ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<gu_timeval_diff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    %rsi,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rbx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    %rbx,%rax
  sub    %rdx,%rax
  imul   $0xf4240,%rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rdx,%rbx
  sub    %rax,%rbx
  pxor   %xmm1,%xmm1
  cvtsi2sd %rbx,%xmm1
  movsd  D4(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_infered: byte[8]

tool: {"D4": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

assistant: ```c
double gu_timeval_diff(struct timeval *left, struct timeval *right) {
  register long long diff = left->tv_sec;
  diff = ((diff - right->tv_sec) * 1000000LL) + left->tv_usec - right->tv_usec;
  return (((double)diff) * 1.0e-06);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<gu_timeval_diff>:
  endbr64
  mov    (%rdi),%rax
  sub    (%rsi),%rax
  imul   $0xf4240,%rax,%rax
  add    0x8(%rdi),%rax
  sub    0x8(%rsi),%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  mulsd  D7(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: byte[8]

tool: {"D7": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

assistant: ```c
double gu_timeval_diff(struct timeval *left, struct timeval *right) {
  register long long diff = left->tv_sec;
  diff = ((diff - right->tv_sec) * 1000000LL) + left->tv_usec - right->tv_usec;
  return (((double)diff) * 1.0e-06);
}
```
user: Translate the below assembly code into C programming language:

```asm
<gu_timeval_diff>:
  endbr64
  mov    (%rdi),%rax
  sub    (%rsi),%rax
  pxor   %xmm0,%xmm0
  imul   $0xf4240,%rax,%rax
  add    0x8(%rdi),%rax
  sub    0x8(%rsi),%rax
  cvtsi2sd %rax,%xmm0
  mulsd  D3(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_infered: byte[8]

tool: {"D3": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

assistant: ```c
double gu_timeval_diff(struct timeval *left, struct timeval *right) {
  register long long diff = left->tv_sec;
  diff = ((diff - right->tv_sec) * 1000000LL) + left->tv_usec - right->tv_usec;
  return (((double)diff) * 1.0e-06);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<gu_timeval_diff>:
  endbr64
  mov    (%rdi),%rax
  sub    (%rsi),%rax
  pxor   %xmm0,%xmm0
  imul   $0xf4240,%rax,%rax
  add    0x8(%rdi),%rax
  sub    0x8(%rsi),%rax
  cvtsi2sd %rax,%xmm0
  mulsd  D5(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D5
type_guessed: byte[8]

tool: {"D5": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

assistant: ```c
double gu_timeval_diff(struct timeval *left, struct timeval *right) {
  register long long diff = left->tv_sec;
  diff = ((diff - right->tv_sec) * 1000000LL) + left->tv_usec - right->tv_usec;
  return (((double)diff) * 1.0e-06);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vec_scale>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    D8(%rip),%r12
  mov    $0x0,%ebx
  jmp    L6
L23:
  movslq %ebx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movslq %ebx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm1
  movslq %ebx,%rax
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  movslq %ebx,%rax
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movq   %r12,%xmm3
  comisd %xmm3,%xmm1
  jbe    L18
  movq   %xmm1,%r12
L18:
  add    $0x2,%ebx
L6:
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  cmp    %eax,%ebx
  jl     L23
  movq   %r12,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D8
type_guessed: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
double vec_scale(int no, double *vp) {
  register int i;
  register double scale, tmp;
  scale = 0;
  for (i = 0; i < 2 * no; i += 2) {
    tmp = ((vp[i]) * (vp[i])) + ((vp[i + 1]) * (vp[i + 1]));
    if (scale < tmp)
      scale = tmp;
  }
  return sqrt(scale);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vec_scale>:
  endbr64
  add    %edi,%edi
  test   %edi,%edi
  jle    L32
  mov    %rsi,%rax
  sub    $0x1,%edi
  shr    %edi
  mov    %edi,%edi
  shl    $0x4,%rdi
  lea    0x10(%rsi,%rdi,1),%rdx
  pxor   %xmm0,%xmm0
L15:
  movsd  (%rax),%xmm1
  movsd  0x8(%rax),%xmm2
  mulsd  %xmm1,%xmm1
  mulsd  %xmm2,%xmm2
  addsd  %xmm2,%xmm1
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L15
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L19
L23:
  sqrtsd %xmm0,%xmm0
  ret
L32:
  pxor   %xmm0,%xmm0
  jmp    L23
L19:
  sub    $0x8,%rsp
  call   <sqrt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
double vec_scale(int no, double *vp) {
  register int i;
  register double scale, tmp;
  scale = 0;
  for (i = 0; i < 2 * no; i += 2) {
    tmp = ((vp[i]) * (vp[i])) + ((vp[i + 1]) * (vp[i + 1]));
    if (scale < tmp)
      scale = tmp;
  }
  return sqrt(scale);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vec_scale>:
  endbr64
  add    %edi,%edi
  test   %edi,%edi
  jle    L26
  sub    $0x1,%edi
  pxor   %xmm3,%xmm3
  shr    %edi
  movapd %xmm3,%xmm0
  shl    $0x4,%rdi
  lea    0x10(%rsi,%rdi,1),%rax
L25:
  movsd  (%rsi),%xmm1
  movsd  0x8(%rsi),%xmm2
  add    $0x10,%rsi
  mulsd  %xmm1,%xmm1
  mulsd  %xmm2,%xmm2
  addsd  %xmm2,%xmm1
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  cmp    %rsi,%rax
  jne    L25
  ucomisd %xmm1,%xmm3
  ja     L22
  sqrtsd %xmm0,%xmm0
  ret
  nopl   0x0(%rax)
L26:
  pxor   %xmm0,%xmm0
  sqrtsd %xmm0,%xmm0
  ret
L22:
  jmp    1050 <sqrt@plt>
```

assistant: ```c
double vec_scale(int no, double *vp) {
  register int i;
  register double scale, tmp;
  scale = 0;
  for (i = 0; i < 2 * no; i += 2) {
    tmp = ((vp[i]) * (vp[i])) + ((vp[i + 1]) * (vp[i + 1]));
    if (scale < tmp)
      scale = tmp;
  }
  return sqrt(scale);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vec_scale>:
  endbr64
  add    %edi,%edi
  test   %edi,%edi
  jle    L12
  sub    $0x1,%edi
  pxor   %xmm3,%xmm3
  shr    %edi
  movapd %xmm3,%xmm0
  shl    $0x4,%rdi
  lea    0x10(%rsi,%rdi,1),%rax
L6:
  movsd  (%rsi),%xmm2
  movsd  0x8(%rsi),%xmm1
  add    $0x10,%rsi
  mulsd  %xmm1,%xmm1
  mulsd  %xmm2,%xmm2
  addsd  %xmm2,%xmm1
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  cmp    %rsi,%rax
  jne    L6
  ucomisd %xmm1,%xmm3
  ja     L9
  sqrtsd %xmm0,%xmm0
  ret
  nopl   0x0(%rax)
L12:
  pxor   %xmm0,%xmm0
  sqrtsd %xmm0,%xmm0
  ret
L9:
  jmp    1050 <sqrt@plt>
```

assistant: ```c
double vec_scale(int no, double *vp) {
  register int i;
  register double scale, tmp;
  scale = 0;
  for (i = 0; i < 2 * no; i += 2) {
    tmp = ((vp[i]) * (vp[i])) + ((vp[i + 1]) * (vp[i + 1]));
    if (scale < tmp)
      scale = tmp;
  }
  return sqrt(scale);
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L15
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L34
L15:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <scan@plt>
  test   %rax,%rax
  jne    L17
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L34
L17:
  mov    $0x0,%eax
L34:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D16
data_type: string
name: extract_symbolic_data
arguments: label: D7
data_type: string

tool: D16: scanner: no input file.


tool: D7: Can't read input file!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc <= 1) {
    printf("scanner: no input file.\n");
    return -1;
  }
  if (scan(argv[1]) == 0) {
    printf("Can\'t read input file!\n");
    return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L22
  mov    0x8(%rsi),%rdi
  call   <scan@plt>
  test   %rax,%rax
  je     L3
  mov    $0x0,%eax
L10:
  add    $0x8,%rsp
  ret
L22:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L10
L3:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L10
```

tool_calls: name: retrieve_data
arguments: label: D7
data_type: string
name: retrieve_data
arguments: label: D17
data_type: string

tool: D7: scanner: no input file.


tool: D17: Can't read input file!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc <= 1) {
    printf("scanner: no input file.\n");
    return -1;
  }
  if (scan(argv[1]) == 0) {
    printf("Can\'t read input file!\n");
    return -1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L30
  mov    0x8(%rsi),%rdi
  call   <scan@plt>
  test   %rax,%rax
  je     L19
  xor    %eax,%eax
L21:
  add    $0x8,%rsp
  ret
L30:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L21
L19:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L21
```

tool_calls: name: parse_data
arguments: data_tag: D18
type_infered: string
name: parse_data
arguments: data_tag: D4
type_infered: string

tool: D18: scanner: no input file.


tool: D4: Can't read input file!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc <= 1) {
    printf("scanner: no input file.\n");
    return -1;
  }
  if (scan(argv[1]) == 0) {
    printf("Can\'t read input file!\n");
    return -1;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L6
  mov    0x8(%rsi),%rdi
  call   <scan@plt>
  test   %rax,%rax
  je     L24
  xor    %eax,%eax
L0:
  add    $0x8,%rsp
  ret
L6:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L0
L24:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L0
```

tool_calls: name: unpack_data
arguments: label: D4
data_type: string
name: unpack_data
arguments: label: D0
data_type: string

tool: D4: scanner: no input file.


tool: D0: Can't read input file!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc <= 1) {
    printf("scanner: no input file.\n");
    return -1;
  }
  if (scan(argv[1]) == 0) {
    printf("Can\'t read input file!\n");
    return -1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<andProduct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L7
L15:
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x18(%rbp)
  movl   $0x80000000,-0x14(%rbp)
  jmp    L62
L29:
  mov    -0x20(%rbp),%eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jne    L36
  mov    -0xc(%rbp),%eax
  or     %eax,-0x18(%rbp)
  sarl   -0x14(%rbp)
L62:
  cmpl   $0x0,-0x14(%rbp)
  jg     L29
  jmp    L68
L36:
  nop
L68:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L7:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jne    L15
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D21
type: string
name: retrieve_data
arguments: tag: D25
type: string
name: retrieve_data
arguments: tag: D27
type: string

tool: D21: %d

tool: D25: %u %u

tool: D27: %u


assistant: ```c
int andProduct(void) {
  int numTestCases;
  scanf("%d", &numTestCases);
  while (numTestCases--) {
    uint low, high;
    scanf("%u %u", &low, &high);
    uint sharedBits = 0;
    for (uint bitIter = 1 << 31; bitIter > 0; bitIter >>= 1) {
      uint lowBitValue = low & bitIter;
      uint highBitValue = high & bitIter;
      if (lowBitValue != highBitValue) {
        break;
      } else {
        sharedBits |= highBitValue;
      }
    }
    printf("%u\n", sharedBits);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<andProduct>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L3
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D24(%rip),%rbp
  lea    D1(%rip),%rbx
L20:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L20
L3:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D7
data_type: string
name: disassemble_data
arguments: data_label: D24
data_type: string
name: disassemble_data
arguments: data_label: D1
data_type: string

tool: D7: %d

tool: D24: %u %u

tool: D1: %u


assistant: ```c
int andProduct(void) {
  int numTestCases;
  scanf("%d", &numTestCases);
  while (numTestCases--) {
    uint low, high;
    scanf("%u %u", &low, &high);
    uint sharedBits = 0;
    for (uint bitIter = 1 << 31; bitIter > 0; bitIter >>= 1) {
      uint lowBitValue = low & bitIter;
      uint highBitValue = high & bitIter;
      if (lowBitValue != highBitValue) {
        break;
      } else {
        sharedBits |= highBitValue;
      }
    }
    printf("%u\n", sharedBits);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<andProduct>:
  endbr64
  push   %r13
  lea    D1(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L3
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D15(%rip),%rbp
  lea    D3(%rip),%rbx
  cs nopw 0x0(%rax,%rax,1)
L21:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L21
L3:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
type_guessed: string
name: disassemble_data
arguments: data_label: D15
type_guessed: string
name: disassemble_data
arguments: data_label: D3
type_guessed: string

tool: D1: %d

tool: D15: %u %u

tool: D3: %u


assistant: ```c
int andProduct(void) {
  int numTestCases;
  scanf("%d", &numTestCases);
  while (numTestCases--) {
    uint low, high;
    scanf("%u %u", &low, &high);
    uint sharedBits = 0;
    for (uint bitIter = 1 << 31; bitIter > 0; bitIter >>= 1) {
      uint lowBitValue = low & bitIter;
      uint highBitValue = high & bitIter;
      if (lowBitValue != highBitValue) {
        break;
      } else {
        sharedBits |= highBitValue;
      }
    }
    printf("%u\n", sharedBits);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<andProduct>:
  endbr64
  push   %r13
  lea    D3(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L6
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D10(%rip),%rbp
  lea    D29(%rip),%rbx
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L9
L6:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
datatype: string
name: retrieve_data
arguments: data_tag: D10
datatype: string
name: retrieve_data
arguments: data_tag: D29
datatype: string

tool: D3: %d

tool: D10: %u %u

tool: D29: %u


assistant: ```c
int andProduct(void) {
  int numTestCases;
  scanf("%d", &numTestCases);
  while (numTestCases--) {
    uint low, high;
    scanf("%u %u", &low, &high);
    uint sharedBits = 0;
    for (uint bitIter = 1 << 31; bitIter > 0; bitIter >>= 1) {
      uint lowBitValue = low & bitIter;
      uint highBitValue = high & bitIter;
      if (lowBitValue != highBitValue) {
        break;
      } else {
        sharedBits |= highBitValue;
      }
    }
    printf("%u\n", sharedBits);
  }
  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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L5
L14:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <possible_trees@plt>
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L5:
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L14
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
type_infered: string
name: extract_symbolic_data
arguments: label: D5
type_infered: string
name: extract_symbolic_data
arguments: label: D15
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string

tool: D14: Enter a tree size N to see how many trees exist such that
the inorder traversal yields [0, 1, ... N-1]

tool: D5: > 

tool: D15: %u


tool: D19: %zu

assistant: ```c
int main(void) {
  printf("Enter a tree size N to see how many trees exist such that\n"
         "the inorder traversal yields [0, 1, ... N-1]\n");
  printf("> ");
  size_t tree_sz;
  while (scanf("%zu", &tree_sz) == 1) {
    printf("%u\n", possible_trees(tree_sz));
    printf("> ");
  }
  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   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D47(%rip),%rdi
  call   <puts@plt>
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rbp
  lea    D16(%rip),%rbx
  lea    D21(%rip),%r13
  lea    D2(%rip),%r12
  jmp    L25
L19:
  mov    (%rsp),%rdi
  call   <possible_trees@plt>
  mov    %eax,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L25:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L19
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D47
type_guessed: string
name: disassemble_data
arguments: data_tag: D2
type_guessed: string
name: disassemble_data
arguments: data_tag: D16
type_guessed: string
name: disassemble_data
arguments: data_tag: D21
type_guessed: string

tool: D47: Enter a tree size N to see how many trees exist such that
the inorder traversal yields [0, 1, ... N-1]

tool: D2: > 

tool: D16: %zu

tool: D21: %u


assistant: ```c
int main(void) {
  printf("Enter a tree size N to see how many trees exist such that\n"
         "the inorder traversal yields [0, 1, ... N-1]\n");
  printf("> ");
  size_t tree_sz;
  while (scanf("%zu", &tree_sz) == 1) {
    printf("%u\n", possible_trees(tree_sz));
    printf("> ");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%rdi
  lea    D1(%rip),%r13
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D33(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <puts@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L28
  xchg   %ax,%ax
L16:
  mov    (%rsp),%rdi
  call   <possible_trees@plt>
  mov    %r13,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L28:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L16
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
data_type: string
name: retrieve_data
arguments: data_tag: D1
data_type: string
name: retrieve_data
arguments: data_tag: D28
data_type: string
name: retrieve_data
arguments: data_tag: D33
data_type: string

tool: D0: Enter a tree size N to see how many trees exist such that
the inorder traversal yields [0, 1, ... N-1]

tool: D1: %u


tool: D28: > 

tool: D33: %zu

assistant: ```c
int main(void) {
  printf("Enter a tree size N to see how many trees exist such that\n"
         "the inorder traversal yields [0, 1, ... N-1]\n");
  printf("> ");
  size_t tree_sz;
  while (scanf("%zu", &tree_sz) == 1) {
    printf("%u\n", possible_trees(tree_sz));
    printf("> ");
  }
  return 0;
}
```