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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <struct_by_value_17_x@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  struct_by_value_17_x();
  exit(0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <struct_by_value_17_x@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  struct_by_value_17_x();
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <struct_by_value_17_x@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  struct_by_value_17_x();
  exit(0);
}
```
user: What is the c source code of the assembly code below:

```asm
<ytht_crypt_genpasswd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  lea    D22(%rip),%rax
  jmp    L6
L4:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  mov    %ebx,%eax
  cltd
  idiv   %esi
  mov    %edx,%ecx
  mov    %ecx,%eax
  mov    %eax,%edi
  call   <srand@plt>
  lea    -0x2c(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <random@plt>
  mov    %eax,%ecx
  lea    -0x22(%rbp),%rax
  mov    $0x3,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <to64@plt>
  mov    -0x28(%rbp),%eax
  lea    -0x22(%rbp),%rdx
  lea    0x3(%rdx),%rcx
  mov    $0x3,%edx
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <to64@plt>
  mov    -0x2c(%rbp),%eax
  lea    -0x22(%rbp),%rdx
  lea    0x6(%rdx),%rcx
  mov    $0x2,%edx
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <to64@plt>
  movb   $0x0,-0x1a(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x22(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <ytht_crypt_crypt1@plt>
L6:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D22
type_infered: string

tool: 

assistant: ```c
char *ytht_crypt_genpasswd(char *pw) {
  char salt[10];
  static char pwbuf[68];
  struct timeval tv;
  if (strlen(pw) == 0)
    return "";
  srand(time(0) % getpid());
  gettimeofday(&tv, 0);
  to64(&salt[0], random(), 3);
  to64(&salt[3], tv.tv_usec, 3);
  to64(&salt[6], tv.tv_sec, 2);
  salt[8] = '\0';
  strcpy(pwbuf, pw);
  return ytht_crypt_crypt1(pwbuf, salt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ytht_crypt_genpasswd>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rax
  cmpb   $0x0,(%rdi)
  jne    L10
L15:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    %rdi,%rbx
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%ecx
  mov    %ebp,%eax
  cltd
  idiv   %ecx
  mov    %edx,%edi
  call   <srand@plt>
  lea    0x4(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <random@plt>
  mov    %eax,%esi
  lea    0xe(%rsp),%rbp
  mov    $0x3,%edx
  mov    %rbp,%rdi
  call   <to64@plt>
  lea    0x11(%rsp),%rdi
  mov    $0x3,%edx
  mov    0x8(%rsp),%esi
  call   <to64@plt>
  lea    0x14(%rsp),%rdi
  mov    $0x2,%edx
  mov    0x4(%rsp),%esi
  call   <to64@plt>
  movb   $0x0,0x16(%rsp)
  mov    %rbx,%rsi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ytht_crypt_crypt1@plt>
  jmp    L15
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string

tool: 

assistant: ```c
char *ytht_crypt_genpasswd(char *pw) {
  char salt[10];
  static char pwbuf[68];
  struct timeval tv;
  if (strlen(pw) == 0)
    return "";
  srand(time(0) % getpid());
  gettimeofday(&tv, 0);
  to64(&salt[0], random(), 3);
  to64(&salt[3], tv.tv_usec, 3);
  to64(&salt[6], tv.tv_sec, 2);
  salt[8] = '\0';
  strcpy(pwbuf, pw);
  return ytht_crypt_crypt1(pwbuf, salt);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ytht_crypt_genpasswd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  lea    D3(%rip),%rax
  je     L8
  mov    %rdi,%rbp
  xor    %edi,%edi
  lea    0xe(%rsp),%r12
  call   <time@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    %edx,%edi
  call   <srand@plt>
  lea    0x4(%rsp),%rdi
  xor    %esi,%esi
  call   <gettimeofday@plt>
  xor    %eax,%eax
  call   <random@plt>
  mov    $0x3,%edx
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <to64@plt>
  mov    0x8(%rsp),%esi
  lea    0x11(%rsp),%rdi
  mov    $0x3,%edx
  call   <to64@plt>
  mov    0x4(%rsp),%esi
  lea    0x14(%rsp),%rdi
  mov    $0x2,%edx
  call   <to64@plt>
  mov    %rbp,%rsi
  lea    D7(%rip),%rbp
  movb   $0x0,0x16(%rsp)
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ytht_crypt_crypt1@plt>
L8:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
type_guessed: string

tool: 

assistant: ```c
char *ytht_crypt_genpasswd(char *pw) {
  char salt[10];
  static char pwbuf[68];
  struct timeval tv;
  if (strlen(pw) == 0)
    return "";
  srand(time(0) % getpid());
  gettimeofday(&tv, 0);
  to64(&salt[0], random(), 3);
  to64(&salt[3], tv.tv_usec, 3);
  to64(&salt[6], tv.tv_sec, 2);
  salt[8] = '\0';
  strcpy(pwbuf, pw);
  return ytht_crypt_crypt1(pwbuf, salt);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ytht_crypt_genpasswd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  lea    D17(%rip),%rax
  je     L13
  mov    %rdi,%rbp
  xor    %edi,%edi
  lea    0xe(%rsp),%r12
  call   <time@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    %edx,%edi
  call   <srand@plt>
  lea    0x4(%rsp),%rdi
  xor    %esi,%esi
  call   <gettimeofday@plt>
  xor    %eax,%eax
  call   <random@plt>
  mov    $0x3,%edx
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <to64@plt>
  mov    0x8(%rsp),%esi
  lea    0x11(%rsp),%rdi
  mov    $0x3,%edx
  call   <to64@plt>
  mov    0x4(%rsp),%esi
  lea    0x14(%rsp),%rdi
  mov    $0x2,%edx
  call   <to64@plt>
  mov    %rbp,%rsi
  lea    D15(%rip),%rbp
  movb   $0x0,0x16(%rsp)
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ytht_crypt_crypt1@plt>
L13:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
type: string

tool: 

assistant: ```c
char *ytht_crypt_genpasswd(char *pw) {
  char salt[10];
  static char pwbuf[68];
  struct timeval tv;
  if (strlen(pw) == 0)
    return "";
  srand(time(0) % getpid());
  gettimeofday(&tv, 0);
  to64(&salt[0], random(), 3);
  to64(&salt[3], tv.tv_usec, 3);
  to64(&salt[6], tv.tv_sec, 2);
  salt[8] = '\0';
  strcpy(pwbuf, pw);
  return ytht_crypt_crypt1(pwbuf, salt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_chars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <mblen@plt>
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L31
L29:
  mov    -0x30(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    -0x28(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <mblen@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L21
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L31:
  mov    -0x8(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L29
  jmp    L5
L21:
  nop
L5:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t count_chars(const char *mbs_begin, size_t mbs_len) {
  mblen(NULL, 0);
  size_t chars = 0;
  size_t i = 0;
  while (i < mbs_len) {
    int len = mblen(mbs_begin + i, mbs_len - i);
    if (len < 1)
      break;
    i += len;
    chars++;
  }
  return chars;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<count_chars>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <mblen@plt>
  test   %rbp,%rbp
  je     L38
  mov    $0x0,%ebx
  mov    $0x0,%r12d
L2:
  mov    %rbp,%rsi
  sub    %rbx,%rsi
  lea    0x0(%r13,%rbx,1),%rdi
  call   <mblen@plt>
  test   %eax,%eax
  jle    L34
  cltq
  add    %rax,%rbx
  add    $0x1,%r12
  cmp    %rbx,%rbp
  ja     L2
L34:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L38:
  mov    %rbp,%r12
  jmp    L34
```

assistant: ```c
size_t count_chars(const char *mbs_begin, size_t mbs_len) {
  mblen(NULL, 0);
  size_t chars = 0;
  size_t i = 0;
  while (i < mbs_len) {
    int len = mblen(mbs_begin + i, mbs_len - i);
    if (len < 1)
      break;
    i += len;
    chars++;
  }
  return chars;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_chars>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x8,%rsp
  call   <mblen@plt>
  test   %rbx,%rbx
  je     L26
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  jmp    L38
  nopw   0x0(%rax,%rax,1)
L14:
  cltq
  add    $0x1,%r12
  add    %rax,%r13
  cmp    %r13,%rbx
  jbe    L42
L38:
  mov    %rbx,%rsi
  lea    0x0(%rbp,%r13,1),%rdi
  sub    %r13,%rsi
  call   <mblen@plt>
  test   %eax,%eax
  jg     L14
L42:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  xor    %r12d,%r12d
  jmp    L42
```

assistant: ```c
size_t count_chars(const char *mbs_begin, size_t mbs_len) {
  mblen(NULL, 0);
  size_t chars = 0;
  size_t i = 0;
  while (i < mbs_len) {
    int len = mblen(mbs_begin + i, mbs_len - i);
    if (len < 1)
      break;
    i += len;
    chars++;
  }
  return chars;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_chars>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x8,%rsp
  call   <mblen@plt>
  test   %rbx,%rbx
  je     L47
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L38:
  cltq
  add    $0x1,%r12
  add    %rax,%r13
  cmp    %r13,%rbx
  jbe    L28
L40:
  mov    %rbx,%rsi
  lea    0x0(%rbp,%r13,1),%rdi
  sub    %r13,%rsi
  call   <mblen@plt>
  test   %eax,%eax
  jg     L38
L28:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L47:
  xor    %r12d,%r12d
  jmp    L28
```

assistant: ```c
size_t count_chars(const char *mbs_begin, size_t mbs_len) {
  mblen(NULL, 0);
  size_t chars = 0;
  size_t i = 0;
  while (i < mbs_len) {
    int len = mblen(mbs_begin + i, mbs_len - i);
    if (len < 1)
      break;
    i += len;
    chars++;
  }
  return chars;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<intCmpFn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L22
  mov    $0x0,%eax
  jmp    L12
L22:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L33
  mov    $0xffffffff,%eax
  jmp    L12
L33:
  mov    $0x1,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int intCmpFn(const void *p1, const void *p2) {
  int v1, v2;
  v1 = *((int *)p1);
  v2 = *((int *)p2);
  if (v1 == v2)
    return 0;
  return (v1 < v2) ? -1 : +1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<intCmpFn>:
  endbr64
  mov    (%rdi),%ecx
  mov    (%rsi),%edx
  mov    $0x0,%eax
  cmp    %edx,%ecx
  je     L5
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L5:
  ret
```

assistant: ```c
int intCmpFn(const void *p1, const void *p2) {
  int v1, v2;
  v1 = *((int *)p1);
  v2 = *((int *)p2);
  if (v1 == v2)
    return 0;
  return (v1 < v2) ? -1 : +1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<intCmpFn>:
  endbr64
  mov    (%rsi),%edx
  xor    %eax,%eax
  cmp    %edx,(%rdi)
  je     L3
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L3:
  ret
```

assistant: ```c
int intCmpFn(const void *p1, const void *p2) {
  int v1, v2;
  v1 = *((int *)p1);
  v2 = *((int *)p2);
  if (v1 == v2)
    return 0;
  return (v1 < v2) ? -1 : +1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<intCmpFn>:
  endbr64
  mov    (%rsi),%edx
  xor    %eax,%eax
  cmp    %edx,(%rdi)
  je     L9
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L9:
  ret
```

assistant: ```c
int intCmpFn(const void *p1, const void *p2) {
  int v1, v2;
  v1 = *((int *)p1);
  v2 = *((int *)p2);
  if (v1 == v2)
    return 0;
  return (v1 < v2) ? -1 : +1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fim>:
  endbr64
  push   %rbx
  mov    (%rdi),%rbx
  mov    $0x0,%esi
  jmp    L0
L128:
  cmp    $0x58,%r10b
  setne  %al
  movzbl %al,%eax
  add    $0x1,%eax
  jmp    L60
L74:
  cmp    $0x58,%r11b
  setne  %al
  movzbl %al,%eax
  add    $0x1,%eax
  jmp    L60
L76:
  add    $0x1,%rsi
  cmp    $0x3,%rsi
  je     L29
L0:
  mov    (%rdi,%rsi,8),%r9
  movzbl (%r9),%r10d
  movzbl (%rbx,%rsi,1),%r11d
  mov    $0x0,%eax
  mov    $0x0,%ecx
  mov    $0x0,%edx
L3:
  cmp    %r10b,(%r9,%rax,1)
  sete   %r8b
  movzbl %r8b,%r8d
  add    %r8d,%edx
  mov    (%rdi,%rax,8),%r8
  cmp    %r11b,(%r8,%rsi,1)
  sete   %r8b
  movzbl %r8b,%r8d
  add    %r8d,%ecx
  cmp    $0x3,%edx
  je     L128
  cmp    $0x3,%ecx
  je     L74
  add    $0x1,%rax
  cmp    $0x3,%rax
  jne    L3
  jmp    L76
L29:
  movzbl (%rbx),%eax
  mov    0x8(%rdi),%rdx
  movzbl 0x1(%rdx),%edx
  cmp    %al,%dl
  je     L124
L69:
  movzbl 0x2(%rbx),%ecx
  mov    $0x0,%eax
  cmp    %cl,%dl
  je     L101
L60:
  pop    %rbx
  ret
L124:
  mov    0x10(%rdi),%rcx
  cmp    %al,0x2(%rcx)
  jne    L69
  cmp    $0x58,%al
  setne  %al
  movzbl %al,%eax
  add    $0x1,%eax
  jmp    L60
L101:
  mov    0x10(%rdi),%rdx
  cmp    %cl,(%rdx)
  jne    L60
  cmp    $0x58,%cl
  setne  %al
  movzbl %al,%eax
  add    $0x1,%eax
  jmp    L60
```

assistant: ```c
int fim(char **mat) {
  int i, j, final = 0, cont1 = 0, cont2 = 0;
  char arm1, arm2;
  for (i = 0; i < 3; i++) {
    arm1 = mat[i][0];
    arm2 = mat[0][i];
    for (j = 0; j < 3; j++) {
      if (mat[i][j] == arm1) {
        cont1++;
      }
      if (mat[j][i] == arm2) {
        cont2++;
      }
      if (cont1 == 3) {
        if (arm1 == 'X')
          return 1;
        else
          return 2;
      }
      if (cont2 == 3) {
        if (arm2 == 'X')
          return 1;
        else
          return 2;
      }
    }
    cont1 = 0;
    cont2 = 0;
  }
  arm1 = mat[0][0];
  if (arm1 == mat[1][1] && arm1 == mat[2][2]) {
    if (arm1 == 'X')
      return 1;
    else
      return 2;
  }
  arm1 = mat[0][2];
  if (arm1 == mat[1][1] && arm1 == mat[2][0]) {
    if (arm1 == 'X')
      return 1;
    else
      return 2;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fim>:
  endbr64
  push   %rbp
  mov    %rdi,%r11
  xor    %esi,%esi
  push   %rbx
  mov    (%rdi),%rbp
  mov    %rbp,%rbx
L20:
  movzbl (%rbx),%r9d
  movzbl 0x0(%rbp,%rsi,1),%r10d
  xor    %eax,%eax
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %r10d,%r8d
  mov    %r9d,%edi
L89:
  cmp    %dil,%r9b
  sete   %dil
  movzbl %dil,%edi
  add    %edi,%edx
  xor    %edi,%edi
  cmp    %r8b,%r10b
  sete   %dil
  add    %edi,%ecx
  cmp    $0x3,%edx
  je     L38
  cmp    $0x3,%ecx
  je     L24
  add    $0x1,%rax
  cmp    $0x3,%rax
  je     L22
  mov    (%r11,%rax,8),%r8
  movzbl (%rbx,%rax,1),%edi
  movzbl (%r8,%rsi,1),%r8d
  jmp    L89
  nopw   0x0(%rax,%rax,1)
L22:
  add    $0x1,%rsi
  cmp    $0x3,%rsi
  je     L49
  mov    (%r11,%rsi,8),%rbx
  jmp    L20
L49:
  mov    0x8(%r11),%rax
  movzbl 0x1(%rax),%edx
  cmp    0x0(%rbp),%dl
  je     L18
  xor    %eax,%eax
  cmp    0x2(%rbp),%dl
  je     L66
L53:
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    0x10(%r11),%rax
  cmp    %dl,0x2(%rax)
  je     L0
  xor    %eax,%eax
  cmp    0x2(%rbp),%dl
  jne    L53
L66:
  mov    0x10(%r11),%rcx
  cmp    %dl,(%rcx)
  jne    L53
L0:
  xor    %eax,%eax
  cmp    $0x58,%dl
  pop    %rbx
  pop    %rbp
  setne  %al
  add    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L38:
  xor    %eax,%eax
  cmp    $0x58,%r9b
  pop    %rbx
  pop    %rbp
  setne  %al
  add    $0x1,%eax
  ret
  nop
L24:
  xor    %eax,%eax
  cmp    $0x58,%r10b
  pop    %rbx
  pop    %rbp
  setne  %al
  add    $0x1,%eax
  ret
```

assistant: ```c
int fim(char **mat) {
  int i, j, final = 0, cont1 = 0, cont2 = 0;
  char arm1, arm2;
  for (i = 0; i < 3; i++) {
    arm1 = mat[i][0];
    arm2 = mat[0][i];
    for (j = 0; j < 3; j++) {
      if (mat[i][j] == arm1) {
        cont1++;
      }
      if (mat[j][i] == arm2) {
        cont2++;
      }
      if (cont1 == 3) {
        if (arm1 == 'X')
          return 1;
        else
          return 2;
      }
      if (cont2 == 3) {
        if (arm2 == 'X')
          return 1;
        else
          return 2;
      }
    }
    cont1 = 0;
    cont2 = 0;
  }
  arm1 = mat[0][0];
  if (arm1 == mat[1][1] && arm1 == mat[2][2]) {
    if (arm1 == 'X')
      return 1;
    else
      return 2;
  }
  arm1 = mat[0][2];
  if (arm1 == mat[1][1] && arm1 == mat[2][0]) {
    if (arm1 == 'X')
      return 1;
    else
      return 2;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fim>:
  endbr64
  push   %rbp
  xor    %r11d,%r11d
  push   %rbx
  mov    (%rdi),%rdx
  mov    0x8(%rdi),%r8
  mov    0x10(%rdi),%r10
  movzbl (%rdx),%eax
  movzbl 0x1(%rdx),%r9d
  movzbl (%r8),%edi
  movzbl 0x2(%rdx),%ecx
  cmp    %al,%r9b
  movzbl (%r10),%esi
  sete   %r11b
  xor    %ebx,%ebx
  add    $0x2,%r11d
  cmp    %dil,%al
  sete   %bl
  add    $0x2,%ebx
  cmp    %cl,%al
  je     L101
  cmp    %sil,%al
  je     L195
L180:
  movzbl 0x1(%r8),%edx
  xor    %r11d,%r11d
  movzbl 0x2(%r8),%r8d
  movzbl 0x1(%r10),%ebp
  cmp    %dil,%dl
  sete   %r11b
  xor    %ebx,%ebx
  add    $0x2,%r11d
  cmp    %dl,%r9b
  sete   %bl
  add    $0x2,%ebx
  cmp    %dil,%r8b
  je     L85
  cmp    %bpl,%r9b
  je     L41
L152:
  xor    %edi,%edi
  cmp    %sil,%bpl
  movzbl 0x2(%r10),%r9d
  sete   %dil
  add    $0x2,%edi
  cmp    %cl,%r8b
  sete   %r8b
  movzbl %r8b,%r8d
  add    $0x2,%r8d
  cmp    %sil,%r9b
  je     L252
  cmp    %cl,%r9b
  je     L69
L10:
  cmp    %dl,%al
  je     L98
  xor    %eax,%eax
  cmp    %cl,%dl
  je     L129
L61:
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L252:
  cmp    %cl,%sil
  je     L92
  cmp    $0x3,%edi
  jne    L10
L200:
  mov    %esi,%eax
L161:
  cmp    $0x58,%al
  pop    %rbx
  pop    %rbp
  setne  %al
  movzbl %al,%eax
  add    $0x1,%eax
  ret
  xchg   %ax,%ax
L223:
  cmp    $0x3,%r11d
  je     L161
L195:
  cmp    $0x3,%ebx
  jne    L180
  jmp    L161
  nopl   0x0(%rax)
L85:
  cmp    %bpl,%r9b
  jne    L65
  cmp    $0x3,%r11d
  je     L265
L41:
  cmp    $0x3,%ebx
  jne    L152
  mov    %r9d,%eax
  jmp    L161
  nopl   0x0(%rax)
L101:
  cmp    %sil,%al
  je     L223
  cmp    $0x3,%r11d
  jne    L180
  jmp    L161
  nopl   0x0(%rax)
L98:
  cmp    %r9b,%al
  je     L161
  xor    %eax,%eax
  cmp    %cl,%dl
  jne    L61
L129:
  cmp    %sil,%dl
  jne    L61
  xor    %eax,%eax
  cmp    $0x58,%dl
  setne  %al
  add    $0x1,%eax
  jmp    L61
L92:
  cmp    $0x3,%edi
  je     L200
L69:
  cmp    $0x3,%r8d
  jne    L10
  mov    %ecx,%eax
  jmp    L161
  nopw   0x0(%rax,%rax,1)
L65:
  cmp    $0x3,%r11d
  jne    L152
L265:
  mov    %edi,%eax
  jmp    L161
```

assistant: ```c
int fim(char **mat) {
  int i, j, final = 0, cont1 = 0, cont2 = 0;
  char arm1, arm2;
  for (i = 0; i < 3; i++) {
    arm1 = mat[i][0];
    arm2 = mat[0][i];
    for (j = 0; j < 3; j++) {
      if (mat[i][j] == arm1) {
        cont1++;
      }
      if (mat[j][i] == arm2) {
        cont2++;
      }
      if (cont1 == 3) {
        if (arm1 == 'X')
          return 1;
        else
          return 2;
      }
      if (cont2 == 3) {
        if (arm2 == 'X')
          return 1;
        else
          return 2;
      }
    }
    cont1 = 0;
    cont2 = 0;
  }
  arm1 = mat[0][0];
  if (arm1 == mat[1][1] && arm1 == mat[2][2]) {
    if (arm1 == 'X')
      return 1;
    else
      return 2;
  }
  arm1 = mat[0][2];
  if (arm1 == mat[1][1] && arm1 == mat[2][0]) {
    if (arm1 == 'X')
      return 1;
    else
      return 2;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L1:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x12(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  cmpb   $0x0,-0x12(%rbp)
  jne    L9
  movzbl -0x12(%rbp),%eax
  movzbl -0x11(%rbp),%edx
  sub    %edx,%eax
  jmp    L29
L9:
  movzbl -0x12(%rbp),%eax
  cmp    -0x11(%rbp),%al
  je     L1
  movzbl -0x12(%rbp),%eax
  movzbl -0x11(%rbp),%edx
  sub    %edx,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
int strcmp(const char *p1, const char *p2) {
  const unsigned char *s1 = (const unsigned char *)p1;
  const unsigned char *s2 = (const unsigned char *)p2;
  unsigned char c1, c2;
  do {
    c1 = (unsigned char)*s1++;
    c2 = (unsigned char)*s2++;
    if (c1 == 0)
      return c1 - c2;
  } while (c1 == c2);
  return c1 - c2;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strcmp>:
  endbr64
  mov    $0x0,%eax
L1:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rsi,%rax,1),%ecx
  test   %dl,%dl
  je     L17
  add    $0x1,%rax
  cmp    %cl,%dl
  je     L1
  movzbl %dl,%eax
  movzbl %cl,%ecx
  sub    %ecx,%eax
  ret
L17:
  movzbl %cl,%eax
  neg    %eax
  ret
```

assistant: ```c
int strcmp(const char *p1, const char *p2) {
  const unsigned char *s1 = (const unsigned char *)p1;
  const unsigned char *s2 = (const unsigned char *)p2;
  unsigned char c1, c2;
  do {
    c1 = (unsigned char)*s1++;
    c2 = (unsigned char)*s2++;
    if (c1 == 0)
      return c1 - c2;
  } while (c1 == c2);
  return c1 - c2;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcmp>:
  endbr64
  xor    %eax,%eax
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L5:
  add    $0x1,%rax
  cmp    %cl,%dl
  jne    L7
L28:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rsi,%rax,1),%ecx
  test   %dl,%dl
  jne    L5
  movzbl %cl,%eax
  neg    %eax
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int strcmp(const char *p1, const char *p2) {
  const unsigned char *s1 = (const unsigned char *)p1;
  const unsigned char *s2 = (const unsigned char *)p2;
  unsigned char c1, c2;
  do {
    c1 = (unsigned char)*s1++;
    c2 = (unsigned char)*s2++;
    if (c1 == 0)
      return c1 - c2;
  } while (c1 == c2);
  return c1 - c2;
}
```