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
<Start_Description>:
  endbr64
  lea    D6(%rip),%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type: string

tool: D6: The `Start` class can be implemented by types which provide an abstract notion of a started and stopped state. This can be real processes such as `Thread`, or something like `File` where the on/off correspond to if the file is open or not.

The main nicety of the `Start` class is that it allows use of the `with` macro which performs the `start` function at the opening of a scope block and the `stop` function at the end.

assistant: ```c
const char *Start_Description(void) {
  return "The `Start` class can be implemented by types which provide an abstract "
         "notion of a started and stopped state. This can be real processes such "
         "as `Thread`, or something like `File` where the on/off correspond to "
         "if the file is open or not."
         "\n\n"
         "The main nicety of the `Start` class is that it allows use of the `with` "
         "macro which performs the `start` function at the opening of a scope block "
         "and the `stop` function at the end.";
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xlisten>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <listen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int xlisten(int sockfd, int backlog) {
  int ret = listen(sockfd, backlog);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<xlisten>:
  endbr64
  sub    $0x8,%rsp
  call   <listen@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int xlisten(int sockfd, int backlog) {
  int ret = listen(sockfd, backlog);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xlisten>:
  endbr64
  jmp    1050 <listen@plt>
```

assistant: ```c
int xlisten(int sockfd, int backlog) {
  int ret = listen(sockfd, backlog);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<xlisten>:
  endbr64
  jmp    1050 <listen@plt>
```

assistant: ```c
int xlisten(int sockfd, int backlog) {
  int ret = listen(sockfd, backlog);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dwrap_callback_ctrl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <BIO_next@plt>
  mov    %eax,%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  mov    %ecx,%edi
  call   <BIO_callback_ctrl@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long dwrap_callback_ctrl(BIO *b, int cmd, bio_info_cb *fp) {
  long ret;
  ret = BIO_callback_ctrl(BIO_next(b), cmd, fp);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dwrap_callback_ctrl>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebx
  mov    %rdx,%rbp
  call   <BIO_next@plt>
  mov    %eax,%edi
  mov    %rbp,%rdx
  mov    %ebx,%esi
  call   <BIO_callback_ctrl@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
long dwrap_callback_ctrl(BIO *b, int cmd, bio_info_cb *fp) {
  long ret;
  ret = BIO_callback_ctrl(BIO_next(b), cmd, fp);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dwrap_callback_ctrl>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  call   <BIO_next@plt>
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %ebp,%esi
  mov    %eax,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <BIO_callback_ctrl@plt>
```

assistant: ```c
long dwrap_callback_ctrl(BIO *b, int cmd, bio_info_cb *fp) {
  long ret;
  ret = BIO_callback_ctrl(BIO_next(b), cmd, fp);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dwrap_callback_ctrl>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  call   <BIO_next@plt>
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %ebp,%esi
  mov    %eax,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <BIO_callback_ctrl@plt>
```

assistant: ```c
long dwrap_callback_ctrl(BIO *b, int cmd, bio_info_cb *fp) {
  long ret;
  ret = BIO_callback_ctrl(BIO_next(b), cmd, fp);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  add    $0x3686696a,%rax
  xor    $0xffffffffce5d000b,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L25
L18:
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rax
  shl    $0x4,%rax
  and    $0x1f0,%eax
  or     %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,-0x10(%rbp,%rax,8)
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x2,%rax
  and    $0x3c,%eax
  or     %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,-0x10(%rbp,%rax,8)
  addq   $0x1,-0x18(%rbp)
L25:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L18
  mov    -0x10(%rbp),%rax
  or     $0x61c775a,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[local1] & 31UL) << 4UL;
      state[local1] |= (state[0UL] & 15UL) << 2UL;
      local1++;
    }
    output[0UL] = state[0UL] | 102528858UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  add    $0x3686696a,%rax
  xor    $0xffffffffce5d000b,%rax
  mov    0x8(%rdi),%rdi
  test   %rdi,%rdi
  je     L3
  mov    $0x0,%ecx
L12:
  mov    %rax,%rdx
  shl    $0x4,%rdx
  and    $0x1f0,%edx
  or     %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  and    $0x3c,%edx
  or     %rdx,%rax
  add    $0x1,%rcx
  cmp    %rdi,%rcx
  jne    L12
L3:
  or     $0x61c775a,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[local1] & 31UL) << 4UL;
      state[local1] |= (state[0UL] & 15UL) << 2UL;
      local1++;
    }
    output[0UL] = state[0UL] | 102528858UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rdi
  add    $0x3686696a,%rax
  xor    $0xffffffffce5d000b,%rax
  test   %rdi,%rdi
  je     L0
  xor    %ecx,%ecx
  xchg   %ax,%ax
L4:
  mov    %rax,%rdx
  add    $0x1,%rcx
  shl    $0x4,%rdx
  and    $0x1f0,%edx
  or     %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  and    $0x3c,%edx
  or     %rdx,%rax
  cmp    %rdi,%rcx
  jne    L4
L0:
  or     $0x61c775a,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[local1] & 31UL) << 4UL;
      state[local1] |= (state[0UL] & 15UL) << 2UL;
      local1++;
    }
    output[0UL] = state[0UL] | 102528858UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rdi
  add    $0x3686696a,%rax
  xor    $0xffffffffce5d000b,%rax
  test   %rdi,%rdi
  je     L1
  xor    %ecx,%ecx
  xchg   %ax,%ax
L17:
  mov    %rax,%rdx
  add    $0x1,%rcx
  shl    $0x4,%rdx
  and    $0x1f0,%edx
  or     %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  and    $0x3c,%edx
  or     %rdx,%rax
  cmp    %rdi,%rcx
  jne    L17
L1:
  or     $0x61c775a,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[local1] & 31UL) << 4UL;
      state[local1] |= (state[0UL] & 15UL) << 2UL;
      local1++;
    }
    output[0UL] = state[0UL] | 102528858UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<pattern>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%ebx
  mov    $0x0,%r12d
  jmp    L27
L23:
  mov    %r12d,%eax
  add    $0x1,%eax
  mov    %eax,%r12d
L20:
  movsbl %r12b,%eax
  and    $0x7f,%eax
  mov    %eax,%edi
  call   <isprint@plt>
  test   %eax,%eax
  je     L23
  mov    %r12d,%ecx
  mov    %ecx,%eax
  add    $0x1,%eax
  mov    %eax,%r12d
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  and    $0x7f,%ecx
  mov    %ecx,%edx
  mov    %dl,(%rax)
L27:
  mov    %ebx,%eax
  lea    -0x1(%rax),%ebx
  test   %eax,%eax
  jg     L20
  nop
  nop
  add    $0x10,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void pattern(char *cp, int cnt) {
  register char c;
  register int cnt1;
  cnt1 = cnt;
  c = 0;
  while (cnt1-- > 0) {
    while (!isprint((c & 0x7F)))
      c++;
    *cp++ = (c++ & 0x7F);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<pattern>:
  endbr64
  test   %esi,%esi
  jle    L1
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%esi
  lea    (%rdi,%rsi,1),%r13
  mov    $0x0,%ebx
  jmp    L0
L11:
  add    $0x1,%ebx
L0:
  mov    %ebx,%ebp
  and    $0x7f,%ebp
  mov    %ebx,%edi
  and    $0x7f,%edi
  call   <isprint@plt>
  test   %eax,%eax
  je     L11
  add    $0x1,%ebx
  add    $0x1,%r12
  mov    %bpl,-0x1(%r12)
  cmp    %r13,%r12
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  ret
```

assistant: ```c
void pattern(char *cp, int cnt) {
  register char c;
  register int cnt1;
  cnt1 = cnt;
  c = 0;
  while (cnt1-- > 0) {
    while (!isprint((c & 0x7F)))
      c++;
    *cp++ = (c++ & 0x7F);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<pattern>:
  endbr64
  test   %esi,%esi
  jle    L26
  push   %r13
  movslq %esi,%rsi
  push   %r12
  lea    (%rdi,%rsi,1),%r13
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L24:
  mov    %ebx,%edi
  mov    %ebx,%ebp
  add    $0x1,%ebx
  and    $0x7f,%edi
  and    $0x7f,%ebp
  call   <isprint@plt>
  test   %eax,%eax
  je     L24
  add    $0x1,%r12
  mov    %bpl,-0x1(%r12)
  cmp    %r12,%r13
  jne    L24
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  ret
```

assistant: ```c
void pattern(char *cp, int cnt) {
  register char c;
  register int cnt1;
  cnt1 = cnt;
  c = 0;
  while (cnt1-- > 0) {
    while (!isprint((c & 0x7F)))
      c++;
    *cp++ = (c++ & 0x7F);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<pattern>:
  endbr64
  test   %esi,%esi
  jle    L26
  push   %r13
  movslq %esi,%rsi
  push   %r12
  lea    (%rdi,%rsi,1),%r13
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L14:
  mov    %ebx,%edi
  mov    %ebx,%ebp
  add    $0x1,%ebx
  and    $0x7f,%edi
  and    $0x7f,%ebp
  call   <isprint@plt>
  test   %eax,%eax
  je     L14
  add    $0x1,%r12
  mov    %bpl,-0x1(%r12)
  cmp    %r12,%r13
  jne    L14
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  ret
```

assistant: ```c
void pattern(char *cp, int cnt) {
  register char c;
  register int cnt1;
  cnt1 = cnt;
  c = 0;
  while (cnt1-- > 0) {
    while (!isprint((c & 0x7F)))
      c++;
    *cp++ = (c++ & 0x7F);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<posix_spawnattr_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int posix_spawnattr_init(posix_spawnattr_t *attr) {
  memset(attr, '\0', sizeof(*attr));
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<posix_spawnattr_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int posix_spawnattr_init(posix_spawnattr_t *attr) {
  memset(attr, '\0', sizeof(*attr));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<posix_spawnattr_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edx
  xor    %esi,%esi
  call   <memset@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int posix_spawnattr_init(posix_spawnattr_t *attr) {
  memset(attr, '\0', sizeof(*attr));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<posix_spawnattr_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edx
  xor    %esi,%esi
  call   <memset@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int posix_spawnattr_init(posix_spawnattr_t *attr) {
  memset(attr, '\0', sizeof(*attr));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %eax,%edi
  call   <validate@plt>
  test   %rax,%rax
  je     L20
  mov    $0x0,%eax
  call   <init_solution@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0xffffffff,-0x10(%rbp)
  jmp    L15
L64:
  movl   $0xffffffff,-0xc(%rbp)
  jmp    L13
L87:
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
L13:
  addl   $0x1,-0xc(%rbp)
  cmpl   $0x8,-0xc(%rbp)
  jle    L87
L15:
  addl   $0x1,-0x10(%rbp)
  cmpl   $0x8,-0x10(%rbp)
  jle    L64
  mov    -0x8(%rbp),%rax
  mov    $0x30,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <solve@plt>
  movl   $0xffffffff,-0x10(%rbp)
  jmp    L14
L11:
  movl   $0xffffffff,-0xc(%rbp)
  jmp    L57
L75:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    $0x1,%edx
  lea    D5(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L57:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L75
  mov    $0x1,%edx
  lea    D14(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L14:
  addl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L11
  jmp    L23
L20:
  mov    $0x6,%edx
  lea    D8(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L23:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
type: string
name: read_data
arguments: data_label: D14
type: string
name: read_data
arguments: data_label: D8
type: string

tool: D5:  

tool: D14: 


tool: D8: Error


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int j;
  char **solution;
  ;
  if (validate(argc, argv, 0, 0)) {
    solution = init_solution();
    i = -1;
    while (++i < 9) {
      j = -1;
      while (++j < 9)
        solution[i][j] = argv[i + 1][j];
    }
    solve(solution, 0, 0, '0');
    i = -1;
    while (solution[++i]) {
      j = -1;
      while (solution[i][++j]) {
        write(1, &solution[i][j], 1);
        write(1, " ", 1);
      };
      write(1, "\n", 1);
    }
  } else
    write(1, "Error\n", 6);
  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
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  mov    $0x0,%ecx
  mov    $0x0,%edx
  call   <validate@plt>
  test   %rax,%rax
  je     L85
  mov    $0x0,%eax
  call   <init_solution@plt>
  mov    %rax,%rbp
  lea    0x8(%rbx),%rdi
  mov    %rax,%rsi
  lea    0x48(%rax),%r8
L14:
  mov    $0x0,%eax
L34:
  mov    (%rdi),%rdx
  movzbl (%rdx,%rax,1),%ecx
  mov    (%rsi),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L34
  add    $0x8,%rdi
  add    $0x8,%rsi
  cmp    %rsi,%r8
  jne    L14
  mov    $0x30,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <solve@plt>
  lea    D23(%rip),%r12
  lea    D24(%rip),%r13
  cmpq   $0x0,0x0(%rbp)
  jne    L16
  jmp    L37
L82:
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    %rbx,%rsi
  add    0x0(%rbp),%rsi
  add    $0x1,%rbx
  cmpb   $0x0,(%rsi)
  jne    L82
L44:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  add    $0x8,%rbp
  cmpq   $0x0,0x0(%rbp)
  je     L37
L16:
  mov    0x0(%rbp),%rsi
  mov    $0x1,%ebx
  cmpb   $0x0,(%rsi)
  jne    L82
  jmp    L44
L85:
  mov    $0x6,%edx
  lea    D18(%rip),%rsi
  mov    $0x1,%edi
  call   <write@plt>
L37:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D23
type: string
name: extract_symbolic_data
arguments: label: D24
type: string
name: extract_symbolic_data
arguments: label: D18
type: string

tool: D23:  

tool: D24: 


tool: D18: Error


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int j;
  char **solution;
  ;
  if (validate(argc, argv, 0, 0)) {
    solution = init_solution();
    i = -1;
    while (++i < 9) {
      j = -1;
      while (++j < 9)
        solution[i][j] = argv[i + 1][j];
    }
    solve(solution, 0, 0, '0');
    i = -1;
    while (solution[++i]) {
      j = -1;
      while (solution[i][++j]) {
        write(1, &solution[i][j], 1);
        write(1, " ", 1);
      };
      write(1, "\n", 1);
    }
  } else
    write(1, "Error\n", 6);
  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   %r13
  xor    %ecx,%ecx
  xor    %edx,%edx
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <validate@plt>
  test   %rax,%rax
  je     L9
  xor    %eax,%eax
  call   <init_solution@plt>
  lea    0x8(%rbx),%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  lea    0x48(%rax),%r8
  nopl   0x0(%rax,%rax,1)
L70:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L63:
  mov    (%rdi),%rdx
  movzbl (%rdx,%rax,1),%ecx
  mov    (%rsi),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L63
  add    $0x8,%rsi
  add    $0x8,%rdi
  cmp    %rsi,%r8
  jne    L70
  xor    %esi,%esi
  mov    $0x30,%ecx
  xor    %edx,%edx
  mov    %rbp,%rdi
  call   <solve@plt>
  mov    0x0(%rbp),%rsi
  lea    D12(%rip),%r13
  lea    D25(%rip),%r12
  test   %rsi,%rsi
  je     L12
  nopl   0x0(%rax)
L69:
  cmpb   $0x0,(%rsi)
  mov    $0x1,%ebx
  je     L67
  nopw   0x0(%rax,%rax,1)
L71:
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
  mov    %r12,%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
  mov    0x0(%rbp),%rsi
  add    %rbx,%rsi
  add    $0x1,%rbx
  cmpb   $0x0,(%rsi)
  jne    L71
L67:
  mov    %r13,%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  add    $0x8,%rbp
  call   <write@plt>
  mov    0x0(%rbp),%rsi
  test   %rsi,%rsi
  jne    L69
L12:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  mov    $0x6,%edx
  lea    D9(%rip),%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L12
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
type_guessed: string
name: extract_symbolic_data
arguments: label: D25
type_guessed: string
name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D12: 


tool: D25:  

tool: D9: Error


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int j;
  char **solution;
  ;
  if (validate(argc, argv, 0, 0)) {
    solution = init_solution();
    i = -1;
    while (++i < 9) {
      j = -1;
      while (++j < 9)
        solution[i][j] = argv[i + 1][j];
    }
    solve(solution, 0, 0, '0');
    i = -1;
    while (solution[++i]) {
      j = -1;
      while (solution[i][++j]) {
        write(1, &solution[i][j], 1);
        write(1, " ", 1);
      };
      write(1, "\n", 1);
    }
  } else
    write(1, "Error\n", 6);
  return (0);
}
```