ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  je     L17
  mov    (%rsi),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  mov    0x8(%rsi),%rsi
  xor    %r12d,%r12d
  mov    $0xa,%edx
  xor    %r13d,%r13d
  mov    %r12d,%edi
  xor    %ebp,%ebp
  call   <mpz_init_set_str@plt>
  mov    0x10(%rbx),%rsi
  mov    $0xa,%edx
  mov    %r13d,%edi
  call   <mpz_init_set_str@plt>
  mov    %ebp,%edi
  call   <mpz_init@plt>
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <mpz_mul@plt>
  mov    %ebp,%esi
  lea    D13(%rip),%rdi
  call   <gmp_printf@plt>
  xor    %eax,%eax
  jmp    L9
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
datatype: string
name: extract_symbolic_data
arguments: data_tag: D13
datatype: string

tool: D5: Usage: %s <number> <number>


tool: D13: Product is: %Zd


assistant: ```c
int main(int argc, char **argv) {
  mpz_t a, b, p;
  if (argc != 3) {
    printf("Usage: %s <number> <number>\n", argv[0]);
    return 1;
  }
  mpz_init_set_str(a, argv[1], 10);
  mpz_init_set_str(b, argv[2], 10);
  mpz_init(p);
  mpz_mul(p, a, b);
  gmp_printf("Product is: %Zd\n", p);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  je     L9
  mov    (%rsi),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  mov    0x8(%rsi),%rsi
  xor    %r12d,%r12d
  mov    $0xa,%edx
  xor    %r13d,%r13d
  mov    %r12d,%edi
  xor    %ebp,%ebp
  call   <mpz_init_set_str@plt>
  mov    0x10(%rbx),%rsi
  mov    $0xa,%edx
  mov    %r13d,%edi
  call   <mpz_init_set_str@plt>
  mov    %ebp,%edi
  call   <mpz_init@plt>
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <mpz_mul@plt>
  mov    %ebp,%esi
  lea    D0(%rip),%rdi
  call   <gmp_printf@plt>
  xor    %eax,%eax
  jmp    L8
```

tool_calls: name: retrieve_data
arguments: data_label: D1
data_type: string
name: retrieve_data
arguments: data_label: D0
data_type: string

tool: D1: Usage: %s <number> <number>


tool: D0: Product is: %Zd


assistant: ```c
int main(int argc, char **argv) {
  mpz_t a, b, p;
  if (argc != 3) {
    printf("Usage: %s <number> <number>\n", argv[0]);
    return 1;
  }
  mpz_init_set_str(a, argv[1], 10);
  mpz_init_set_str(b, argv[2], 10);
  mpz_init(p);
  mpz_mul(p, a, b);
  gmp_printf("Product is: %Zd\n", p);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L1
L3:
  addq   $0x1,-0x8(%rbp)
L1:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t size = 0;
  while (str[size] != '\0') {
    ++size;
  }
  return size;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L7
  mov    $0x0,%eax
L17:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L17
  ret
L7:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t size = 0;
  while (str[size] != '\0') {
    ++size;
  }
  return size;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L8
  nopl   0x0(%rax,%rax,1)
L15:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L15
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t size = 0;
  while (str[size] != '\0') {
    ++size;
  }
  return size;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L8
  nopl   0x0(%rax,%rax,1)
L10:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L10
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t size = 0;
  while (str[size] != '\0') {
    ++size;
  }
  return size;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<remove_comments>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    $0x25,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L63
L22:
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L65
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jbe    L79
  mov    -0x10(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  je     L79
L65:
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    $0x1,%eax
  jmp    L64
L79:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L29
  mov    $0x1,%eax
  jmp    L64
L29:
  mov    -0x8(%rbp),%rax
  mov    $0x25,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x10(%rbp)
L63:
  cmpq   $0x0,-0x10(%rbp)
  jne    L22
  mov    $0x0,%eax
L64:
  leave
  ret
```

assistant: ```c
int remove_comments(char *s, char *com) {
  char *p = NULL;
  char *q = NULL;
  com[0] = '\0';
  p = strchr(s, '%');
  while (p != NULL) {
    if (p == s || ((p > s) && *(p - 1) != '\\')) {
      strcpy(com, p);
      *p = '\0';
      return 1;
    }
    q = p + 1;
    if (*q == '\0')
      return 1;
    p = strchr(q, '%');
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<remove_comments>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  movb   $0x0,(%rsi)
  mov    $0x25,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L38
  mov    %rax,%rbx
  cmp    %rax,%rbp
  jne    L41
  jmp    L100
L71:
  lea    0x1(%rbx),%rdi
  cmpb   $0x0,0x1(%rbx)
  je     L72
  mov    $0x25,%esi
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L9
  cmp    %rbx,%rbp
  je     L100
L41:
  cmp    %rbx,%rbp
  jae    L71
  cmpb   $0x5c,-0x1(%rbx)
  je     L71
L100:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  movb   $0x0,(%rbx)
  mov    $0x1,%eax
L108:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  mov    $0x0,%eax
  jmp    L108
L38:
  mov    $0x0,%eax
  jmp    L108
L72:
  mov    $0x1,%eax
  jmp    L108
```

assistant: ```c
int remove_comments(char *s, char *com) {
  char *p = NULL;
  char *q = NULL;
  com[0] = '\0';
  p = strchr(s, '%');
  while (p != NULL) {
    if (p == s || ((p > s) && *(p - 1) != '\\')) {
      strcpy(com, p);
      *p = '\0';
      return 1;
    }
    q = p + 1;
    if (*q == '\0')
      return 1;
    p = strchr(q, '%');
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<remove_comments>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movb   $0x0,(%rsi)
  mov    $0x25,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L25
  mov    %rax,%rbx
  cmp    %rax,%rbp
  jne    L51
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L70:
  mov    $0x25,%esi
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L25
  cmp    %rbx,%rbp
  je     L0
L51:
  cmp    %rbx,%rbp
  jae    L42
  cmpb   $0x5c,-0x1(%rbx)
  jne    L0
L42:
  cmpb   $0x0,0x1(%rbx)
  lea    0x1(%rbx),%rdi
  jne    L70
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  movb   $0x0,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int remove_comments(char *s, char *com) {
  char *p = NULL;
  char *q = NULL;
  com[0] = '\0';
  p = strchr(s, '%');
  while (p != NULL) {
    if (p == s || ((p > s) && *(p - 1) != '\\')) {
      strcpy(com, p);
      *p = '\0';
      return 1;
    }
    q = p + 1;
    if (*q == '\0')
      return 1;
    p = strchr(q, '%');
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<remove_comments>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movb   $0x0,(%rsi)
  mov    $0x25,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L41
  mov    %rax,%rbx
  cmp    %rax,%rbp
  jne    L14
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L5:
  mov    $0x25,%esi
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L41
  cmp    %rbx,%rbp
  je     L6
L14:
  cmp    %rbx,%rbp
  jae    L12
  cmpb   $0x5c,-0x1(%rbx)
  jne    L6
L12:
  cmpb   $0x0,0x1(%rbx)
  lea    0x1(%rbx),%rdi
  jne    L5
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  movb   $0x0,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int remove_comments(char *s, char *com) {
  char *p = NULL;
  char *q = NULL;
  com[0] = '\0';
  p = strchr(s, '%');
  while (p != NULL) {
    if (p == s || ((p > s) && *(p - 1) != '\\')) {
      strcpy(com, p);
      *p = '\0';
      return 1;
    }
    q = p + 1;
    if (*q == '\0')
      return 1;
    p = strchr(q, '%');
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<backtrack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    %r9,-0x48(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <remaining_moves@plt>
  cmp    $0x1,%eax
  jle    L33
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2a,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jl     L52
  mov    -0x40(%rbp),%rax
  mov    (%rax),%ebx
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edi
  call   <floor@plt>
  lea    (%rbx,%rax,1),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x48(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x48(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L32
L30:
  mov    -0x24(%rbp),%eax
  imul   %eax,%eax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x48(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  addl   $0x1,-0x18(%rbp)
L32:
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L30
L52:
  movl   $0x0,-0x14(%rbp)
  jmp    L45
L24:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x14(%rbp)
L45:
  mov    -0x24(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L24
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
L33:
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void backtrack(int n, int index, char *board, int *top, int *max, char **stack) {
  if (remaining_moves(n, board) > 1) {
    board[index] = '*';
    if (*top >= *max) {
      *max = *max + floor(*max / 2);
      stack = realloc(stack, (*max) * sizeof(char *));
      for (int i = *top; i < (*max); i++) {
        stack[i] = malloc((n * n) * sizeof(char));
      }
    }
    for (int i = 0; i < n * n; i++) {
      stack[*top][i] = board[i];
    }
    *top += 1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<backtrack>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%r14d
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %r8,%r15
  mov    %r9,%rbp
  mov    %rdx,%rsi
  call   <remaining_moves@plt>
  cmp    $0x1,%eax
  jle    L48
  movslq %r14d,%r14
  movb   $0x2a,0x0(%r13,%r14,1)
  mov    (%r15),%r14d
  cmp    %r14d,(%r12)
  jge    L53
L64:
  imul   %ebx,%ebx
  test   %ebx,%ebx
  jle    L17
  mov    %ebx,%ebx
  mov    $0x0,%eax
L35:
  movzbl 0x0(%r13,%rax,1),%ecx
  movslq (%r12),%rdx
  mov    0x0(%rbp,%rdx,8),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L35
L17:
  addl   $0x1,(%r12)
L48:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  mov    %r14d,%edi
  shr    $0x1f,%edi
  add    %r14d,%edi
  sar    %edi
  call   <floor@plt>
  add    %eax,%r14d
  mov    %r14d,(%r15)
  lea    0x0(,%r14,8),%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  mov    (%r12),%r14d
  cmp    (%r15),%r14d
  jge    L64
  mov    %ebx,%eax
  imul   %ebx,%eax
  mov    %eax,0xc(%rsp)
  movslq %r14d,%r14
L16:
  mov    0xc(%rsp),%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp,%r14,8)
  add    $0x1,%r14
  cmp    %r14d,(%r15)
  jg     L16
  jmp    L64
```

assistant: ```c
void backtrack(int n, int index, char *board, int *top, int *max, char **stack) {
  if (remaining_moves(n, board) > 1) {
    board[index] = '*';
    if (*top >= *max) {
      *max = *max + floor(*max / 2);
      stack = realloc(stack, (*max) * sizeof(char *));
      for (int i = *top; i < (*max); i++) {
        stack[i] = malloc((n * n) * sizeof(char));
      }
    }
    for (int i = 0; i < n * n; i++) {
      stack[*top][i] = board[i];
    }
    *top += 1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<backtrack>:
  endbr64
  push   %r15
  mov    %r9,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  movslq %esi,%rbx
  mov    %rdx,%rsi
  sub    $0x8,%rsp
  call   <remaining_moves@plt>
  cmp    $0x1,%eax
  jle    L35
  movb   $0x2a,0x0(%rbp,%rbx,1)
  movslq (%r12),%rax
  imul   %r13d,%r13d
  mov    (%r14),%ebx
  cmp    %ebx,%eax
  jge    L63
L77:
  test   %r13d,%r13d
  je     L52
  movslq %r13d,%rsi
  xor    %edx,%edx
  jmp    L50
  nopl   0x0(%rax)
L76:
  movslq (%r12),%rax
L50:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    (%r15,%rax,8),%rax
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L76
  mov    (%r12),%eax
L52:
  add    $0x1,%eax
  mov    %eax,(%r12)
L35:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L63:
  mov    %ebx,%edi
  shr    $0x1f,%edi
  add    %ebx,%edi
  sar    %edi
  call   <floor@plt>
  mov    %r15,%rdi
  add    %eax,%ebx
  mov    %ebx,(%r14)
  lea    0x0(,%rbx,8),%esi
  call   <realloc@plt>
  mov    %rax,%r15
  movslq (%r12),%rax
  cmp    (%r14),%eax
  jge    L77
  movslq %eax,%rbx
  nopl   0x0(%rax)
L29:
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%r15,%rbx,8)
  add    $0x1,%rbx
  cmp    %ebx,(%r14)
  jg     L29
  movslq (%r12),%rax
  jmp    L77
```

assistant: ```c
void backtrack(int n, int index, char *board, int *top, int *max, char **stack) {
  if (remaining_moves(n, board) > 1) {
    board[index] = '*';
    if (*top >= *max) {
      *max = *max + floor(*max / 2);
      stack = realloc(stack, (*max) * sizeof(char *));
      for (int i = *top; i < (*max); i++) {
        stack[i] = malloc((n * n) * sizeof(char));
      }
    }
    for (int i = 0; i < n * n; i++) {
      stack[*top][i] = board[i];
    }
    *top += 1;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<backtrack>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %r9,%rbp
  push   %rbx
  movslq %esi,%rbx
  mov    %rdx,%rsi
  sub    $0x18,%rsp
  call   <remaining_moves@plt>
  cmp    $0x1,%eax
  jle    L1
  movb   $0x2a,(%r12,%rbx,1)
  movslq 0x0(%r13),%rbx
  imul   %r15d,%r15d
  mov    (%r14),%edx
  cmp    %edx,%ebx
  jge    L58
L22:
  test   %r15d,%r15d
  je     L59
  mov    0x0(%rbp,%rbx,8),%rax
  movzbl (%r12),%edx
  movslq %r15d,%rsi
  mov    %dl,(%rax)
  mov    $0x1,%eax
  cmp    $0x1,%r15d
  je     L64
  nopl   0x0(%rax)
L51:
  movslq 0x0(%r13),%rdx
  movzbl (%r12,%rax,1),%ecx
  mov    0x0(%rbp,%rdx,8),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L51
L64:
  mov    0x0(%r13),%ebx
L59:
  add    $0x1,%ebx
  mov    %ebx,0x0(%r13)
L1:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L58:
  mov    %edx,%edi
  mov    %edx,0xc(%rsp)
  shr    $0x1f,%edi
  add    %edx,%edi
  sar    %edi
  call   <floor@plt>
  mov    0xc(%rsp),%edx
  mov    %rbp,%rdi
  add    %eax,%edx
  mov    %edx,(%r14)
  lea    0x0(,%rdx,8),%esi
  call   <realloc@plt>
  movslq 0x0(%r13),%rbx
  mov    %rax,%rbp
  cmp    (%r14),%ebx
  jge    L22
  nopw   0x0(%rax,%rax,1)
L25:
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    %ebx,(%r14)
  jg     L25
  movslq 0x0(%r13),%rbx
  jmp    L22
```

assistant: ```c
void backtrack(int n, int index, char *board, int *top, int *max, char **stack) {
  if (remaining_moves(n, board) > 1) {
    board[index] = '*';
    if (*top >= *max) {
      *max = *max + floor(*max / 2);
      stack = realloc(stack, (*max) * sizeof(char *));
      for (int i = *top; i < (*max); i++) {
        stack[i] = malloc((n * n) * sizeof(char));
      }
    }
    for (int i = 0; i < n * n; i++) {
      stack[*top][i] = board[i];
    }
    *top += 1;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<countDigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L22
  mov    $0x1,%eax
  jmp    L30
L22:
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L0:
  addl   $0x1,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movabs $0xcccccccccccccccd,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,-0x18(%rbp)
L35:
  cmpq   $0x0,-0x18(%rbp)
  jne    L0
  mov    -0x4(%rbp),%eax
L30:
  pop    %rbp
  ret
```

assistant: ```c
int countDigits(unsigned long n) {
  if (n == 0) {
    return (1);
  }
  int count = 0;
  while (n != 0) {
    count++;
    n /= 10;
  }
  return (count);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<countDigits>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%ecx
  test   %rdi,%rdi
  je     L15
  mov    $0x0,%ecx
  movabs $0xcccccccccccccccd,%rdi
L19:
  add    $0x1,%ecx
  mov    %rdx,%rsi
  mov    %rdx,%rax
  mul    %rdi
  shr    $0x3,%rdx
  cmp    $0x9,%rsi
  ja     L19
L15:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int countDigits(unsigned long n) {
  if (n == 0) {
    return (1);
  }
  int count = 0;
  while (n != 0) {
    count++;
    n /= 10;
  }
  return (count);
}
```
user: What is the c source code of the assembly code below:

```asm
<countDigits>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%r8d
  test   %rdi,%rdi
  je     L7
  movabs $0xcccccccccccccccd,%rsi
  xor    %r8d,%r8d
  nop
L13:
  mov    %rdx,%rax
  mov    %rdx,%rcx
  add    $0x1,%r8d
  mul    %rsi
  shr    $0x3,%rdx
  cmp    $0x9,%rcx
  ja     L13
L7:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int countDigits(unsigned long n) {
  if (n == 0) {
    return (1);
  }
  int count = 0;
  while (n != 0) {
    count++;
    n /= 10;
  }
  return (count);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<countDigits>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%r8d
  test   %rdi,%rdi
  je     L18
  movabs $0xcccccccccccccccd,%rsi
  xor    %r8d,%r8d
  nop
L10:
  mov    %rdx,%rax
  mov    %rdx,%rcx
  add    $0x1,%r8d
  mul    %rsi
  shr    $0x3,%rdx
  cmp    $0x9,%rcx
  ja     L10
L18:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int countDigits(unsigned long n) {
  if (n == 0) {
    return (1);
  }
  int count = 0;
  while (n != 0) {
    count++;
    n /= 10;
  }
  return (count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<durationtolong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L142
L150:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  lea    -0x18(%rbp),%rcx
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  je     L62
  cmpl   $0x0,-0x1c(%rbp)
  je     L142
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  test   %eax,%eax
  je     L124
  test   %eax,%eax
  js     L142
  cmp    $0x79,%eax
  jg     L142
  cmp    $0x4d,%eax
  jl     L142
  sub    $0x4d,%eax
  cmp    $0x2c,%eax
  ja     L142
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D6(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D6(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
L124:
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,-0x20(%rbp)
  jmp    L142
  mov    -0x1c(%rbp),%eax
  imul   $0x3c,%eax,%eax
  add    %eax,-0x20(%rbp)
  jmp    L142
  mov    -0x1c(%rbp),%eax
  imul   $0xe10,%eax,%eax
  add    %eax,-0x20(%rbp)
  jmp    L142
  mov    -0x1c(%rbp),%eax
  imul   $0x15180,%eax,%eax
  add    %eax,-0x20(%rbp)
  jmp    L142
  mov    -0x1c(%rbp),%eax
  imul   $0x93a80,%eax,%eax
  add    %eax,-0x20(%rbp)
  jmp    L142
  mov    -0x1c(%rbp),%eax
  imul   $0x278d00,%eax,%eax
  add    %eax,-0x20(%rbp)
  jmp    L142
  mov    -0x1c(%rbp),%eax
  imul   $0x1e187e0,%eax,%eax
  add    %eax,-0x20(%rbp)
L142:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L150
  jmp    L84
L62:
  nop
L84:
  mov    -0x20(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L70
  call   <__stack_chk_fail@plt>
L70:
  leave
  ret
```

assistant: ```c
int durationtolong(const char *string) {
  int total = 0;
  int current = 0;
  char *ch = (char *)string, *och;
  while (*ch) {
    och = ch;
    current = strtol(ch, &ch, 10);
    if (och == ch)
      break;
    if (current) {
      switch (*ch++) {
      case '\0':
        ch--;
      case 's':
        total += current;
        break;
      case 'm':
        total += (current * 60);
        break;
      case 'h':
        total += (current * 3600);
        break;
      case 'd':
        total += (current * 86400);
        break;
      case 'w':
        total += (current * 604800);
        break;
      case 'M':
        total += (current * 2592000);
        break;
      case 'y':
        total += (current * 31557600);
      }
    }
  }
  return total;
}
```