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
<ez_writer_write_track>:
  endbr64
  push   %r12
  and    $0xe,%esi
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x2,%edx
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %sil,0x7(%rsp)
  lea    0x6(%rsp),%rsi
  movb   $0x1b,0x6(%rsp)
  call   <serial_port_write@plt>
  test   %eax,%eax
  jne    L4
L34:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L4:
  xor    %esi,%esi
  mov    %rbx,%rdx
  mov    %rbp,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  je     L37
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
L37:
  mov    $0x1,%eax
  test   %rbx,%rbx
  je     L34
  lea    -0x2(%rbx),%rdx
  lea    0x1(%rbp),%rsi
  mov    %r12,%rdi
  call   <serial_port_write@plt>
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  jmp    L34
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool ez_writer_write_track(struct serial_port *sport, unsigned track, const char *trackdata, size_t len) {
  char track_begin[] = {'\x1b', track & (((1) << ((1) & (1 | 2 | 3))) | ((1) << ((2) & (1 | 2 | 3))) | ((1) << ((3) & (1 | 2 | 3))))};
  if (!serial_port_write(sport, track_begin, sizeof track_begin / sizeof track_begin[0]))
    return (false);
  if (memchr(trackdata, '\0', len) != NULL)
    len = strlen(trackdata);
  if (len == 0)
    return (true);
  if (!serial_port_write(sport, trackdata + 1, len - 2))
    return (false);
  return (true);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fso>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x4(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%r8d
  lea    D36(%rip),%rax
  mov    %rax,%rcx
  lea    D29(%rip),%rax
  mov    %rax,%rdx
  lea    D33(%rip),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x10,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D36
type: string
name: retrieve_data
arguments: label: D29
type: string
name: retrieve_data
arguments: label: D33
type: string
name: retrieve_data
arguments: label: D16
type: string

tool: D36: 16:29:08

tool: D29: Mar 15 2022

tool: D33: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c

tool: D16: %s: %s %s This is a shared functio - 3asd - %f


assistant: ```c
int fso(int a) {
  printf("%s: %s %s This is a shared functio - 3asd - %f\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c", "Mar 15 2022", "16:29:08", a * 1.0);
  return 16;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fso>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%r8d
  lea    D1(%rip),%rcx
  lea    D16(%rip),%rdx
  lea    D37(%rip),%rsi
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    $0x10,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type_guessed: string
name: unpack_data
arguments: tag: D16
type_guessed: string
name: unpack_data
arguments: tag: D37
type_guessed: string
name: unpack_data
arguments: tag: D27
type_guessed: string

tool: D1: 16:29:08

tool: D16: Mar 15 2022

tool: D37: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c

tool: D27: %s: %s %s This is a shared functio - 3asd - %f


assistant: ```c
int fso(int a) {
  printf("%s: %s %s This is a shared functio - 3asd - %f\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c", "Mar 15 2022", "16:29:08", a * 1.0);
  return 16;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fso>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%r8d
  lea    D16(%rip),%rcx
  lea    D39(%rip),%rdx
  lea    D25(%rip),%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x10,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D16
datatype: string
name: parse_data
arguments: label: D39
datatype: string
name: parse_data
arguments: label: D25
datatype: string
name: parse_data
arguments: label: D7
datatype: string

tool: D16: 16:29:08

tool: D39: Mar 15 2022

tool: D25: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c

tool: D7: %s: %s %s This is a shared functio - 3asd - %f


assistant: ```c
int fso(int a) {
  printf("%s: %s %s This is a shared functio - 3asd - %f\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c", "Mar 15 2022", "16:29:08", a * 1.0);
  return 16;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fso>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%r8d
  lea    D31(%rip),%rcx
  lea    D8(%rip),%rdx
  lea    D16(%rip),%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x10,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D31
type: string
name: read_data
arguments: tag: D8
type: string
name: read_data
arguments: tag: D16
type: string
name: read_data
arguments: tag: D4
type: string

tool: D31: 16:29:08

tool: D8: Mar 15 2022

tool: D16: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c

tool: D4: %s: %s %s This is a shared functio - 3asd - %f


assistant: ```c
int fso(int a) {
  printf("%s: %s %s This is a shared functio - 3asd - %f\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/jarogames/gregory/refs/heads/master/concepts/cpp/Funq.c", "Mar 15 2022", "16:29:08", a * 1.0);
  return 16;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<shellsort2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L48
L68:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L48:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L68
  jmp    L37
L10:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L44
L83:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L24
L19:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  sub    %eax,-0xc(%rbp)
L24:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L6
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jl     L19
L6:
  addl   $0x1,-0x10(%rbp)
L44:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L83
L37:
  cmpl   $0x0,-0x8(%rbp)
  jg     L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void shellsort2(int v[], int n) {
  int i, j, temp, h;
  h = 1;
  while (h < n / 3)
    h = 3 * h + 1;
  while (h >= 1) {
    for (i = h; i < n; i++) {
      for (j = i; j >= h && v[i] < v[j - h]; j -= h) {
        temp = v[j];
        v[j] = v[j - h];
        v[j - h] = temp;
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<shellsort2>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  movslq %esi,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  cmp    $0x5,%esi
  jle    L105
  mov    $0x1,%r9d
L15:
  lea    0x1(%r9,%r9,2),%r9d
  cmp    %eax,%r9d
  jl     L15
  test   %r9d,%r9d
  jle    L51
L11:
  movslq %r9d,%rbp
  lea    (%rdi,%rbp,4),%r12
  mov    %rbp,%rax
  neg    %rax
  lea    (%rdi,%rax,4),%rcx
  mov    %r9d,%r10d
  neg    %r10d
  movslq %r10d,%r10
L74:
  mov    %r12,%r8
  mov    %rbp,%r11
  cmp    %r9d,%ebx
  jg     L108
  jmp    L74
L100:
  add    $0x1,%r11
  add    $0x4,%r8
  cmp    %r11d,%ebx
  jle    L74
L108:
  cmp    %r11d,%r9d
  jg     L100
  mov    %r11,%rax
L96:
  mov    (%rcx,%rax,4),%edx
  cmp    %edx,(%r8)
  jge    L100
  mov    (%rdi,%rax,4),%esi
  mov    %edx,(%rdi,%rax,4)
  mov    %esi,(%rcx,%rax,4)
  add    %r10,%rax
  cmp    %eax,%r9d
  jle    L96
  jmp    L100
L105:
  mov    $0x1,%r9d
  jmp    L11
L51:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void shellsort2(int v[], int n) {
  int i, j, temp, h;
  h = 1;
  while (h < n / 3)
    h = 3 * h + 1;
  while (h >= 1) {
    for (i = h; i < n; i++) {
      for (j = i; j >= h && v[i] < v[j - h]; j -= h) {
        temp = v[j];
        v[j] = v[j - h];
        v[j - h] = temp;
      }
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<shellsort2>:
  endbr64
  movslq %esi,%rax
  push   %r12
  mov    $0x1,%r9d
  push   %rbp
  push   %rbx
  mov    %rax,%rbx
  imul   $0x55555556,%rax,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  cmp    $0x5,%ebx
  jle    L37
  nopl   0x0(%rax,%rax,1)
L60:
  lea    0x1(%r9,%r9,2),%r9d
  cmp    %eax,%r9d
  jl     L60
L37:
  movslq %r9d,%rbp
  mov    %r9d,%r10d
  mov    %rdi,%rcx
  lea    0x0(,%rbp,4),%rax
  neg    %r10d
  lea    (%rdi,%rax,1),%r12
  sub    %rax,%rcx
  movslq %r10d,%r10
  nopl   0x0(%rax,%rax,1)
L21:
  mov    %r12,%r8
  mov    %rbp,%r11
  cmp    %r9d,%ebx
  jle    L21
  nopl   0x0(%rax,%rax,1)
L49:
  cmp    %r11d,%r9d
  jg     L13
  mov    %r11,%rax
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L25:
  mov    (%rdi,%rax,4),%esi
  mov    %edx,(%rdi,%rax,4)
  mov    %esi,(%rcx,%rax,4)
  add    %r10,%rax
  cmp    %eax,%r9d
  jg     L13
L40:
  mov    (%rcx,%rax,4),%edx
  cmp    %edx,(%r8)
  jl     L25
L13:
  add    $0x1,%r11
  add    $0x4,%r8
  cmp    %r11d,%ebx
  jg     L49
  jmp    L21
```

assistant: ```c
void shellsort2(int v[], int n) {
  int i, j, temp, h;
  h = 1;
  while (h < n / 3)
    h = 3 * h + 1;
  while (h >= 1) {
    for (i = h; i < n; i++) {
      for (j = i; j >= h && v[i] < v[j - h]; j -= h) {
        temp = v[j];
        v[j] = v[j - h];
        v[j - h] = temp;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<shellsort2>:
  endbr64
  movslq %esi,%rax
  push   %r12
  mov    $0x1,%r9d
  push   %rbp
  push   %rbx
  mov    %rax,%rbx
  imul   $0x55555556,%rax,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  cmp    $0x5,%ebx
  jle    L19
  nopl   0x0(%rax,%rax,1)
L20:
  lea    0x1(%r9,%r9,2),%r9d
  cmp    %eax,%r9d
  jl     L20
L19:
  movslq %r9d,%rbp
  mov    %r9d,%r10d
  mov    %rdi,%rcx
  lea    0x0(,%rbp,4),%rax
  neg    %r10d
  lea    (%rdi,%rax,1),%r12
  sub    %rax,%rcx
  movslq %r10d,%r10
  nopl   0x0(%rax,%rax,1)
L65:
  mov    %r12,%r8
  mov    %rbp,%r11
  cmp    %r9d,%ebx
  jle    L65
  nopl   0x0(%rax,%rax,1)
L49:
  mov    %r11,%rax
  jmp    L31
  nopl   (%rax)
L22:
  mov    (%rdi,%rax,4),%esi
  mov    %edx,(%rdi,%rax,4)
  mov    %esi,(%rcx,%rax,4)
  add    %r10,%rax
  cmp    %eax,%r9d
  jg     L30
L31:
  mov    (%rcx,%rax,4),%edx
  cmp    (%r8),%edx
  jg     L22
L30:
  add    $0x1,%r11
  add    $0x4,%r8
  cmp    %r11d,%ebx
  jg     L49
  jmp    L65
```

assistant: ```c
void shellsort2(int v[], int n) {
  int i, j, temp, h;
  h = 1;
  while (h < n / 3)
    h = 3 * h + 1;
  while (h >= 1) {
    for (i = h; i < n; i++) {
      for (j = i; j >= h && v[i] < v[j - h]; j -= h) {
        temp = v[j];
        v[j] = v[j - h];
        v[j - h] = temp;
      }
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mul8bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  sub    $0xffffff80,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  sar    $0x8,%eax
  pop    %rbp
  ret
```

assistant: ```c
int mul8bit(int a, int b) {
  int t = a * b + 128;
  return ((t + (t >> 8)) >> 8);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mul8bit>:
  endbr64
  imul   %esi,%edi
  sub    $0xffffff80,%edi
  mov    %edi,%eax
  sar    $0x8,%eax
  add    %edi,%eax
  sar    $0x8,%eax
  ret
```

assistant: ```c
int mul8bit(int a, int b) {
  int t = a * b + 128;
  return ((t + (t >> 8)) >> 8);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mul8bit>:
  endbr64
  imul   %esi,%edi
  sub    $0xffffff80,%edi
  mov    %edi,%eax
  sar    $0x8,%eax
  add    %edi,%eax
  sar    $0x8,%eax
  ret
```

assistant: ```c
int mul8bit(int a, int b) {
  int t = a * b + 128;
  return ((t + (t >> 8)) >> 8);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<mul8bit>:
  endbr64
  imul   %esi,%edi
  sub    $0xffffff80,%edi
  mov    %edi,%eax
  sar    $0x8,%eax
  add    %edi,%eax
  sar    $0x8,%eax
  ret
```

assistant: ```c
int mul8bit(int a, int b) {
  int t = a * b + 128;
  return ((t + (t >> 8)) >> 8);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x1,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  jmp    L66
L46:
  mov    -0x48(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L70
  addq   $0x1,-0x40(%rbp)
  jmp    L124
L70:
  mov    -0x38(%rbp),%rax
  imul   -0x8(%rbp),%rax
  mov    %rax,-0x38(%rbp)
L124:
  addq   $0x1,-0x30(%rbp)
L66:
  mov    -0x30(%rbp),%rax
  cmp    -0x58(%rbp),%rax
  jb     L46
  cmpq   $0x0,-0x40(%rbp)
  jne    L18
  mov    -0x38(%rbp),%rax
  jmp    L32
L18:
  mov    $0x0,%eax
L32:
  mov    %rax,-0x28(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L97
L80:
  mov    -0x20(%rbp),%rax
  sub    -0x58(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  mov    -0x48(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L19
  subq   $0x1,-0x40(%rbp)
  jmp    L112
L19:
  mov    -0x38(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x18(%rbp)
  mov    %rax,-0x38(%rbp)
L112:
  cmpq   $0x0,-0x10(%rbp)
  jne    L111
  addq   $0x1,-0x40(%rbp)
  jmp    L15
L111:
  mov    -0x38(%rbp),%rax
  imul   -0x10(%rbp),%rax
  mov    %rax,-0x38(%rbp)
L15:
  cmpq   $0x0,-0x40(%rbp)
  jne    L5
  mov    -0x38(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jbe    L5
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x28(%rbp)
L5:
  addq   $0x1,-0x20(%rbp)
L97:
  mov    -0x20(%rbp),%rax
  cmp    -0x50(%rbp),%rax
  jb     L80
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t solve(const char *number, size_t n, uint64_t k) {
  uint64_t zeroes = 0;
  uint64_t x = 1;
  for (size_t i = 0; i < k; ++i) {
    uint64_t d = number[i] - '0';
    if (d == 0)
      zeroes += 1;
    else
      x *= d;
  }
  uint64_t result = zeroes ? 0 : x;
  for (size_t i = k; i < n; ++i) {
    uint64_t d0 = number[i - k] - '0';
    uint64_t d1 = number[i] - '0';
    if (d0 == 0)
      zeroes -= 1;
    else
      x /= d0;
    if (d1 == 0)
      zeroes += 1;
    else
      x *= d1;
    if (!zeroes && x > result)
      result = x;
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<solve>:
  endbr64
  mov    %rdi,%r11
  mov    %rdx,%r9
  test   %rdx,%rdx
  je     L99
  mov    %rdi,%rcx
  lea    (%rdi,%rdx,1),%rdi
  mov    $0x1,%eax
  mov    $0x0,%r8d
  jmp    L128
L40:
  imul   %rdx,%rax
L122:
  add    $0x1,%rcx
  cmp    %rdi,%rcx
  je     L68
L128:
  movsbl (%rcx),%edx
  sub    $0x30,%edx
  movslq %edx,%rdx
  test   %rdx,%rdx
  jne    L40
  add    $0x1,%r8
  jmp    L122
L68:
  test   %r8,%r8
  mov    $0x0,%edi
  cmove  %rax,%rdi
L113:
  cmp    %rsi,%r9
  jae    L61
  mov    %r11,%r10
  add    %rsi,%r11
  sub    %r9,%r11
  jmp    L76
L99:
  mov    %rdx,%r8
  mov    $0x1,%eax
  mov    $0x1,%edi
  jmp    L113
L73:
  mov    $0x0,%edx
  div    %rcx
  jmp    L102
L58:
  imul   %rsi,%rax
L30:
  mov    %rdi,%rdx
  cmp    %rdi,%rax
  cmovae %rax,%rdi
  test   %r8,%r8
  cmovne %rdx,%rdi
  add    $0x1,%r10
  cmp    %r10,%r11
  je     L61
L76:
  movsbl (%r10),%ecx
  sub    $0x30,%ecx
  movslq %ecx,%rcx
  movsbl (%r10,%r9,1),%esi
  sub    $0x30,%esi
  movslq %esi,%rsi
  test   %rcx,%rcx
  jne    L73
  sub    $0x1,%r8
L102:
  test   %rsi,%rsi
  jne    L58
  add    $0x1,%r8
  jmp    L30
L61:
  mov    %rdi,%rax
  ret
```

assistant: ```c
uint64_t solve(const char *number, size_t n, uint64_t k) {
  uint64_t zeroes = 0;
  uint64_t x = 1;
  for (size_t i = 0; i < k; ++i) {
    uint64_t d = number[i] - '0';
    if (d == 0)
      zeroes += 1;
    else
      x *= d;
  }
  uint64_t result = zeroes ? 0 : x;
  for (size_t i = k; i < n; ++i) {
    uint64_t d0 = number[i - k] - '0';
    uint64_t d1 = number[i] - '0';
    if (d0 == 0)
      zeroes -= 1;
    else
      x /= d0;
    if (d1 == 0)
      zeroes += 1;
    else
      x *= d1;
    if (!zeroes && x > result)
      result = x;
  }
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<solve>:
  endbr64
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  mov    %rdx,%r9
  test   %rdx,%rdx
  je     L54
  mov    %rdi,%rcx
  lea    (%rdx,%rdi,1),%rsi
  mov    $0x1,%eax
  xor    %edi,%edi
  jmp    L44
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rcx
  add    $0x1,%rdi
  cmp    %rcx,%rsi
  je     L105
L44:
  movsbl (%rcx),%edx
  sub    $0x30,%edx
  movslq %edx,%rdx
  test   %rdx,%rdx
  je     L1
  add    $0x1,%rcx
  imul   %rdx,%rax
  cmp    %rcx,%rsi
  jne    L44
L105:
  xor    %r8d,%r8d
  test   %rdi,%rdi
  mov    %r9,%r10
  cmove  %rax,%r8
L22:
  cmp    %r10,%rbx
  jbe    L38
  sub    %r9,%r10
  add    %r11,%r10
  add    %rbx,%r11
  sub    %r9,%r11
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L70:
  sub    $0x1,%rdi
  test   %rsi,%rsi
  jne    L4
L41:
  add    $0x1,%rdi
L3:
  cmp    %r8,%rax
  mov    %r8,%rdx
  cmovae %rax,%r8
  test   %rdi,%rdi
  cmovne %rdx,%r8
  add    $0x1,%r10
  cmp    %r10,%r11
  je     L38
L69:
  movsbl (%r10),%ecx
  movsbl (%r10,%r9,1),%esi
  sub    $0x30,%ecx
  sub    $0x30,%esi
  movslq %ecx,%rcx
  movslq %esi,%rsi
  test   %rcx,%rcx
  je     L70
  xor    %edx,%edx
  div    %rcx
  test   %rsi,%rsi
  je     L41
L4:
  imul   %rsi,%rax
  jmp    L3
  nopl   0x0(%rax)
L38:
  mov    %r8,%rax
  pop    %rbx
  ret
  nopl   (%rax)
L54:
  xor    %r10d,%r10d
  xor    %edi,%edi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  jmp    L22
```

assistant: ```c
uint64_t solve(const char *number, size_t n, uint64_t k) {
  uint64_t zeroes = 0;
  uint64_t x = 1;
  for (size_t i = 0; i < k; ++i) {
    uint64_t d = number[i] - '0';
    if (d == 0)
      zeroes += 1;
    else
      x *= d;
  }
  uint64_t result = zeroes ? 0 : x;
  for (size_t i = k; i < n; ++i) {
    uint64_t d0 = number[i - k] - '0';
    uint64_t d1 = number[i] - '0';
    if (d0 == 0)
      zeroes -= 1;
    else
      x /= d0;
    if (d1 == 0)
      zeroes += 1;
    else
      x *= d1;
    if (!zeroes && x > result)
      result = x;
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<solve>:
  endbr64
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  mov    %rdx,%r9
  test   %rdx,%rdx
  je     L62
  mov    %rdi,%rcx
  lea    (%rdi,%rdx,1),%rsi
  mov    $0x1,%eax
  xor    %edi,%edi
  jmp    L110
  nopw   0x0(%rax,%rax,1)
L9:
  add    $0x1,%rcx
  add    $0x1,%rdi
  cmp    %rsi,%rcx
  je     L82
L110:
  movsbl (%rcx),%edx
  sub    $0x30,%edx
  movslq %edx,%rdx
  test   %rdx,%rdx
  je     L9
  add    $0x1,%rcx
  imul   %rdx,%rax
  cmp    %rsi,%rcx
  jne    L110
L82:
  xor    %r8d,%r8d
  test   %rdi,%rdi
  mov    %r9,%r10
  cmove  %rax,%r8
L41:
  cmp    %r10,%rbx
  jbe    L111
  sub    %r9,%r10
  add    %r11,%r10
  add    %rbx,%r11
  sub    %r9,%r11
  jmp    L96
  nopl   0x0(%rax,%rax,1)
L87:
  sub    $0x1,%rdi
  test   %rsi,%rsi
  jne    L34
L127:
  add    $0x1,%rdi
L118:
  cmp    %r8,%rax
  mov    %r8,%rdx
  cmovae %rax,%r8
  test   %rdi,%rdi
  cmovne %rdx,%r8
  add    $0x1,%r10
  cmp    %r10,%r11
  je     L111
L96:
  movsbl (%r10),%ecx
  movsbl (%r10,%r9,1),%esi
  sub    $0x30,%ecx
  sub    $0x30,%esi
  movslq %ecx,%rcx
  movslq %esi,%rsi
  test   %rcx,%rcx
  je     L87
  xor    %edx,%edx
  div    %rcx
  test   %rsi,%rsi
  je     L127
L34:
  imul   %rsi,%rax
  jmp    L118
  nopl   0x0(%rax)
L111:
  mov    %r8,%rax
  pop    %rbx
  ret
  nopl   (%rax)
L62:
  xor    %r10d,%r10d
  xor    %edi,%edi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  jmp    L41
```

assistant: ```c
uint64_t solve(const char *number, size_t n, uint64_t k) {
  uint64_t zeroes = 0;
  uint64_t x = 1;
  for (size_t i = 0; i < k; ++i) {
    uint64_t d = number[i] - '0';
    if (d == 0)
      zeroes += 1;
    else
      x *= d;
  }
  uint64_t result = zeroes ? 0 : x;
  for (size_t i = k; i < n; ++i) {
    uint64_t d0 = number[i - k] - '0';
    uint64_t d1 = number[i] - '0';
    if (d0 == 0)
      zeroes -= 1;
    else
      x /= d0;
    if (d1 == 0)
      zeroes += 1;
    else
      x *= d1;
    if (!zeroes && x > result)
      result = x;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<locatesubstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L57
L58:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L9
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L48
L44:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L54
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L7
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L44
  jmp    L7
L54:
  nop
L7:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x18(%rbp),%rax
  jmp    L8
L9:
  addq   $0x1,-0x18(%rbp)
L57:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
  mov    $0x0,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
char *locatesubstr(char *str1, char *str2) {
  char *p1, *p2;
  for (; *str1; str1++)
    if (*str1 == *str2) {
      for (p1 = str1, p2 = str2; *p2 && *p1; p1++, p2++)
        if (*p1 != *p2)
          break;
      if (*p2 == 0)
        return str1;
    }
  return 0;
}
```