ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<parse_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L3:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L20
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L3
L20:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int parse_int(const char *text, int *nread) {
  int n = 0;
  const char *s = text;
  while ((int)*s >= (int)'0' && (int)*s <= (int)'9') {
    int d = (int)*s - (int)'0';
    n = 10 * n + d;
    s++;
  }
  *nread = s - text;
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_int>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L6
  mov    %rdi,%rdx
  mov    $0x0,%ecx
L19:
  lea    (%rcx,%rcx,4),%ecx
  movsbl %al,%eax
  lea    -0x30(%rax,%rcx,2),%ecx
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  lea    -0x30(%rax),%r8d
  cmp    $0x9,%r8b
  jbe    L19
L21:
  sub    %rdi,%rdx
  mov    %edx,(%rsi)
  mov    %ecx,%eax
  ret
L6:
  mov    %rdi,%rdx
  mov    $0x0,%ecx
  jmp    L21
```

assistant: ```c
int parse_int(const char *text, int *nread) {
  int n = 0;
  const char *s = text;
  while ((int)*s >= (int)'0' && (int)*s <= (int)'9') {
    int d = (int)*s - (int)'0';
    n = 10 * n + d;
    s++;
  }
  *nread = s - text;
  return n;
}
```
user: What is the c source code of the assembly code below:

```asm
<parse_int>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L14
  mov    %rdi,%rdx
  xor    %r8d,%r8d
  nopl   (%rax)
L16:
  sub    $0x30,%eax
  lea    (%r8,%r8,4),%ecx
  add    $0x1,%rdx
  movsbl %al,%eax
  lea    (%rax,%rcx,2),%r8d
  movzbl (%rdx),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  jbe    L16
  sub    %edi,%edx
  mov    %r8d,%eax
  mov    %edx,(%rsi)
  ret
  nopl   (%rax)
L14:
  xor    %edx,%edx
  xor    %r8d,%r8d
  mov    %edx,(%rsi)
  mov    %r8d,%eax
  ret
```

assistant: ```c
int parse_int(const char *text, int *nread) {
  int n = 0;
  const char *s = text;
  while ((int)*s >= (int)'0' && (int)*s <= (int)'9') {
    int d = (int)*s - (int)'0';
    n = 10 * n + d;
    s++;
  }
  *nread = s - text;
  return n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<parse_int>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L11
  mov    %rdi,%rdx
  xor    %r8d,%r8d
  nopl   (%rax)
L12:
  sub    $0x30,%eax
  lea    (%r8,%r8,4),%ecx
  add    $0x1,%rdx
  movsbl %al,%eax
  lea    (%rax,%rcx,2),%r8d
  movzbl (%rdx),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  jbe    L12
  sub    %edi,%edx
  mov    %r8d,%eax
  mov    %edx,(%rsi)
  ret
  nopl   (%rax)
L11:
  xor    %edx,%edx
  xor    %r8d,%r8d
  mov    %edx,(%rsi)
  mov    %r8d,%eax
  ret
```

assistant: ```c
int parse_int(const char *text, int *nread) {
  int n = 0;
  const char *s = text;
  while ((int)*s >= (int)'0' && (int)*s <= (int)'9') {
    int d = (int)*s - (int)'0';
    n = 10 * n + d;
    s++;
  }
  *nread = s - text;
  return n;
}
```
user: What is the c source code of the assembly code below?

```asm
<vga_writer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  movsbl -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <vga_put_char@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *vga_writer(void *ud, char c) {
  (*(int *)ud)++;
  vga_put_char(c);
  return ud;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vga_writer>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  addl   $0x1,(%rdi)
  movsbl %sil,%edi
  call   <vga_put_char@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
void *vga_writer(void *ud, char c) {
  (*(int *)ud)++;
  vga_put_char(c);
  return ud;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<vga_writer>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  addl   $0x1,(%rdi)
  movsbl %sil,%edi
  call   <vga_put_char@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *vga_writer(void *ud, char c) {
  (*(int *)ud)++;
  vga_put_char(c);
  return ud;
}
```
user: Translate the below assembly code into C programming language:

```asm
<vga_writer>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  addl   $0x1,(%rdi)
  movsbl %sil,%edi
  call   <vga_put_char@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *vga_writer(void *ud, char c) {
  (*(int *)ud)++;
  vga_put_char(c);
  return ud;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ExactSol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  movsd  %xmm0,-0x28(%rbp)
  movsd  %xmm1,-0x30(%rbp)
  mov    %rdi,-0x38(%rbp)
  mov    $0x1,%edi
  call   <atan@plt>
  shl    $0x2,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  movsd  -0x28(%rbp),%xmm0
  subsd  -0x30(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <cos@plt>
  movq   %xmm0,%rax
  mov    -0x38(%rbp),%rdx
  mov    %rax,(%rdx)
  movsd  -0x18(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  movsd  -0x28(%rbp),%xmm0
  addsd  -0x30(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    -0x38(%rbp),%rdx
  lea    0x8(%rdx),%rbx
  mov    %eax,%edi
  call   <cos@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void ExactSol(double x, double t, double w[2]) {
  double pi = 4 * atan(1.);
  w[0] = cos(2 * pi * (x - t));
  w[1] = cos(2 * pi * (x + t));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ExactSol>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  movsd  %xmm1,0x10(%rsp)
  mov    %rdi,%rbx
  mov    $0x1,%edi
  call   <atan@plt>
  shl    $0x2,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  movsd  0x8(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  movsd  %xmm2,0x18(%rsp)
  mulsd  %xmm2,%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movsd  %xmm0,0x8(%rbx)
  add    $0x20,%rsp
  pop    %rbx
  ret
```

assistant: ```c
void ExactSol(double x, double t, double w[2]) {
  double pi = 4 * atan(1.);
  w[0] = cos(2 * pi * (x - t));
  w[1] = cos(2 * pi * (x + t));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ExactSol>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x1,%edi
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  movsd  %xmm1,0x10(%rsp)
  call   <atan@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  shl    $0x2,%eax
  cvtsi2sd %eax,%xmm1
  addsd  %xmm1,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm1,0x18(%rsp)
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movsd  %xmm0,0x8(%rbx)
  add    $0x20,%rsp
  pop    %rbx
  ret
```

assistant: ```c
void ExactSol(double x, double t, double w[2]) {
  double pi = 4 * atan(1.);
  w[0] = cos(2 * pi * (x - t));
  w[1] = cos(2 * pi * (x + t));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ExactSol>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x1,%edi
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  movsd  %xmm1,0x10(%rsp)
  call   <atan@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  shl    $0x2,%eax
  cvtsi2sd %eax,%xmm1
  addsd  %xmm1,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm1,0x18(%rsp)
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movsd  %xmm0,0x8(%rbx)
  add    $0x20,%rsp
  pop    %rbx
  ret
```

assistant: ```c
void ExactSol(double x, double t, double w[2]) {
  double pi = 4 * atan(1.);
  w[0] = cos(2 * pi * (x - t));
  w[1] = cos(2 * pi * (x + t));
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  test   %eax,%eax
  jns    L41
  movl   $0x0,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  neg    %eax
  mov    %eax,-0x20(%rbp)
L41:
  mov    -0x20(%rbp),%eax
  cmp    $0x1,%eax
  je     L19
  mov    -0x20(%rbp),%eax
  cmp    $0x2,%eax
  jne    L89
L19:
  movl   $0x1,-0x14(%rbp)
  jmp    L49
L89:
  movl   $0x1,-0xc(%rbp)
  jmp    L3
L55:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L55
L49:
  cmpl   $0x0,-0x10(%rbp)
  jne    L29
  mov    -0x20(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L29
  negl   -0x14(%rbp)
L29:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
type_guessed: string
name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D13: %d

tool: D6: %d


assistant: ```c
int main() {
  int number;
  int first = 0;
  int second = 1;
  int fibonacci = 0;
  int key = 1;
  scanf("%d", &number);
  if (number < 0) {
    key = 0;
    number *= -1;
  }
  if (number == 1 || number == 2) {
    fibonacci = 1;
  } else {
    for (int i = 1; i < number; i++) {
      fibonacci = first + second;
      first = second;
      second = fibonacci;
    }
  }
  if (key == 0 && number % 2 == 0) {
    fibonacci *= -1;
  }
  printf("%d\n", fibonacci);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  test   %ecx,%ecx
  js     L89
  mov    %ecx,%r8d
  lea    -0x1(%rcx),%eax
  mov    $0x1,%esi
  cmp    $0x1,%eax
  jbe    L39
  mov    $0x1,%r9d
  mov    $0x0,%esi
  cmp    $0x1,%ecx
  jle    L39
L67:
  mov    $0x1,%eax
  mov    $0x1,%esi
  mov    $0x0,%edx
L50:
  mov    %esi,%edi
  add    %edx,%esi
  add    $0x1,%eax
  mov    %edi,%edx
  cmp    %eax,%ecx
  jne    L50
  test   %r9d,%r9d
  jne    L39
L33:
  test   $0x1,%r8b
  jne    L39
  neg    %esi
L39:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L89:
  mov    %ecx,%eax
  neg    %eax
  mov    %eax,0x4(%rsp)
  mov    %eax,%r8d
  mov    %ecx,%edx
  not    %edx
  cmp    $0x1,%edx
  jbe    L86
  mov    $0x0,%esi
  cmp    $0xffffffff,%ecx
  jge    L33
  mov    %eax,%ecx
  mov    $0x0,%r9d
  jmp    L67
L86:
  mov    $0x1,%esi
  jmp    L33
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: string
name: parse_data
arguments: tag: D1
datatype: string

tool: D9: %d

tool: D1: %d


assistant: ```c
int main() {
  int number;
  int first = 0;
  int second = 1;
  int fibonacci = 0;
  int key = 1;
  scanf("%d", &number);
  if (number < 0) {
    key = 0;
    number *= -1;
  }
  if (number == 1 || number == 2) {
    fibonacci = 1;
  } else {
    for (int i = 1; i < number; i++) {
      fibonacci = first + second;
      first = second;
      second = fibonacci;
    }
  }
  if (key == 0 && number % 2 == 0) {
    fibonacci *= -1;
  }
  printf("%d\n", fibonacci);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D12(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  mov    %ecx,%r8d
  test   %ecx,%ecx
  js     L117
  lea    -0x1(%rcx),%eax
  mov    $0x1,%esi
  cmp    $0x1,%eax
  jbe    L118
  mov    $0x1,%r9d
  xor    %esi,%esi
  cmp    $0x1,%ecx
  jle    L118
L110:
  mov    $0x1,%eax
  mov    $0x1,%esi
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L57:
  mov    %esi,%edi
  add    $0x1,%eax
  add    %edx,%esi
  mov    %edi,%edx
  cmp    %ecx,%eax
  jl     L57
  test   %r9d,%r9d
  jne    L118
L113:
  and    $0x1,%r8b
  jne    L118
  neg    %esi
L118:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L85
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L117:
  neg    %r8d
  mov    %r8d,0x4(%rsp)
  cmp    $0xfffffffd,%ecx
  ja     L72
  add    $0x1,%ecx
  je     L112
  mov    %r8d,%ecx
  xor    %r9d,%r9d
  jmp    L110
L72:
  mov    $0x1,%esi
  jmp    L113
L112:
  xor    %esi,%esi
  jmp    L113
L85:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D12
type: string
name: parse_data
arguments: tag: D15
type: string

tool: D12: %d

tool: D15: %d


assistant: ```c
int main() {
  int number;
  int first = 0;
  int second = 1;
  int fibonacci = 0;
  int key = 1;
  scanf("%d", &number);
  if (number < 0) {
    key = 0;
    number *= -1;
  }
  if (number == 1 || number == 2) {
    fibonacci = 1;
  } else {
    for (int i = 1; i < number; i++) {
      fibonacci = first + second;
      first = second;
      second = fibonacci;
    }
  }
  if (key == 0 && number % 2 == 0) {
    fibonacci *= -1;
  }
  printf("%d\n", fibonacci);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D8(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  mov    %ecx,%r8d
  test   %ecx,%ecx
  js     L37
  lea    -0x1(%rcx),%eax
  mov    $0x1,%esi
  cmp    $0x1,%eax
  jbe    L105
  mov    $0x1,%r9d
  xor    %esi,%esi
  cmp    $0x1,%ecx
  jle    L105
L29:
  mov    $0x1,%eax
  mov    $0x1,%esi
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L56:
  mov    %esi,%edi
  add    $0x1,%eax
  add    %edx,%esi
  mov    %edi,%edx
  cmp    %ecx,%eax
  jl     L56
  test   %r9d,%r9d
  jne    L105
L40:
  and    $0x1,%r8b
  jne    L105
  neg    %esi
L105:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L37:
  neg    %r8d
  mov    %r8d,0x4(%rsp)
  cmp    $0xfffffffd,%ecx
  ja     L64
  add    $0x1,%ecx
  je     L112
  mov    %r8d,%ecx
  xor    %r9d,%r9d
  jmp    L29
L64:
  mov    $0x1,%esi
  jmp    L40
L112:
  xor    %esi,%esi
  jmp    L40
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D18
type_infered: string

tool: D8: %d

tool: D18: %d


assistant: ```c
int main() {
  int number;
  int first = 0;
  int second = 1;
  int fibonacci = 0;
  int key = 1;
  scanf("%d", &number);
  if (number < 0) {
    key = 0;
    number *= -1;
  }
  if (number == 1 || number == 2) {
    fibonacci = 1;
  } else {
    for (int i = 1; i < number; i++) {
      fibonacci = first + second;
      first = second;
      second = fibonacci;
    }
  }
  if (key == 0 && number % 2 == 0) {
    fibonacci *= -1;
  }
  printf("%d\n", fibonacci);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<percentages>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,%rax
  mov    %rcx,%rdx
  mov    %r8,-0x48(%rbp)
  mov    $0x0,%r12d
  mov    -0x48(%rbp),%r14
  mov    $0x0,%ebx
  jmp    L19
L10:
  mov    (%rax),%rcx
  mov    (%rdx),%rsi
  sub    %rsi,%rcx
  mov    %rcx,%r13
  test   %r13,%r13
  jns    L29
  mov    (%rax),%rcx
  mov    %ecx,%esi
  mov    (%rdx),%rcx
  mov    %ecx,%edi
  sub    %edi,%esi
  mov    %esi,%ecx
  movslq %ecx,%r13
L29:
  mov    %r14,%rcx
  lea    0x8(%rcx),%r14
  mov    %r13,(%rcx)
  mov    (%rcx),%rcx
  add    %rcx,%r12
  mov    %rax,%rsi
  lea    0x8(%rsi),%rax
  mov    %rdx,%rcx
  lea    0x8(%rcx),%rdx
  mov    (%rsi),%rsi
  mov    %rsi,(%rcx)
  add    $0x1,%ebx
L19:
  cmp    -0x34(%rbp),%ebx
  jl     L10
  test   %r12,%r12
  jne    L39
  mov    $0x1,%r12d
L39:
  mov    %r12,%rax
  shr    $0x3f,%rax
  add    %r12,%rax
  sar    %rax
  mov    %rax,-0x28(%rbp)
  mov    $0x0,%ebx
  jmp    L4
L20:
  mov    -0x48(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x48(%rbp)
  mov    (%rax),%rax
  imul   $0x3e8,%rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  cqto
  idiv   %r12
  mov    %rax,%rcx
  mov    -0x40(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    %ecx,%edx
  mov    %edx,(%rax)
  add    $0x1,%ebx
L4:
  cmp    -0x34(%rbp),%ebx
  jl     L20
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

assistant: ```c
long percentages(cnt, out, new, old, diffs)
int cnt;
int *out;
register long *new;
register long *old;
long *diffs;
{
  register int i;
  register long change;
  register long total_change;
  register long *dp;
  long half_total;
  total_change = 0;
  dp = diffs;
  for (i = 0; i < cnt; i++) {
    if ((change = *new - *old) < 0) {
      change = (int)((unsigned long)*new - (unsigned long)*old);
    }
    total_change += (*dp++ = change);
    *old++ = *new ++;
  }
  if (total_change == 0) {
    total_change = 1;
  }
  half_total = total_change / 2l;
  for (i = 0; i < cnt; i++) {
    *out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
  }
  return (total_change);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<percentages>:
  endbr64
  test   %edi,%edi
  jle    L2
  push   %rbx
  mov    %rsi,%r9
  mov    %rdx,%r11
  mov    %edi,%ebx
  mov    $0x0,%edx
  mov    $0x0,%r10d
L13:
  mov    (%r11,%rdx,8),%rax
  mov    %eax,%esi
  sub    (%rcx,%rdx,8),%esi
  movslq %esi,%rsi
  sub    (%rcx,%rdx,8),%rax
  cmovs  %rsi,%rax
  mov    %rax,(%r8,%rdx,8)
  add    %rax,%r10
  mov    (%r11,%rdx,8),%rax
  mov    %rax,(%rcx,%rdx,8)
  add    $0x1,%rdx
  cmp    %rbx,%rdx
  jne    L13
  mov    $0x1,%esi
  test   %r10,%r10
  je     L35
  mov    %r10,%rax
  shr    $0x3f,%rax
  add    %r10,%rax
  mov    %r10,%rsi
  sar    %rax
  mov    %rax,%r10
L35:
  mov    $0x0,%ecx
L4:
  imul   $0x3e8,(%r8,%rcx,8),%rax
  add    %r10,%rax
  cqto
  idiv   %rsi
  mov    %eax,(%r9,%rcx,4)
  add    $0x1,%rcx
  cmp    %ecx,%edi
  jg     L4
  mov    %rsi,%rax
  pop    %rbx
  ret
L2:
  mov    $0x1,%esi
  mov    %rsi,%rax
  ret
```

assistant: ```c
long percentages(cnt, out, new, old, diffs)
int cnt;
int *out;
register long *new;
register long *old;
long *diffs;
{
  register int i;
  register long change;
  register long total_change;
  register long *dp;
  long half_total;
  total_change = 0;
  dp = diffs;
  for (i = 0; i < cnt; i++) {
    if ((change = *new - *old) < 0) {
      change = (int)((unsigned long)*new - (unsigned long)*old);
    }
    total_change += (*dp++ = change);
    *old++ = *new ++;
  }
  if (total_change == 0) {
    total_change = 1;
  }
  half_total = total_change / 2l;
  for (i = 0; i < cnt; i++) {
    *out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
  }
  return (total_change);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<percentages>:
  endbr64
  mov    %rsi,%r9
  mov    %rdx,%r11
  test   %edi,%edi
  jle    L4
  movslq %edi,%r10
  xor    %edx,%edx
  xor    %edi,%edi
  nopl   0x0(%rax)
L12:
  mov    (%r11,%rdx,8),%rax
  mov    %eax,%esi
  sub    (%rcx,%rdx,8),%esi
  sub    (%rcx,%rdx,8),%rax
  movslq %esi,%rsi
  cmovs  %rsi,%rax
  mov    %rax,(%r8,%rdx,8)
  add    %rax,%rdi
  mov    (%r11,%rdx,8),%rax
  mov    %rax,(%rcx,%rdx,8)
  add    $0x1,%rdx
  cmp    %rdx,%r10
  jne    L12
  mov    $0x1,%r11d
  test   %rdi,%rdi
  je     L37
  mov    %rdi,%rax
  mov    %rdi,%r11
  shr    $0x3f,%rax
  add    %rdi,%rax
  sar    %rax
  mov    %rax,%rdi
L37:
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L24:
  imul   $0x3e8,(%r8,%rcx,8),%rax
  add    %rdi,%rax
  cqto
  idiv   %r11
  mov    %eax,(%r9,%rcx,4)
  add    $0x1,%rcx
  cmp    %rcx,%r10
  jne    L24
  mov    %r11,%rax
  ret
  nopl   0x0(%rax)
L4:
  mov    $0x1,%r11d
  mov    %r11,%rax
  ret
```

assistant: ```c
long percentages(cnt, out, new, old, diffs)
int cnt;
int *out;
register long *new;
register long *old;
long *diffs;
{
  register int i;
  register long change;
  register long total_change;
  register long *dp;
  long half_total;
  total_change = 0;
  dp = diffs;
  for (i = 0; i < cnt; i++) {
    if ((change = *new - *old) < 0) {
      change = (int)((unsigned long)*new - (unsigned long)*old);
    }
    total_change += (*dp++ = change);
    *old++ = *new ++;
  }
  if (total_change == 0) {
    total_change = 1;
  }
  half_total = total_change / 2l;
  for (i = 0; i < cnt; i++) {
    *out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
  }
  return (total_change);
}
```
user: Translate the below assembly code into C programming language:

```asm
<percentages>:
  endbr64
  test   %edi,%edi
  jle    L42
  mov    %rsi,%r9
  push   %rbx
  mov    %rdx,%r10
  mov    %rcx,%r11
  movslq %edi,%rdi
  xor    %eax,%eax
  xor    %esi,%esi
  jmp    L33
  nop
L23:
  mov    %rbx,(%r8,%rax,8)
  add    %rbx,%rsi
L49:
  mov    (%r10,%rax,8),%rdx
  mov    %rdx,(%r11,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdi
  je     L28
L33:
  mov    (%r10,%rax,8),%rdx
  mov    (%r11,%rax,8),%rcx
  mov    %rdx,%rbx
  sub    %rcx,%rbx
  jns    L23
  sub    %ecx,%edx
  movslq %edx,%rdx
  mov    %rdx,(%r8,%rax,8)
  add    %rdx,%rsi
  jmp    L49
  cs nopw 0x0(%rax,%rax,1)
L28:
  mov    $0x1,%r10d
  test   %rsi,%rsi
  je     L63
  mov    %rsi,%rax
  mov    %rsi,%r10
  shr    $0x3f,%rax
  add    %rsi,%rax
  sar    %rax
  mov    %rax,%rsi
L63:
  xor    %ecx,%ecx
L38:
  imul   $0x3e8,(%r8,%rcx,8),%rax
  add    %rsi,%rax
  cqto
  idiv   %r10
  mov    %eax,(%r9,%rcx,4)
  add    $0x1,%rcx
  cmp    %rcx,%rdi
  jne    L38
  mov    %r10,%rax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L42:
  mov    $0x1,%eax
  ret
```

assistant: ```c
long percentages(cnt, out, new, old, diffs)
int cnt;
int *out;
register long *new;
register long *old;
long *diffs;
{
  register int i;
  register long change;
  register long total_change;
  register long *dp;
  long half_total;
  total_change = 0;
  dp = diffs;
  for (i = 0; i < cnt; i++) {
    if ((change = *new - *old) < 0) {
      change = (int)((unsigned long)*new - (unsigned long)*old);
    }
    total_change += (*dp++ = change);
    *old++ = *new ++;
  }
  if (total_change == 0) {
    total_change = 1;
  }
  half_total = total_change / 2l;
  for (i = 0; i < cnt; i++) {
    *out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
  }
  return (total_change);
}
```