ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x1000,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  call   <Mem_Init@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r15
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r13
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r14
  setne  %dil
  call   <assert@plt>
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  nopl   0x0(%rax)
L4:
  mov    $0x1f4,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  jne    L4
  mov    %r15,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r14,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r13,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <Mem_Free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0x9c4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  xor    %edi,%edi
  call   <exit@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  void *ptr[7];
  ptr[0] = Mem_Alloc(500);
  assert(ptr[0] != NULL);
  ptr[1] = Mem_Alloc(500);
  assert(ptr[1] != NULL);
  ptr[2] = Mem_Alloc(500);
  assert(ptr[2] != NULL);
  ptr[3] = Mem_Alloc(500);
  assert(ptr[3] != NULL);
  ptr[4] = Mem_Alloc(500);
  assert(ptr[4] != NULL);
  ptr[5] = Mem_Alloc(500);
  assert(ptr[5] != NULL);
  ptr[6] = Mem_Alloc(500);
  assert(ptr[6] != NULL);
  while (Mem_Alloc(500) != NULL)
    ;
  assert(Mem_Free(ptr[1]) == 0);
  assert(Mem_Free(ptr[5]) == 0);
  assert(Mem_Free(ptr[2]) == 0);
  assert(Mem_Free(ptr[4]) == 0);
  assert(Mem_Free(ptr[3]) == 0);
  ptr[2] = Mem_Alloc(2500);
  assert(ptr[2] != NULL);
  exit(0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <addSum@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <subSum@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <divSum@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <multiSum@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D12
type_guessed: string
name: parse_data
arguments: tag: D13
type_guessed: string
name: parse_data
arguments: tag: D29
type_guessed: string
name: parse_data
arguments: tag: D39
type_guessed: string

tool: D12: The addition of 2 and 3 is: %i


tool: D13: The subtraction of 2 and 3 is: %i


tool: D29: The division of 2 and 3 is: %f


tool: D39: The multiplication of 2 and 3 is: %i


assistant: ```c
int main() {
  int resultAdd = addSum(2, 3);
  printf("The addition of 2 and 3 is: %i\n", resultAdd);
  int resultSub = subSum(2, 3);
  printf("The subtraction of 2 and 3 is: %i\n", resultSub);
  double resultDiv = divSum(2, 3);
  printf("The division of 2 and 3 is: %f\n", resultDiv);
  int resultMulti = multiSum(2, 3);
  printf("The multiplication of 2 and 3 is: %i\n", resultMulti);
  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    $0x3,%esi
  mov    $0x2,%edi
  call   <addSum@plt>
  mov    %eax,%esi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <subSum@plt>
  mov    %eax,%esi
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <divSum@plt>
  cvttsd2si %xmm0,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <multiSum@plt>
  mov    %eax,%esi
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D33
data_type: string
name: disassemble_data
arguments: data_label: D31
data_type: string
name: disassemble_data
arguments: data_label: D15
data_type: string
name: disassemble_data
arguments: data_label: D35
data_type: string

tool: D33: The addition of 2 and 3 is: %i


tool: D31: The subtraction of 2 and 3 is: %i


tool: D15: The division of 2 and 3 is: %f


tool: D35: The multiplication of 2 and 3 is: %i


assistant: ```c
int main() {
  int resultAdd = addSum(2, 3);
  printf("The addition of 2 and 3 is: %i\n", resultAdd);
  int resultSub = subSum(2, 3);
  printf("The subtraction of 2 and 3 is: %i\n", resultSub);
  double resultDiv = divSum(2, 3);
  printf("The division of 2 and 3 is: %f\n", resultDiv);
  int resultMulti = multiSum(2, 3);
  printf("The multiplication of 2 and 3 is: %i\n", resultMulti);
  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    $0x3,%esi
  mov    $0x2,%edi
  call   <addSum@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <subSum@plt>
  lea    D32(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <divSum@plt>
  lea    D22(%rip),%rdi
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <multiSum@plt>
  lea    D25(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
datatype: string
name: disassemble_data
arguments: data_tag: D32
datatype: string
name: disassemble_data
arguments: data_tag: D22
datatype: string
name: disassemble_data
arguments: data_tag: D25
datatype: string

tool: D6: The addition of 2 and 3 is: %i


tool: D32: The subtraction of 2 and 3 is: %i


tool: D22: The division of 2 and 3 is: %f


tool: D25: The multiplication of 2 and 3 is: %i


assistant: ```c
int main() {
  int resultAdd = addSum(2, 3);
  printf("The addition of 2 and 3 is: %i\n", resultAdd);
  int resultSub = subSum(2, 3);
  printf("The subtraction of 2 and 3 is: %i\n", resultSub);
  double resultDiv = divSum(2, 3);
  printf("The division of 2 and 3 is: %f\n", resultDiv);
  int resultMulti = multiSum(2, 3);
  printf("The multiplication of 2 and 3 is: %i\n", resultMulti);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <addSum@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <subSum@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <divSum@plt>
  lea    D38(%rip),%rdi
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <multiSum@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string
name: retrieve_data
arguments: tag: D0
type: string
name: retrieve_data
arguments: tag: D38
type: string
name: retrieve_data
arguments: tag: D13
type: string

tool: D6: The addition of 2 and 3 is: %i


tool: D0: The subtraction of 2 and 3 is: %i


tool: D38: The division of 2 and 3 is: %f


tool: D13: The multiplication of 2 and 3 is: %i


assistant: ```c
int main() {
  int resultAdd = addSum(2, 3);
  printf("The addition of 2 and 3 is: %i\n", resultAdd);
  int resultSub = subSum(2, 3);
  printf("The subtraction of 2 and 3 is: %i\n", resultSub);
  double resultDiv = divSum(2, 3);
  printf("The division of 2 and 3 is: %f\n", resultDiv);
  int resultMulti = multiSum(2, 3);
  printf("The multiplication of 2 and 3 is: %i\n", resultMulti);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compare_ints_min>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  setg   %al
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%ecx
  setl   %al
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int compare_ints_min(const void *a, const void *b) {
  const int *da = (const int *)a;
  const int *db = (const int *)b;
  return (*da > *db) - (*da < *db);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<compare_ints_min>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  setg   %al
  movzbl %al,%eax
  setl   %dl
  movzbl %dl,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int compare_ints_min(const void *a, const void *b) {
  const int *da = (const int *)a;
  const int *db = (const int *)b;
  return (*da > *db) - (*da < *db);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<compare_ints_min>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  setg   %al
  setl   %dl
  movzbl %dl,%edx
  movzbl %al,%eax
  sub    %edx,%eax
  ret
```

assistant: ```c
int compare_ints_min(const void *a, const void *b) {
  const int *da = (const int *)a;
  const int *db = (const int *)b;
  return (*da > *db) - (*da < *db);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<compare_ints_min>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  setg   %al
  setl   %dl
  movzbl %dl,%edx
  movzbl %al,%eax
  sub    %edx,%eax
  ret
```

assistant: ```c
int compare_ints_min(const void *a, const void *b) {
  const int *da = (const int *)a;
  const int *db = (const int *)b;
  return (*da > *db) - (*da < *db);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x9,-0x8(%rbp)
  movl   $0x3,-0x4(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <printBits@plt>
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printBits@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printBits@plt>
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  or     -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printBits@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  xor    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printBits@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  not    %eax
  mov    %eax,%edi
  call   <printBits@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D11
data_type: string
name: retrieve_data
arguments: tag: D47
data_type: string
name: retrieve_data
arguments: tag: D45
data_type: string
name: retrieve_data
arguments: tag: D23
data_type: string
name: retrieve_data
arguments: tag: D58
data_type: string
name: retrieve_data
arguments: tag: D50
data_type: string

tool: D11: a:


tool: D47: b:


tool: D45: a&b:


tool: D23: a|b:


tool: D58: a^b:


tool: D50: ~a:


assistant: ```c
void test1() {
  unsigned int a = 9;
  unsigned int b = 3;
  printf("a:\n");
  printBits(a);
  printf("b:\n");
  printBits(b);
  printf("a&b:\n");
  printBits(a & b);
  printf("a|b:\n");
  printBits(a | b);
  printf("a^b:\n");
  printBits(a ^ b);
  printf("~a:\n");
  printBits(~a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test1>:
  endbr64
  sub    $0x8,%rsp
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    $0x9,%edi
  call   <printBits@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <printBits@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <printBits@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    $0xb,%edi
  call   <printBits@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%edi
  call   <printBits@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0xfffffff6,%edi
  call   <printBits@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D21
type_infered: string
name: extract_symbolic_data
arguments: tag: D5
type_infered: string
name: extract_symbolic_data
arguments: tag: D4
type_infered: string
name: extract_symbolic_data
arguments: tag: D39
type_infered: string
name: extract_symbolic_data
arguments: tag: D9
type_infered: string
name: extract_symbolic_data
arguments: tag: D8
type_infered: string

tool: D21: a:


tool: D5: b:


tool: D4: a&b:


tool: D39: a|b:


tool: D9: a^b:


tool: D8: ~a:


assistant: ```c
void test1() {
  unsigned int a = 9;
  unsigned int b = 3;
  printf("a:\n");
  printBits(a);
  printf("b:\n");
  printBits(b);
  printf("a&b:\n");
  printBits(a & b);
  printf("a|b:\n");
  printBits(a | b);
  printf("a^b:\n");
  printBits(a ^ b);
  printf("~a:\n");
  printBits(~a);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test1>:
  endbr64
  sub    $0x8,%rsp
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    $0x9,%edi
  call   <printBits@plt>
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <printBits@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <printBits@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    $0xb,%edi
  call   <printBits@plt>
  lea    D50(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%edi
  call   <printBits@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0xfffffff6,%edi
  add    $0x8,%rsp
  jmp    1060 <printBits@plt>
```

tool_calls: name: read_data
arguments: tag: D39
data_type: string
name: read_data
arguments: tag: D33
data_type: string
name: read_data
arguments: tag: D52
data_type: string
name: read_data
arguments: tag: D32
data_type: string
name: read_data
arguments: tag: D50
data_type: string
name: read_data
arguments: tag: D13
data_type: string

tool: D39: a:


tool: D33: b:


tool: D52: a&b:


tool: D32: a|b:


tool: D50: a^b:


tool: D13: ~a:


assistant: ```c
void test1() {
  unsigned int a = 9;
  unsigned int b = 3;
  printf("a:\n");
  printBits(a);
  printf("b:\n");
  printBits(b);
  printf("a&b:\n");
  printBits(a & b);
  printf("a|b:\n");
  printBits(a | b);
  printf("a^b:\n");
  printBits(a ^ b);
  printf("~a:\n");
  printBits(~a);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<test1>:
  endbr64
  sub    $0x8,%rsp
  lea    D51(%rip),%rdi
  call   <printf@plt>
  mov    $0x9,%edi
  call   <printBits@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <printBits@plt>
  lea    D56(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <printBits@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    $0xb,%edi
  call   <printBits@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%edi
  call   <printBits@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0xfffffff6,%edi
  add    $0x8,%rsp
  jmp    1060 <printBits@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D51
type_infered: string
name: disassemble_data
arguments: data_label: D16
type_infered: string
name: disassemble_data
arguments: data_label: D56
type_infered: string
name: disassemble_data
arguments: data_label: D24
type_infered: string
name: disassemble_data
arguments: data_label: D32
type_infered: string
name: disassemble_data
arguments: data_label: D12
type_infered: string

tool: D51: a:


tool: D16: b:


tool: D56: a&b:


tool: D24: a|b:


tool: D32: a^b:


tool: D12: ~a:


assistant: ```c
void test1() {
  unsigned int a = 9;
  unsigned int b = 3;
  printf("a:\n");
  printBits(a);
  printf("b:\n");
  printBits(b);
  printf("a&b:\n");
  printBits(a & b);
  printf("a|b:\n");
  printBits(a | b);
  printf("a^b:\n");
  printBits(a ^ b);
  printf("~a:\n");
  printBits(~a);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<simpsons>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  fldt   0x20(%rbp)
  fldt   0x10(%rbp)
  fsubrp %st,%st(1)
  fildl  -0x2c(%rbp)
  fdivrp %st,%st(1)
  fstpt  -0x10(%rbp)
  push   0x18(%rbp)
  push   0x10(%rbp)
  mov    -0x28(%rbp),%rax
  call   *%rax
  add    $0x10,%rsp
  fstpt  -0x40(%rbp)
  push   0x28(%rbp)
  push   0x20(%rbp)
  mov    -0x28(%rbp),%rax
  call   *%rax
  add    $0x10,%rsp
  fldt   -0x40(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x20(%rbp)
  jmp    L11
L8:
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x40(%rbp)
  fildl  -0x40(%rbp)
  fldt   -0x10(%rbp)
  fmulp  %st,%st(1)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  mov    -0x28(%rbp),%rax
  call   *%rax
  add    $0x10,%rsp
  fldt   0xe8e(%rip)        # 2000 <_fini+0xe00>
  fmulp  %st,%st(1)
  fstpt  -0x40(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0x30(%rbp)
  fildl  -0x30(%rbp)
  fldt   -0x10(%rbp)
  fmulp  %st,%st(1)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  mov    -0x28(%rbp),%rax
  call   *%rax
  add    $0x10,%rsp
  fadd   %st(0),%st
  fldt   -0x40(%rbp)
  faddp  %st,%st(1)
  fldt   -0x20(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x20(%rbp)
  subl   $0x2,-0x2c(%rbp)
L11:
  cmpl   $0x2,-0x2c(%rbp)
  jne    L8
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x40(%rbp)
  fildl  -0x40(%rbp)
  fldt   -0x10(%rbp)
  fmulp  %st,%st(1)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  mov    -0x28(%rbp),%rax
  call   *%rax
  add    $0x10,%rsp
  fldt   0xe24(%rip)        # 2000 <_fini+0xe00>
  fmulp  %st,%st(1)
  fldt   -0x20(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x20(%rbp)
  fldt   -0x10(%rbp)
  fldt   0xe21(%rip)        # 2010 <_fini+0xe10>
  fdivrp %st,%st(1)
  fldt   -0x20(%rbp)
  fmulp  %st,%st(1)
  fstpt  -0x20(%rbp)
  fldt   -0x20(%rbp)
  leave
  ret
```

assistant: ```c
long double simpsons(long double (*f)(long double), long double a, long double b, int n) {
  long double h = (b - a) / n;
  long double sum = f(a) + f(b);
  while (n - 2) {
    sum += (4 * f((n - 1) * h) + 2 * f((n - 2) * h));
    n = n - 2;
  }
  sum += 4 * f((n - 1) * h);
  return sum *= (h / 3);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<simpsons>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbp
  fldt   0x70(%rsp)
  mov    %esi,%ebx
  fldt   0x80(%rsp)
  fsub   %st(1),%st
  mov    %esi,0x10(%rsp)
  fildl  0x10(%rsp)
  fdivrp %st,%st(1)
  fstpt  0x10(%rsp)
  fstpt  (%rsp)
  call   *%rdi
  fstpt  0x20(%rsp)
  push   0x88(%rsp)
  push   0x88(%rsp)
  call   *%rbp
  fldt   0x30(%rsp)
  faddp  %st,%st(1)
  fstpt  0x30(%rsp)
  add    $0x20,%rsp
  cmp    $0x2,%ebx
  je     L9
L10:
  lea    -0x1(%rbx),%eax
  mov    %eax,0x20(%rsp)
  fildl  0x20(%rsp)
  fldt   (%rsp)
  fmulp  %st,%st(1)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   *%rbp
  fstpt  0x30(%rsp)
  sub    $0x2,%ebx
  mov    %ebx,0x4c(%rsp)
  fildl  0x4c(%rsp)
  fldt   0x10(%rsp)
  fmulp  %st,%st(1)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   *%rbp
  fldt   0x40(%rsp)
  fmuls  0xe6e(%rip)        # 2000 <_fini+0xe2c>
  fxch   %st(1)
  fadd   %st(0),%st
  faddp  %st,%st(1)
  fldt   0x30(%rsp)
  faddp  %st,%st(1)
  fstpt  0x30(%rsp)
  add    $0x20,%rsp
  cmp    $0x2,%ebx
  jne    L10
L9:
  push   0x8(%rsp)
  push   0x8(%rsp)
  call   *%rbp
  fmuls  0xe45(%rip)        # 2000 <_fini+0xe2c>
  fldt   0x20(%rsp)
  faddp  %st,%st(1)
  fldt   0x10(%rsp)
  fdivs  0xe39(%rip)        # 2004 <_fini+0xe30>
  fmulp  %st,%st(1)
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
long double simpsons(long double (*f)(long double), long double a, long double b, int n) {
  long double h = (b - a) / n;
  long double sum = f(a) + f(b);
  while (n - 2) {
    sum += (4 * f((n - 1) * h) + 2 * f((n - 2) * h));
    n = n - 2;
  }
  sum += 4 * f((n - 1) * h);
  return sum *= (h / 3);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<simpsons>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x68,%rsp
  fldt   0x80(%rsp)
  mov    %esi,0x10(%rsp)
  fldt   0x90(%rsp)
  fsub   %st(1),%st
  fildl  0x10(%rsp)
  fdivrp %st,%st(1)
  fstpt  0x10(%rsp)
  fstpt  (%rsp)
  call   *%rdi
  fstpt  0x20(%rsp)
  push   0x98(%rsp)
  push   0x98(%rsp)
  call   *%rbp
  fldt   0x30(%rsp)
  add    $0x20,%rsp
  faddp  %st,%st(1)
  cmp    $0x2,%ebx
  je     L19
  movss  D1(%rip),%xmm0
  movss  %xmm0,0x4c(%rsp)
  nopw   0x0(%rax,%rax,1)
L21:
  lea    -0x1(%rbx),%eax
  fstpt  0x30(%rsp)
  sub    $0x10,%rsp
  sub    $0x2,%ebx
  mov    %eax,0x20(%rsp)
  fildl  0x20(%rsp)
  fldt   0x10(%rsp)
  fmulp  %st,%st(1)
  fstpt  (%rsp)
  call   *%rbp
  sub    $0x10,%rsp
  fstpt  0x40(%rsp)
  mov    %ebx,0x30(%rsp)
  fildl  0x30(%rsp)
  fldt   0x20(%rsp)
  fmulp  %st,%st(1)
  fstpt  (%rsp)
  call   *%rbp
  fldt   0x40(%rsp)
  fmuls  0x6c(%rsp)
  fxch   %st(1)
  fadd   %st(0),%st
  faddp  %st,%st(1)
  fldt   0x50(%rsp)
  add    $0x20,%rsp
  faddp  %st,%st(1)
  cmp    $0x2,%ebx
  jne    L21
L1:
  fstpt  0x10(%rsp)
  push   0x8(%rsp)
  push   0x8(%rsp)
  call   *%rbp
  fmuls  0x5c(%rsp)
  fldt   0x20(%rsp)
  faddp  %st,%st(1)
  fldt   0x10(%rsp)
  fdivs  0xe19(%rip)        # 2004 <_fini+0xdfc>
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  fmulp  %st,%st(1)
  ret
  nopl   0x0(%rax)
L19:
  movss  D1(%rip),%xmm1
  movss  %xmm1,0x4c(%rsp)
  jmp    L1
```

tool_calls: name: retrieve_data
arguments: label: D1
type_guessed: byte[4]

tool: {"D1": [0x00,0x00,0x80,0x40]}

assistant: ```c
long double simpsons(long double (*f)(long double), long double a, long double b, int n) {
  long double h = (b - a) / n;
  long double sum = f(a) + f(b);
  while (n - 2) {
    sum += (4 * f((n - 1) * h) + 2 * f((n - 2) * h));
    n = n - 2;
  }
  sum += 4 * f((n - 1) * h);
  return sum *= (h / 3);
}
```