ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mk_addr>:
  endbr64
  push   %r12
  lea    D18(%rip),%r12
  mov    %edi,%edx
  lea    D17(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D17
type_infered: string

tool: D17: levels/level%d.txt

assistant: ```c
char *mk_addr(int level) {
  static char buffer[100];
  sprintf(buffer,
          "levels/level"
          "%d"
          ".txt",
          level);
  return buffer;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mk_addr>:
  endbr64
  push   %r12
  lea    D9(%rip),%r12
  mov    %edi,%edx
  lea    D17(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D17
type: string

tool: D17: levels/level%d.txt

assistant: ```c
char *mk_addr(int level) {
  static char buffer[100];
  sprintf(buffer,
          "levels/level"
          "%d"
          ".txt",
          level);
  return buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<load_noinstall_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0xa,%rax
  mov    %rax,%rdi
  call   <lt_malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x14(%rbp),%eax
  add    $0x5,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D7: .libs

assistant: ```c
char *load_noinstall_path(char const *arg, int pathlen) {
  char *newarg, *expanded_path;
  int newpathlen;
  newarg = (char *)lt_malloc(strlen(arg) + 10);
  strcpy(newarg, arg);
  newarg[pathlen] = 0;
  newpathlen = pathlen;
  strcat(newarg, ".libs");
  newpathlen += sizeof(".libs") - 1;
  newarg[newpathlen] = 0;
  expanded_path = newarg;
  return expanded_path;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<load_noinstall_path>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  call   <strlen@plt>
  lea    0xa(%rax),%rdi
  call   <lt_malloc@plt>
  mov    %rax,%rbp
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movslq %ebx,%rax
  movb   $0x0,0x0(%rbp,%rax,1)
  lea    D8(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  add    $0x5,%ebx
  movslq %ebx,%rbx
  movb   $0x0,0x0(%rbp,%rbx,1)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type: string

tool: D8: .libs

assistant: ```c
char *load_noinstall_path(char const *arg, int pathlen) {
  char *newarg, *expanded_path;
  int newpathlen;
  newarg = (char *)lt_malloc(strlen(arg) + 10);
  strcpy(newarg, arg);
  newarg[pathlen] = 0;
  newpathlen = pathlen;
  strcat(newarg, ".libs");
  newpathlen += sizeof(".libs") - 1;
  newarg[newpathlen] = 0;
  expanded_path = newarg;
  return expanded_path;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<load_noinstall_path>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  call   <strlen@plt>
  lea    0xa(%rax),%rdi
  call   <lt_malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  movslq %ebx,%rax
  add    $0x5,%ebx
  mov    %r12,%rdi
  movb   $0x0,(%r12,%rax,1)
  movslq %ebx,%rbx
  lea    D5(%rip),%rsi
  call   <strcat@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
datatype: string

tool: D5: .libs

assistant: ```c
char *load_noinstall_path(char const *arg, int pathlen) {
  char *newarg, *expanded_path;
  int newpathlen;
  newarg = (char *)lt_malloc(strlen(arg) + 10);
  strcpy(newarg, arg);
  newarg[pathlen] = 0;
  newpathlen = pathlen;
  strcat(newarg, ".libs");
  newpathlen += sizeof(".libs") - 1;
  newarg[newpathlen] = 0;
  expanded_path = newarg;
  return expanded_path;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<load_noinstall_path>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  call   <strlen@plt>
  lea    0xa(%rax),%rdi
  call   <lt_malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  movslq %ebx,%rax
  add    $0x5,%ebx
  mov    %r12,%rdi
  movb   $0x0,(%r12,%rax,1)
  movslq %ebx,%rbx
  lea    D7(%rip),%rsi
  call   <strcat@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
type_guessed: string

tool: D7: .libs

assistant: ```c
char *load_noinstall_path(char const *arg, int pathlen) {
  char *newarg, *expanded_path;
  int newpathlen;
  newarg = (char *)lt_malloc(strlen(arg) + 10);
  strcpy(newarg, arg);
  newarg[pathlen] = 0;
  newpathlen = pathlen;
  strcat(newarg, ".libs");
  newpathlen += sizeof(".libs") - 1;
  newarg[newpathlen] = 0;
  expanded_path = newarg;
  return expanded_path;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x3,%esi
  lea    D0(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x6,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x10,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0xffffff84,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0xc,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0xffffffde,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
data_type: string

tool: D0: %d


assistant: ```c
int main() {
  printf("%d\n", 1 + 2);
  printf("%d\n", 2 - 1);
  printf("%d\n", 2 * 3);
  printf("%d\n", 4 / 2);
  printf("%d\n", 4 % 3);
  printf("%d\n", 1 & 2);
  printf("%d\n", 1 | 2);
  printf("%d\n", 1 ^ 2);
  printf("%d\n", 1 && 0);
  printf("%d\n", 1 || 0);
  printf("%d\n", 5 == 5);
  printf("%d\n", 1 != 1);
  printf("%d\n", 3 < 4);
  printf("%d\n", 3 > 4);
  printf("%d\n", 3 <= 4);
  printf("%d\n", 3 >= 4);
  printf("%d\n", 1 << 4);
  printf("%d\n", 16 >> 4);
  printf("%d\n", !0);
  printf("%d\n", ~123);
  printf("%d\n", +12);
  printf("%d\n", -34);
  int i = 0, *a = &i;
  printf("%d\n", ++i);
  printf("%d\n", --i);
  printf("%d\n", i++);
  printf("%d\n", i--);
  printf("%d\n", *a);
  i += 10;
  i -= 10;
  i *= 10;
  i /= 10;
  i %= 10;
  i <<= 10;
  i >>= 10;
  i &= 10;
  i |= 10;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x6,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x10,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xffffff84,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xc,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xffffffde,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
datatype: string

tool: D6: %d


assistant: ```c
int main() {
  printf("%d\n", 1 + 2);
  printf("%d\n", 2 - 1);
  printf("%d\n", 2 * 3);
  printf("%d\n", 4 / 2);
  printf("%d\n", 4 % 3);
  printf("%d\n", 1 & 2);
  printf("%d\n", 1 | 2);
  printf("%d\n", 1 ^ 2);
  printf("%d\n", 1 && 0);
  printf("%d\n", 1 || 0);
  printf("%d\n", 5 == 5);
  printf("%d\n", 1 != 1);
  printf("%d\n", 3 < 4);
  printf("%d\n", 3 > 4);
  printf("%d\n", 3 <= 4);
  printf("%d\n", 3 >= 4);
  printf("%d\n", 1 << 4);
  printf("%d\n", 16 >> 4);
  printf("%d\n", !0);
  printf("%d\n", ~123);
  printf("%d\n", +12);
  printf("%d\n", -34);
  int i = 0, *a = &i;
  printf("%d\n", ++i);
  printf("%d\n", --i);
  printf("%d\n", i++);
  printf("%d\n", i--);
  printf("%d\n", *a);
  i += 10;
  i -= 10;
  i *= 10;
  i /= 10;
  i %= 10;
  i <<= 10;
  i >>= 10;
  i &= 10;
  i |= 10;
  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   %rbp
  lea    D4(%rip),%rbp
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x6,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x10,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xffffff84,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xc,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xffffffde,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
type: string

tool: D4: %d


assistant: ```c
int main() {
  printf("%d\n", 1 + 2);
  printf("%d\n", 2 - 1);
  printf("%d\n", 2 * 3);
  printf("%d\n", 4 / 2);
  printf("%d\n", 4 % 3);
  printf("%d\n", 1 & 2);
  printf("%d\n", 1 | 2);
  printf("%d\n", 1 ^ 2);
  printf("%d\n", 1 && 0);
  printf("%d\n", 1 || 0);
  printf("%d\n", 5 == 5);
  printf("%d\n", 1 != 1);
  printf("%d\n", 3 < 4);
  printf("%d\n", 3 > 4);
  printf("%d\n", 3 <= 4);
  printf("%d\n", 3 >= 4);
  printf("%d\n", 1 << 4);
  printf("%d\n", 16 >> 4);
  printf("%d\n", !0);
  printf("%d\n", ~123);
  printf("%d\n", +12);
  printf("%d\n", -34);
  int i = 0, *a = &i;
  printf("%d\n", ++i);
  printf("%d\n", --i);
  printf("%d\n", i++);
  printf("%d\n", i--);
  printf("%d\n", *a);
  i += 10;
  i -= 10;
  i *= 10;
  i /= 10;
  i %= 10;
  i <<= 10;
  i >>= 10;
  i &= 10;
  i |= 10;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<checkRow>:
  endbr64
  mov    %rdi,%r8
  mov    %esi,%edi
  mov    %r8,%rdx
  mov    $0x0,%ecx
  mov    $0x0,%r9d
  jmp    L160
L133:
  add    $0x1,%ecx
  add    $0x3,%rdx
  cmp    $0x2,%ecx
  jg     L103
  test   %al,%al
  jne    L103
L160:
  mov    %r9d,%eax
  cmp    %dil,(%rdx)
  jne    L133
  cmp    %dil,0x1(%rdx)
  jne    L133
  cmp    %dil,0x2(%rdx)
  sete   %al
  jmp    L133
L103:
  test   %al,%al
  jne    L66
  mov    $0x1,%edx
  jmp    L38
L11:
  mov    %edx,%ecx
  add    $0x1,%rdx
  cmp    $0x2,%ecx
  jg     L66
L38:
  cmp    %dil,-0x1(%r8,%rdx,1)
  jne    L11
  lea    0x2(%rdx),%ecx
  movslq %ecx,%rcx
  cmp    %dil,(%r8,%rcx,1)
  jne    L11
  lea    0x5(%rdx),%ecx
  movslq %ecx,%rcx
  cmp    %dil,(%r8,%rcx,1)
  jne    L11
  mov    $0x1,%eax
L66:
  cmp    %sil,(%r8)
  je     L153
  cmp    %sil,0x2(%r8)
  je     L24
L173:
  movsbl %al,%eax
  ret
L153:
  cmp    %sil,0x4(%r8)
  jne    L173
  cmp    %sil,0x8(%r8)
  sete   %dl
  add    %edx,%eax
  cmp    0x2(%r8),%sil
  jne    L173
  jmp    L53
L24:
  cmp    %sil,0x4(%r8)
  jne    L173
L53:
  cmp    %sil,0x6(%r8)
  sete   %dl
  add    %edx,%eax
  jmp    L173
```

assistant: ```c
int checkRow(char *playground, char user) {
  int p;
  char won = 0;
  for (p = 0; p < 3 && won == 0; p++) {
    if ((playground[p * 3 + 0] == user) && (playground[p * 3 + 1] == user) && (playground[p * 3 + 2] == user))
      won += 1;
  }
  for (p = 0; p < 3 && won == 0; p++) {
    if ((playground[3 * 0 + p] == user) && (playground[3 * 1 + p] == user) && (playground[3 * 2 + p] == user))
      won += 1;
  }
  if ((playground[3 * 0 + 0] == user) && (playground[3 * 1 + 1] == user) && (playground[3 * 2 + 2] == user))
    won += 1;
  if ((playground[3 * 0 + 2] == user) && (playground[3 * 1 + 1] == user) && (playground[3 * 2 + 0] == user))
    won += 1;
  return won;
}
```
user: Translate the below assembly code into C programming language:

```asm
<checkRow>:
  endbr64
  mov    %esi,%r8d
  mov    %rdi,%rdx
  xor    %ecx,%ecx
L110:
  mov    $0x1,%r9d
  xor    %eax,%eax
  cmp    %r8b,(%rdx)
  je     L23
L34:
  add    $0x1,%ecx
  add    $0x3,%rdx
  cmp    $0x2,%ecx
  jg     L77
L64:
  test   %r9b,%r9b
  jne    L110
L116:
  movzbl 0x2(%rdi),%edx
  cmp    %sil,(%rdi)
  je     L159
  cmp    %dl,%sil
  je     L121
L118:
  movsbl %al,%eax
  ret
L23:
  cmp    %r8b,0x1(%rdx)
  jne    L34
  cmp    %r8b,0x2(%rdx)
  sete   %al
  setne  %r9b
  add    $0x1,%ecx
  add    $0x3,%rdx
  cmp    $0x2,%ecx
  jle    L64
  nopl   (%rax)
L77:
  test   %r9b,%r9b
  je     L116
  mov    $0x1,%edx
  mov    %r9d,%ecx
  xor    %eax,%eax
  cmp    %r8b,-0x1(%rdi,%rdx,1)
  je     L16
  cs nopw 0x0(%rax,%rax,1)
L82:
  cmp    $0x2,%edx
  setle  %r10b
  add    $0x1,%rdx
  test   %cl,%r10b
  je     L116
  mov    %r9d,%ecx
  xor    %eax,%eax
  cmp    %r8b,-0x1(%rdi,%rdx,1)
  jne    L82
L16:
  cmp    %r8b,0x2(%rdi,%rdx,1)
  jne    L82
  cmp    %r8b,0x5(%rdi,%rdx,1)
  sete   %al
  setne  %cl
  jmp    L82
L121:
  cmp    %sil,0x4(%rdi)
  jne    L118
L87:
  cmp    %sil,0x6(%rdi)
  sete   %dl
  add    %edx,%eax
  movsbl %al,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L159:
  cmp    %sil,0x4(%rdi)
  jne    L118
  cmp    %sil,0x8(%rdi)
  sete   %cl
  add    %ecx,%eax
  cmp    %dl,%sil
  jne    L118
  jmp    L87
```

assistant: ```c
int checkRow(char *playground, char user) {
  int p;
  char won = 0;
  for (p = 0; p < 3 && won == 0; p++) {
    if ((playground[p * 3 + 0] == user) && (playground[p * 3 + 1] == user) && (playground[p * 3 + 2] == user))
      won += 1;
  }
  for (p = 0; p < 3 && won == 0; p++) {
    if ((playground[3 * 0 + p] == user) && (playground[3 * 1 + p] == user) && (playground[3 * 2 + p] == user))
      won += 1;
  }
  if ((playground[3 * 0 + 0] == user) && (playground[3 * 1 + 1] == user) && (playground[3 * 2 + 2] == user))
    won += 1;
  if ((playground[3 * 0 + 2] == user) && (playground[3 * 1 + 1] == user) && (playground[3 * 2 + 0] == user))
    won += 1;
  return won;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<checkRow>:
  endbr64
  movzbl (%rdi),%edx
  cmp    %dl,%sil
  je     L102
  movzbl 0x2(%rdi),%ecx
L238:
  cmp    0x3(%rdi),%sil
  je     L32
L236:
  cmp    %sil,0x6(%rdi)
  je     L202
L198:
  cmp    %dl,%sil
  je     L128
L75:
  cmp    0x1(%rdi),%sil
  je     L120
L23:
  xor    %eax,%eax
  cmp    %sil,0x2(%rdi)
  je     L46
L40:
  mov    %ecx,%r8d
  cmp    %dl,%sil
  je     L86
  cmp    %cl,%sil
  je     L123
L257:
  movsbl %al,%eax
  ret
  nopl   0x0(%rax)
L128:
  cmp    %sil,0x3(%rdi)
  jne    L75
  cmp    %sil,0x6(%rdi)
  mov    %ecx,%r8d
  sete   %al
  jne    L75
  nopw   0x0(%rax,%rax,1)
L86:
  cmp    %dl,0x4(%rdi)
  jne    L257
  cmp    %dl,0x8(%rdi)
  sete   %cl
  add    %ecx,%eax
  cmp    %dl,%r8b
  jne    L257
L241:
  cmp    %sil,0x6(%rdi)
  sete   %dl
  add    %edx,%eax
  movsbl %al,%eax
  ret
  nop
L32:
  cmp    0x4(%rdi),%sil
  jne    L236
  cmp    0x5(%rdi),%sil
  sete   %al
  jne    L236
  jmp    L40
  nopl   (%rax)
L102:
  cmp    0x1(%rdi),%sil
  movzbl 0x2(%rdi),%ecx
  jne    L238
  cmp    %cl,%sil
  mov    %esi,%r8d
  sete   %al
  je     L86
  jmp    L238
  cs nopw 0x0(%rax,%rax,1)
L123:
  cmp    %sil,0x4(%rdi)
  je     L241
  movsbl %al,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L120:
  cmp    0x4(%rdi),%sil
  jne    L23
  cmp    %sil,0x7(%rdi)
  sete   %al
  je     L40
  jmp    L23
  nopl   0x0(%rax)
L202:
  cmp    %sil,0x7(%rdi)
  jne    L198
  cmp    %sil,0x8(%rdi)
  sete   %al
  je     L40
  jmp    L198
  nopl   0x0(%rax)
L46:
  cmp    %sil,0x5(%rdi)
  jne    L40
  cmp    %sil,0x8(%rdi)
  sete   %al
  jmp    L40
```

assistant: ```c
int checkRow(char *playground, char user) {
  int p;
  char won = 0;
  for (p = 0; p < 3 && won == 0; p++) {
    if ((playground[p * 3 + 0] == user) && (playground[p * 3 + 1] == user) && (playground[p * 3 + 2] == user))
      won += 1;
  }
  for (p = 0; p < 3 && won == 0; p++) {
    if ((playground[3 * 0 + p] == user) && (playground[3 * 1 + p] == user) && (playground[3 * 2 + p] == user))
      won += 1;
  }
  if ((playground[3 * 0 + 0] == user) && (playground[3 * 1 + 1] == user) && (playground[3 * 2 + 2] == user))
    won += 1;
  if ((playground[3 * 0 + 2] == user) && (playground[3 * 1 + 1] == user) && (playground[3 * 2 + 0] == user))
    won += 1;
  return won;
}
```
user: Translate the below assembly code into C programming language:

```asm
<genrand_res53>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <genrand_int32@plt>
  sar    $0x5,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <genrand_int32@plt>
  sar    $0x6,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  js     L33
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L9
L33:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L9:
  movsd  D9(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  mov    -0x8(%rbp),%rax
  test   %rax,%rax
  js     L16
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L29
L16:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L29:
  addsd  %xmm0,%xmm1
  movsd  D0(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type_infered: f64
name: parse_data
arguments: label: D0
type_infered: byte[8]

tool: D9: 67108864

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0xa0,0x3c]}

assistant: ```c
double genrand_res53(void) {
  unsigned long a = genrand_int32() >> 5, b = genrand_int32() >> 6;
  return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<genrand_res53>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <genrand_int32@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <genrand_int32@plt>
  sar    $0x5,%ebx
  movslq %ebx,%rdx
  test   %rdx,%rdx
  js     L26
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
L33:
  mulsd  D14(%rip),%xmm0
  sar    $0x6,%eax
  cltq
  test   %rax,%rax
  js     L39
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
L36:
  addsd  %xmm1,%xmm0
  mulsd  D17(%rip),%xmm0
  pop    %rbx
  ret
L26:
  mov    %rdx,%rcx
  shr    %rcx
  and    $0x1,%edx
  or     %rdx,%rcx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rcx,%xmm0
  addsd  %xmm0,%xmm0
  jmp    L33
L39:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm1,%xmm1
  cvtsi2sd %rdx,%xmm1
  addsd  %xmm1,%xmm1
  jmp    L36
```

tool_calls: name: read_data
arguments: data_label: D14
data_type: f64
name: read_data
arguments: data_label: D17
data_type: byte[8]

tool: D14: 67108864

tool: {"D17": [0x00,0x00,0x00,0x00,0x00,0x00,0xa0,0x3c]}

assistant: ```c
double genrand_res53(void) {
  unsigned long a = genrand_int32() >> 5, b = genrand_int32() >> 6;
  return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<genrand_res53>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <genrand_int32@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <genrand_int32@plt>
  sar    $0x5,%ebx
  movslq %ebx,%rdx
  test   %rdx,%rdx
  js     L5
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
L8:
  mulsd  D0(%rip),%xmm0
  sar    $0x6,%eax
  cltq
  test   %rax,%rax
  js     L21
  pxor   %xmm1,%xmm1
  pop    %rbx
  cvtsi2sd %rax,%xmm1
  addsd  %xmm1,%xmm0
  mulsd  D14(%rip),%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %rdx,%rcx
  and    $0x1,%edx
  pxor   %xmm0,%xmm0
  shr    %rcx
  or     %rdx,%rcx
  cvtsi2sd %rcx,%xmm0
  addsd  %xmm0,%xmm0
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L21:
  mov    %rax,%rdx
  and    $0x1,%eax
  pxor   %xmm1,%xmm1
  pop    %rbx
  shr    %rdx
  or     %rax,%rdx
  cvtsi2sd %rdx,%xmm1
  addsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  mulsd  D14(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
type_infered: f64
name: read_data
arguments: data_label: D14
type_infered: byte[8]

tool: D0: 67108864

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0xa0,0x3c]}

assistant: ```c
double genrand_res53(void) {
  unsigned long a = genrand_int32() >> 5, b = genrand_int32() >> 6;
  return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
```