ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D17
type: string
name: read_data
arguments: tag: D33
type: string
name: read_data
arguments: tag: D54
type: string
name: read_data
arguments: tag: D3
type: string
name: read_data
arguments: tag: D45
type: string
name: read_data
arguments: tag: D86
type: string
name: read_data
arguments: tag: D75
type: string
name: read_data
arguments: tag: D82
type: string
name: read_data
arguments: tag: D23
type: string

tool: D17: Hello world.


tool: D33: Hello susumu.


tool: D54: Added new branch.


tool: D3: Hello universe.


tool: D45: Hello moon.


tool: D86: Hello Mario.


tool: D75: Hello Vince.


tool: D82: Hello Elephant.


tool: D23: Hello Tiger.


assistant: ```c
int main() {
  printf("Hello world.\n");
  printf("Hello susumu.\n");
  printf("Added new branch.\n");
  printf("Hello universe.\n");
  printf("Hello moon.\n");
  printf("Hello Mario.\n");
  printf("Hello Vince.\n");
  printf("Hello Elephant.\n");
  printf("Hello Tiger.\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D82(%rip),%rdi
  call   <printf@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D53(%rip),%rdi
  call   <printf@plt>
  lea    D43(%rip),%rdi
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
datatype: string
name: unpack_data
arguments: tag: D82
datatype: string
name: unpack_data
arguments: tag: D29
datatype: string
name: unpack_data
arguments: tag: D24
datatype: string
name: unpack_data
arguments: tag: D53
datatype: string
name: unpack_data
arguments: tag: D43
datatype: string
name: unpack_data
arguments: tag: D52
datatype: string
name: unpack_data
arguments: tag: D32
datatype: string
name: unpack_data
arguments: tag: D10
datatype: string

tool: D5: Hello world.


tool: D82: Hello susumu.


tool: D29: Added new branch.


tool: D24: Hello universe.


tool: D53: Hello moon.


tool: D43: Hello Mario.


tool: D52: Hello Vince.


tool: D32: Hello Elephant.


tool: D10: Hello Tiger.


assistant: ```c
int main() {
  printf("Hello world.\n");
  printf("Hello susumu.\n");
  printf("Added new branch.\n");
  printf("Hello universe.\n");
  printf("Hello moon.\n");
  printf("Hello Mario.\n");
  printf("Hello Vince.\n");
  printf("Hello Elephant.\n");
  printf("Hello Tiger.\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D44(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D87(%rip),%rdi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D83(%rip),%rdi
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D44
datatype: string
name: retrieve_data
arguments: data_tag: D9
datatype: string
name: retrieve_data
arguments: data_tag: D25
datatype: string
name: retrieve_data
arguments: data_tag: D87
datatype: string
name: retrieve_data
arguments: data_tag: D32
datatype: string
name: retrieve_data
arguments: data_tag: D20
datatype: string
name: retrieve_data
arguments: data_tag: D3
datatype: string
name: retrieve_data
arguments: data_tag: D83
datatype: string
name: retrieve_data
arguments: data_tag: D50
datatype: string

tool: D44: Hello world.


tool: D9: Hello susumu.


tool: D25: Added new branch.


tool: D87: Hello universe.


tool: D32: Hello moon.


tool: D20: Hello Mario.


tool: D3: Hello Vince.


tool: D83: Hello Elephant.


tool: D50: Hello Tiger.


assistant: ```c
int main() {
  printf("Hello world.\n");
  printf("Hello susumu.\n");
  printf("Added new branch.\n");
  printf("Hello universe.\n");
  printf("Hello moon.\n");
  printf("Hello Mario.\n");
  printf("Hello Vince.\n");
  printf("Hello Elephant.\n");
  printf("Hello Tiger.\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D82(%rip),%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    D51(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D15
type_guessed: string
name: disassemble_data
arguments: data_label: D47
type_guessed: string
name: disassemble_data
arguments: data_label: D3
type_guessed: string
name: disassemble_data
arguments: data_label: D82
type_guessed: string
name: disassemble_data
arguments: data_label: D26
type_guessed: string
name: disassemble_data
arguments: data_label: D6
type_guessed: string
name: disassemble_data
arguments: data_label: D51
type_guessed: string
name: disassemble_data
arguments: data_label: D39
type_guessed: string
name: disassemble_data
arguments: data_label: D25
type_guessed: string

tool: D15: Hello world.


tool: D47: Hello susumu.


tool: D3: Added new branch.


tool: D82: Hello universe.


tool: D26: Hello moon.


tool: D6: Hello Mario.


tool: D51: Hello Vince.


tool: D39: Hello Elephant.


tool: D25: Hello Tiger.


assistant: ```c
int main() {
  printf("Hello world.\n");
  printf("Hello susumu.\n");
  printf("Added new branch.\n");
  printf("Hello universe.\n");
  printf("Hello moon.\n");
  printf("Hello Mario.\n");
  printf("Hello Vince.\n");
  printf("Hello Elephant.\n");
  printf("Hello Tiger.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<EMACPowerManagementStatusGet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x2c,%eax
  cltq
  mov    (%rax),%eax
  and    $0x61,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t EMACPowerManagementStatusGet(uint32_t ui32Base) {
  ;
  return ((*((volatile uint32_t *)(ui32Base + 0x0000002C))) & (0x00000040 | 0x00000020 | 0x00000001));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<EMACPowerManagementStatusGet>:
  endbr64
  add    $0x2c,%edi
  movslq %edi,%rdi
  mov    (%rdi),%eax
  and    $0x61,%eax
  ret
```

assistant: ```c
uint32_t EMACPowerManagementStatusGet(uint32_t ui32Base) {
  ;
  return ((*((volatile uint32_t *)(ui32Base + 0x0000002C))) & (0x00000040 | 0x00000020 | 0x00000001));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<EMACPowerManagementStatusGet>:
  endbr64
  add    $0x2c,%edi
  movslq %edi,%rdi
  mov    (%rdi),%eax
  and    $0x61,%eax
  ret
```

assistant: ```c
uint32_t EMACPowerManagementStatusGet(uint32_t ui32Base) {
  ;
  return ((*((volatile uint32_t *)(ui32Base + 0x0000002C))) & (0x00000040 | 0x00000020 | 0x00000001));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<EMACPowerManagementStatusGet>:
  endbr64
  add    $0x2c,%edi
  movslq %edi,%rdi
  mov    (%rdi),%eax
  and    $0x61,%eax
  ret
```

assistant: ```c
uint32_t EMACPowerManagementStatusGet(uint32_t ui32Base) {
  ;
  return ((*((volatile uint32_t *)(ui32Base + 0x0000002C))) & (0x00000040 | 0x00000020 | 0x00000001));
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
L27:
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  D27(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L29
  movsd  -0x10(%rbp),%xmm0
  addsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  jmp    L0
L29:
  movsd  -0x10(%rbp),%xmm0
  subsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
L0:
  movsd  D12(%rip),%xmm0
  comisd -0x8(%rbp),%xmm0
  ja     L35
  addl   $0x1,-0x14(%rbp)
  jmp    L27
L35:
  nop
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D27
type_guessed: f64
name: disassemble_data
arguments: data_tag: D12
type_guessed: byte[8]
name: disassemble_data
arguments: data_tag: D1
type_guessed: string

tool: D27: 1

tool: {"D12": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D1: %.6f


assistant: ```c
int main(void) {
  int i;
  double pi = 0;
  for (i = 0;; i++) {
    double temp = 1.0 / (i * 2 + 1);
    if (i % 2 == 0)
      pi += temp;
    else
      pi -= temp;
    if (temp < 1e-6)
      break;
  }
  printf("%.6f\n", pi);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%eax
  pxor   %xmm1,%xmm1
  mov    $0x0,%edx
  movsd  D27(%rip),%xmm4
  movsd  D22(%rip),%xmm3
  jmp    L6
L26:
  addsd  %xmm0,%xmm1
L24:
  add    $0x2,%eax
  comisd %xmm0,%xmm3
  ja     L4
  add    $0x1,%edx
L6:
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movapd %xmm4,%xmm0
  divsd  %xmm2,%xmm0
  test   $0x1,%dl
  je     L26
  subsd  %xmm0,%xmm1
  jmp    L24
L4:
  movapd %xmm1,%xmm0
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D27
type_guessed: f64
name: parse_data
arguments: data_label: D22
type_guessed: byte[8]
name: parse_data
arguments: data_label: D10
type_guessed: string

tool: D27: 1

tool: {"D22": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D10: %.6f


assistant: ```c
int main(void) {
  int i;
  double pi = 0;
  for (i = 0;; i++) {
    double temp = 1.0 / (i * 2 + 1);
    if (i % 2 == 0)
      pi += temp;
    else
      pi -= temp;
    if (temp < 1e-6)
      break;
  }
  printf("%.6f\n", pi);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%eax
  pxor   %xmm1,%xmm1
  xor    %edx,%edx
  movsd  D29(%rip),%xmm4
  movsd  D11(%rip),%xmm3
  jmp    L37
  nopl   (%rax)
L13:
  add    $0x2,%eax
  comisd %xmm0,%xmm3
  addsd  %xmm0,%xmm1
  ja     L19
L24:
  add    $0x1,%edx
L37:
  pxor   %xmm2,%xmm2
  movapd %xmm4,%xmm0
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  test   $0x1,%dl
  je     L13
  add    $0x2,%eax
  comisd %xmm0,%xmm3
  subsd  %xmm0,%xmm1
  jbe    L24
L19:
  movapd %xmm1,%xmm0
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D29
type_guessed: f64
name: retrieve_data
arguments: tag: D11
type_guessed: byte[8]
name: retrieve_data
arguments: tag: D1
type_guessed: string

tool: D29: 1

tool: {"D11": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D1: %.6f


assistant: ```c
int main(void) {
  int i;
  double pi = 0;
  for (i = 0;; i++) {
    double temp = 1.0 / (i * 2 + 1);
    if (i % 2 == 0)
      pi += temp;
    else
      pi -= temp;
    if (temp < 1e-6)
      break;
  }
  printf("%.6f\n", pi);
  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    $0x1,%eax
  pxor   %xmm1,%xmm1
  xor    %edx,%edx
  movsd  D25(%rip),%xmm4
  movsd  D3(%rip),%xmm3
  jmp    L30
  nopl   (%rax)
L23:
  add    $0x2,%eax
  comisd %xmm0,%xmm3
  addsd  %xmm0,%xmm1
  ja     L17
L9:
  add    $0x1,%edx
L30:
  pxor   %xmm2,%xmm2
  movapd %xmm4,%xmm0
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  test   $0x1,%dl
  je     L23
  add    $0x2,%eax
  comisd %xmm0,%xmm3
  subsd  %xmm0,%xmm1
  jbe    L9
L17:
  movapd %xmm1,%xmm0
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D25
data_type: f64
name: disassemble_data
arguments: data_tag: D3
data_type: byte[8]
name: disassemble_data
arguments: data_tag: D8
data_type: string

tool: D25: 1

tool: {"D3": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D8: %.6f


assistant: ```c
int main(void) {
  int i;
  double pi = 0;
  for (i = 0;; i++) {
    double temp = 1.0 / (i * 2 + 1);
    if (i % 2 == 0)
      pi += temp;
    else
      pi -= temp;
    if (temp < 1e-6)
      break;
  }
  printf("%.6f\n", pi);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reorderLevel>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    %r8d,%eax
  mov    %al,-0x2c(%rbp)
  cmpb   $0x0,-0x2c(%rbp)
  jne    L17
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L85
L17:
  mov    $0x1,%eax
  jmp    L81
L85:
  mov    $0x0,%eax
L81:
  mov    %al,-0x2c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L27:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jg     L65
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L56
  movsbl -0x2c(%rbp),%r8d
  mov    -0x18(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rsi
  mov    -0x20(%rbp),%rcx
  add    %rcx,%rsi
  mov    -0x14(%rbp),%ecx
  lea    0x1(%rcx),%edi
  mov    %eax,%ecx
  call   <reorderLevel>
  sub    $0x1,%eax
  add    %eax,-0x4(%rbp)
L56:
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L27
  jmp    L41
L65:
  nop
L41:
  cmpb   $0x0,-0x2c(%rbp)
  je     L69
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
L69:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int reorderLevel(int level, char *pszText, const int *plevel, int cch, char fReverse) {
  int ich;
  fReverse = fReverse || ((level)&1);
  for (ich = 0; ich < cch; ich++) {
    if (plevel[ich] < level) {
      break;
    } else if (plevel[ich] > level) {
      ich += reorderLevel(level + 1, pszText + ich, plevel + ich, cch - ich, fReverse) - 1;
    }
  }
  if (fReverse) {
    reverse(pszText, ich);
  }
  return ich;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reorderLevel>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %rsi,0x8(%rsp)
  mov    %rdx,%r13
  mov    %ecx,%r12d
  test   %r8b,%r8b
  jne    L37
  mov    %edi,%r14d
  and    $0x1,%r14d
  test   %ecx,%ecx
  jle    L94
L44:
  mov    $0x0,%ebx
  movsbl %r14b,%r15d
  jmp    L52
L4:
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jle    L39
L52:
  movslq %ebx,%rsi
  lea    0x0(%r13,%rsi,4),%rdx
  mov    (%rdx),%eax
  cmp    %ebp,%eax
  jl     L39
  jle    L4
  mov    %r12d,%ecx
  sub    %ebx,%ecx
  add    0x8(%rsp),%rsi
  lea    0x1(%rbp),%edi
  mov    %r15d,%r8d
  call   <reorderLevel>
  lea    -0x1(%rbx,%rax,1),%ebx
  jmp    L4
L94:
  mov    $0x0,%ebx
L39:
  test   %r14b,%r14b
  jne    L35
L76:
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  mov    $0x1,%r14d
  mov    $0x0,%ebx
  test   %ecx,%ecx
  jg     L44
L35:
  mov    %ebx,%esi
  mov    0x8(%rsp),%rdi
  call   <reverse@plt>
  jmp    L76
```

assistant: ```c
int reorderLevel(int level, char *pszText, const int *plevel, int cch, char fReverse) {
  int ich;
  fReverse = fReverse || ((level)&1);
  for (ich = 0; ich < cch; ich++) {
    if (plevel[ich] < level) {
      break;
    } else if (plevel[ich] > level) {
      ich += reorderLevel(level + 1, pszText + ich, plevel + ich, cch - ich, fReverse) - 1;
    }
  }
  if (fReverse) {
    reverse(pszText, ich);
  }
  return ich;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reorderLevel>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x18,%rsp
  test   %r8b,%r8b
  je     L91
  mov    $0x1,%r15d
  xor    %r12d,%r12d
  test   %ecx,%ecx
  jle    L65
L99:
  xor    %r12d,%r12d
  lea    0x1(%rbp),%edi
L76:
  movslq %r12d,%rsi
  lea    0x0(%r13,%rsi,4),%rdx
  cmp    %ebp,(%rdx)
  jl     L96
  jg     L45
L69:
  add    $0x1,%r12d
  cmp    %r12d,%ebx
  jg     L76
L96:
  test   %r15d,%r15d
  jne    L65
L30:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L45:
  mov    %ebx,%ecx
  add    %r14,%rsi
  mov    %r15d,%r8d
  mov    %edi,0xc(%rsp)
  sub    %r12d,%ecx
  call   <reorderLevel>
  mov    0xc(%rsp),%edi
  lea    -0x1(%r12,%rax,1),%r12d
  jmp    L69
  nopw   0x0(%rax,%rax,1)
L91:
  mov    %edi,%r15d
  and    $0x1,%r15d
  test   %ecx,%ecx
  jg     L99
  xor    %r12d,%r12d
  jmp    L96
L65:
  mov    %r12d,%esi
  mov    %r14,%rdi
  call   <reverse@plt>
  jmp    L30
```

assistant: ```c
int reorderLevel(int level, char *pszText, const int *plevel, int cch, char fReverse) {
  int ich;
  fReverse = fReverse || ((level)&1);
  for (ich = 0; ich < cch; ich++) {
    if (plevel[ich] < level) {
      break;
    } else if (plevel[ich] > level) {
      ich += reorderLevel(level + 1, pszText + ich, plevel + ich, cch - ich, fReverse) - 1;
    }
  }
  if (fReverse) {
    reverse(pszText, ich);
  }
  return ich;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<boxesForGauss>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x28(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x24(%rbp),%xmm1
  movsd  D19(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x24(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sdl -0x28(%rbp),%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  D15(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <floor@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L37
  subl   $0x1,-0x14(%rbp)
L37:
  mov    -0x14(%rbp),%eax
  add    $0x2,%eax
  mov    %eax,-0x10(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x24(%rbp),%xmm1
  movsd  D19(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x24(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x28(%rbp),%eax
  imul   -0x14(%rbp),%eax
  imul   -0x14(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  mov    -0x28(%rbp),%eax
  imul   -0x14(%rbp),%eax
  shl    $0x2,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  mov    -0x28(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    $0x0,%eax
  sub    %edx,%eax
  shl    $0x2,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <round@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L43
L17:
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jge    L18
  mov    -0x14(%rbp),%eax
  jmp    L47
L18:
  mov    -0x10(%rbp),%eax
L47:
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,(%rdx)
  addl   $0x1,-0x18(%rbp)
L43:
  mov    -0x18(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L17
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D19
type_guessed: f64
name: unpack_data
arguments: data_tag: D15
type_guessed: byte[8]

tool: D19: 12

tool: {"D15": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void boxesForGauss(int sigma, int *sizes, int n) {
  int i;
  int wl = (int)floor(sqrt(((double)12 * sigma * sigma / n) + 1));
  if (wl % 2 == 0)
    wl--;
  int wu = wl + 2;
  double mIdeal = ((double)12 * sigma * sigma - n * wl * wl - 4 * n * wl - 3 * n) / (-4 * wl - 4);
  int m = (int)round(mIdeal);
  for (i = 0; i < n; i++)
    sizes[i] = (int)(i < m ? wl : wu);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<boxesForGauss>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbp
  mov    %edx,%r13d
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
  movapd %xmm0,%xmm1
  mulsd  D1(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %edx,%xmm1
  movsd  %xmm0,0x8(%rsp)
  divsd  %xmm1,%xmm0
  addsd  D13(%rip),%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <floor@plt>
  mov    %rax,%rbx
  and    $0x1,%eax
  cmp    $0x1,%eax
  sbb    $0x0,%ebx
  mov    %ebx,%eax
  imul   %r13d,%eax
  mov    %eax,%edx
  imul   %ebx,%edx
  pxor   %xmm1,%xmm1
  cvtsi2sd %edx,%xmm1
  movsd  0x8(%rsp),%xmm0
  subsd  %xmm1,%xmm0
  shl    $0x2,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  lea    0x0(%r13,%r13,2),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  mov    %ebx,%eax
  not    %eax
  shl    $0x2,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  call   <round@plt>
  test   %r13d,%r13d
  jle    L7
  lea    0x2(%rbx),%r12d
  mov    %r13d,%esi
  mov    $0x0,%edx
L1:
  cmp    %edx,%eax
  mov    %r12d,%ecx
  cmovg  %ebx,%ecx
  mov    %ecx,0x0(%rbp,%rdx,4)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L1
L7:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: f64
name: read_data
arguments: data_label: D13
type_guessed: byte[8]

tool: D1: 12

tool: {"D13": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void boxesForGauss(int sigma, int *sizes, int n) {
  int i;
  int wl = (int)floor(sqrt(((double)12 * sigma * sigma / n) + 1));
  if (wl % 2 == 0)
    wl--;
  int wu = wl + 2;
  double mIdeal = ((double)12 * sigma * sigma - n * wl * wl - 4 * n * wl - 3 * n) / (-4 * wl - 4);
  int m = (int)round(mIdeal);
  for (i = 0; i < n; i++)
    sizes[i] = (int)(i < m ? wl : wu);
}
```