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
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  foo();
  foo();
  foo();
  foo();
  foo();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  foo();
  foo();
  foo();
  foo();
  foo();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x160,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x140(%rbp)
  movl   $0x1,-0x13c(%rbp)
  movl   $0x2,-0x138(%rbp)
  movl   $0x6,-0x134(%rbp)
  movl   $0x18,-0x130(%rbp)
  movl   $0x78,-0x12c(%rbp)
  movl   $0x2d0,-0x128(%rbp)
  movl   $0x13b0,-0x124(%rbp)
  movl   $0x9d80,-0x120(%rbp)
  movl   $0x58980,-0x11c(%rbp)
  movl   $0x0,-0x148(%rbp)
  movl   $0x2,-0x14c(%rbp)
  jmp    L19
L95:
  mov    -0x14c(%rbp),%eax
  mov    %eax,-0x144(%rbp)
  movl   $0x0,-0x154(%rbp)
  movl   $0x0,-0x150(%rbp)
  jmp    L39
L68:
  mov    -0x150(%rbp),%eax
  cltq
  movl   $0xffffffff,-0x110(%rbp,%rax,4)
  addl   $0x1,-0x150(%rbp)
L39:
  cmpl   $0x40,-0x150(%rbp)
  jle    L68
  jmp    L63
L81:
  mov    -0x154(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x154(%rbp)
  cltq
  mov    -0x144(%rbp),%edx
  mov    %edx,-0x110(%rbp,%rax,4)
  lea    -0x140(%rbp),%rdx
  mov    -0x144(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <f_digs@plt>
  mov    %eax,-0x144(%rbp)
  cmpl   $0x3c,-0x154(%rbp)
  jle    L63
  movl   $0x0,-0x154(%rbp)
  jmp    L0
L63:
  lea    -0x110(%rbp),%rcx
  mov    -0x144(%rbp),%eax
  mov    $0x41,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <in@plt>
  test   %eax,%eax
  je     L81
L0:
  cmpl   $0x3c,-0x154(%rbp)
  jne    L70
  addl   $0x1,-0x148(%rbp)
L70:
  addl   $0x1,-0x14c(%rbp)
L19:
  cmpl   $0xf423f,-0x14c(%rbp)
  jle    L95
  mov    -0x148(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D9: counter: %d


assistant: ```c
int main() {
  int f[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
  int non_repeats;
  int arr[65];
  int i, n, counter;
  int temp_n;
  counter = 0;
  for (n = 2; n < 1000000; n++) {
    temp_n = n;
    non_repeats = 0;
    for (i = 0; i < 65; i++) {
      arr[i] = -1;
    }
    while (!in(temp_n, arr, 65)) {
      arr[non_repeats++] = temp_n;
      temp_n = f_digs(temp_n, f);
      if (non_repeats > 60) {
        non_repeats = 0;
        break;
      }
    }
    if (non_repeats == 60)
      counter++;
  }
  printf("counter: %d\n", counter);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  movl   $0x2,0x18(%rsp)
  movl   $0x6,0x1c(%rsp)
  movl   $0x18,0x20(%rsp)
  movl   $0x78,0x24(%rsp)
  movl   $0x2d0,0x28(%rsp)
  movl   $0x13b0,0x2c(%rsp)
  movl   $0x9d80,0x30(%rsp)
  movl   $0x58980,0x34(%rsp)
  movl   $0x0,0xc(%rsp)
  mov    $0x2,%r15d
  lea    0x144(%rsp),%r12
  jmp    L56
L16:
  cmp    $0x3c,%ebp
  sete   %al
  movzbl %al,%eax
  add    %eax,0xc(%rsp)
L15:
  add    $0x1,%r15d
  cmp    $0xf4240,%r15d
  je     L30
L56:
  lea    0x40(%rsp),%rax
L41:
  movl   $0xffffffff,(%rax)
  add    $0x4,%rax
  cmp    %r12,%rax
  jne    L41
  mov    %r15d,%ebx
  mov    $0x0,%ebp
  lea    0x40(%rsp),%r13
L10:
  mov    $0x41,%edx
  mov    %r13,%rsi
  mov    %ebx,%edi
  call   <in@plt>
  test   %eax,%eax
  jne    L16
  mov    %ebx,0x0(%r13,%rbp,4)
  lea    0x10(%rsp),%rsi
  mov    %ebx,%edi
  call   <f_digs@plt>
  mov    %eax,%ebx
  add    $0x1,%rbp
  cmp    $0x3d,%rbp
  jne    L10
  jmp    L15
L30:
  mov    0xc(%rsp),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x158,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D4
data_type: string

tool: D4: counter: %d


assistant: ```c
int main() {
  int f[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
  int non_repeats;
  int arr[65];
  int i, n, counter;
  int temp_n;
  counter = 0;
  for (n = 2; n < 1000000; n++) {
    temp_n = n;
    non_repeats = 0;
    for (i = 0; i < 65; i++) {
      arr[i] = -1;
    }
    while (!in(temp_n, arr, 65)) {
      arr[non_repeats++] = temp_n;
      temp_n = f_digs(temp_n, f);
      if (non_repeats > 60) {
        non_repeats = 0;
        break;
      }
    }
    if (non_repeats == 60)
      counter++;
  }
  printf("counter: %d\n", counter);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    $0x2,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  movabs $0x100000001,%rax
  movl   $0x0,0xc(%rsp)
  lea    0x40(%rsp),%rbx
  lea    0x10(%rsp),%r12
  mov    %rax,0x10(%rsp)
  movabs $0x600000002,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x7800000018,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x13b0000002d0,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x5898000009d80,%rax
  mov    %rax,0x30(%rsp)
  nopl   0x0(%rax)
L40:
  mov    $0x20,%ecx
  mov    %rbx,%rdi
  mov    %r13d,%r14d
  xor    %r15d,%r15d
  mov    $0xffffffffffffffff,%rax
  rep stos %rax,%es:(%rdi)
  movl   $0xffffffff,(%rdi)
  jmp    L33
L11:
  mov    %r14d,%edi
  mov    %r12,%rsi
  mov    %r14d,(%rbx,%r15,4)
  add    $0x1,%r15
  call   <f_digs@plt>
  mov    %eax,%r14d
  cmp    $0x3d,%r15
  je     L21
L33:
  mov    $0x41,%edx
  mov    %rbx,%rsi
  mov    %r14d,%edi
  call   <in@plt>
  test   %eax,%eax
  je     L11
  xor    %eax,%eax
  cmp    $0x3c,%r15d
  sete   %al
  add    %eax,0xc(%rsp)
L21:
  add    $0x1,%r13d
  cmp    $0xf4240,%r13d
  jne    L40
  mov    0xc(%rsp),%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x158,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D3
datatype: string

tool: D3: counter: %d


assistant: ```c
int main() {
  int f[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
  int non_repeats;
  int arr[65];
  int i, n, counter;
  int temp_n;
  counter = 0;
  for (n = 2; n < 1000000; n++) {
    temp_n = n;
    non_repeats = 0;
    for (i = 0; i < 65; i++) {
      arr[i] = -1;
    }
    while (!in(temp_n, arr, 65)) {
      arr[non_repeats++] = temp_n;
      temp_n = f_digs(temp_n, f);
      if (non_repeats > 60) {
        non_repeats = 0;
        break;
      }
    }
    if (non_repeats == 60)
      counter++;
  }
  printf("counter: %d\n", counter);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    $0x2,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x158,%rsp
  movdqa D28(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  movl   $0x0,0xc(%rsp)
  lea    0x40(%rsp),%rbx
  movaps %xmm0,0x10(%rsp)
  movdqa D1(%rip),%xmm0
  lea    0x10(%rsp),%r12
  mov    %rax,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    $0x20,%ecx
  mov    %rbx,%rdi
  mov    %r13d,%r14d
  xor    %r15d,%r15d
  mov    $0xffffffffffffffff,%rax
  rep stos %rax,%es:(%rdi)
  movl   $0xffffffff,(%rdi)
  jmp    L48
L31:
  mov    %r14d,%edi
  mov    %r12,%rsi
  mov    %r14d,(%rbx,%r15,4)
  add    $0x1,%r15
  call   <f_digs@plt>
  mov    %eax,%r14d
  cmp    $0x3d,%r15
  je     L35
L48:
  mov    $0x41,%edx
  mov    %rbx,%rsi
  mov    %r14d,%edi
  call   <in@plt>
  test   %eax,%eax
  je     L31
  xor    %eax,%eax
  cmp    $0x3c,%r15d
  sete   %al
  add    %eax,0xc(%rsp)
L35:
  add    $0x1,%r13d
  cmp    $0xf4240,%r13d
  jne    L9
  mov    0xc(%rsp),%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x158,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D28
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D8
datatype: byte[8]
name: extract_symbolic_data
arguments: label: D1
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D21
datatype: string

tool: {"D28": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D8": [0x80,0x9d,0x00,0x00,0x80,0x89,0x05,0x00]}

tool: {"D1": [0x18,0x00,0x00,0x00,0x78,0x00,0x00,0x00,0xd0,0x02,0x00,0x00,0xb0,0x13,0x00,0x00]}

tool: D21: counter: %d


assistant: ```c
int main() {
  int f[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
  int non_repeats;
  int arr[65];
  int i, n, counter;
  int temp_n;
  counter = 0;
  for (n = 2; n < 1000000; n++) {
    temp_n = n;
    non_repeats = 0;
    for (i = 0; i < 65; i++) {
      arr[i] = -1;
    }
    while (!in(temp_n, arr, 65)) {
      arr[non_repeats++] = temp_n;
      temp_n = f_digs(temp_n, f);
      if (non_repeats > 60) {
        non_repeats = 0;
        break;
      }
    }
    if (non_repeats == 60)
      counter++;
  }
  printf("counter: %d\n", counter);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <init_rqcounter@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  init_rqcounter();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <init_rqcounter@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  init_rqcounter();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <init_rqcounter@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  init_rqcounter();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unzip>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    $0x8,%edx
  call   <read_bits@plt>
  cmp    $0x1f,%eax
  jne    L62
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  cmp    $0x8b,%eax
  jne    L55
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  cmp    $0x8,%al
  jne    L116
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  mov    %eax,%r12d
  mov    $0x20,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  test   $0x4,%r12b
  jne    L114
L93:
  test   $0x8,%r12b
  je     L138
L97:
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  test   %eax,%eax
  jne    L97
L138:
  test   $0x10,%r12b
  je     L81
L47:
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  test   %eax,%eax
  jne    L47
L81:
  test   $0x2,%r12b
  jne    L77
L50:
  mov    %r13,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <inflate@plt>
L39:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L114:
  mov    $0x10,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  mov    %eax,%eax
  shl    $0x3,%rax
  add    %rax,(%rbx)
  jmp    L93
L77:
  mov    $0x10,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <read_bits@plt>
  jmp    L50
L62:
  mov    $0xffffffff,%eax
  jmp    L39
L55:
  mov    $0xffffffff,%eax
  jmp    L39
L116:
  mov    $0xffffffff,%eax
  jmp    L39
```

assistant: ```c
int unzip(const unsigned char *data, long *p, void (*callback)(unsigned char d)) {
  unsigned char cm, flg;
  if (read_bits(data, p, 8) != 0x1f || read_bits(data, p, 8) != 0x8b) {
    return -1;
  }
  cm = read_bits(data, p, 8);
  if (cm != 0x8) {
    return -1;
  }
  flg = read_bits(data, p, 8);
  if (flg & 0xe0)
    ;
  read_bits(data, p, 32);
  read_bits(data, p, 8);
  read_bits(data, p, 8);
  if (flg & 0x4) {
    unsigned int xlen;
    xlen = read_bits(data, p, 16);
    *p += ((long)xlen) * 8;
  }
  if (flg & 0x8) {
    while (read_bits(data, p, 8))
      ;
  }
  if (flg & 0x10) {
    while (read_bits(data, p, 8))
      ;
  }
  if (flg & 0x2)
    read_bits(data, p, 16);
  return inflate(data, p, callback);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<unzip>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  mov    $0x8,%edx
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <read_bits@plt>
  cmp    $0x1f,%eax
  jne    L41
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  cmp    $0x8b,%eax
  jne    L41
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  cmp    $0x8,%al
  jne    L41
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  mov    $0x20,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <read_bits@plt>
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  test   $0x4,%bl
  jne    L110
L82:
  test   $0x8,%bl
  je     L33
  nopl   0x0(%rax)
L29:
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  test   %eax,%eax
  jne    L29
L33:
  test   $0x10,%bl
  jne    L116
L108:
  and    $0x2,%ebx
  jne    L119
L9:
  add    $0x8,%rsp
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <inflate@plt>
  cs nopw 0x0(%rax,%rax,1)
L116:
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  test   %eax,%eax
  jne    L116
  jmp    L108
  cs nopw 0x0(%rax,%rax,1)
L110:
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  mov    %eax,%eax
  shl    $0x3,%rax
  add    %rax,0x0(%rbp)
  jmp    L82
  nop
L119:
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L41:
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int unzip(const unsigned char *data, long *p, void (*callback)(unsigned char d)) {
  unsigned char cm, flg;
  if (read_bits(data, p, 8) != 0x1f || read_bits(data, p, 8) != 0x8b) {
    return -1;
  }
  cm = read_bits(data, p, 8);
  if (cm != 0x8) {
    return -1;
  }
  flg = read_bits(data, p, 8);
  if (flg & 0xe0)
    ;
  read_bits(data, p, 32);
  read_bits(data, p, 8);
  read_bits(data, p, 8);
  if (flg & 0x4) {
    unsigned int xlen;
    xlen = read_bits(data, p, 16);
    *p += ((long)xlen) * 8;
  }
  if (flg & 0x8) {
    while (read_bits(data, p, 8))
      ;
  }
  if (flg & 0x10) {
    while (read_bits(data, p, 8))
      ;
  }
  if (flg & 0x2)
    read_bits(data, p, 16);
  return inflate(data, p, callback);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<unzip>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  mov    $0x8,%edx
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <read_bits@plt>
  cmp    $0x1f,%eax
  jne    L0
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  cmp    $0x8b,%eax
  jne    L0
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  cmp    $0x8,%al
  jne    L0
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  mov    $0x20,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <read_bits@plt>
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  test   $0x4,%bl
  jne    L110
L86:
  test   $0x8,%bl
  je     L71
  nopl   0x0(%rax)
L17:
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  test   %eax,%eax
  jne    L17
L71:
  test   $0x10,%bl
  jne    L88
L16:
  and    $0x2,%ebx
  jne    L118
L107:
  add    $0x8,%rsp
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <inflate@plt>
  cs nopw 0x0(%rax,%rax,1)
L88:
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  test   %eax,%eax
  jne    L88
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L110:
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  mov    %eax,%eax
  shl    $0x3,%rax
  add    %rax,0x0(%rbp)
  jmp    L86
  nop
L118:
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <read_bits@plt>
  jmp    L107
  nopw   0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int unzip(const unsigned char *data, long *p, void (*callback)(unsigned char d)) {
  unsigned char cm, flg;
  if (read_bits(data, p, 8) != 0x1f || read_bits(data, p, 8) != 0x8b) {
    return -1;
  }
  cm = read_bits(data, p, 8);
  if (cm != 0x8) {
    return -1;
  }
  flg = read_bits(data, p, 8);
  if (flg & 0xe0)
    ;
  read_bits(data, p, 32);
  read_bits(data, p, 8);
  read_bits(data, p, 8);
  if (flg & 0x4) {
    unsigned int xlen;
    xlen = read_bits(data, p, 16);
    *p += ((long)xlen) * 8;
  }
  if (flg & 0x8) {
    while (read_bits(data, p, 8))
      ;
  }
  if (flg & 0x10) {
    while (read_bits(data, p, 8))
      ;
  }
  if (flg & 0x2)
    read_bits(data, p, 16);
  return inflate(data, p, callback);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<SizeWMov>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L14
  mov    $0x5,%eax
  jmp    L39
L14:
  cmpl   $0xff,-0x4(%rbp)
  jg     L20
  mov    $0x6,%eax
  jmp    L39
L20:
  mov    $0x9,%eax
L39:
  pop    %rbp
  ret
```

assistant: ```c
int SizeWMov(int x) {
  if (x == 0)
    return 5;
  else if (x < 256)
    return 6;
  else
    return 9;
}
```
user: What is the c source code of the assembly code below?

```asm
<SizeWMov>:
  endbr64
  mov    $0x5,%eax
  test   %edi,%edi
  je     L2
  cmp    $0x100,%edi
  setge  %al
  movzbl %al,%eax
  lea    0x6(%rax,%rax,2),%eax
L2:
  ret
```

assistant: ```c
int SizeWMov(int x) {
  if (x == 0)
    return 5;
  else if (x < 256)
    return 6;
  else
    return 9;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<SizeWMov>:
  endbr64
  mov    $0x5,%eax
  test   %edi,%edi
  je     L4
  xor    %eax,%eax
  cmp    $0x100,%edi
  setge  %al
  lea    0x6(%rax,%rax,2),%eax
L4:
  ret
```

assistant: ```c
int SizeWMov(int x) {
  if (x == 0)
    return 5;
  else if (x < 256)
    return 6;
  else
    return 9;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<SizeWMov>:
  endbr64
  mov    $0x5,%eax
  test   %edi,%edi
  je     L0
  xor    %eax,%eax
  cmp    $0x100,%edi
  setge  %al
  lea    0x6(%rax,%rax,2),%eax
L0:
  ret
```

assistant: ```c
int SizeWMov(int x) {
  if (x == 0)
    return 5;
  else if (x < 256)
    return 6;
  else
    return 9;
}
```
user: Translate the below assembly code into C programming language:

```asm
<p1evl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x28(%rbp)
  mov    %rdi,-0x30(%rbp)
  mov    %esi,-0x34(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movsd  (%rax),%xmm0
  movsd  -0x28(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
L7:
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x28(%rbp),%xmm1
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movsd  (%rax),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L7
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double p1evl(double x, double *coef, int N) {
  double ans;
  double *p;
  int i;
  p = coef;
  ans = x + *p++;
  i = N - 1;
  do {
    ans = ans * x + *p++;
  } while (--i);
  return ans;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<p1evl>:
  endbr64
  movapd %xmm0,%xmm1
  lea    0x8(%rdi),%rax
  addsd  (%rdi),%xmm0
  lea    -0x2(%rsi),%edx
  lea    0x10(%rdi,%rdx,8),%rdx
L7:
  mulsd  %xmm1,%xmm0
  add    $0x8,%rax
  addsd  -0x8(%rax),%xmm0
  cmp    %rdx,%rax
  jne    L7
  ret
```

assistant: ```c
double p1evl(double x, double *coef, int N) {
  double ans;
  double *p;
  int i;
  p = coef;
  ans = x + *p++;
  i = N - 1;
  do {
    ans = ans * x + *p++;
  } while (--i);
  return ans;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<p1evl>:
  endbr64
  movapd %xmm0,%xmm1
  movsd  (%rdi),%xmm0
  lea    -0x2(%rsi),%edx
  lea    0x8(%rdi),%rax
  lea    0x10(%rdi,%rdx,8),%rdx
  addsd  %xmm1,%xmm0
  nopl   0x0(%rax)
L9:
  mulsd  %xmm1,%xmm0
  add    $0x8,%rax
  addsd  -0x8(%rax),%xmm0
  cmp    %rdx,%rax
  jne    L9
  ret
```

assistant: ```c
double p1evl(double x, double *coef, int N) {
  double ans;
  double *p;
  int i;
  p = coef;
  ans = x + *p++;
  i = N - 1;
  do {
    ans = ans * x + *p++;
  } while (--i);
  return ans;
}
```
user: What is the c source code of the assembly code below?

```asm
<p1evl>:
  endbr64
  movapd %xmm0,%xmm1
  movsd  (%rdi),%xmm0
  lea    -0x2(%rsi),%edx
  lea    0x8(%rdi),%rax
  lea    0x10(%rdi,%rdx,8),%rdx
  addsd  %xmm1,%xmm0
  nopl   0x0(%rax)
L6:
  mulsd  %xmm1,%xmm0
  add    $0x8,%rax
  addsd  -0x8(%rax),%xmm0
  cmp    %rax,%rdx
  jne    L6
  ret
```

assistant: ```c
double p1evl(double x, double *coef, int N) {
  double ans;
  double *p;
  int i;
  p = coef;
  ans = x + *p++;
  i = N - 1;
  do {
    ans = ans * x + *p++;
  } while (--i);
  return ans;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mov_reg64_preg64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    $0x48,%edi
  call   <put8@plt>
  mov    $0x8b,%edi
  call   <put8@plt>
  mov    -0x4(%rbp),%eax
  shl    $0x3,%eax
  or     -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <put8@plt>
  nop
  leave
  ret
```

assistant: ```c
void mov_reg64_preg64(int reg1, int reg2) {
  put8(0x48);
  put8(0x8B);
  put8((reg1 << 3) | reg2);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mov_reg64_preg64>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    $0x48,%edi
  call   <put8@plt>
  mov    $0x8b,%edi
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  or     %ebp,%edi
  call   <put8@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void mov_reg64_preg64(int reg1, int reg2) {
  put8(0x48);
  put8(0x8B);
  put8((reg1 << 3) | reg2);
}
```