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
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <is_prime@plt>
  cmp    $0x1,%eax
  je     L11
  lea    D0(%rip),%rdi
  call   <printf@plt>
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L11:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  jmp    L6
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D24
type_guessed: string
name: read_data
arguments: data_label: D10
type_guessed: string
name: read_data
arguments: data_label: D0
type_guessed: string
name: read_data
arguments: data_label: D19
type_guessed: string

tool: D24: Enter number: 


tool: D10: %d

tool: D0: 0

tool: D19: 1

assistant: ```c
int main() {
  int n, prime;
  printf("Enter number: \n");
  scanf("%d", &n);
  prime = is_prime(n);
  if (prime == 1) {
    printf("1");
  } else {
    printf("0");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D1(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D18(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <is_prime@plt>
  cmp    $0x1,%eax
  je     L16
  lea    D22(%rip),%rdi
  call   <printf@plt>
L20:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L16:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D1
type_infered: string
name: retrieve_data
arguments: tag: D18
type_infered: string
name: retrieve_data
arguments: tag: D22
type_infered: string
name: retrieve_data
arguments: tag: D24
type_infered: string

tool: D1: Enter number: 


tool: D18: %d

tool: D22: 0

tool: D24: 1

assistant: ```c
int main() {
  int n, prime;
  printf("Enter number: \n");
  scanf("%d", &n);
  prime = is_prime(n);
  if (prime == 1) {
    printf("1");
  } else {
    printf("0");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D36(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D11(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <is_prime@plt>
  cmp    $0x1,%eax
  je     L17
  lea    D20(%rip),%rdi
  call   <printf@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L17:
  lea    D39(%rip),%rdi
  call   <printf@plt>
  jmp    L9
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D36
type: string
name: disassemble_data
arguments: data_tag: D11
type: string
name: disassemble_data
arguments: data_tag: D20
type: string
name: disassemble_data
arguments: data_tag: D39
type: string

tool: D36: Enter number: 


tool: D11: %d

tool: D20: 0

tool: D39: 1

assistant: ```c
int main() {
  int n, prime;
  printf("Enter number: \n");
  scanf("%d", &n);
  prime = is_prime(n);
  if (prime == 1) {
    printf("1");
  } else {
    printf("0");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<add>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int add(int i, int j) {
  int a = i + j;
  return (a);
}
```
user: What is the c source code of the assembly code below:

```asm
<add>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int add(int i, int j) {
  int a = i + j;
  return (a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<add>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int add(int i, int j) {
  int a = i + j;
  return (a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<enough_space>:
  endbr64
  mov    %rdx,%rax
  mov    0x4(%rdx),%r8d
  test   %esi,%esi
  jg     L129
  mov    $0x0,%ecx
  test   %r8d,%r8d
  jne    L119
L129:
  mov    (%rax),%edx
  cmp    $0x1,%esi
  jg     L143
  mov    $0x0,%ecx
  test   %edx,%edx
  jne    L119
L143:
  mov    0x10(%rax),%r9d
  cmp    $0x5,%esi
  jle    L124
  mov    $0x0,%ecx
  test   %r9d,%r9d
  jne    L119
L124:
  mov    0xc(%rax),%r10d
  cmp    $0x6,%esi
  jle    L94
  mov    $0x0,%ecx
  test   %r10d,%r10d
  jne    L119
L94:
  cmp    $0x5,%edi
  jle    L85
  mov    %r8d,%esi
  or     %edx,%esi
  or     %r9d,%esi
  or     %r10d,%esi
  or     0x8(%rax),%esi
  mov    $0x0,%ecx
  test   $0x10,%sil
  jne    L119
  cmp    $0x6,%edi
  jle    L29
  or     %r8d,%edx
  or     %r9d,%edx
  or     %r10d,%edx
  or     0x8(%rax),%edx
  test   $0x8,%dl
  sete   %cl
  movzbl %cl,%ecx
L119:
  mov    %ecx,%eax
  ret
L85:
  cmp    $0x1,%edi
  jg     L145
  mov    %r8d,%esi
  or     %edx,%esi
  or     %r9d,%esi
  or     %r10d,%esi
  or     0x8(%rax),%esi
  mov    $0x0,%ecx
  test   $0x1,%sil
  jne    L119
L145:
  test   %edi,%edi
  jg     L19
  or     %r8d,%edx
  or     %r9d,%edx
  or     %r10d,%edx
  or     0x8(%rax),%edx
  test   $0x2,%dl
  sete   %cl
  movzbl %cl,%ecx
  jmp    L119
L19:
  mov    $0x1,%ecx
  jmp    L119
L29:
  mov    $0x1,%ecx
  jmp    L119
```

assistant: ```c
int enough_space(int x, int y, uint8_t *piece) {
  if (piece[1] && (y < 1))
    return 0;
  if (piece[0] && (y < 2))
    return 0;
  if (piece[4] && (y > 5))
    return 0;
  if (piece[3] && (y > 6))
    return 0;
  if (x > 5)
    if ((piece[0] & 16) | (piece[1] & 16) | (piece[2] & 16) | (piece[3] & 16) | (piece[4] & 16))
      return 0;
  if (x > 6)
    if ((piece[0] & 8) | (piece[1] & 8) | (piece[2] & 8) | (piece[3] & 8) | (piece[4] & 8))
      return 0;
  if (x < 2)
    if ((piece[0] & 1) | (piece[1] & 1) | (piece[2] & 1) | (piece[3] & 1) | (piece[4] & 1))
      return 0;
  if (x < 1)
    if ((piece[0] & 2) | (piece[1] & 2) | (piece[2] & 2) | (piece[3] & 2) | (piece[4] & 2))
      return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<enough_space>:
  endbr64
  mov    0x4(%rdx),%eax
  mov    %edi,%ecx
  test   %esi,%esi
  jg     L38
  xor    %r8d,%r8d
  test   %eax,%eax
  jne    L24
L38:
  mov    (%rdx),%edi
  cmp    $0x1,%esi
  jg     L100
  xor    %r8d,%r8d
  test   %edi,%edi
  jne    L24
L100:
  mov    0x10(%rdx),%r9d
  cmp    $0x5,%esi
  jle    L99
  xor    %r8d,%r8d
  test   %r9d,%r9d
  jne    L24
L99:
  mov    0xc(%rdx),%r10d
  cmp    $0x6,%esi
  jle    L0
  xor    %r8d,%r8d
  test   %r10d,%r10d
  jne    L24
L0:
  cmp    $0x5,%ecx
  jle    L130
  or     %edi,%eax
  xor    %r8d,%r8d
  or     %r9d,%eax
  or     %r10d,%eax
  or     0x8(%rdx),%eax
  test   $0x10,%al
  jne    L24
  cmp    $0x6,%ecx
  je     L27
  xor    %r8d,%r8d
  test   $0x8,%al
  sete   %r8b
L24:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L130:
  mov    $0x1,%r8d
  cmp    $0x1,%ecx
  jg     L24
  or     %edi,%eax
  xor    %r8d,%r8d
  or     %r9d,%eax
  or     %r10d,%eax
  or     0x8(%rdx),%eax
  test   $0x1,%al
  jne    L24
  cmp    $0x1,%ecx
  je     L27
  xor    %r8d,%r8d
  test   $0x2,%al
  sete   %r8b
  mov    %r8d,%eax
  ret
L27:
  mov    $0x1,%r8d
  jmp    L24
```

assistant: ```c
int enough_space(int x, int y, uint8_t *piece) {
  if (piece[1] && (y < 1))
    return 0;
  if (piece[0] && (y < 2))
    return 0;
  if (piece[4] && (y > 5))
    return 0;
  if (piece[3] && (y > 6))
    return 0;
  if (x > 5)
    if ((piece[0] & 16) | (piece[1] & 16) | (piece[2] & 16) | (piece[3] & 16) | (piece[4] & 16))
      return 0;
  if (x > 6)
    if ((piece[0] & 8) | (piece[1] & 8) | (piece[2] & 8) | (piece[3] & 8) | (piece[4] & 8))
      return 0;
  if (x < 2)
    if ((piece[0] & 1) | (piece[1] & 1) | (piece[2] & 1) | (piece[3] & 1) | (piece[4] & 1))
      return 0;
  if (x < 1)
    if ((piece[0] & 2) | (piece[1] & 2) | (piece[2] & 2) | (piece[3] & 2) | (piece[4] & 2))
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<enough_space>:
  endbr64
  mov    0x4(%rdx),%eax
  mov    %edi,%ecx
  test   %esi,%esi
  jg     L123
  xor    %r8d,%r8d
  test   %eax,%eax
  jne    L110
L123:
  mov    (%rdx),%edi
  cmp    $0x1,%esi
  jg     L113
  xor    %r8d,%r8d
  test   %edi,%edi
  jne    L110
L113:
  mov    0x10(%rdx),%r9d
  cmp    $0x5,%esi
  jle    L96
  xor    %r8d,%r8d
  test   %r9d,%r9d
  jne    L110
L96:
  mov    0xc(%rdx),%r10d
  cmp    $0x6,%esi
  jle    L130
  xor    %r8d,%r8d
  test   %r10d,%r10d
  jne    L110
L130:
  cmp    $0x5,%ecx
  jle    L111
  or     %edi,%eax
  xor    %r8d,%r8d
  or     %r9d,%eax
  or     %r10d,%eax
  or     0x8(%rdx),%eax
  test   $0x10,%al
  jne    L110
  cmp    $0x6,%ecx
  je     L40
  xor    %r8d,%r8d
  test   $0x8,%al
  sete   %r8b
L110:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L111:
  mov    $0x1,%r8d
  cmp    $0x1,%ecx
  jg     L110
  or     %edi,%eax
  xor    %r8d,%r8d
  or     %r9d,%eax
  or     %r10d,%eax
  or     0x8(%rdx),%eax
  test   $0x1,%al
  jne    L110
  cmp    $0x1,%ecx
  je     L40
  xor    %r8d,%r8d
  test   $0x2,%al
  sete   %r8b
  mov    %r8d,%eax
  ret
L40:
  mov    $0x1,%r8d
  jmp    L110
```

assistant: ```c
int enough_space(int x, int y, uint8_t *piece) {
  if (piece[1] && (y < 1))
    return 0;
  if (piece[0] && (y < 2))
    return 0;
  if (piece[4] && (y > 5))
    return 0;
  if (piece[3] && (y > 6))
    return 0;
  if (x > 5)
    if ((piece[0] & 16) | (piece[1] & 16) | (piece[2] & 16) | (piece[3] & 16) | (piece[4] & 16))
      return 0;
  if (x > 6)
    if ((piece[0] & 8) | (piece[1] & 8) | (piece[2] & 8) | (piece[3] & 8) | (piece[4] & 8))
      return 0;
  if (x < 2)
    if ((piece[0] & 1) | (piece[1] & 1) | (piece[2] & 1) | (piece[3] & 1) | (piece[4] & 1))
      return 0;
  if (x < 1)
    if ((piece[0] & 2) | (piece[1] & 2) | (piece[2] & 2) | (piece[3] & 2) | (piece[4] & 2))
      return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x54(%rbp)
  movl   $0x0,-0x50(%rbp)
  movl   $0x0,-0x4c(%rbp)
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x40(%rbp),%rax
  lea    D29(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0x58(%rbp)
  jmp    L34
L9:
  addl   $0x1,-0x50(%rbp)
  cmpl   $0xa,-0x58(%rbp)
  jne    L1
  addl   $0x1,-0x54(%rbp)
L1:
  cmpl   $0x20,-0x58(%rbp)
  jne    L22
  addl   $0x1,-0x4c(%rbp)
L22:
  mov    -0x58(%rbp),%ecx
  mov    -0x58(%rbp),%edx
  mov    -0x58(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0x58(%rbp)
L34:
  cmpl   $0xffffffff,-0x58(%rbp)
  jne    L9
  mov    -0x50(%rbp),%ecx
  mov    -0x4c(%rbp),%edx
  mov    -0x54(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D43
type_guessed: string
name: parse_data
arguments: data_tag: D4
type_guessed: string
name: parse_data
arguments: data_tag: D29
type_guessed: string
name: parse_data
arguments: data_tag: D8
type_guessed: string
name: parse_data
arguments: data_tag: D21
type_guessed: string

tool: D43: Digite o nome do arquivo: 

tool: D4: %s

tool: D29: r

tool: D8: %i %c %x


tool: D21: L=%i P=%i B=%i


assistant: ```c
int main() {
  FILE *arquivo;
  int c;
  char fnome[51];
  int cl = 0;
  int cb = 0;
  int cp = 0;
  printf("Digite o nome do arquivo: ");
  scanf("%s", fnome);
  arquivo = fopen(fnome, "r");
  c = getc(arquivo);
  while (c != -1) {
    cb++;
    if (c == '\n')
      cl = cl + 1;
    if (c == ' ')
      cp = cp + 1;
    printf("%i %c %x\n", c, c, c);
    c = getc(arquivo);
  }
  printf("L=%i P=%i B=%i\n", cl, cp, cb);
  fclose(arquivo);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  lea    D43(%rip),%rsi
  mov    %rbx,%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <getc@plt>
  cmp    $0xffffffff,%eax
  je     L28
  mov    %eax,%esi
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  mov    $0x0,%r14d
  lea    D1(%rip),%r13
  jmp    L4
L58:
  add    $0x1,%r14d
L55:
  mov    %esi,%ecx
  mov    %esi,%edx
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <getc@plt>
  mov    %eax,%esi
  cmp    $0xffffffff,%eax
  je     L12
L4:
  add    $0x1,%ebx
  cmp    $0xa,%esi
  je     L58
  cmp    $0x20,%esi
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  jmp    L55
L28:
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  mov    $0x0,%r14d
L12:
  mov    %ebx,%ecx
  mov    %r12d,%edx
  mov    %r14d,%esi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D10
data_type: string
name: parse_data
arguments: label: D2
data_type: string
name: parse_data
arguments: label: D43
data_type: string
name: parse_data
arguments: label: D1
data_type: string
name: parse_data
arguments: label: D33
data_type: string

tool: D10: Digite o nome do arquivo: 

tool: D2: %s

tool: D43: r

tool: D1: %i %c %x


tool: D33: L=%i P=%i B=%i


assistant: ```c
int main() {
  FILE *arquivo;
  int c;
  char fnome[51];
  int cl = 0;
  int cb = 0;
  int cp = 0;
  printf("Digite o nome do arquivo: ");
  scanf("%s", fnome);
  arquivo = fopen(fnome, "r");
  c = getc(arquivo);
  while (c != -1) {
    cb++;
    if (c == '\n')
      cl = cl + 1;
    if (c == ' ')
      cp = cp + 1;
    printf("%i %c %x\n", c, c, c);
    c = getc(arquivo);
  }
  printf("L=%i P=%i B=%i\n", cl, cp, cb);
  fclose(arquivo);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r14
  lea    D47(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D36(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  lea    D9(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <getc@plt>
  cmp    $0xffffffff,%eax
  je     L49
  mov    %eax,%esi
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  lea    D37(%rip),%rbx
  jmp    L2
  nopl   0x0(%rax)
L35:
  xor    %eax,%eax
  cmp    $0x20,%esi
  sete   %al
  add    %eax,%r13d
L59:
  mov    %esi,%ecx
  mov    %esi,%edx
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <getc@plt>
  mov    %eax,%esi
  cmp    $0xffffffff,%eax
  je     L4
L2:
  add    $0x1,%r12d
  cmp    $0xa,%esi
  jne    L35
  add    $0x1,%r14d
  jmp    L59
L49:
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  nopl   0x0(%rax,%rax,1)
L4:
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %r14d,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D47
datatype: string
name: retrieve_data
arguments: label: D36
datatype: string
name: retrieve_data
arguments: label: D9
datatype: string
name: retrieve_data
arguments: label: D37
datatype: string
name: retrieve_data
arguments: label: D34
datatype: string

tool: D47: Digite o nome do arquivo: 

tool: D36: %s

tool: D9: r

tool: D37: %i %c %x


tool: D34: L=%i P=%i B=%i


assistant: ```c
int main() {
  FILE *arquivo;
  int c;
  char fnome[51];
  int cl = 0;
  int cb = 0;
  int cp = 0;
  printf("Digite o nome do arquivo: ");
  scanf("%s", fnome);
  arquivo = fopen(fnome, "r");
  c = getc(arquivo);
  while (c != -1) {
    cb++;
    if (c == '\n')
      cl = cl + 1;
    if (c == ' ')
      cp = cp + 1;
    printf("%i %c %x\n", c, c, c);
    c = getc(arquivo);
  }
  printf("L=%i P=%i B=%i\n", cl, cp, cb);
  fclose(arquivo);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r14
  lea    D25(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  lea    D44(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <getc@plt>
  cmp    $0xffffffff,%eax
  je     L32
  mov    %eax,%esi
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  lea    D36(%rip),%rbx
  jmp    L33
  nopl   0x0(%rax)
L55:
  xor    %eax,%eax
  cmp    $0x20,%esi
  sete   %al
  add    %eax,%r13d
L56:
  mov    %esi,%ecx
  mov    %esi,%edx
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <getc@plt>
  mov    %eax,%esi
  cmp    $0xffffffff,%eax
  je     L58
L33:
  add    $0x1,%r12d
  cmp    $0xa,%esi
  jne    L55
  add    $0x1,%r14d
  jmp    L56
L32:
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  nopl   0x0(%rax,%rax,1)
L58:
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %r14d,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D25
type: string
name: disassemble_data
arguments: data_label: D19
type: string
name: disassemble_data
arguments: data_label: D44
type: string
name: disassemble_data
arguments: data_label: D36
type: string
name: disassemble_data
arguments: data_label: D17
type: string

tool: D25: Digite o nome do arquivo: 

tool: D19: %s

tool: D44: r

tool: D36: %i %c %x


tool: D17: L=%i P=%i B=%i


assistant: ```c
int main() {
  FILE *arquivo;
  int c;
  char fnome[51];
  int cl = 0;
  int cb = 0;
  int cp = 0;
  printf("Digite o nome do arquivo: ");
  scanf("%s", fnome);
  arquivo = fopen(fnome, "r");
  c = getc(arquivo);
  while (c != -1) {
    cb++;
    if (c == '\n')
      cl = cl + 1;
    if (c == ' ')
      cp = cp + 1;
    printf("%i %c %x\n", c, c, c);
    c = getc(arquivo);
  }
  printf("L=%i P=%i B=%i\n", cl, cp, cb);
  fclose(arquivo);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<_scale_val>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  movsd  %xmm2,-0x18(%rbp)
  movsd  %xmm3,-0x20(%rbp)
  movsd  %xmm4,-0x28(%rbp)
  movsd  -0x8(%rbp),%xmm0
  subsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x28(%rbp),%xmm0
  subsd  -0x20(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  subsd  -0x10(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double _scale_val(double val, double oldlo, double oldhi, double newlo, double newhi) {
  val -= oldlo;
  val *= (newhi - newlo) / (oldhi - oldlo);
  val += newlo;
  return val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_scale_val>:
  endbr64
  subsd  %xmm1,%xmm0
  subsd  %xmm3,%xmm4
  subsd  %xmm1,%xmm2
  divsd  %xmm2,%xmm4
  mulsd  %xmm0,%xmm4
  addsd  %xmm4,%xmm3
  movapd %xmm3,%xmm0
  ret
```

assistant: ```c
double _scale_val(double val, double oldlo, double oldhi, double newlo, double newhi) {
  val -= oldlo;
  val *= (newhi - newlo) / (oldhi - oldlo);
  val += newlo;
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_scale_val>:
  endbr64
  subsd  %xmm3,%xmm4
  subsd  %xmm1,%xmm2
  subsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm4
  mulsd  %xmm0,%xmm4
  addsd  %xmm4,%xmm3
  movapd %xmm3,%xmm0
  ret
```

assistant: ```c
double _scale_val(double val, double oldlo, double oldhi, double newlo, double newhi) {
  val -= oldlo;
  val *= (newhi - newlo) / (oldhi - oldlo);
  val += newlo;
  return val;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_scale_val>:
  endbr64
  subsd  %xmm3,%xmm4
  subsd  %xmm1,%xmm2
  subsd  %xmm1,%xmm0
  divsd  %xmm2,%xmm4
  mulsd  %xmm0,%xmm4
  addsd  %xmm4,%xmm3
  movapd %xmm3,%xmm0
  ret
```

assistant: ```c
double _scale_val(double val, double oldlo, double oldhi, double newlo, double newhi) {
  val -= oldlo;
  val *= (newhi - newlo) / (oldhi - oldlo);
  val += newlo;
  return val;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<niggli_reduce>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  call   <initialize@plt>
  mov    $0x0,%eax
  call   <step0@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L41:
  mov    $0x0,%eax
  call   <step1@plt>
  mov    $0x0,%eax
  call   <step2@plt>
  test   %rax,%rax
  jne    L56
  mov    $0x0,%eax
  call   <step3@plt>
  mov    $0x0,%eax
  call   <step4@plt>
  mov    $0x0,%eax
  call   <step5@plt>
  test   %rax,%rax
  jne    L40
  mov    $0x0,%eax
  call   <step6@plt>
  test   %rax,%rax
  jne    L2
  mov    $0x0,%eax
  call   <step7@plt>
  test   %rax,%rax
  jne    L16
  mov    $0x0,%eax
  call   <step8@plt>
  test   %rax,%rax
  je     L49
  jmp    L31
L56:
  nop
  jmp    L31
L40:
  nop
  jmp    L31
L2:
  nop
  jmp    L31
L16:
  nop
L31:
  addl   $0x1,-0x4(%rbp)
L0:
  cmpl   $0x9,-0x4(%rbp)
  jle    L41
  jmp    L93
L49:
  nop
L93:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <finalize@plt>
  nop
  leave
  ret
```

assistant: ```c
void niggli_reduce(double *lattice_, const double eps_) {
  int i;
  initialize(lattice_, eps_);
  step0();
  for (i = 0; i < 10; i++) {
    if (step1()) {
      ;
      ;
      ;
    }
    if (step2()) {
      ;
      ;
      ;
      continue;
    }
    if (step3()) {
      ;
      ;
      ;
    }
    if (step4()) {
      ;
      ;
      ;
    }
    if (step5()) {
      ;
      ;
      ;
      continue;
    }
    if (step6()) {
      ;
      ;
      ;
      continue;
    }
    if (step7()) {
      ;
      ;
      ;
      continue;
    }
    if (step8()) {
      ;
      ;
      ;
      continue;
    }
    break;
  }
  finalize(lattice_);
}
```