%rbx),%edi
  call   <ft_isspace@plt>
  test   %rax,%rax
  jne    L12
  movzbl (%rbx),%eax
  cmp    $0x2b,%al
  je     L39
  mov    $0x0,%esi
  cmp    $0x2d,%al
  je     L17
L54:
  movzbl (%rbx),%edx
  lea    -0x30(%rdx),%ecx
  mov    $0x0,%eax
  cmp    $0x9,%cl
  ja     L35
L8:
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  movsbl %dl,%edx
  sub    $0x30,%edx
  sub    %edx,%eax
  add    $0x1,%rbx
  movzbl (%rbx),%edx
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L8
L35:
  mov    %eax,%edx
  neg    %edx
  test   %esi,%esi
  cmove  %edx,%eax
  pop    %rbx
  ret
L39:
  add    $0x1,%rbx
  mov    $0x0,%esi
  jmp    L54
L17:
  add    $0x1,%rbx
  mov    $0x1,%esi
  jmp    L54
```

assistant: ```c
int ft_atoi(const char *str) {
  int negative;
  int value;
  negative = 0;
  value = 0;
  while (ft_isspace(*str))
    str++;
  if (*str == '+')
    str++;
  else if (*str == '-') {
    negative = 1;
    str++;
  }
  while ((*str <= '9') && (*str >= '0')) {
    value = (value * 10) - (*str - '0');
    str++;
  }
  if (!negative)
    value = -value;
  return (value);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L31:
  add    $0x1,%rbx
L34:
  movsbl (%rbx),%edi
  call   <ft_isspace@plt>
  test   %rax,%rax
  jne    L31
  movzbl (%rbx),%eax
  cmp    $0x2b,%al
  je     L22
  cmp    $0x2d,%al
  je     L50
  lea    -0x30(%rax),%edx
  xor    %edi,%edi
  cmp    $0x9,%dl
  ja     L84
L77:
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L40:
  sub    $0x30,%eax
  add    $0x1,%rbx
  lea    (%r8,%r8,4),%edx
  movsbl %al,%ecx
  movzbl (%rbx),%eax
  add    %edx,%edx
  mov    %edx,%r8d
  lea    -0x30(%rax),%esi
  sub    %ecx,%r8d
  cmp    $0x9,%sil
  jbe    L40
  sub    %edx,%ecx
  test   %edi,%edi
  pop    %rbx
  cmove  %ecx,%r8d
  mov    %r8d,%eax
  ret
  nop
L50:
  movzbl 0x1(%rbx),%eax
  lea    0x1(%rbx),%rcx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L84
  mov    %rcx,%rbx
  mov    $0x1,%edi
  jmp    L77
  nopw   0x0(%rax,%rax,1)
L22:
  movzbl 0x1(%rbx),%eax
  xor    %edi,%edi
  add    $0x1,%rbx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L77
L84:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ft_atoi(const char *str) {
  int negative;
  int value;
  negative = 0;
  value = 0;
  while (ft_isspace(*str))
    str++;
  if (*str == '+')
    str++;
  else if (*str == '-') {
    negative = 1;
    str++;
  }
  while ((*str <= '9') && (*str >= '0')) {
    value = (value * 10) - (*str - '0');
    str++;
  }
  if (!negative)
    value = -value;
  return (value);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L36:
  add    $0x1,%rbx
L12:
  movsbl (%rbx),%edi
  call   <ft_isspace@plt>
  test   %rax,%rax
  jne    L36
  movzbl (%rbx),%eax
  cmp    $0x2b,%al
  je     L55
  cmp    $0x2d,%al
  je     L49
  lea    -0x30(%rax),%edx
  xor    %edi,%edi
  cmp    $0x9,%dl
  ja     L65
L80:
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L68:
  sub    $0x30,%eax
  add    $0x1,%rbx
  lea    (%r8,%r8,4),%edx
  movsbl %al,%ecx
  movzbl (%rbx),%eax
  add    %edx,%edx
  mov    %edx,%r8d
  lea    -0x30(%rax),%esi
  sub    %ecx,%r8d
  cmp    $0x9,%sil
  jbe    L68
  sub    %edx,%ecx
  test   %edi,%edi
  pop    %rbx
  cmove  %ecx,%r8d
  mov    %r8d,%eax
  ret
  nop
L49:
  movzbl 0x1(%rbx),%eax
  lea    0x1(%rbx),%rcx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L65
  mov    %rcx,%rbx
  mov    $0x1,%edi
  jmp    L80
  nopw   0x0(%rax,%rax,1)
L55:
  movzbl 0x1(%rbx),%eax
  xor    %edi,%edi
  add    $0x1,%rbx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L80
L65:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ft_atoi(const char *str) {
  int negative;
  int value;
  negative = 0;
  value = 0;
  while (ft_isspace(*str))
    str++;
  if (*str == '+')
    str++;
  else if (*str == '-') {
    negative = 1;
    str++;
  }
  while ((*str <= '9') && (*str >= '0')) {
    value = (value * 10) - (*str - '0');
    str++;
  }
  if (!negative)
    value = -value;
  return (value);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L14:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L14
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D12
datatype: string
name: parse_data
arguments: data_label: D5
datatype: string

tool: D12: There were %d arguments provided:


tool: D5: argument #%d	-	%s


assistant: ```c
int main(int arg_count, char *arg_list[]) {
  int i;
  printf("There were %d arguments provided:\n", arg_count);
  for (i = 0; i < arg_count; i++)
    printf("argument #%d\t-\t%s\n", i, arg_list[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %edi,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L19
  mov    %ebp,%ebp
  mov    $0x0,%ebx
  lea    D3(%rip),%r13
L6:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L6
L19:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
datatype: string
name: unpack_data
arguments: data_label: D3
datatype: string

tool: D5: There were %d arguments provided:


tool: D3: argument #%d	-	%s


assistant: ```c
int main(int arg_count, char *arg_list[]) {
  int i;
  printf("There were %d arguments provided:\n", arg_count);
  for (i = 0; i < arg_count; i++)
    printf("argument #%d\t-\t%s\n", i, arg_list[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D16(%rip),%rdi
  push   %rbx
  mov    %ebp,%esi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L1
  xor    %ebx,%ebx
  lea    D3(%rip),%r13
  nopl   0x0(%rax)
L3:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L3
L1:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: label: D16
type: string
name: read_data
arguments: label: D3
type: string

tool: D16: There were %d arguments provided:


tool: D3: argument #%d	-	%s


assistant: ```c
int main(int arg_count, char *arg_list[]) {
  int i;
  printf("There were %d arguments provided:\n", arg_count);
  for (i = 0; i < arg_count; i++)
    printf("argument #%d\t-\t%s\n", i, arg_list[i]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D12(%rip),%rdi
  push   %rbx
  mov    %ebp,%esi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L13
  xor    %ebx,%ebx
  lea    D3(%rip),%r13
  nopl   0x0(%rax)
L9:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L9
L13:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D12
data_type: string
name: extract_symbolic_data
arguments: tag: D3
data_type: string

tool: D12: There were %d arguments provided:


tool: D3: argument #%d	-	%s


assistant: ```c
int main(int arg_count, char *arg_list[]) {
  int i;
  printf("There were %d arguments provided:\n", arg_count);
  for (i = 0; i < arg_count; i++)
    printf("argument #%d\t-\t%s\n", i, arg_list[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<has_diff_genotype>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D18(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L53
L31:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L72
  mov    -0x8(%rbp),%rax
  lea    D17(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L72
  mov    -0x28(%rbp),%rax
  lea    D17(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L28
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L72
L28:
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L72
  movl   $0x1,-0x10(%rbp)
  jmp    L45
L72:
  addl   $0x1,-0xc(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
L53:
  cmpq   $0x0,-0x8(%rbp)
  jne    L31
L45:
  mov    -0x10(%rbp),%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D18
type: string
name: read_data
arguments: label: D17
type: string

tool: D18: 	

tool: D17: N

assistant: ```c
int has_diff_genotype(char *line, int start_colN, char *refBase) {
  int isDiff = 0;
  char *part;
  part = strtok(line, "\t");
  int i = 0;
  while (part != NULL) {
    if (i < start_colN || strcmp(part, "N") == 0) {
      ;
    } else if (strcmp(refBase, "N") == 0) {
      strcpy(refBase, part);
    } else if (strcmp(refBase, part) != 0) {
      isDiff = 1;
      break;
    }
    i++;
    part = strtok(NULL, "\t");
  }
  return isDiff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<has_diff_genotype>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r12d
  mov    %rdx,%r15
  lea    D2(%rip),%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L69
  mov    %rax,%rbx
  mov    $0x0,%ebp
  lea    D22(%rip),%r14
  lea    D2(%rip),%r13
  jmp    L63
L78:
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L64
L57:
  add    $0x1,%ebp
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L88
L63:
  cmp    %ebp,%r12d
  jg     L57
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L57
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L78
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  jmp    L57
L88:
  mov    $0x0,%eax
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L69:
  mov    $0x0,%eax
  jmp    L6
L64:
  mov    $0x1,%eax
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
data_type: string
name: extract_symbolic_data
arguments: label: D22
data_type: string

tool: D2: 	

tool: D22: N

assistant: ```c
int has_diff_genotype(char *line, int start_colN, char *refBase) {
  int isDiff = 0;
  char *part;
  part = strtok(line, "\t");
  int i = 0;
  while (part != NULL) {
    if (i < start_colN || strcmp(part, "N") == 0) {
      ;
    } else if (strcmp(refBase, "N") == 0) {
      strcpy(refBase, part);
    } else if (strcmp(refBase, part) != 0) {
      isDiff = 1;
      break;
    }
    i++;
    part = strtok(NULL, "\t");
  }
  return isDiff;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<has_diff_genotype>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  mov    %esi,%r12d
  mov    %r13,%rsi
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strtok@plt>
  test   %rax,%rax
  je     L36
  mov    %rax,%rbp
  xor    %ebx,%ebx
  lea    D10(%rip),%r14
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %r13,%rsi
  xor    %edi,%edi
  add    $0x1,%ebx
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L36
L19:
  cmp    %ebx,%r12d
  jg     L32
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L32
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <strcmp@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  test   %rax,%rax
  jne    L15
  call   <strcpy@plt>
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L15:
  call   <strcmp@plt>
  test   %rax,%rax
  je     L32
  mov    $0x1,%eax
  jmp    L87
  nopl   0x0(%rax)
L36:
  xor    %eax,%eax
L87:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string

tool: D1: 	

tool: D10: N

assistant: ```c
int has_diff_genotype(char *line, int start_colN, char *refBase) {
  int isDiff = 0;
  char *part;
  part = strtok(line, "\t");
  int i = 0;
  while (part != NULL) {
    if (i < start_colN || strcmp(part, "N") == 0) {
      ;
    } else if (strcmp(refBase, "N") == 0) {
      strcpy(refBase, part);
    } else if (strcmp(refBase, part) != 0) {
      isDiff = 1;
      break;
    }
    i++;
    part = strtok(NULL, "\t");
  }
  return isDiff;
}
```
user: Translate the below assembly code into C programming language:

```asm
<has_diff_genotype>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  mov    %esi,%r12d
  mov    %r13,%rsi
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strtok@plt>
  test   %rax,%rax
  je     L26
  mov    %rax,%rbp
  xor    %ebx,%ebx
  lea    D16(%rip),%r14
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %r13,%rsi
  xor    %edi,%edi
  add    $0x1,%ebx
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L26
L18:
  cmp    %ebx,%r12d
  jg     L27
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L27
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <strcmp@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  test   %rax,%rax
  jne    L67
  call   <strcpy@plt>
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L67:
  call   <strcmp@plt>
  test   %rax,%rax
  je     L27
  mov    $0x1,%eax
  jmp    L0
  nopl   0x0(%rax)
L26:
  xor    %eax,%eax
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: label: D15
type: string
name: disassemble_data
arguments: label: D16
type: string

tool: D15: 	

tool: D16: N

assistant: ```c
int has_diff_genotype(char *line, int start_colN, char *refBase) {
  int isDiff = 0;
  char *part;
  part = strtok(line, "\t");
  int i = 0;
  while (part != NULL) {
    if (i < start_colN || strcmp(part, "N") == 0) {
      ;
    } else if (strcmp(refBase, "N") == 0) {
      strcpy(refBase, part);
    } else if (strcmp(refBase, part) != 0) {
      isDiff = 1;
      break;
    }
    i++;
    part = strtok(NULL, "\t");
  }
  return isDiff;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xor_encode_uint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L11:
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%eax
  mov    $0x0,%edx
  divq   -0x38(%rbp)
  mov    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  xor    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    %rax,-0x28(%rbp)
  ja     L11
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void xor_encode_uint(unsigned int *enc, const unsigned int *dec, size_t len, const unsigned int *key, size_t keylen) {
  unsigned int i = 0;
  while (i < len) {
    enc[i] = dec[i] ^ key[i % keylen];
    i++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xor_encode_uint>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  mov    %rdx,%r9
  test   %rdx,%rdx
  je     L17
  mov    $0x0,%edi
  mov    $0x0,%esi
L9:
  mov    %rsi,%rax
  mov    $0x0,%edx
  div    %r8
  mov    (%rcx,%rdx,4),%eax
  xor    (%r11,%rsi,4),%eax
  mov    %eax,(%r10,%rsi,4)
  add    $0x1,%edi
  mov    %edi,%esi
  cmp    %r9,%rsi
  jb     L9
L17:
  ret
```

assistant: ```c
void xor_encode_uint(unsigned int *enc, const unsigned int *dec, size_t len, const unsigned int *key, size_t keylen) {
  unsigned int i = 0;
  while (i < len) {
    enc[i] = dec[i] ^ key[i % keylen];
    i++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xor_encode_uint>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  mov    %rdx,%r9
  test   %rdx,%rdx
  je     L4
  xor    %edi,%edi
  xor    %esi,%esi
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    %rsi,%rax
  xor    %edx,%edx
  div    %r8
  mov    (%rcx,%rdx,4),%eax
  xor    (%r11,%rsi,4),%eax
  mov    %eax,(%r10,%rsi,4)
  lea    0x1(%rdi),%esi
  mov    %rsi,%rdi
  cmp    %r9,%rsi
  jb     L10
L4:
  ret
```

assistant: ```c
void xor_encode_uint(unsigned int *enc, const unsigned int *dec, size_t len, const unsigned int *key, size_t keylen) {
  unsigned int i = 0;
  while (i < len) {
    enc[i] = dec[i] ^ key[i % keylen];
    i++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xor_encode_uint>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%r11
  mov    %rdx,%r9
  test   %rdx,%rdx
  je     L4
  xor    %edi,%edi
  xor    %esi,%esi
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %rsi,%rax
  xor    %edx,%edx
  div    %r8
  mov    (%rcx,%rdx,4),%eax
  xor    (%r11,%rsi,4),%eax
  mov    %eax,(%r10,%rsi,4)
  lea    0x1(%rdi),%esi
  mov    %rsi,%rdi
  cmp    %r9,%rsi
  jb     L5
L4:
  ret
```

assistant: ```c
void xor_encode_uint(unsigned int *enc, const unsigned int *dec, size_t len, const unsigned int *key, size_t keylen) {
  unsigned int i = 0;
  while (i < len) {
    enc[i] = dec[i] ^ key[i % keylen];
    i++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<reverseWords>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L39
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L7
L62:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L27
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reverse@plt>
  jmp    L7
L27:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L7
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L62
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reverse@plt>
  jmp    L34
L39:
  nop
L34:
  leave
  ret
```

assistant: ```c
void reverseWords(char *str) {
  if (!str) {
    return;
  }
  char *start = str;
  char *temp = str;
  while (*temp) {
    temp++;
    if (*temp == '\0') {
      reverse(start, temp - 1);
    } else if (*temp == ' ') {
      reverse(start, temp - 1);
      start = temp + 1;
    }
  }
  reverse(str, temp - 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<reverseWords>:
  endbr64
  test   %rdi,%rdi
  je     L33
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  cmpb   $0x0,(%rdi)
  je     L41
  mov    %rdi,%rsi
  mov    %rdi,%r12
  jmp    L67
L18:
  mov    %r12,%rdi
  call   <reverse@plt>
L42:
  cmpb   $0x0,(%rbx)
  je     L47
L39:
  mov    %rbx,%rsi
L67:
  lea    0x1(%rsi),%rbx
  movzbl 0x1(%rsi),%eax
  test   %al,%al
  je     L18
  cmp    $0x20,%al
  jne    L39
  mov    %r12,%rdi
  call   <reverse@plt>
  lea    0x1(%rbx),%r12
  jmp    L42
L41:
  mov    %rdi,%rbx
L47:
  lea    -0x1(%rbx),%rsi
  mov    %rbp,%rdi
  call   <reverse@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  ret
```

assistant: ```c
void reverseWords(char *str) {
  if (!str) {
    return;
  }
  char *start = str;
  char *temp = str;
  while (*temp) {
    temp++;
    if (*temp == '\0') {
      reverse(start, temp - 1);
    } else if (*temp == ' ') {
      reverse(start, temp - 1);
      start = temp + 1;
    }
  }
  reverse(str, temp - 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverseWords>:
  endbr64
  test   %rdi,%rdi
  je     L9
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmpb   $0x0,(%rdi)
  je     L54
  mov    %rdi,%rbp
  mov    %rdi,%r13
  jmp    L72
  nopl   0x0(%rax,%rax,1)
L34:
  cmp    $0x20,%al
  je     L17
L69:
  mov    %rbx,%rbp
L72:
  movzbl 0x1(%rbp),%eax
  lea    0x1(%rbp),%rbx
  test   %al,%al
  jne    L34
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <reverse@plt>
L45:
  cmpb   $0x0,(%rbx)
  jne    L69
L6:
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <reverse@plt>
L17:
  mov    %r13,%rdi
  mov    %rbp,%rsi
  lea    0x2(%rbp),%r13
  call   <reverse@plt>
  jmp    L45
  nopl   0x0(%rax)
L9:
  ret
  nopl   0x0(%rax)
L54:
  lea    -0x1(%rdi),%rbp
  jmp    L6
```

assistant: ```c
void reverseWords(char *str) {
  if (!str) {
    return;
  }
  char *start = str;
  char *temp = str;
  while (*temp) {
    temp++;
    if (*temp == '\0') {
      reverse(start, temp - 1);
    } else if (*temp == ' ') {
      reverse(start, temp - 1);
      start = temp + 1;
    }
  }
  reverse(str, temp - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverseWords>:
  endbr64
  test   %rdi,%rdi
  je     L55
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmpb   $0x0,(%rdi)
  je     L46
  mov    %rdi,%rbp
  mov    %rdi,%r13
  jmp    L60
  nopl   0x0(%rax,%rax,1)
L52:
  cmp    $0x20,%al
  je     L77
L59:
  mov    %rbx,%rbp
L60:
  movzbl 0x1(%rbp),%eax
  lea    0x1(%rbp),%rbx
  test   %al,%al
  jne    L52
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <reverse@plt>
L0:
  cmpb   $0x0,(%rbx)
  jne    L59
L64:
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <reverse@plt>
L77:
  mov    %r13,%rdi
  mov    %rbp,%rsi
  lea    0x2(%rbp),%r13
  call   <reverse@plt>
  jmp    L0
  nopl   0x0(%rax)
L55:
  ret
  nopl   0x0(%rax)
L46:
  lea    -0x1(%rdi),%rbp
  jmp    L64
```

assistant: ```c
void reverseWords(char *str) {
  if (!str) {
    return;
  }
  char *start = str;
  char *temp = str;
  while (*temp) {
    temp++;
    if (*temp == '\0') {
      reverse(start, temp - 1);
    } else if (*temp == ' ') {
      reverse(start, temp - 1);
      start = temp + 1;
    }
  }
  reverse(str, temp - 1);
}
```