ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_print_comb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x0,-0x10(%rbp)
  jmp    L29
L15:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  imul   $0xffffff9c,%eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  imul   $0xffffff9c,%eax,%edx
  mov    -0x10(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   $0xfffffff6,%eax,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L23
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L23
  mov    -0xc(%rbp),%eax
  add    $0x30,%eax
  mov    %al,-0x13(%rbp)
  mov    -0x8(%rbp),%eax
  add    $0x30,%eax
  mov    %al,-0x12(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x30,%eax
  mov    %al,-0x11(%rbp)
  movsbl -0x13(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  movsbl -0x12(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  movsbl -0x11(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  cmpl   $0x314,-0x10(%rbp)
  jg     L23
  mov    $0x2c,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
L23:
  addl   $0x1,-0x10(%rbp)
L29:
  cmpl   $0x315,-0x10(%rbp)
  jle    L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void ft_print_comb(void) {
  int c, d, u, i;
  char c2;
  char d2;
  char u2;
  i = 0;
  while (i < 790) {
    c = (i / 100);
    d = ((i - (c * 100)) / 10);
    u = (i - (c * 100)) - (d * 10);
    if (c < d && d < u) {
      c2 = c + '0';
      d2 = d + '0';
      u2 = u + '0';
      putchar(c2);
      putchar(d2);
      putchar(u2);
      if (i < 789) {
        putchar(',');
        putchar(' ');
      }
    }
    i++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ft_print_comb>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%r12d
  mov    $0x0,%ebp
  jmp    L13
L24:
  cmp    $0x315,%r12d
  jg     L28
L5:
  add    $0x1,%ebp
  add    $0x1,%r12d
L13:
  movslq %ebp,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0xffffff9c,%eax,%edx
  add    %ebp,%edx
  movslq %edx,%rbx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%ebx
  imul   $0xfffffff6,%ebx,%r14d
  add    %edx,%r14d
  cmp    %ebx,%eax
  jge    L24
  cmp    %r14d,%ebx
  jge    L24
  lea    0x30(%rax),%edi
  movsbl %dil,%edi
  call   <putchar@plt>
  lea    0x30(%rbx),%edi
  movsbl %dil,%edi
  call   <putchar@plt>
  lea    0x30(%r14),%edi
  movsbl %dil,%edi
  call   <putchar@plt>
  cmp    $0x314,%ebp
  jg     L24
  mov    $0x2c,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L5
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void ft_print_comb(void) {
  int c, d, u, i;
  char c2;
  char d2;
  char u2;
  i = 0;
  while (i < 790) {
    c = (i / 100);
    d = ((i - (c * 100)) / 10);
    u = (i - (c * 100)) - (d * 10);
    if (c < d && d < u) {
      c2 = c + '0';
      d2 = d + '0';
      u2 = u + '0';
      putchar(c2);
      putchar(d2);
      putchar(u2);
      if (i < 789) {
        putchar(',');
        putchar(' ');
      }
    }
    i++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_print_comb>:
  endbr64
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L44:
  add    $0x1,%ebp
  cmp    $0x316,%ebp
  je     L20
L25:
  mov    %ebp,%edi
  imul   $0x51eb851f,%rdi,%rdi
  shr    $0x25,%rdi
  imul   $0xffffff9c,%edi,%eax
  add    %ebp,%eax
  movslq %eax,%rbx
  cltd
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  sub    %edx,%ebx
  imul   $0xfffffff6,%ebx,%edx
  lea    (%rdx,%rax,1),%r12d
  cmp    %ebx,%edi
  jge    L44
  cmp    %r12d,%ebx
  jge    L44
  add    $0x30,%edi
  movzbl %dil,%edi
  call   <putchar@plt>
  lea    0x30(%rbx),%edi
  movzbl %dil,%edi
  call   <putchar@plt>
  lea    0x30(%r12),%edx
  movsbl %dl,%edi
  call   <putchar@plt>
  cmp    $0x315,%ebp
  je     L20
  mov    $0x2c,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    $0x316,%ebp
  jne    L25
L20:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void ft_print_comb(void) {
  int c, d, u, i;
  char c2;
  char d2;
  char u2;
  i = 0;
  while (i < 790) {
    c = (i / 100);
    d = ((i - (c * 100)) / 10);
    u = (i - (c * 100)) - (d * 10);
    if (c < d && d < u) {
      c2 = c + '0';
      d2 = d + '0';
      u2 = u + '0';
      putchar(c2);
      putchar(d2);
      putchar(u2);
      if (i < 789) {
        putchar(',');
        putchar(' ');
      }
    }
    i++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_print_comb>:
  endbr64
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L48:
  add    $0x1,%ebp
  cmp    $0x316,%ebp
  je     L20
L16:
  mov    %ebp,%edi
  imul   $0x51eb851f,%rdi,%rdi
  shr    $0x25,%rdi
  imul   $0xffffff9c,%edi,%eax
  add    %ebp,%eax
  movslq %eax,%rbx
  cltd
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  sub    %edx,%ebx
  imul   $0xfffffff6,%ebx,%edx
  lea    (%rdx,%rax,1),%r12d
  cmp    %ebx,%edi
  jge    L48
  cmp    %r12d,%ebx
  jge    L48
  add    $0x30,%edi
  movzbl %dil,%edi
  call   <putchar@plt>
  lea    0x30(%rbx),%edi
  movzbl %dil,%edi
  call   <putchar@plt>
  lea    0x30(%r12),%edx
  movsbl %dl,%edi
  call   <putchar@plt>
  cmp    $0x315,%ebp
  je     L20
  mov    $0x2c,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    $0x316,%ebp
  jne    L16
L20:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void ft_print_comb(void) {
  int c, d, u, i;
  char c2;
  char d2;
  char u2;
  i = 0;
  while (i < 790) {
    c = (i / 100);
    d = ((i - (c * 100)) / 10);
    u = (i - (c * 100)) - (d * 10);
    if (c < d && d < u) {
      c2 = c + '0';
      d2 = d + '0';
      u2 = u + '0';
      putchar(c2);
      putchar(d2);
      putchar(u2);
      if (i < 789) {
        putchar(',');
        putchar(' ');
      }
    }
    i++;
  }
}
```
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
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    D55(%rip),%rdi
  call   <strlen@plt>
  shl    $0x3,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L61:
  cmp    %rdx,%rsp
  je     L64
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L61
L64:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L8
  orq    $0x0,-0x8(%rsp,%rax,1)
L8:
  mov    %rsp,%r12
  lea    D55(%rip),%rbx
  mov    %rbx,%rsi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D55(%rip),%r13
  jmp    L33
L49:
  lea    0x0(,%rbx,8),%edx
  movsbl 0x0(%r13,%rbx,1),%edi
  mov    %r12,%rsi
  call   <batata@plt>
  add    $0x1,%rbx
L33:
  mov    %r13,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L49
  lea    D51(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D55(%rip),%r13
  lea    D9(%rip),%r14
  jmp    L24
L35:
  movsbl (%r12,%rbx,1),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
L24:
  mov    %r13,%rdi
  call   <strlen@plt>
  shl    $0x3,%eax
  cmp    %ebx,%eax
  jg     L35
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L73
  mov    $0x0,%eax
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L73:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D55
type: string
name: extract_symbolic_data
arguments: label: D10
type: string
name: extract_symbolic_data
arguments: label: D19
type: string
name: extract_symbolic_data
arguments: label: D51
type: string
name: extract_symbolic_data
arguments: label: D9
type: string

tool: D55: joao doidjo

tool: D10: [original] %s


tool: D19: tamanho do texto original %d


tool: D51: [resultado] 

tool: D9: %d

assistant: ```c
int main() {
  char *texto = "joao doidjo";
  char resultado[strlen(texto) * 8];
  printf("[original] %s\n", texto);
  printf("tamanho do texto original %d\n", (int)strlen(texto));
  for (int i = 0; i < strlen(texto); i++)
    batata(texto[i], resultado, i * 8);
  printf("[resultado] ");
  for (int i = 0; i < strlen(texto) * 8; i++)
    printf("%d", (int)resultado[i]);
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  shl    $0x3,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L82
L10:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L10
L82:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L73
L22:
  mov    %r12,%rsi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <printf@plt>
  mov    %r12,%rdi
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    D23(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L0
  nopl   0x0(%rax)
L15:
  movsbl (%r12,%rbx,1),%edi
  lea    0x0(,%rbx,8),%edx
  mov    %r13,%rsi
  add    $0x1,%rbx
  call   <batata@plt>
L0:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L15
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    D46(%rip),%r14
  jmp    L76
  nopl   (%rax)
L21:
  movsbl 0x0(%r13,%rbx,1),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
L76:
  mov    %r12,%rdi
  call   <strlen@plt>
  shl    $0x3,%eax
  cmp    %ebx,%eax
  jg     L21
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L73:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L22
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D10
type_infered: string
name: retrieve_data
arguments: data_label: D27
type_infered: string
name: retrieve_data
arguments: data_label: D23
type_infered: string
name: retrieve_data
arguments: data_label: D47
type_infered: string
name: retrieve_data
arguments: data_label: D46
type_infered: string

tool: D10: joao doidjo

tool: D27: [original] %s


tool: D23: tamanho do texto original %d


tool: D47: [resultado] 

tool: D46: %d

assistant: ```c
int main() {
  char *texto = "joao doidjo";
  char resultado[strlen(texto) * 8];
  printf("[original] %s\n", texto);
  printf("tamanho do texto original %d\n", (int)strlen(texto));
  for (int i = 0; i < strlen(texto); i++)
    batata(texto[i], resultado, i * 8);
  printf("[resultado] ");
  for (int i = 0; i < strlen(texto) * 8; i++)
    printf("%d", (int)resultado[i]);
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  lea    D19(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  shl    $0x3,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L54
L15:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L15
L54:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L65
L20:
  mov    %r12,%rsi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <printf@plt>
  mov    %r12,%rdi
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    D27(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L70
  nopl   0x0(%rax)
L8:
  movsbl (%r12,%rbx,1),%edi
  lea    0x0(,%rbx,8),%edx
  mov    %r13,%rsi
  add    $0x1,%rbx
  call   <batata@plt>
L70:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L8
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    D39(%rip),%r14
  jmp    L23
  nopl   (%rax)
L55:
  movsbl 0x0(%r13,%rbx,1),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
L23:
  mov    %r12,%rdi
  call   <strlen@plt>
  shl    $0x3,%eax
  cmp    %ebx,%eax
  jg     L55
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L65:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L20
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D19
data_type: string
name: read_data
arguments: data_tag: D40
data_type: string
name: read_data
arguments: data_tag: D27
data_type: string
name: read_data
arguments: data_tag: D3
data_type: string
name: read_data
arguments: data_tag: D39
data_type: string

tool: D19: joao doidjo

tool: D40: [original] %s


tool: D27: tamanho do texto original %d


tool: D3: [resultado] 

tool: D39: %d

assistant: ```c
int main() {
  char *texto = "joao doidjo";
  char resultado[strlen(texto) * 8];
  printf("[original] %s\n", texto);
  printf("tamanho do texto original %d\n", (int)strlen(texto));
  for (int i = 0; i < strlen(texto); i++)
    batata(texto[i], resultado, i * 8);
  printf("[resultado] ");
  for (int i = 0; i < strlen(texto) * 8; i++)
    printf("%d", (int)resultado[i]);
  printf("\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<prtab>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L8:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L8
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type_infered: string

tool: D1: 	

assistant: ```c
void prtab(int nr_tab) {
  int i;
  for (i = 0; i < nr_tab; i++) {
    printf("\t");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<prtab>:
  endbr64
  test   %edi,%edi
  jle    L18
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    $0x0,%ebx
  lea    D2(%rip),%r12
L3:
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
data_type: string

tool: D2: 	

assistant: ```c
void prtab(int nr_tab) {
  int i;
  for (i = 0; i < nr_tab; i++) {
    printf("\t");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<prtab>:
  endbr64
  test   %edi,%edi
  jle    L19
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L17
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L19:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type_infered: string

tool: D5: 	

assistant: ```c
void prtab(int nr_tab) {
  int i;
  for (i = 0; i < nr_tab; i++) {
    printf("\t");
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<prtab>:
  endbr64
  test   %edi,%edi
  jle    L14
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L14:
  ret
```

tool_calls: name: read_data
arguments: label: D6
data_type: string

tool: D6: 	

assistant: ```c
void prtab(int nr_tab) {
  int i;
  for (i = 0; i < nr_tab; i++) {
    printf("\t");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<merc_get_type>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1780,-0x4(%rbp)
  jle    L71
  cmpl   $0x178a,-0x4(%rbp)
  jg     L71
  mov    $0x0,%eax
  jmp    L15
L71:
  cmpl   $0x178a,-0x4(%rbp)
  jle    L32
  cmpl   $0x1794,-0x4(%rbp)
  jg     L32
  mov    $0x1,%eax
  jmp    L15
L32:
  cmpl   $0x1794,-0x4(%rbp)
  jle    L0
  cmpl   $0x179e,-0x4(%rbp)
  jg     L0
  mov    $0x2,%eax
  jmp    L15
L0:
  mov    $0xffffffff,%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int merc_get_type(int class_) {
  if (class_ >= 6017 && class_ <= 6026)
    return 0;
  if (class_ >= 6027 && class_ <= 6036)
    return 1;
  if (class_ >= 6037 && class_ <= 6046)
    return 2;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<merc_get_type>:
  endbr64
  lea    -0x1781(%rdi),%edx
  mov    $0x0,%eax
  cmp    $0x9,%edx
  jbe    L12
  lea    -0x178b(%rdi),%edx
  mov    $0x1,%eax
  cmp    $0x9,%edx
  jbe    L12
  sub    $0x1795,%edi
  cmp    $0xa,%edi
  sbb    %eax,%eax
  and    $0x3,%eax
  sub    $0x1,%eax
L12:
  ret
```

assistant: ```c
int merc_get_type(int class_) {
  if (class_ >= 6017 && class_ <= 6026)
    return 0;
  if (class_ >= 6027 && class_ <= 6036)
    return 1;
  if (class_ >= 6037 && class_ <= 6046)
    return 2;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<merc_get_type>:
  endbr64
  lea    -0x1781(%rdi),%edx
  xor    %eax,%eax
  cmp    $0x9,%edx
  jbe    L3
  lea    -0x178b(%rdi),%edx
  mov    $0x1,%eax
  cmp    $0x9,%edx
  jbe    L3
  sub    $0x1795,%edi
  cmp    $0xa,%edi
  sbb    %eax,%eax
  and    $0x3,%eax
  sub    $0x1,%eax
L3:
  ret
```

assistant: ```c
int merc_get_type(int class_) {
  if (class_ >= 6017 && class_ <= 6026)
    return 0;
  if (class_ >= 6027 && class_ <= 6036)
    return 1;
  if (class_ >= 6037 && class_ <= 6046)
    return 2;
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<merc_get_type>:
  endbr64
  lea    -0x1781(%rdi),%edx
  xor    %eax,%eax
  cmp    $0x9,%edx
  jbe    L11
  lea    -0x178b(%rdi),%edx
  mov    $0x1,%eax
  cmp    $0x9,%edx
  jbe    L11
  sub    $0x1795,%edi
  cmp    $0xa,%edi
  sbb    %eax,%eax
  and    $0x3,%eax
  sub    $0x1,%eax
L11:
  ret
```

assistant: ```c
int merc_get_type(int class_) {
  if (class_ >= 6017 && class_ <= 6026)
    return 0;
  if (class_ >= 6027 && class_ <= 6036)
    return 1;
  if (class_ >= 6037 && class_ <= 6046)
    return 2;
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_pow2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movq   $0x1,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L31
L10:
  mov    -0x28(%rbp),%rax
  and    -0x8(%rbp),%rax
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L0
  addl   $0x1,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L0:
  shlq   -0x8(%rbp)
  addl   $0x1,-0x14(%rbp)
L31:
  cmpq   $0x0,-0x8(%rbp)
  jne    L10
  cmpl   $0x1,-0x18(%rbp)
  je     L16
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  or     -0x10(%rbp),%eax
  jmp    L38
L16:
  mov    -0x10(%rbp),%eax
L38:
  pop    %rbp
  ret
```

assistant: ```c
int is_pow2(size_t x) {
  size_t y = 1;
  int cnt = 0, pos = 0, is, posr = 0;
  while (y) {
    is = (x & y) > 0;
    if (is) {
      cnt++;
      posr = pos;
    }
    y <<= 1;
    pos++;
  }
  return cnt == 1 ? posr : posr | INT_MIN;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_pow2>:
  endbr64
  mov    $0x0,%esi
  mov    $0x0,%eax
  mov    $0x0,%ecx
  mov    $0x1,%edx
  jmp    L3
L45:
  add    %rdx,%rdx
  add    $0x1,%eax
  cmp    $0x40,%eax
  je     L2
L3:
  test   %rdx,%rdi
  je     L45
  add    $0x1,%ecx
  mov    %eax,%esi
  jmp    L45
L2:
  cmp    $0x1,%ecx
  je     L24
  mov    D1(%rip),%rax
  or     (%rax),%esi
L24:
  mov    %esi,%eax
  ret
```

assistant: ```c
int is_pow2(size_t x) {
  size_t y = 1;
  int cnt = 0, pos = 0, is, posr = 0;
  while (y) {
    is = (x & y) > 0;
    if (is) {
      cnt++;
      posr = pos;
    }
    y <<= 1;
    pos++;
  }
  return cnt == 1 ? posr : posr | INT_MIN;
}
```
user: What is the c source code of the assembly code below?

```asm
<is_pow2>:
  endbr64
  xor    %r8d,%r8d
  xor    %eax,%eax
  xor    %ecx,%ecx
  mov    $0x1,%edx
L4:
  test   %rdx,%rdi
  je     L17
  add    $0x1,%ecx
  mov    %eax,%r8d
L17:
  add    $0x1,%eax
  add    %rdx,%rdx
  cmp    $0x40,%eax
  jne    L4
  cmp    $0x1,%ecx
  je     L25
  mov    D1(%rip),%rax
  or     (%rax),%r8d
L25:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_pow2(size_t x) {
  size_t y = 1;
  int cnt = 0, pos = 0, is, posr = 0;
  while (y) {
    is = (x & y) > 0;
    if (is) {
      cnt++;
      posr = pos;
    }
    y <<= 1;
    pos++;
  }
  return cnt == 1 ? posr : posr | INT_MIN;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<is_pow2>:
  endbr64
  xor    %r8d,%r8d
  xor    %eax,%eax
  xor    %ecx,%ecx
  mov    $0x1,%edx
L9:
  test   %rdi,%rdx
  je     L11
  add    $0x1,%ecx
  mov    %eax,%r8d
L11:
  add    $0x1,%eax
  add    %rdx,%rdx
  cmp    $0x40,%eax
  jne    L9
  cmp    $0x1,%ecx
  je     L6
  mov    D3(%rip),%rax
  or     (%rax),%r8d
L6:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_pow2(size_t x) {
  size_t y = 1;
  int cnt = 0, pos = 0, is, posr = 0;
  while (y) {
    is = (x & y) > 0;
    if (is) {
      cnt++;
      posr = pos;
    }
    y <<= 1;
    pos++;
  }
  return cnt == 1 ? posr : posr | INT_MIN;
}
```