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
<atan_66s>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  D18(%rip),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x8(%rbp),%xmm1
  movsd  D3(%rip),%xmm0
  addsd  %xmm1,%xmm0
  mulsd  -0x18(%rbp),%xmm0
  movsd  D21(%rip),%xmm1
  addsd  -0x8(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D18
type: f64
name: parse_data
arguments: label: D3
type: f64
name: parse_data
arguments: label: D21
type: f64

tool: D18: 0.4378497304

tool: D3: 1.6867629106

tool: D21: 1.6867633134

assistant: ```c
double atan_66s(double x) {
  static const double c1 = 1.6867629106;
  static const double c2 = 0.4378497304;
  static const double c3 = 1.6867633134;
  double x2;
  x2 = x * x;
  return (x * (c1 + x2 * c2) / (c3 + x2));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atan_66s>:
  endbr64
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  movapd %xmm2,%xmm1
  mulsd  D9(%rip),%xmm1
  addsd  D21(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  addsd  D23(%rip),%xmm2
  divsd  %xmm2,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_guessed: f64
name: extract_symbolic_data
arguments: tag: D21
type_guessed: f64
name: extract_symbolic_data
arguments: tag: D23
type_guessed: f64

tool: D9: 0.4378497304

tool: D21: 1.6867629106

tool: D23: 1.6867633134

assistant: ```c
double atan_66s(double x) {
  static const double c1 = 1.6867629106;
  static const double c2 = 0.4378497304;
  static const double c3 = 1.6867633134;
  double x2;
  x2 = x * x;
  return (x * (c1 + x2 * c2) / (c3 + x2));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atan_66s>:
  endbr64
  movapd %xmm0,%xmm2
  movsd  D18(%rip),%xmm1
  mulsd  %xmm0,%xmm2
  mulsd  %xmm2,%xmm1
  addsd  D0(%rip),%xmm2
  addsd  D6(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D18
type_guessed: f64
name: retrieve_data
arguments: data_label: D0
type_guessed: f64
name: retrieve_data
arguments: data_label: D6
type_guessed: f64

tool: D18: 0.4378497304

tool: D0: 1.6867633134

tool: D6: 1.6867629106

assistant: ```c
double atan_66s(double x) {
  static const double c1 = 1.6867629106;
  static const double c2 = 0.4378497304;
  static const double c3 = 1.6867633134;
  double x2;
  x2 = x * x;
  return (x * (c1 + x2 * c2) / (c3 + x2));
}
```
user: What is the c source code of the assembly code below?

```asm
<atan_66s>:
  endbr64
  movapd %xmm0,%xmm2
  movsd  D14(%rip),%xmm1
  mulsd  %xmm0,%xmm2
  mulsd  %xmm2,%xmm1
  addsd  D24(%rip),%xmm2
  addsd  D3(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D14
datatype: f64
name: parse_data
arguments: label: D24
datatype: f64
name: parse_data
arguments: label: D3
datatype: f64

tool: D14: 0.4378497304

tool: D24: 1.6867633134

tool: D3: 1.6867629106

assistant: ```c
double atan_66s(double x) {
  static const double c1 = 1.6867629106;
  static const double c2 = 0.4378497304;
  static const double c3 = 1.6867633134;
  double x2;
  x2 = x * x;
  return (x * (c1 + x2 * c2) / (c3 + x2));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Saturate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L39
L29:
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L67:
  mov    -0x10(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  cwtl
  mov    %eax,-0x8(%rbp)
  cmpl   $0x7ff,-0x8(%rbp)
  jle    L58
  movl   $0x7ff,-0x8(%rbp)
  jmp    L61
L58:
  cmpl   $0xfffff800,-0x8(%rbp)
  jge    L61
  movl   $0xfffff800,-0x8(%rbp)
L61:
  mov    -0x10(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dx,(%rax)
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x10(%rbp)
L34:
  cmpl   $0x7,-0x4(%rbp)
  jle    L67
L39:
  cmpl   $0x3f,-0x10(%rbp)
  jle    L29
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L69
  mov    -0x18(%rbp),%rax
  add    $0x7e,%rax
  movzwl (%rax),%edx
  mov    -0x18(%rbp),%rax
  add    $0x7e,%rax
  xor    $0x1,%edx
  mov    %dx,(%rax)
  nop
L69:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Saturate(short *Block_Ptr) {
  int i;
  int sum;
  int val;
  int j;
  {
    sum = 0;
    i = 0;
    while (i < 64) {
      j = 0;
      while (j < 8) {
        val = (int)*(Block_Ptr + i);
        if (val > 2047) {
          val = 2047;
        } else if (val < -2048) {
          val = -2048;
        }
        *(Block_Ptr + i) = (short)val;
        sum += val;
        j++;
        i++;
      }
    }
    if ((sum & 1) == 0) {
      *(Block_Ptr + 63) = (short)((int)*(Block_Ptr + 63) ^ 1);
    }
    return;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<Saturate>:
  endbr64
  lea    0x10(%rdi),%r8
  mov    $0x0,%esi
  mov    $0x0,%r9d
L26:
  lea    -0x10(%r8),%rdx
L6:
  movswl (%rdx),%eax
  mov    $0xfffff800,%ecx
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  mov    $0x7ff,%ecx
  cmp    %ecx,%eax
  cmovg  %ecx,%eax
  mov    %ax,(%rdx)
  add    %eax,%esi
  add    $0x2,%rdx
  cmp    %r8,%rdx
  jne    L6
  add    $0x8,%r9d
  add    $0x10,%r8
  cmp    $0x40,%r9d
  jne    L26
  test   $0x1,%sil
  jne    L1
  xorw   $0x1,0x7e(%rdi)
L1:
  ret
```

assistant: ```c
void Saturate(short *Block_Ptr) {
  int i;
  int sum;
  int val;
  int j;
  {
    sum = 0;
    i = 0;
    while (i < 64) {
      j = 0;
      while (j < 8) {
        val = (int)*(Block_Ptr + i);
        if (val > 2047) {
          val = 2047;
        } else if (val < -2048) {
          val = -2048;
        }
        *(Block_Ptr + i) = (short)val;
        sum += val;
        j++;
        i++;
      }
    }
    if ((sum & 1) == 0) {
      *(Block_Ptr + 63) = (short)((int)*(Block_Ptr + 63) ^ 1);
    }
    return;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Saturate>:
  endbr64
  lea    0x10(%rdi),%r8
  xor    %esi,%esi
  xor    %r9d,%r9d
  nopl   (%rax)
L28:
  lea    -0x10(%r8),%rdx
  nopl   0x0(%rax)
L18:
  movswl (%rdx),%eax
  mov    $0xfffff800,%ecx
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  mov    $0x7ff,%ecx
  cmp    %ecx,%eax
  cmovg  %ecx,%eax
  add    $0x2,%rdx
  mov    %ax,-0x2(%rdx)
  add    %eax,%esi
  cmp    %r8,%rdx
  jne    L18
  add    $0x8,%r9d
  lea    0x10(%rdx),%r8
  cmp    $0x40,%r9d
  jne    L28
  and    $0x1,%esi
  jne    L21
  xorw   $0x1,0x7e(%rdi)
L21:
  ret
```

assistant: ```c
void Saturate(short *Block_Ptr) {
  int i;
  int sum;
  int val;
  int j;
  {
    sum = 0;
    i = 0;
    while (i < 64) {
      j = 0;
      while (j < 8) {
        val = (int)*(Block_Ptr + i);
        if (val > 2047) {
          val = 2047;
        } else if (val < -2048) {
          val = -2048;
        }
        *(Block_Ptr + i) = (short)val;
        sum += val;
        j++;
        i++;
      }
    }
    if ((sum & 1) == 0) {
      *(Block_Ptr + 63) = (short)((int)*(Block_Ptr + 63) ^ 1);
    }
    return;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<checkpass>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L41
  cmpq   $0x0,-0x10(%rbp)
  jne    L30
L41:
  mov    $0x0,%eax
  jmp    L47
L30:
  cmpq   $0x0,-0x18(%rbp)
  je     L31
  mov    -0x18(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L31
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <checkpass_pam@plt>
  jmp    L47
L31:
  mov    $0x0,%eax
L47:
  leave
  ret
```

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

tool: D5: pam

assistant: ```c
int checkpass(const char *user, const char *pass, const char *challenge) {
  if (!user || !pass)
    return 0;
  if (challenge) {
    if (strcmp(challenge, "pam") == 0)
      return checkpass_pam(user, pass);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<checkpass>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdx,%rdi
  test   %rbx,%rbx
  sete   %al
  test   %rsi,%rsi
  sete   %dl
  or     %dl,%al
  jne    L33
  mov    %rsi,%rbp
  test   %rdi,%rdi
  je     L33
  lea    D7(%rip),%rsi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L37
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <checkpass_pam@plt>
  jmp    L37
L33:
  mov    $0x0,%eax
L37:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: string

tool: D7: pam

assistant: ```c
int checkpass(const char *user, const char *pass, const char *challenge) {
  if (!user || !pass)
    return 0;
  if (challenge) {
    if (strcmp(challenge, "pam") == 0)
      return checkpass_pam(user, pass);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<checkpass>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rdx,%rdi
  sub    $0x8,%rsp
  test   %rbp,%rbp
  sete   %al
  test   %rsi,%rsi
  sete   %dl
  or     %dl,%al
  jne    L15
  test   %rdi,%rdi
  je     L15
  mov    %rsi,%r12
  lea    D0(%rip),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L22
L15:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L22:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <checkpass_pam@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
datatype: string

tool: D0: pam

assistant: ```c
int checkpass(const char *user, const char *pass, const char *challenge) {
  if (!user || !pass)
    return 0;
  if (challenge) {
    if (strcmp(challenge, "pam") == 0)
      return checkpass_pam(user, pass);
  }
  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
<checkpass>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rdx,%rdi
  sub    $0x8,%rsp
  test   %rbp,%rbp
  sete   %al
  test   %rsi,%rsi
  sete   %dl
  or     %dl,%al
  jne    L3
  test   %rdi,%rdi
  je     L3
  mov    %rsi,%r12
  lea    D6(%rip),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L17
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L17:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <checkpass_pam@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
data_type: string

tool: D6: pam

assistant: ```c
int checkpass(const char *user, const char *pass, const char *challenge) {
  if (!user || !pass)
    return 0;
  if (challenge) {
    if (strcmp(challenge, "pam") == 0)
      return checkpass_pam(user, pass);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L26
L20:
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <rotr@plt>
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jg     L20
  mov    -0x10(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D11
type_guessed: string
name: disassemble_data
arguments: tag: D49
type_guessed: string

tool: D11: %d

tool: D49: 
 %d %d %d

assistant: ```c
void main() {
  int a, b, c, n;
  scanf("%d", &a);
  scanf("%d", &b);
  scanf("%d", &c);
  scanf("%d", &n);
  while (n > 0) {
    rotr(&a, &b, &c);
    n--;
  }
  printf("\n %d %d %d", a, b, c);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D13(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L19
  lea    0x10(%rsp),%rbx
L1:
  lea    0xc(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %rbx,%rdx
  call   <rotr@plt>
  mov    0x14(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x14(%rsp)
  test   %eax,%eax
  jg     L1
L19:
  mov    0x10(%rsp),%ecx
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x20,%rsp
  pop    %rbx
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D13
data_type: string
name: unpack_data
arguments: tag: D12
data_type: string

tool: D13: %d

tool: D12: 
 %d %d %d

assistant: ```c
void main() {
  int a, b, c, n;
  scanf("%d", &a);
  scanf("%d", &b);
  scanf("%d", &c);
  scanf("%d", &n);
  while (n > 0) {
    rotr(&a, &b, &c);
    n--;
  }
  printf("\n %d %d %d", a, b, c);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  lea    D19(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%r12
  lea    0xc(%rsp),%rbp
  mov    %r12,%rsi
  lea    0x10(%rsp),%rbx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L12
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <rotr@plt>
  mov    0x14(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x14(%rsp)
  test   %eax,%eax
  jg     L18
L12:
  mov    0x10(%rsp),%ecx
  mov    0xc(%rsp),%edx
  lea    D14(%rip),%rdi
  mov    0x8(%rsp),%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D19
type: string
name: read_data
arguments: tag: D14
type: string

tool: D19: %d

tool: D14: 
 %d %d %d

assistant: ```c
void main() {
  int a, b, c, n;
  scanf("%d", &a);
  scanf("%d", &b);
  scanf("%d", &c);
  scanf("%d", &n);
  while (n > 0) {
    rotr(&a, &b, &c);
    n--;
  }
  printf("\n %d %d %d", a, b, c);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D19(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%r12
  lea    0xc(%rsp),%rbp
  mov    %r12,%rsi
  lea    0x10(%rsp),%rbx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L6
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <rotr@plt>
  mov    0x14(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x14(%rsp)
  test   %eax,%eax
  jg     L27
L6:
  mov    0x10(%rsp),%ecx
  mov    0xc(%rsp),%edx
  lea    D12(%rip),%rdi
  mov    0x8(%rsp),%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
type: string
name: parse_data
arguments: data_tag: D12
type: string

tool: D19: %d

tool: D12: 
 %d %d %d

assistant: ```c
void main() {
  int a, b, c, n;
  scanf("%d", &a);
  scanf("%d", &b);
  scanf("%d", &c);
  scanf("%d", &n);
  while (n > 0) {
    rotr(&a, &b, &c);
    n--;
  }
  printf("\n %d %d %d", a, b, c);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D37(%rip),%xmm0
  movsd  %xmm0,-0x50(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x48(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x40(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  D49(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  D46(%rip),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  D1(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D17(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  lea    -0x50(%rbp),%rax
  mov    %rax,-0x60(%rbp)
  lea    -0x50(%rbp),%rax
  add    $0x40,%rax
  mov    %rax,-0x58(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x68(%rbp)
  jmp    L26
L19:
  mov    -0x60(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  -0x68(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x68(%rbp)
  addq   $0x8,-0x60(%rbp)
L26:
  mov    -0x60(%rbp),%rax
  cmp    -0x58(%rbp),%rax
  jb     L19
  movsd  -0x68(%rbp),%xmm0
  ucomisd D32(%rip),%xmm0
  setnp  %al
  mov    $0x0,%edx
  movsd  -0x68(%rbp),%xmm0
  ucomisd D32(%rip),%xmm0
  cmovne %edx,%eax
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D37
type_infered: f64
name: retrieve_data
arguments: data_label: D49
type_infered: f32[2]
name: retrieve_data
arguments: data_label: D46
type_infered: f32[2]
name: retrieve_data
arguments: data_label: D1
type_infered: byte[8]
name: retrieve_data
arguments: data_label: D17
type_infered: byte[8]
name: retrieve_data
arguments: data_label: D32
type_infered: f64

tool: D37: 2393736.541207228

tool: {"D49": [0.0,11.177374839782715]}

tool: {"D46": [0.0,12.11850643157959]}

tool: {"D1": [0x00,0x00,0x00,0x80,0x8a,0x5f,0x4a,0x41]}

tool: {"D17": [0x00,0x00,0x00,0x80,0xd4,0x6c,0x51,0x41]}

tool: D32: 12345

assistant: ```c
int main(int argc, char *argv[]) {
  const double numbers[8] = {2.39373654120722785592079162598E6, 0, 0, 0, 1234567, 2345678, 3456789, 4567890};
  double result;
  const double *ptr = numbers, *end = numbers + 8;
  int i;
  result = 0.0;
  while (ptr < end) {
    result += *ptr;
    ptr++;
  }
  return (result == 12345.0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    D13(%rip),%rax
  mov    %rax,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  mov    D20(%rip),%rax
  mov    %rax,0x20(%rsp)
  mov    D0(%rip),%rax
  mov    %rax,0x28(%rsp)
  mov    D12(%rip),%rax
  mov    %rax,0x30(%rsp)
  mov    D1(%rip),%rax
  mov    %rax,0x38(%rsp)
  mov    %rsp,%rax
  pxor   %xmm0,%xmm0
  lea    0x40(%rsp),%rdx
L3:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L3
  ucomisd D23(%rip),%xmm0
  setnp  %al
  movzbl %al,%eax
  mov    $0x0,%edx
  cmovne %edx,%eax
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x58,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
data_type: f64
name: parse_data
arguments: data_label: D20
data_type: f32[2]
name: parse_data
arguments: data_label: D0
data_type: f32[2]
name: parse_data
arguments: data_label: D12
data_type: byte[8]
name: parse_data
arguments: data_label: D1
data_type: byte[8]
name: parse_data
arguments: data_label: D23
data_type: f64

tool: D13: 2393736.541207228

tool: {"D20": [0.0,11.177374839782715]}

tool: {"D0": [0.0,12.11850643157959]}

tool: {"D12": [0x00,0x00,0x00,0x80,0x8a,0x5f,0x4a,0x41]}

tool: {"D1": [0x00,0x00,0x00,0x80,0xd4,0x6c,0x51,0x41]}

tool: D23: 12345

assistant: ```c
int main(int argc, char *argv[]) {
  const double numbers[8] = {2.39373654120722785592079162598E6, 0, 0, 0, 1234567, 2345678, 3456789, 4567890};
  double result;
  const double *ptr = numbers, *end = numbers + 8;
  int i;
  result = 0.0;
  while (ptr < end) {
    result += *ptr;
    ptr++;
  }
  return (result == 12345.0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  movsd  D41(%rip),%xmm1
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    D31(%rip),%rax
  lea    0x40(%rsp),%rdx
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %rax,0x20(%rsp)
  mov    D12(%rip),%rax
  movq   $0x0,0x18(%rsp)
  mov    %rax,0x28(%rsp)
  mov    D19(%rip),%rax
  movsd  %xmm1,(%rsp)
  mov    %rax,0x30(%rsp)
  mov    D49(%rip),%rax
  mov    %rax,0x38(%rsp)
  mov    %rsp,%rax
  jmp    L28
  xchg   %ax,%ax
L23:
  movsd  (%rax),%xmm1
L28:
  add    $0x8,%rax
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L23
  xor    %eax,%eax
  ucomisd D36(%rip),%xmm0
  mov    $0x0,%edx
  setnp  %al
  cmovne %edx,%eax
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L14
  add    $0x58,%rsp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D41
type_guessed: f64
name: disassemble_data
arguments: tag: D31
type_guessed: f32[2]
name: disassemble_data
arguments: tag: D12
type_guessed: f32[2]
name: disassemble_data
arguments: tag: D19
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D49
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D36
type_guessed: f64

tool: D41: 2393736.541207228

tool: {"D31": [0.0,11.177374839782715]}

tool: {"D12": [0.0,12.11850643157959]}

tool: {"D19": [0x00,0x00,0x00,0x80,0x8a,0x5f,0x4a,0x41]}

tool: {"D49": [0x00,0x00,0x00,0x80,0xd4,0x6c,0x51,0x41]}

tool: D36: 12345

assistant: ```c
int main(int argc, char *argv[]) {
  const double numbers[8] = {2.39373654120722785592079162598E6, 0, 0, 0, 1234567, 2345678, 3456789, 4567890};
  double result;
  const double *ptr = numbers, *end = numbers + 8;
  int i;
  result = 0.0;
  while (ptr < end) {
    result += *ptr;
    ptr++;
  }
  return (result == 12345.0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  const double numbers[8] = {2.39373654120722785592079162598E6, 0, 0, 0, 1234567, 2345678, 3456789, 4567890};
  double result;
  const double *ptr = numbers, *end = numbers + 8;
  int i;
  result = 0.0;
  while (ptr < end) {
    result += *ptr;
    ptr++;
  }
  return (result == 12345.0);
}
```