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

```asm
<mergeKLists>:
  endbr64
  mov    %rdi,%rax
  mov    (%rdi),%rdi
  cmp    $0x1,%esi
  jle    L13
  push   %rbp
  lea    -0x2(%rsi),%edx
  push   %rbx
  lea    0x10(%rax,%rdx,8),%rbp
  lea    0x8(%rax),%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L15:
  mov    (%rbx),%rsi
  add    $0x8,%rbx
  call   <mergeTwoLists@plt>
  mov    %rax,%rdi
  cmp    %rbp,%rbx
  jne    L15
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %rdi,%rax
  ret
```

assistant: ```c
struct ListNode *mergeKLists(struct ListNode **lists, int listsSize) {
  struct ListNode *result = lists[0];
  int i = 1;
  for (i = 1; i < listsSize; i++) {
    result = mergeTwoLists(result, lists[i]);
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strnatcmp0>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  mov    %rsi,0x18(%rsp)
  mov    %edx,%r14d
  mov    %rdi,%r13
  mov    %rsi,%r12
  jmp    L11
L59:
  mov    %ebp,%edi
  call   <nat_isdigit@plt>
  test   %rax,%rax
  je     L113
  cmp    $0x30,%ebx
  je     L78
  cmp    $0x30,%ebp
  je     L78
  mov    %r12,%rsi
  mov    0x8(%rsp),%rdi
  call   <compare_right@plt>
  test   %eax,%eax
  je     L113
  jmp    L110
L78:
  mov    %r15,%rsi
  mov    0x8(%rsp),%rdi
  call   <compare_left@plt>
  test   %eax,%eax
  je     L113
  jmp    L110
L9:
  test   %r14d,%r14d
  je     L75
  mov    0x18(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  call   <strcasecmp@plt>
  jmp    L110
L75:
  mov    0x18(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  call   <strcmp@plt>
  jmp    L110
L99:
  cmp    %ebp,%ebx
  jl     L131
  add    $0x1,%r13
  add    $0x1,%r12
  cmp    %ebp,%ebx
  jg     L150
L11:
  mov    %r13,0x8(%rsp)
  movsbl 0x0(%r13),%edi
  call   <to_int@plt>
  mov    %eax,%ebx
  mov    %r12,%r15
  movsbl (%r12),%edi
  call   <to_int@plt>
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <nat_isdigit@plt>
  test   %rax,%rax
  jne    L59
L113:
  mov    %ebx,%eax
  or     %ebp,%eax
  je     L9
  test   %r14d,%r14d
  je     L99
  mov    %ebx,%edi
  call   <nat_unify_case@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <nat_unify_case@plt>
  mov    %eax,%ebp
  jmp    L99
L150:
  mov    $0x1,%eax
  jmp    L110
L131:
  mov    $0xffffffff,%eax
L110:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int strnatcmp0(char const *a, char const *b, int fold_case) {
  int ai, bi;
  int ca, cb;
  int fractional, result;
  ;
  ai = bi = 0;
  while (1) {
    ca = to_int(a[ai]);
    cb = to_int(b[bi]);
    if (nat_isdigit(ca) && nat_isdigit(cb)) {
      fractional = (ca == '0' || cb == '0');
      if (fractional) {
        if ((result = compare_left(a + ai, b + bi)) != 0)
          return result;
      } else {
        if ((result = compare_right(a + ai, b + bi)) != 0)
          return result;
      }
    }
    if (!ca && !cb) {
      if (fold_case)
        return strcasecmp(a, b);
      else
        return strcmp(a, b);
    }
    if (fold_case) {
      ca = nat_unify_case(ca);
      cb = nat_unify_case(cb);
    }
    if (ca < cb)
      return -1;
    else if (ca > cb)
      return +1;
    ++ai;
    ++bi;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strnatcmp0>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  jmp    L117
  nopl   (%rax)
L53:
  mov    %ebp,%eax
  or     %r14d,%eax
  je     L44
L7:
  test   %r13d,%r13d
  jne    L72
  cmp    %r14d,%ebp
  jl     L123
L81:
  add    $0x1,%r12
  add    $0x1,%rbx
  cmp    %r14d,%ebp
  jg     L55
L117:
  movsbl (%r12),%edi
  call   <to_int@plt>
  movsbl (%rbx),%edi
  mov    %eax,%ebp
  call   <to_int@plt>
  mov    %ebp,%edi
  mov    %eax,%r14d
  call   <nat_isdigit@plt>
  test   %rax,%rax
  je     L53
  mov    %r14d,%edi
  call   <nat_isdigit@plt>
  test   %rax,%rax
  je     L53
  mov    %rbx,%rsi
  mov    %r12,%rdi
  cmp    $0x30,%ebp
  je     L25
  cmp    $0x30,%r14d
  je     L25
  call   <compare_right@plt>
  test   %eax,%eax
  je     L53
L126:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L72:
  mov    %ebp,%edi
  call   <nat_unify_case@plt>
  mov    %r14d,%edi
  mov    %eax,%ebp
  call   <nat_unify_case@plt>
  mov    %eax,%r14d
  cmp    %r14d,%ebp
  jge    L81
L123:
  add    $0x18,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L25:
  call   <compare_left@plt>
  test   %eax,%eax
  jne    L126
  mov    %ebp,%eax
  or     %r14d,%eax
  jne    L7
  nopl   0x0(%rax)
L44:
  mov    0x8(%rsp),%rdi
  mov    %r15,%rsi
  test   %r13d,%r13d
  je     L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <strcasecmp@plt>
L55:
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L17:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10e0 <strcmp@plt>
```

assistant: ```c
int strnatcmp0(char const *a, char const *b, int fold_case) {
  int ai, bi;
  int ca, cb;
  int fractional, result;
  ;
  ai = bi = 0;
  while (1) {
    ca = to_int(a[ai]);
    cb = to_int(b[bi]);
    if (nat_isdigit(ca) && nat_isdigit(cb)) {
      fractional = (ca == '0' || cb == '0');
      if (fractional) {
        if ((result = compare_left(a + ai, b + bi)) != 0)
          return result;
      } else {
        if ((result = compare_right(a + ai, b + bi)) != 0)
          return result;
      }
    }
    if (!ca && !cb) {
      if (fold_case)
        return strcasecmp(a, b);
      else
        return strcmp(a, b);
    }
    if (fold_case) {
      ca = nat_unify_case(ca);
      cb = nat_unify_case(cb);
    }
    if (ca < cb)
      return -1;
    else if (ca > cb)
      return +1;
    ++ai;
    ++bi;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strnatcmp0>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  jmp    L129
  nopl   (%rax)
L89:
  mov    %ebp,%eax
  or     %r14d,%eax
  je     L124
L88:
  test   %r13d,%r13d
  jne    L53
  cmp    %r14d,%ebp
  jl     L52
L128:
  add    $0x1,%r12
  add    $0x1,%rbx
  cmp    %r14d,%ebp
  jg     L27
L129:
  movsbl (%r12),%edi
  call   <to_int@plt>
  movsbl (%rbx),%edi
  mov    %eax,%ebp
  call   <to_int@plt>
  mov    %ebp,%edi
  mov    %eax,%r14d
  call   <nat_isdigit@plt>
  test   %rax,%rax
  je     L89
  mov    %r14d,%edi
  call   <nat_isdigit@plt>
  test   %rax,%rax
  je     L89
  mov    %rbx,%rsi
  mov    %r12,%rdi
  cmp    $0x30,%ebp
  je     L112
  cmp    $0x30,%r14d
  je     L112
  call   <compare_right@plt>
  test   %eax,%eax
  je     L89
L29:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L53:
  mov    %ebp,%edi
  call   <nat_unify_case@plt>
  mov    %r14d,%edi
  mov    %eax,%ebp
  call   <nat_unify_case@plt>
  mov    %eax,%r14d
  cmp    %r14d,%ebp
  jge    L128
L52:
  add    $0x18,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L112:
  call   <compare_left@plt>
  test   %eax,%eax
  jne    L29
  mov    %ebp,%eax
  or     %r14d,%eax
  jne    L88
  nopl   0x0(%rax)
L124:
  mov    0x8(%rsp),%rdi
  mov    %r15,%rsi
  test   %r13d,%r13d
  je     L28
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <strcasecmp@plt>
L27:
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L28:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10e0 <strcmp@plt>
```

assistant: ```c
int strnatcmp0(char const *a, char const *b, int fold_case) {
  int ai, bi;
  int ca, cb;
  int fractional, result;
  ;
  ai = bi = 0;
  while (1) {
    ca = to_int(a[ai]);
    cb = to_int(b[bi]);
    if (nat_isdigit(ca) && nat_isdigit(cb)) {
      fractional = (ca == '0' || cb == '0');
      if (fractional) {
        if ((result = compare_left(a + ai, b + bi)) != 0)
          return result;
      } else {
        if ((result = compare_right(a + ai, b + bi)) != 0)
          return result;
      }
    }
    if (!ca && !cb) {
      if (fold_case)
        return strcasecmp(a, b);
      else
        return strcmp(a, b);
    }
    if (fold_case) {
      ca = nat_unify_case(ca);
      cb = nat_unify_case(cb);
    }
    if (ca < cb)
      return -1;
    else if (ca > cb)
      return +1;
    ++ai;
    ++bi;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L100
L85:
  movl   $0x0,-0x24(%rbp)
  lea    -0x1e(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x13(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L63
L13:
  mov    -0x28(%rbp),%eax
  cltq
  movzbl -0x1e(%rbp,%rax,1),%eax
  cmp    $0x3f,%al
  je     L108
  mov    -0x28(%rbp),%eax
  cltq
  movzbl -0x13(%rbp,%rax,1),%eax
  cmp    $0x3f,%al
  je     L108
  mov    -0x28(%rbp),%eax
  cltq
  movzbl -0x1e(%rbp,%rax,1),%edx
  mov    -0x28(%rbp),%eax
  cltq
  movzbl -0x13(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L108
  movl   $0x1,-0x24(%rbp)
  jmp    L20
L108:
  addl   $0x1,-0x28(%rbp)
L63:
  lea    -0x1e(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x28(%rbp)
  jl     L13
L20:
  cmpl   $0x0,-0x24(%rbp)
  je     L102
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L100
L102:
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L100:
  mov    -0x2c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  test   %eax,%eax
  jne    L85
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L109
  call   <__stack_chk_fail@plt>
L109:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D38
datatype: string
name: unpack_data
arguments: data_tag: D5
datatype: string
name: unpack_data
arguments: data_tag: D11
datatype: string
name: unpack_data
arguments: data_tag: D39
datatype: string

tool: D38: %d

tool: D5: %s

tool: D11: No


tool: D39: Yes


assistant: ```c
int main() {
  int T, i, f;
  char x[11], y[11];
  scanf("%d", &T);
  while (T--) {
    f = 0;
    scanf("%s", x);
    scanf("%s", y);
    for (i = 0; i < strlen(x); i++) {
      if (x[i] != '?' && y[i] != '?' && x[i] != y[i]) {
        f = 1;
        break;
      }
    }
    if (f)
      printf("No\n");
    else
      printf("Yes\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L40
  lea    0x12(%rsp),%rbp
  lea    D31(%rip),%r12
  lea    D17(%rip),%r14
  lea    D13(%rip),%r13
  jmp    L23
L54:
  add    $0x1,%rbx
L62:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L15
  movzbl (%rbx,%rbp,1),%eax
  cmp    $0x3f,%al
  je     L54
  movzbl 0x1d(%rsp,%rbx,1),%edx
  cmp    %dl,%al
  je     L54
  cmp    $0x3f,%dl
  je     L54
  mov    %r13,%rdi
  call   <printf@plt>
L19:
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L40
L23:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    0x1d(%rsp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x0,%ebx
  jmp    L62
L40:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L91
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L15:
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L19
L91:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D30
data_type: string
name: extract_symbolic_data
arguments: label: D31
data_type: string
name: extract_symbolic_data
arguments: label: D17
data_type: string
name: extract_symbolic_data
arguments: label: D13
data_type: string

tool: D30: %d

tool: D31: %s

tool: D17: Yes


tool: D13: No


assistant: ```c
int main() {
  int T, i, f;
  char x[11], y[11];
  scanf("%d", &T);
  while (T--) {
    f = 0;
    scanf("%s", x);
    scanf("%s", y);
    for (i = 0; i < strlen(x); i++) {
      if (x[i] != '?' && y[i] != '?' && x[i] != y[i]) {
        f = 1;
        break;
      }
    }
    if (f)
      printf("No\n");
    else
      printf("Yes\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  lea    D24(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%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     L44
  lea    0x12(%rsp),%r15
  lea    D32(%rip),%r12
  lea    0x1d(%rsp),%rbp
  lea    D35(%rip),%r14
  lea    D7(%rip),%r13
  nopl   (%rax)
L40:
  mov    %r15,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  jmp    L27
  cs nopw 0x0(%rax,%rax,1)
L21:
  movzbl (%r15,%rbx,1),%eax
  cmp    $0x3f,%al
  je     L10
  movzbl 0x0(%rbp,%rbx,1),%edx
  cmp    $0x3f,%dl
  je     L10
  cmp    %dl,%al
  jne    L15
L10:
  add    $0x1,%rbx
L27:
  mov    %r15,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L21
  mov    %r14,%rdi
  call   <printf@plt>
L52:
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L40
L44:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L15:
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L52
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D24
data_type: string
name: retrieve_data
arguments: data_tag: D32
data_type: string
name: retrieve_data
arguments: data_tag: D35
data_type: string
name: retrieve_data
arguments: data_tag: D7
data_type: string

tool: D24: %d

tool: D32: %s

tool: D35: Yes


tool: D7: No


assistant: ```c
int main() {
  int T, i, f;
  char x[11], y[11];
  scanf("%d", &T);
  while (T--) {
    f = 0;
    scanf("%s", x);
    scanf("%s", y);
    for (i = 0; i < strlen(x); i++) {
      if (x[i] != '?' && y[i] != '?' && x[i] != y[i]) {
        f = 1;
        break;
      }
    }
    if (f)
      printf("No\n");
    else
      printf("Yes\n");
  }
  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
  lea    D20(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%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     L65
  lea    0x12(%rsp),%r15
  lea    D4(%rip),%r12
  lea    0x1d(%rsp),%rbp
  lea    D33(%rip),%r14
  lea    D15(%rip),%r13
  nopl   (%rax)
L26:
  mov    %r15,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  jmp    L15
  cs nopw 0x0(%rax,%rax,1)
L35:
  movzbl (%r15,%rbx,1),%eax
  cmp    $0x3f,%al
  je     L46
  movzbl 0x0(%rbp,%rbx,1),%edx
  cmp    $0x3f,%dl
  je     L46
  cmp    %dl,%al
  jne    L44
L46:
  add    $0x1,%rbx
L15:
  mov    %r15,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L35
  mov    %r14,%rdi
  call   <printf@plt>
L88:
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L26
L65:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L44:
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L88
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D20
datatype: string
name: parse_data
arguments: label: D4
datatype: string
name: parse_data
arguments: label: D33
datatype: string
name: parse_data
arguments: label: D15
datatype: string

tool: D20: %d

tool: D4: %s

tool: D33: Yes


tool: D15: No


assistant: ```c
int main() {
  int T, i, f;
  char x[11], y[11];
  scanf("%d", &T);
  while (T--) {
    f = 0;
    scanf("%s", x);
    scanf("%s", y);
    for (i = 0; i < strlen(x); i++) {
      if (x[i] != '?' && y[i] != '?' && x[i] != y[i]) {
        f = 1;
        break;
      }
    }
    if (f)
      printf("No\n");
    else
      printf("Yes\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0x8(%rbp),%rax
  movl   $0x2,(%rax)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jne    L10
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L12
L10:
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L12:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string

tool: D6: %d %d


tool: D12: foo

assistant: ```c
int main(void) {
  int *p = (int *)malloc(sizeof(int));
  free(p);
  int *q = (int *)malloc(sizeof(int));
  *p = 1;
  *q = 2;
  if (p == q)
    printf("%d %d\n", *p, *q);
  else
    printf("foo\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x1,(%rbx)
  movl   $0x2,(%rax)
  cmp    %rax,%rbx
  je     L14
  lea    D13(%rip),%rdi
  call   <puts@plt>
L16:
  mov    $0x0,%eax
  pop    %rbx
  ret
L14:
  mov    $0x2,%edx
  mov    (%rbx),%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L16
```

tool_calls: name: unpack_data
arguments: label: D13
data_type: string
name: unpack_data
arguments: label: D16
data_type: string

tool: D13: foo

tool: D16: %d %d


assistant: ```c
int main(void) {
  int *p = (int *)malloc(sizeof(int));
  free(p);
  int *q = (int *)malloc(sizeof(int));
  *p = 1;
  *q = 2;
  if (p == q)
    printf("%d %d\n", *p, *q);
  else
    printf("foo\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <free@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x1,(%rbx)
  movl   $0x2,(%rax)
  cmp    %rax,%rbx
  je     L8
  lea    D14(%rip),%rdi
  call   <puts@plt>
L2:
  xor    %eax,%eax
  pop    %rbx
  ret
L8:
  mov    $0x2,%edx
  mov    $0x2,%esi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L2
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type: string
name: extract_symbolic_data
arguments: tag: D16
type: string

tool: D14: foo

tool: D16: %d %d


assistant: ```c
int main(void) {
  int *p = (int *)malloc(sizeof(int));
  free(p);
  int *q = (int *)malloc(sizeof(int));
  *p = 1;
  *q = 2;
  if (p == q)
    printf("%d %d\n", *p, *q);
  else
    printf("foo\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <free@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x1,(%rbx)
  movl   $0x2,(%rax)
  cmp    %rax,%rbx
  je     L5
  lea    D2(%rip),%rdi
  call   <puts@plt>
L12:
  xor    %eax,%eax
  pop    %rbx
  ret
L5:
  mov    $0x2,%edx
  mov    $0x2,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L12
```

tool_calls: name: parse_data
arguments: tag: D2
type_infered: string
name: parse_data
arguments: tag: D0
type_infered: string

tool: D2: foo

tool: D0: %d %d


assistant: ```c
int main(void) {
  int *p = (int *)malloc(sizeof(int));
  free(p);
  int *q = (int *)malloc(sizeof(int));
  *p = 1;
  *q = 2;
  if (p == q)
    printf("%d %d\n", *p, *q);
  else
    printf("foo\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<quickSortRec>:
  endbr64
  cmp    %edx,%esi
  jge    L74
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%r9d
  mov    %edx,%ebx
  movslq %esi,%rax
  lea    (%rdi,%rax,4),%r10
  mov    %esi,%ecx
  mov    %edx,%eax
  jmp    L62
L3:
  movslq %eax,%rdx
  lea    -0x4(%r12,%rdx,4),%rdx
L21:
  mov    %eax,%ebp
  sub    $0x1,%eax
  mov    %rdx,%r8
  mov    (%rdx),%edi
  sub    $0x4,%rdx
  test   %eax,%eax
  jle    L34
  cmp    %edi,%esi
  jl     L21
L34:
  cmp    %ecx,%eax
  jle    L8
  movslq %ecx,%rdx
  lea    (%r12,%rdx,4),%rdx
  mov    (%rdx),%esi
  mov    %edi,(%rdx)
  mov    %esi,(%r8)
L62:
  mov    (%r10),%esi
  movslq %ecx,%rdx
  lea    (%r12,%rdx,4),%rdx
L26:
  cmp    %esi,(%rdx)
  jg     L3
  add    $0x1,%ecx
  add    $0x4,%rdx
  cmp    %ecx,%ebx
  jg     L26
  jmp    L3
L8:
  cltq
  lea    (%r12,%rax,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%r10)
  mov    %esi,(%rax)
  lea    -0x2(%rbp),%edx
  mov    %r9d,%esi
  mov    %r12,%rdi
  call   <quickSortRec>
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <quickSortRec>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L74:
  ret
```

assistant: ```c
void quickSortRec(int *tab, int min, int max) {
  int pivotIndex;
  int tmp_max;
  int tmp_min;
  int tmp;
  pivotIndex = 0;
  tmp_max = 0;
  tmp_min = 0;
  if (min < max) {
    pivotIndex = min;
    tmp_max = max;
    tmp_min = min;
    while (tmp_min < tmp_max) {
      while ((tab[tmp_min] <= tab[pivotIndex]) && ++tmp_min < max)
        ;
      while (tab[--tmp_max] > tab[pivotIndex] && tmp_max > 0)
        ;
      if (tmp_min < tmp_max) {
        tmp = tab[tmp_min];
        tab[tmp_min] = tab[tmp_max];
        tab[tmp_max] = tmp;
      }
    }
    tmp = tab[pivotIndex];
    tab[pivotIndex] = tab[tmp_max];
    tab[tmp_max] = tmp;
    quickSortRec(tab, min, tmp_max - 1);
    quickSortRec(tab, tmp_max + 1, max);
  }
}
```