ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <sum@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    -0xc(%rbp),%edx
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type: string
name: extract_symbolic_data
arguments: label: D29
type: string
name: extract_symbolic_data
arguments: label: D14
type: string

tool: D9: Enter the number: 

tool: D29: %d

tool: D14: Sum of digits in %d is %d


assistant: ```c
int main() {
  int num, result;
  printf("Enter the number: ");
  scanf("%d", &num);
  result = sum(num);
  printf("Sum of digits in %d is %d\n", num, result);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sum@plt>
  mov    %eax,%edx
  mov    0x4(%rsp),%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D29
type_infered: string
name: disassemble_data
arguments: data_label: D9
type_infered: string
name: disassemble_data
arguments: data_label: D26
type_infered: string

tool: D29: Enter the number: 

tool: D9: %d

tool: D26: Sum of digits in %d is %d


assistant: ```c
int main() {
  int num, result;
  printf("Enter the number: ");
  scanf("%d", &num);
  result = sum(num);
  printf("Sum of digits in %d is %d\n", num, result);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sum@plt>
  mov    0x4(%rsp),%esi
  lea    D7(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
type_infered: string
name: read_data
arguments: tag: D13
type_infered: string
name: read_data
arguments: tag: D7
type_infered: string

tool: D15: Enter the number: 

tool: D13: %d

tool: D7: Sum of digits in %d is %d


assistant: ```c
int main() {
  int num, result;
  printf("Enter the number: ");
  scanf("%d", &num);
  result = sum(num);
  printf("Sum of digits in %d is %d\n", num, result);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D24(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sum@plt>
  mov    0x4(%rsp),%esi
  lea    D4(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D24
data_type: string
name: unpack_data
arguments: tag: D13
data_type: string
name: unpack_data
arguments: tag: D4
data_type: string

tool: D24: Enter the number: 

tool: D13: %d

tool: D4: Sum of digits in %d is %d


assistant: ```c
int main() {
  int num, result;
  printf("Enter the number: ");
  scanf("%d", &num);
  result = sum(num);
  printf("Sum of digits in %d is %d\n", num, result);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_open>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D7(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int _open(const char *buf, int flags, int mode) {
  errno = EIO;
  return (-1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<_open>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int _open(const char *buf, int flags, int mode) {
  errno = EIO;
  return (-1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_open>:
  endbr64
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int _open(const char *buf, int flags, int mode) {
  errno = EIO;
  return (-1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_open>:
  endbr64
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int _open(const char *buf, int flags, int mode) {
  errno = EIO;
  return (-1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type: string

tool: D7: This is exec1;argc = %d


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  printf("This is exec1;argc = %d\n", argc);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
type: string

tool: D4: This is exec1;argc = %d


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  printf("This is exec1;argc = %d\n", argc);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
type: string

tool: D1: This is exec1;argc = %d


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  printf("This is exec1;argc = %d\n", argc);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
type: string

tool: D7: This is exec1;argc = %d


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  printf("This is exec1;argc = %d\n", argc);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<combination_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x28(%rbp)
  jmp    L38
L9:
  movl   $0x0,-0x24(%rbp)
  jmp    L52
L13:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jns    L8
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
L8:
  addl   $0x1,-0x24(%rbp)
L52:
  mov    -0x3c(%rbp),%eax
  sub    -0x24(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L13
  addl   $0x1,-0x28(%rbp)
L38:
  mov    -0x28(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L9
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
void combination_sort(char *combinations[512][22], int variation_number) {
  char temp[22];
  int i, j;
  for (i = 0; i < variation_number; i++) {
    for (j = 0; j < variation_number - j; j++) {
      if (strcmp(combinations[i], combinations[i + 1]) < 0) {
        strcpy(temp, combinations[i]);
        strcpy(combinations[i], combinations[i + 1]);
        strcpy(combinations[i + 1], temp);
      }
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<combination_sort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L0
  mov    %esi,%r13d
  mov    %rdi,%r14
  lea    -0x1(%rsi),%eax
  lea    (%rax,%rax,4),%rdx
  lea    (%rax,%rdx,2),%rax
  shl    $0x4,%rax
  lea    0xb0(%rdi,%rax,1),%rax
  mov    %rax,0x8(%rsp)
  lea    0x10(%rsp),%r15
  jmp    L50
L9:
  add    $0x1,%ebx
  mov    %r13d,%eax
  sub    %ebx,%eax
  cmp    %ebx,%eax
  jle    L46
L25:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jns    L9
  mov    %r12,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  jmp    L9
L46:
  cmp    %r14,0x8(%rsp)
  je     L0
L50:
  mov    %r14,%r12
  add    $0xb0,%r14
  mov    %r14,%rbp
  mov    $0x0,%ebx
  jmp    L25
L0:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void combination_sort(char *combinations[512][22], int variation_number) {
  char temp[22];
  int i, j;
  for (i = 0; i < variation_number; i++) {
    for (j = 0; j < variation_number - j; j++) {
      if (strcmp(combinations[i], combinations[i + 1]) < 0) {
        strcpy(temp, combinations[i]);
        strcpy(combinations[i], combinations[i + 1]);
        strcpy(combinations[i + 1], temp);
      }
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<combination_sort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L2
  lea    -0x1(%rsi),%eax
  mov    %esi,%r13d
  mov    %rdi,%rbp
  mov    %rsp,%r14
  lea    (%rax,%rax,4),%rdx
  lea    (%rax,%rdx,2),%rax
  shl    $0x4,%rax
  lea    0xb0(%rdi,%rax,1),%r15
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %rbp,%r12
  xor    %ebx,%ebx
  add    $0xb0,%rbp
  jmp    L21
  xchg   %ax,%ax
L42:
  add    $0x1,%ebx
  mov    %r13d,%eax
  sub    %ebx,%eax
  cmp    %eax,%ebx
  jge    L67
L21:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jns    L42
  mov    %r12,%rsi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <strcpy@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r13d,%eax
  sub    %ebx,%eax
  cmp    %eax,%ebx
  jl     L21
L67:
  cmp    %rbp,%r15
  jne    L64
L2:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void combination_sort(char *combinations[512][22], int variation_number) {
  char temp[22];
  int i, j;
  for (i = 0; i < variation_number; i++) {
    for (j = 0; j < variation_number - j; j++) {
      if (strcmp(combinations[i], combinations[i + 1]) < 0) {
        strcpy(temp, combinations[i]);
        strcpy(combinations[i], combinations[i + 1]);
        strcpy(combinations[i + 1], temp);
      }
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<combination_sort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L6
  lea    -0x1(%rsi),%eax
  mov    %esi,%r14d
  mov    %rdi,%rbx
  mov    %rsp,%r13
  lea    (%rax,%rax,4),%rdx
  lea    (%rax,%rdx,2),%rax
  shl    $0x4,%rax
  lea    0xb0(%rdi,%rax,1),%r15
  jmp    L59
  nopl   0x0(%rax)
L40:
  cmp    %rbx,%r15
  je     L6
L59:
  mov    %rbx,%rbp
  xor    %r12d,%r12d
  add    $0xb0,%rbx
L34:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jns    L40
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%r12d
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %r14d,%eax
  sub    %r12d,%eax
  cmp    %eax,%r12d
  jl     L34
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L6:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void combination_sort(char *combinations[512][22], int variation_number) {
  char temp[22];
  int i, j;
  for (i = 0; i < variation_number; i++) {
    for (j = 0; j < variation_number - j; j++) {
      if (strcmp(combinations[i], combinations[i + 1]) < 0) {
        strcpy(temp, combinations[i]);
        strcpy(combinations[i], combinations[i + 1]);
        strcpy(combinations[i + 1], temp);
      }
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x3,-0x1c(%rbp)
  jg     L58
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L81
L55:
  cmpl   $0x10,-0xc(%rbp)
  je     L62
  cmpl   $0x10,-0xc(%rbp)
  jg     L58
  cmpl   $0xe,-0xc(%rbp)
  je     L62
  cmpl   $0xe,-0xc(%rbp)
  jg     L58
  cmpl   $0x1,-0xc(%rbp)
  je     L62
  cmpl   $0xb,-0xc(%rbp)
  je     L62
  jmp    L58
L62:
  mov    -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <bar@plt>
  movq   %xmm0,%rax
  movsd  D1(%rip),%xmm0
  movq   %rax,%xmm1
  divsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L58:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L55
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L81:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D25
type_guessed: string
name: disassemble_data
arguments: tag: D1
type_guessed: f64
name: disassemble_data
arguments: tag: D6
type_guessed: string

tool: D25: foo

tool: D1: 1000

tool: D6: %1.2fW

assistant: ```c
void foo(const char *pptr, int caplen) {
  int type;
  const char *tptr;
  if (caplen < 4) {
    (void)printf("foo");
    return;
  }
  while (tptr < pptr) {
    switch (type) {
    case 0x01:
      printf("");
    case 0x0b:
      printf("");
    case 0x0e:
      printf("");
    case 0x10:
      printf("%1.2fW", bar(tptr, caplen) / 1000.0);
    }
  }
  printf("foo");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<foo>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    $0x0,%ebp
  cmp    $0x3,%esi
  jle    L24
L1:
  test   %rbx,%rbx
  je     L41
  test   %rbp,%rbp
  je     L1
  mov    %r12d,%esi
  mov    $0x0,%edi
  call   <bar@plt>
  divsd  D16(%rip),%xmm0
  lea    D18(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L1
L24:
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L4
L41:
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D16
type: f64
name: disassemble_data
arguments: data_label: D18
type: string
name: disassemble_data
arguments: data_label: D25
type: string

tool: D16: 1000

tool: D18: %1.2fW

tool: D25: foo

assistant: ```c
void foo(const char *pptr, int caplen) {
  int type;
  const char *tptr;
  if (caplen < 4) {
    (void)printf("foo");
    return;
  }
  while (tptr < pptr) {
    switch (type) {
    case 0x01:
      printf("");
    case 0x0b:
      printf("");
    case 0x0e:
      printf("");
    case 0x10:
      printf("%1.2fW", bar(tptr, caplen) / 1000.0);
    }
  }
  printf("foo");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  cmp    $0x3,%esi
  jle    L23
  nopl   (%rax)
L39:
  test   %rbx,%rbx
  je     L23
L32:
  test   %bpl,%bpl
  je     L39
  xor    %edi,%edi
  mov    %r12d,%esi
  call   <bar@plt>
  lea    D22(%rip),%rdi
  mov    $0x1,%eax
  divsd  D1(%rip),%xmm0
  call   <printf@plt>
  test   %rbx,%rbx
  jne    L32
L23:
  pop    %rbx
  lea    D14(%rip),%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D22
data_type: string
name: parse_data
arguments: data_label: D1
data_type: f64
name: parse_data
arguments: data_label: D14
data_type: string

tool: D22: %1.2fW

tool: D1: 1000

tool: D14: foo

assistant: ```c
void foo(const char *pptr, int caplen) {
  int type;
  const char *tptr;
  if (caplen < 4) {
    (void)printf("foo");
    return;
  }
  while (tptr < pptr) {
    switch (type) {
    case 0x01:
      printf("");
    case 0x0b:
      printf("");
    case 0x0e:
      printf("");
    case 0x10:
      printf("%1.2fW", bar(tptr, caplen) / 1000.0);
    }
  }
  printf("foo");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%esi
  jle    L15
  mov    %esi,%ebp
  lea    D18(%rip),%r12
  xor    %ebx,%ebx
  test   %rdi,%rdi
  je     L15
  nopl   (%rax)
L29:
  test   %bl,%bl
  je     L29
  xor    %edi,%edi
  mov    %ebp,%esi
  call   <bar@plt>
  mov    %r12,%rdi
  mov    $0x1,%eax
  divsd  D3(%rip),%xmm0
  call   <printf@plt>
  jmp    L29
  nopl   0x0(%rax)
L15:
  pop    %rbx
  lea    D24(%rip),%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D18
type: string
name: extract_symbolic_data
arguments: data_tag: D3
type: f64
name: extract_symbolic_data
arguments: data_tag: D24
type: string

tool: D18: %1.2fW

tool: D3: 1000

tool: D24: foo

assistant: ```c
void foo(const char *pptr, int caplen) {
  int type;
  const char *tptr;
  if (caplen < 4) {
    (void)printf("foo");
    return;
  }
  while (tptr < pptr) {
    switch (type) {
    case 0x01:
      printf("");
    case 0x0b:
      printf("");
    case 0x0e:
      printf("");
    case 0x10:
      printf("%1.2fW", bar(tptr, caplen) / 1000.0);
    }
  }
  printf("foo");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<linenoiseUtf8ReadCode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L180
  mov    $0xffffffffffffffff,%rax
  jmp    L156
L180:
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L26
  mov    -0x8(%rbp),%rax
  jmp    L156
L26:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  test   %al,%al
  jns    L80
  movzbl -0x9(%rbp),%eax
  and    $0xe0,%eax
  cmp    $0xc0,%eax
  jne    L191
  cmpq   $0x1,-0x28(%rbp)
  ja     L116
  mov    $0xffffffffffffffff,%rax
  jmp    L156
L116:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L80
  mov    -0x8(%rbp),%rax
  jmp    L156
L191:
  movzbl -0x9(%rbp),%eax
  and    $0xf0,%eax
  cmp    $0xe0,%eax
  jne    L96
  cmpq   $0x2,-0x28(%rbp)
  ja     L76
  mov    $0xffffffffffffffff,%rax
  jmp    L156
L76:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L80
  mov    -0x8(%rbp),%rax
  jmp    L156
L96:
  movzbl -0x9(%rbp),%eax
  and    $0xf8,%eax
  cmp    $0xf0,%eax
  jne    L171
  cmpq   $0x2,-0x28(%rbp)
  ja     L21
  mov    $0xffffffffffffffff,%rax
  jmp    L156
L21:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x3,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L80
  mov    -0x8(%rbp),%rax
  jmp    L156
L171:
  mov    $0xffffffffffffffff,%rax
  jmp    L156
L80:
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <utf8BytesToCodePoint@plt>
L156:
  leave
  ret
```

assistant: ```c
size_t linenoiseUtf8ReadCode(int fd, char *buf, size_t buf_len, int *cp) {
  if (buf_len < 1)
    return -1;
  size_t nread = read(fd, &buf[0], 1);
  if (nread <= 0)
    return nread;
  unsigned char byte = buf[0];
  if ((byte & 0x80) == 0) {
    ;
  } else if ((byte & 0xE0) == 0xC0) {
    if (buf_len < 2)
      return -1;
    nread = read(fd, &buf[1], 1);
    if (nread <= 0)
      return nread;
  } else if ((byte & 0xF0) == 0xE0) {
    if (buf_len < 3)
      return -1;
    nread = read(fd, &buf[1], 2);
    if (nread <= 0)
      return nread;
  } else if ((byte & 0xF8) == 0xF0) {
    if (buf_len < 3)
      return -1;
    nread = read(fd, &buf[1], 3);
    if (nread <= 0)
      return nread;
  } else {
    return -1;
  }
  return utf8BytesToCodePoint(buf, buf_len, cp);
}
```