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

```asm
<printInfo>:
  endbr64
  mov    %rdi,%rsi
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D5
data_type: string

tool: D5: INFO : %s

assistant: ```c
int printInfo(const char *message) {
  int ret = 0;
  ret = printf("INFO : %s", message);
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<printInfo>:
  endbr64
  mov    %rdi,%rsi
  lea    D1(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string

tool: D1: INFO : %s

assistant: ```c
int printInfo(const char *message) {
  int ret = 0;
  ret = printf("INFO : %s", message);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x1,-0x18(%rbp)
  jmp    L57
L102:
  mov    -0x20(%rbp),%rax
  test   %rax,%rax
  jne    L99
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L57
L99:
  mov    -0x20(%rbp),%rax
  test   %rax,%rax
  jns    L26
  mov    -0x20(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  jne    L26
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L57
L26:
  mov    -0x20(%rbp),%rax
  test   %rax,%rax
  jns    L77
  mov    -0x20(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L77
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L57
L77:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L70
L41:
  mov    -0x18(%rbp),%rax
  imul   -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  cmpq   $0x1,-0x10(%rbp)
  jne    L151
  cmpq   $0x270f,-0x18(%rbp)
  jg     L151
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L152
L151:
  movabs $0x17328cc00,%rax
  cmp    %rax,-0x18(%rbp)
  jle    L189
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L152
L189:
  cmpq   $0x1,-0x10(%rbp)
  jne    L167
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D63(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L167:
  subq   $0x1,-0x10(%rbp)
L70:
  cmpq   $0x0,-0x10(%rbp)
  jg     L41
L152:
  movq   $0x1,-0x18(%rbp)
L57:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L102
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D42
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D68
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D63
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D64
type_guessed: string

tool: D42: Underflow!

tool: D68: Overflow!

tool: D63: %lld


tool: D64: %lld

assistant: ```c
int main() {
  long long a, b, c, ans = 1, i;
  while (scanf("%lld", &a) == 1) {
    if (a == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 != 0) {
      printf("Overflow!\n");
      continue;
    }
    for (i = a; i > 0; i--) {
      ans *= i;
      if (i == 1 && ans < 10000) {
        printf("Underflow!\n");
        break;
      } else if (ans > 6227020800) {
        printf("Overflow!\n");
        break;
      } else if (i == 1) {
        printf("%lld\n", ans);
      }
    }
    ans = 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D44(%rip),%rbp
  lea    D12(%rip),%r12
  movabs $0x17328cc00,%rbx
  jmp    L97
L37:
  mov    %r12,%rdi
  call   <puts@plt>
  jmp    L97
L103:
  test   $0x1,%al
  je     L107
  lea    D23(%rip),%rdi
  call   <puts@plt>
  jmp    L97
L107:
  mov    %r12,%rdi
  call   <puts@plt>
  jmp    L97
L47:
  cmp    %rbx,%rsi
  jg     L90
  cmp    $0x1,%rax
  je     L2
L95:
  sub    $0x1,%rax
  je     L97
  imul   %rax,%rsi
  cmp    $0x1,%rax
  jne    L47
  cmp    $0x270f,%rsi
  jg     L47
L20:
  mov    %r12,%rdi
  call   <puts@plt>
  jmp    L97
L90:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  jmp    L97
L2:
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L97:
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L80
  mov    (%rsp),%rax
  test   %rax,%rax
  je     L37
  js     L103
  cmp    $0x1,%rax
  je     L20
  cmp    %rbx,%rax
  jg     L90
  mov    %rax,%rsi
  jmp    L95
L80:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D44
data_type: string
name: unpack_data
arguments: data_tag: D12
data_type: string
name: unpack_data
arguments: data_tag: D23
data_type: string
name: unpack_data
arguments: data_tag: D1
data_type: string

tool: D44: %lld

tool: D12: Underflow!

tool: D23: Overflow!

tool: D1: %lld


assistant: ```c
int main() {
  long long a, b, c, ans = 1, i;
  while (scanf("%lld", &a) == 1) {
    if (a == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 != 0) {
      printf("Overflow!\n");
      continue;
    }
    for (i = a; i > 0; i--) {
      ans *= i;
      if (i == 1 && ans < 10000) {
        printf("Underflow!\n");
        break;
      } else if (ans > 6227020800) {
        printf("Overflow!\n");
        break;
      } else if (i == 1) {
        printf("%lld\n", ans);
      }
    }
    ans = 1;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r14
  lea    D12(%rip),%r14
  push   %r13
  lea    D22(%rip),%r13
  push   %r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  movabs $0x17328cc00,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L106
L75:
  mov    (%rsp),%rax
  test   %rax,%rax
  je     L6
  js     L139
  cmp    $0x1,%rax
  je     L6
  mov    %rax,%rsi
  cmp    %rbx,%rax
  jg     L94
  nopl   (%rax)
L10:
  sub    $0x1,%rax
  imul   %rax,%rsi
  cmp    $0x1,%rax
  jne    L69
  cmp    $0x270f,%rsi
  jle    L6
L69:
  cmp    %rbx,%rsi
  jg     L94
  cmp    $0x1,%rax
  jne    L10
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L75
L106:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L139:
  test   $0x1,%al
  jne    L94
  nopl   0x0(%rax)
L6:
  lea    D29(%rip),%rdi
  call   <puts@plt>
  jmp    L0
  nopl   0x0(%rax)
L94:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L0
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D12
data_type: string
name: read_data
arguments: data_label: D22
data_type: string
name: read_data
arguments: data_label: D8
data_type: string
name: read_data
arguments: data_label: D29
data_type: string

tool: D12: Overflow!

tool: D22: %lld


tool: D8: %lld

tool: D29: Underflow!

assistant: ```c
int main() {
  long long a, b, c, ans = 1, i;
  while (scanf("%lld", &a) == 1) {
    if (a == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 != 0) {
      printf("Overflow!\n");
      continue;
    }
    for (i = a; i > 0; i--) {
      ans *= i;
      if (i == 1 && ans < 10000) {
        printf("Underflow!\n");
        break;
      } else if (ans > 6227020800) {
        printf("Overflow!\n");
        break;
      } else if (i == 1) {
        printf("%lld\n", ans);
      }
    }
    ans = 1;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r14
  lea    D17(%rip),%r14
  push   %r13
  lea    D23(%rip),%r13
  push   %r12
  push   %rbp
  lea    D20(%rip),%rbp
  push   %rbx
  movabs $0x17328cc00,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nopw   0x0(%rax,%rax,1)
L87:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L131
L97:
  mov    (%rsp),%rax
  test   %rax,%rax
  je     L78
  js     L113
  cmp    $0x1,%rax
  je     L78
  mov    %rax,%rsi
  cmp    %rbx,%rax
  jg     L77
  nopl   (%rax)
L102:
  sub    $0x1,%rax
  imul   %rax,%rsi
  cmp    $0x1,%rax
  jne    L134
  cmp    $0x270f,%rsi
  jle    L78
L134:
  cmp    %rbx,%rsi
  jg     L77
  cmp    $0x1,%rax
  jne    L102
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L97
L131:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L113:
  test   $0x1,%al
  jne    L77
  nopl   0x0(%rax)
L78:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  jmp    L87
  nopl   0x0(%rax)
L77:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L87
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D23
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D20
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D16
type_infered: string

tool: D17: Overflow!

tool: D23: %lld


tool: D20: %lld

tool: D16: Underflow!

assistant: ```c
int main() {
  long long a, b, c, ans = 1, i;
  while (scanf("%lld", &a) == 1) {
    if (a == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 == 0) {
      printf("Underflow!\n");
      continue;
    } else if (a < 0 && a % 2 != 0) {
      printf("Overflow!\n");
      continue;
    }
    for (i = a; i > 0; i--) {
      ans *= i;
      if (i == 1 && ans < 10000) {
        printf("Underflow!\n");
        break;
      } else if (ans > 6227020800) {
        printf("Overflow!\n");
        break;
      } else if (i == 1) {
        printf("%lld\n", ans);
      }
    }
    ans = 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bm_utf8_rune_prev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L25
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L25
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L44
L25:
  mov    $0x0,%eax
  jmp    L0
L44:
  nop
L30:
  subq   $0x1,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L49
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xc0,%eax
  cmp    $0x80,%eax
  je     L30
L49:
  mov    -0x20(%rbp),%rax
  sub    -0x10(%rbp),%rax
L0:
  leave
  ret
```

assistant: ```c
size_t bm_utf8_rune_prev(const char *string, size_t start) {
  assert(string);
  size_t len = strlen(string), i = start;
  if (i == 0 || len < start || !*string)
    return 0;
  while (--i > 0 && (string[i] & 0xc0) == 0x80)
    ;
  return start - i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bm_utf8_rune_prev>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rax,%rcx
  lea    -0x1(%rbp),%rax
  mov    $0x0,%edx
  cmp    %rax,%rcx
  jbe    L23
  cmpb   $0x0,(%rbx)
  je     L23
  mov    %rbp,%rax
L29:
  sub    $0x1,%rax
  je     L3
  movsbl (%rbx,%rax,1),%edx
  and    $0xc0,%edx
  cmp    $0x80,%edx
  je     L29
L3:
  mov    %rbp,%rdx
  sub    %rax,%rdx
L23:
  mov    %rdx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
size_t bm_utf8_rune_prev(const char *string, size_t start) {
  assert(string);
  size_t len = strlen(string), i = start;
  if (i == 0 || len < start || !*string)
    return 0;
  while (--i > 0 && (string[i] & 0xc0) == 0x80)
    ;
  return start - i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bm_utf8_rune_prev>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rbp),%rdx
  mov    %rax,%r8
  xor    %eax,%eax
  cmp    %rdx,%r8
  jbe    L29
  cmpb   $0x0,(%rbx)
  je     L29
  mov    %rbp,%rax
  jmp    L53
  nopl   (%rax)
L44:
  movsbl (%rbx,%rax,1),%edx
  and    $0xc0,%edx
  cmp    $0x80,%edx
  jne    L0
L53:
  sub    $0x1,%rax
  jne    L44
  mov    %rbp,%rax
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  sub    %rax,%rbp
  mov    %rbp,%rax
  jmp    L29
```

assistant: ```c
size_t bm_utf8_rune_prev(const char *string, size_t start) {
  assert(string);
  size_t len = strlen(string), i = start;
  if (i == 0 || len < start || !*string)
    return 0;
  while (--i > 0 && (string[i] & 0xc0) == 0x80)
    ;
  return start - i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bm_utf8_rune_prev>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rbp),%rdx
  mov    %rax,%r8
  xor    %eax,%eax
  cmp    %rdx,%r8
  jbe    L6
  cmpb   $0x0,(%rbx)
  je     L6
  mov    %rbp,%rax
  jmp    L27
  nopl   (%rax)
L17:
  movsbl (%rbx,%rax,1),%edx
  and    $0xc0,%edx
  cmp    $0x80,%edx
  jne    L14
L27:
  sub    $0x1,%rax
  jne    L17
  mov    %rbp,%rax
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  sub    %rax,%rbp
  mov    %rbp,%rax
  jmp    L6
```

assistant: ```c
size_t bm_utf8_rune_prev(const char *string, size_t start) {
  assert(string);
  size_t len = strlen(string), i = start;
  if (i == 0 || len < start || !*string)
    return 0;
  while (--i > 0 && (string[i] & 0xc0) == 0x80)
    ;
  return start - i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<quick_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L15
L47:
  addl   $0x1,-0x10(%rbp)
L74:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jg     L47
  jmp    L64
L73:
  subl   $0x1,-0xc(%rbp)
L64:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L73
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L15
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x10(%rbp)
  subl   $0x1,-0xc(%rbp)
L15:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L74
  mov    -0x1c(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jge    L45
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort>
L45:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L18
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort>
L18:
  nop
  leave
  ret
```

assistant: ```c
void quick_sort(int *A, int left, int right) {
  int i = left;
  int j = right;
  int pivot = A[(left + right) / 2];
  while (i <= j) {
    while (A[i] < pivot) {
      i++;
    }
    while (A[j] > pivot) {
      j--;
    }
    if (i <= j) {
      int temp = A[i];
      A[i] = A[j];
      A[j] = temp;
      i++;
      j--;
    }
  }
  if (left < j) {
    quick_sort(A, left, j);
  }
  if (i < right) {
    quick_sort(A, i, right);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<quick_sort>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r10d
  mov    %edx,%r12d
  lea    (%rsi,%rdx,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cltq
  mov    (%rdi,%rax,4),%esi
  mov    %r12d,%edx
  mov    %r10d,%ebx
  jmp    L90
L126:
  mov    %edi,(%r8)
  mov    %ecx,(%r9)
  add    $0x1,%ebx
  sub    $0x1,%edx
  jmp    L90
L53:
  cmp    %r10d,%edx
  jg     L44
L77:
  cmp    %r12d,%ebx
  jl     L91
L27:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L44:
  mov    %r10d,%esi
  mov    %rbp,%rdi
  call   <quick_sort>
  jmp    L77
L91:
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <quick_sort>
  jmp    L27
L0:
  movslq %edx,%rax
  lea    0x0(%rbp,%rax,4),%r9
  mov    (%r9),%edi
  cmp    %edi,%esi
  jge    L126
L23:
  movslq %edx,%rax
  lea    -0x4(%rbp,%rax,4),%rax
L123:
  sub    $0x1,%edx
  mov    %rax,%r9
  sub    $0x4,%rax
  mov    0x4(%rax),%edi
  cmp    %esi,%edi
  jg     L123
L74:
  cmp    %ebx,%edx
  jge    L126
L90:
  cmp    %edx,%ebx
  jg     L53
  movslq %ebx,%rax
  shl    $0x2,%rax
  lea    0x0(%rbp,%rax,1),%r8
  mov    (%r8),%ecx
  lea    0x4(%rbp,%rax,1),%rax
  cmp    %ecx,%esi
  jle    L0
L73:
  add    $0x1,%ebx
  mov    %rax,%r8
  add    $0x4,%rax
  mov    -0x4(%rax),%ecx
  cmp    %esi,%ecx
  jl     L73
  movslq %edx,%rax
  lea    0x0(%rbp,%rax,4),%r9
  mov    (%r9),%edi
  cmp    %edi,%esi
  jl     L23
  jmp    L74
```

assistant: ```c
void quick_sort(int *A, int left, int right) {
  int i = left;
  int j = right;
  int pivot = A[(left + right) / 2];
  while (i <= j) {
    while (A[i] < pivot) {
      i++;
    }
    while (A[j] > pivot) {
      j--;
    }
    if (i <= j) {
      int temp = A[i];
      A[i] = A[j];
      A[j] = temp;
      i++;
      j--;
    }
  }
  if (left < j) {
    quick_sort(A, left, j);
  }
  if (i < right) {
    quick_sort(A, i, right);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quick_sort>:
  endbr64
  push   %r12
  mov    %esi,%r10d
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
L56:
  lea    (%r10,%rbp,1),%edx
  mov    %r10d,%ebx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  mov    %ebp,%edx
  sar    %eax
  cltq
  mov    (%r12,%rax,4),%edi
  nopl   0x0(%rax,%rax,1)
L115:
  cmp    %edx,%ebx
  jg     L45
L107:
  movslq %ebx,%rax
  movslq %edx,%r11
  shl    $0x2,%rax
  lea    (%r12,%r11,4),%r9
  lea    (%r12,%rax,1),%rcx
  mov    (%r9),%r8d
  lea    0x4(%r12,%rax,1),%rax
  mov    (%rcx),%esi
  cmp    %edi,%esi
  jge    L5
  nopl   0x0(%rax)
L31:
  mov    %rax,%rcx
  mov    (%rax),%esi
  lea    0x4(%rax),%rax
  add    $0x1,%ebx
  cmp    %edi,%esi
  jl     L31
  cmp    %r8d,%edi
  jge    L13
L65:
  lea    -0x4(%r12,%r11,4),%rax
  nopw   0x0(%rax,%rax,1)
L81:
  mov    %rax,%r9
  mov    (%rax),%r8d
  sub    $0x4,%rax
  sub    $0x1,%edx
  cmp    %edi,%r8d
  jg     L81
L13:
  cmp    %ebx,%edx
  jl     L115
  mov    %r8d,(%rcx)
  add    $0x1,%ebx
  sub    $0x1,%edx
  mov    %esi,(%r9)
L27:
  cmp    %edx,%ebx
  jle    L107
L45:
  cmp    %r10d,%edx
  jg     L98
L44:
  cmp    %ebp,%ebx
  jge    L121
  mov    %ebx,%r10d
  jmp    L56
  nop
L5:
  cmp    %r8d,%edi
  jl     L65
  mov    %r8d,(%rcx)
  add    $0x1,%ebx
  sub    $0x1,%edx
  mov    %esi,(%r9)
  jmp    L27
L121:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L98:
  mov    %r10d,%esi
  mov    %r12,%rdi
  call   <quick_sort>
  jmp    L44
```

assistant: ```c
void quick_sort(int *A, int left, int right) {
  int i = left;
  int j = right;
  int pivot = A[(left + right) / 2];
  while (i <= j) {
    while (A[i] < pivot) {
      i++;
    }
    while (A[j] > pivot) {
      j--;
    }
    if (i <= j) {
      int temp = A[i];
      A[i] = A[j];
      A[j] = temp;
      i++;
      j--;
    }
  }
  if (left < j) {
    quick_sort(A, left, j);
  }
  if (i < right) {
    quick_sort(A, i, right);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<g>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L17
  mov    $0x1,%eax
  jmp    L18
L17:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <g>
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <g>
L18:
  leave
  ret
```

assistant: ```c
int g(int x) {
  if (x == 0)
    return 1;
  else
    return g(g(x - 1) - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<g>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  jne    L4
  ret
L4:
  sub    $0x8,%rsp
  sub    $0x1,%edi
  call   <g>
  lea    -0x1(%rax),%edi
  call   <g>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int g(int x) {
  if (x == 0)
    return 1;
  else
    return g(g(x - 1) - 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<g>:
  endbr64
  cmp    $0x7,%edi
  jbe    L27
  mov    %edi,%eax
  sub    $0x8,%eax
  je     L27
  sub    $0x8,%rsp
L24:
  lea    -0x1(%rax),%edi
  call   <g>
  sub    $0x1,%eax
  jne    L24
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int g(int x) {
  if (x == 0)
    return 1;
  else
    return g(g(x - 1) - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<g>:
  endbr64
  cmp    $0x7,%edi
  jbe    L5
  mov    %edi,%eax
  sub    $0x8,%eax
  je     L5
  sub    $0x8,%rsp
L0:
  lea    -0x1(%rax),%edi
  call   <g>
  sub    $0x1,%eax
  jne    L0
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int g(int x) {
  if (x == 0)
    return 1;
  else
    return g(g(x - 1) - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<cvector_dot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L8:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L8
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double cvector_dot(const double *a, const double *b, int n) {
  double sum = 0.0;
  for (int i = 0; i < n; i++) {
    sum += a[i] * b[i];
  }
  return sum;
}
```