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
<userFunctions>:
  endbr64
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  mov    $0x5,%edx
  push   %rbp
  mov    %esi,%ebp
  lea    D14(%rip),%rsi
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <strncmp@plt>
  test   %rax,%rax
  je     L88
L28:
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <writeSock@plt>
  mov    $0x400,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <readSock@plt>
  movsbl (%rsp),%edx
  cmp    $0x22,%dl
  je     L3
  cmp    $0x23,%dl
  je     L49
  cmp    $0x24,%dl
  jne    L67
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <writeArticle@plt>
  jmp    L28
  nop
L3:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <listArticles@plt>
  jmp    L28
  nop
L49:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <readArticle@plt>
  jmp    L28
  nop
L67:
  lea    D3(%rip),%rsi
  mov    %r12,%rdi
  call   <logData@plt>
L17:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L88:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <adminFunctions@plt>
  xor    %eax,%eax
  jmp    L17
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: string
name: parse_data
arguments: data_label: D14
type_guessed: string
name: parse_data
arguments: data_label: D3
type_guessed: string

tool: D1: AA READY!

tool: D14: admin

tool: D3: unknown action %x

assistant: ```c
int userFunctions(FILE *logfile, int sock, char *user) {
  char action[1024];
  size_t len;
  if (0 == strncmp(user, "admin", 5)) {
    adminFunctions(logfile, sock);
    return 0;
  }
  while (1) {
    writeSock(sock, "\x41\x41 READY!", sizeof("\x41\x41 READY!"));
    memset(action, 0, sizeof(action));
    len = readSock(sock, action, sizeof(action));
    if (action[0] == 0x22) {
      listArticles(sock, logfile, action);
    } else if (action[0] == 0x23) {
      readArticle(sock, logfile, action);
    } else if (action[0] == 0x24) {
      writeArticle(sock, logfile, action);
    } else {
      logData(logfile, "unknown action %x", action[0]);
      return;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<common>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_infered: string

tool: D2: hello from common.c


assistant: ```c
void common(void) {
  printf("hello from common.c\n");
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<common>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
type_infered: string

tool: D9: hello from common.c


assistant: ```c
void common(void) {
  printf("hello from common.c\n");
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<common>:
  endbr64
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D5
type_guessed: string

tool: D5: hello from common.c


assistant: ```c
void common(void) {
  printf("hello from common.c\n");
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<common>:
  endbr64
  lea    D9(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D9
datatype: string

tool: D9: hello from common.c


assistant: ```c
void common(void) {
  printf("hello from common.c\n");
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<PR15623>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L9
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <clang_analyzer_eval@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void PR15623(int n) {
  if ((n == 0) != 0) {
    clang_analyzer_eval(n == 0);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<PR15623>:
  endbr64
  test   %edi,%edi
  je     L9
  ret
L9:
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <clang_analyzer_eval@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void PR15623(int n) {
  if ((n == 0) != 0) {
    clang_analyzer_eval(n == 0);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<PR15623>:
  endbr64
  test   %edi,%edi
  je     L5
  ret
  nopl   0x0(%rax)
L5:
  mov    $0x1,%edi
  jmp    1050 <clang_analyzer_eval@plt>
```

assistant: ```c
void PR15623(int n) {
  if ((n == 0) != 0) {
    clang_analyzer_eval(n == 0);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<PR15623>:
  endbr64
  test   %edi,%edi
  je     L5
  ret
  nopl   0x0(%rax)
L5:
  mov    $0x1,%edi
  jmp    1050 <clang_analyzer_eval@plt>
```

assistant: ```c
void PR15623(int n) {
  if ((n == 0) != 0) {
    clang_analyzer_eval(n == 0);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bitops_reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L19
L10:
  mov    -0x18(%rbp),%eax
  sub    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L19:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L10
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int bitops_reverse(int i, int length) {
  uint8_t index;
  int r = 0;
  int b;
  for (index = 0; index < length; index++) {
    b = (i >> (length - index - 1)) & 0x01;
    r |= b << index;
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitops_reverse>:
  endbr64
  test   %esi,%esi
  jle    L6
  mov    $0x0,%r8d
  mov    $0x0,%edx
  lea    -0x1(%rsi),%r9d
L1:
  mov    %r9d,%ecx
  sub    %edx,%ecx
  mov    %edi,%eax
  sar    %cl,%eax
  and    $0x1,%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  or     %eax,%r8d
  add    $0x1,%edx
  cmp    %edx,%esi
  jne    L1
L14:
  mov    %r8d,%eax
  ret
L6:
  mov    $0x0,%r8d
  jmp    L14
```

assistant: ```c
int bitops_reverse(int i, int length) {
  uint8_t index;
  int r = 0;
  int b;
  for (index = 0; index < length; index++) {
    b = (i >> (length - index - 1)) & 0x01;
    r |= b << index;
  }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bitops_reverse>:
  endbr64
  xor    %r8d,%r8d
  test   %esi,%esi
  jle    L1
  xor    %edx,%edx
  lea    -0x1(%rsi),%r9d
  nopl   0x0(%rax)
L6:
  mov    %r9d,%ecx
  mov    %edi,%eax
  sub    %edx,%ecx
  sar    %cl,%eax
  mov    %edx,%ecx
  add    $0x1,%edx
  and    $0x1,%eax
  shl    %cl,%eax
  or     %eax,%r8d
  cmp    %edx,%esi
  jne    L6
L1:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitops_reverse(int i, int length) {
  uint8_t index;
  int r = 0;
  int b;
  for (index = 0; index < length; index++) {
    b = (i >> (length - index - 1)) & 0x01;
    r |= b << index;
  }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bitops_reverse>:
  endbr64
  xor    %r8d,%r8d
  test   %esi,%esi
  jle    L11
  xor    %edx,%edx
  lea    -0x1(%rsi),%r9d
  nopl   0x0(%rax)
L16:
  mov    %r9d,%ecx
  mov    %edi,%eax
  sub    %edx,%ecx
  sar    %cl,%eax
  mov    %edx,%ecx
  add    $0x1,%edx
  and    $0x1,%eax
  shl    %cl,%eax
  or     %eax,%r8d
  cmp    %edx,%esi
  jne    L16
L11:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitops_reverse(int i, int length) {
  uint8_t index;
  int r = 0;
  int b;
  for (index = 0; index < length; index++) {
    b = (i >> (length - index - 1)) & 0x01;
    r |= b << index;
  }
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L14:
  addl   $0x1,-0x4(%rbp)
L15:
  cmpl   $0x9,-0x4(%rbp)
  jle    L14
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D4: %i


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  int local0;
  local0 = 0;
  while (local0 <= 9) {
    local0++;
  }
  printf("%i\n", local0);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_infered: string

tool: D0: %i


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  int local0;
  local0 = 0;
  while (local0 <= 9) {
    local0++;
  }
  printf("%i\n", local0);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type_guessed: string

tool: D4: %i


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  int local0;
  local0 = 0;
  while (local0 <= 9) {
    local0++;
  }
  printf("%i\n", local0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type_guessed: string

tool: D3: %i


assistant: ```c
int main(int argc, char *argv[], char *envp[]) {
  int local0;
  local0 = 0;
  while (local0 <= 9) {
    local0++;
  }
  printf("%i\n", local0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_pow1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  sar    %eax
  mov    %eax,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x1,-0x28(%rbp)
  jne    L15
  mov    -0x24(%rbp),%eax
  jmp    L5
L15:
  mov    -0x18(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_pow1>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_pow1>
  imul   %ebx,%eax
L5:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int _pow1(int N, int M) {
  int P = M >> 1, Q = M - P;
  if (1 == M)
    return N;
  return _pow1(N, P) * _pow1(N, Q);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_pow1>:
  endbr64
  mov    %edi,%eax
  cmp    $0x1,%esi
  je     L6
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %esi,%r13d
  sar    %r13d
  mov    %r13d,%esi
  call   <_pow1>
  mov    %eax,%r12d
  sub    %r13d,%ebx
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <_pow1>
  imul   %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  ret
```

assistant: ```c
int _pow1(int N, int M) {
  int P = M >> 1, Q = M - P;
  if (1 == M)
    return N;
  return _pow1(N, P) * _pow1(N, Q);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverseLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L29:
  addl   $0x1,-0x8(%rbp)
L13:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L29
  subl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L34:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
  subl   $0x1,-0x8(%rbp)
L35:
  cmpl   $0x0,-0x8(%rbp)
  jns    L34
  nop
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type: string

tool: D9: 
%d, %d, %c


assistant: ```c
void reverseLine(char to[], char from[]) {
  int i, j;
  for (i = 0; from[i] != '\0'; i++)
    ;
  i--;
  for (j = 0; i >= 0; ++j, --i) {
    to[j] = from[i];
    printf("\n%d, %d, %c\n", i, j, to[j]);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<reverseLine>:
  endbr64
  cmpb   $0x0,(%rsi)
  je     L24
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    $0x0,%eax
L0:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0x0,(%r12,%rax,1)
  jne    L0
  test   %ebx,%ebx
  js     L8
  movslq %ebx,%rbx
  mov    $0x0,%ebp
  lea    D3(%rip),%r14
L35:
  movzbl (%r12,%rbx,1),%ecx
  mov    %cl,0x0(%r13,%rbp,1)
  movsbl %cl,%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%rbp
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jns    L35
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
type: string

tool: D3: 
%d, %d, %c


assistant: ```c
void reverseLine(char to[], char from[]) {
  int i, j;
  for (i = 0; from[i] != '\0'; i++)
    ;
  i--;
  for (j = 0; i >= 0; ++j, --i) {
    to[j] = from[i];
    printf("\n%d, %d, %c\n", i, j, to[j]);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<reverseLine>:
  endbr64
  cmpb   $0x0,(%rsi)
  je     L16
  push   %r14
  xor    %eax,%eax
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  nopl   0x0(%rax)
L20:
  movslq %eax,%rbx
  add    $0x1,%rax
  cmpb   $0x0,(%r14,%rax,1)
  jne    L20
  xor    %r12d,%r12d
  lea    D2(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L1:
  movsbl (%r14,%rbx,1),%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  sub    $0x1,%rbx
  mov    %cl,0x0(%rbp,%r12,1)
  add    $0x1,%r12
  call   <printf@plt>
  cmp    $0xffffffff,%ebx
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L16:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_infered: string

tool: D2: 
%d, %d, %c


assistant: ```c
void reverseLine(char to[], char from[]) {
  int i, j;
  for (i = 0; from[i] != '\0'; i++)
    ;
  i--;
  for (j = 0; i >= 0; ++j, --i) {
    to[j] = from[i];
    printf("\n%d, %d, %c\n", i, j, to[j]);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<reverseLine>:
  endbr64
  cmpb   $0x0,(%rsi)
  je     L4
  push   %r14
  xor    %eax,%eax
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  nopl   0x0(%rax)
L12:
  movslq %eax,%rbx
  add    $0x1,%rax
  cmpb   $0x0,(%r14,%rax,1)
  jne    L12
  xor    %r12d,%r12d
  lea    D0(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L10:
  movsbl (%r14,%rbx,1),%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  sub    $0x1,%rbx
  mov    %cl,0x0(%rbp,%r12,1)
  add    $0x1,%r12
  call   <printf@plt>
  cmp    $0xffffffff,%ebx
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L4:
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
datatype: string

tool: D0: 
%d, %d, %c


assistant: ```c
void reverseLine(char to[], char from[]) {
  int i, j;
  for (i = 0; from[i] != '\0'; i++)
    ;
  i--;
  for (j = 0; i >= 0; ++j, --i) {
    to[j] = from[i];
    printf("\n%d, %d, %c\n", i, j, to[j]);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<snprint_x8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x8,-0x18(%rbp)
  jmp    L47
L20:
  subl   $0x1,-0x18(%rbp)
  mov    -0x34(%rbp),%eax
  and    $0xf,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x9,-0x14(%rbp)
  jg     L21
  mov    -0x14(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %dl,-0x10(%rbp,%rax,1)
  jmp    L66
L21:
  mov    -0x14(%rbp),%eax
  add    $0x37,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %dl,-0x10(%rbp,%rax,1)
L66:
  sarl   $0x4,-0x34(%rbp)
L47:
  cmpl   $0x0,-0x18(%rbp)
  jne    L20
  jmp    L5
L1:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0x10(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  cmp    %rax,-0x30(%rbp)
  jne    L5
  mov    $0x8,%eax
  jmp    L54
L5:
  mov    -0x18(%rbp),%eax
  cmp    $0x7,%eax
  jbe    L1
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x8,%eax
L54:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
uint8_t snprint_x8(char *buffer, size_t buf_size, uint8_t num) {
  char str[sizeof(num) * 2];
  uint8_t i = sizeof(num) * 2;
  uint8_t tmp;
  while (i) {
    i--;
    tmp = num & 0x0F;
    if (tmp < 10) {
      str[i] = '0' + tmp;
    } else {
      str[i] = 'A' - 10 + tmp;
    }
    num >>= 4;
  }
  while (i <= sizeof(num) * 2 - 1) {
    buffer[i] = str[i];
    i++;
    if (i == buf_size)
      return (sizeof(num) * 2);
  }
  buffer[i] = 0;
  return (sizeof(num) * 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<snprint_x8>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%r8
  mov    %rsi,%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r10
  lea    0x7(%rsp),%rax
  jmp    L28
L25:
  mov    %rcx,%rax
L28:
  mov    %edx,%esi
  and    $0xf,%esi
  lea    0x30(%rsi),%edi
  lea    0x37(%rsi),%ecx
  cmp    $0x9,%esi
  cmovle %edi,%ecx
  mov    %cl,(%rax)
  sar    $0x4,%edx
  lea    -0x1(%rax),%rcx
  cmp    %r10,%rax
  jne    L25
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rcx
L6:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,-0x1(%r8,%rax,1)
  cmp    %r9,%rax
  je     L19
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L6
  movb   $0x0,0x8(%r8)
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    $0x8,%eax
  add    $0x18,%rsp
  ret
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint8_t snprint_x8(char *buffer, size_t buf_size, uint8_t num) {
  char str[sizeof(num) * 2];
  uint8_t i = sizeof(num) * 2;
  uint8_t tmp;
  while (i) {
    i--;
    tmp = num & 0x0F;
    if (tmp < 10) {
      str[i] = '0' + tmp;
    } else {
      str[i] = 'A' - 10 + tmp;
    }
    num >>= 4;
  }
  while (i <= sizeof(num) * 2 - 1) {
    buffer[i] = str[i];
    i++;
    if (i == buf_size)
      return (sizeof(num) * 2);
  }
  buffer[i] = 0;
  return (sizeof(num) * 2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<snprint_x8>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r9
  lea    0x7(%rsp),%rax
  jmp    L32
  nopl   (%rax)
L26:
  mov    %rcx,%rax
L32:
  mov    %edx,%edi
  and    $0xf,%edi
  cmp    $0x9,%edi
  lea    0x30(%rdi),%r8d
  lea    0x37(%rdi),%ecx
  cmovle %r8d,%ecx
  sar    $0x4,%edx
  mov    %cl,(%rax)
  lea    -0x1(%rax),%rcx
  cmp    %rax,%r9
  jne    L26
  sub    $0x1,%rsi
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L9:
  movzbl (%r9,%rax,1),%edx
  mov    %dl,(%r10,%rax,1)
  cmp    %rsi,%rax
  je     L22
  add    $0x1,%rax
  cmp    $0x8,%rax
  jne    L9
  movb   $0x0,0x8(%r10)
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x8,%eax
  add    $0x18,%rsp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint8_t snprint_x8(char *buffer, size_t buf_size, uint8_t num) {
  char str[sizeof(num) * 2];
  uint8_t i = sizeof(num) * 2;
  uint8_t tmp;
  while (i) {
    i--;
    tmp = num & 0x0F;
    if (tmp < 10) {
      str[i] = '0' + tmp;
    } else {
      str[i] = 'A' - 10 + tmp;
    }
    num >>= 4;
  }
  while (i <= sizeof(num) * 2 - 1) {
    buffer[i] = str[i];
    i++;
    if (i == buf_size)
      return (sizeof(num) * 2);
  }
  buffer[i] = 0;
  return (sizeof(num) * 2);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<snprint_x8>:
  endbr64
  mov    %edx,%eax
  and    $0xf,%edx
  mov    %rdi,%rcx
  push   %r12
  cmp    $0xa,%edx
  lea    0x37(%rdx),%edi
  lea    0x30(%rdx),%r8d
  mov    %eax,%edx
  cmovge %edi,%r8d
  sar    $0x4,%edx
  push   %rbp
  and    $0xf,%edx
  push   %rbx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%r9d
  lea    0x30(%rdx),%edi
  mov    %eax,%edx
  cmovge %r9d,%edi
  sar    $0x8,%edx
  and    $0xf,%edx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%r10d
  lea    0x30(%rdx),%r9d
  mov    %eax,%edx
  cmovge %r10d,%r9d
  sar    $0xc,%edx
  and    $0xf,%edx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%r11d
  lea    0x30(%rdx),%r10d
  mov    %eax,%edx
  cmovge %r11d,%r10d
  sar    $0x10,%edx
  and    $0xf,%edx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%ebx
  lea    0x30(%rdx),%r11d
  mov    %eax,%edx
  cmovge %ebx,%r11d
  sar    $0x14,%edx
  and    $0xf,%edx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%ebx
  lea    0x30(%rdx),%ebp
  mov    %eax,%edx
  cmovge %ebx,%ebp
  sar    $0x18,%edx
  and    $0xf,%edx
  cmp    $0xa,%edx
  lea    0x37(%rdx),%r12d
  lea    0x30(%rdx),%ebx
  cmovge %r12d,%ebx
  shr    $0x1c,%eax
  lea    0x30(%rax),%r12d
  lea    0x37(%rax),%edx
  cmp    $0x9,%eax
  mov    %r12d,%eax
  cmovg  %edx,%eax
  mov    %al,(%rcx)
  cmp    $0x1,%rsi
  je     L40
  mov    %bl,0x1(%rcx)
  cmp    $0x2,%rsi
  je     L40
  mov    %bpl,0x2(%rcx)
  cmp    $0x3,%rsi
  je     L40
  mov    %r11b,0x3(%rcx)
  cmp    $0x4,%rsi
  je     L40
  mov    %r10b,0x4(%rcx)
  cmp    $0x5,%rsi
  je     L40
  mov    %r9b,0x5(%rcx)
  cmp    $0x6,%rsi
  je     L40
  mov    %dil,0x6(%rcx)
  cmp    $0x7,%rsi
  je     L40
  mov    %r8b,0x7(%rcx)
  cmp    $0x8,%rsi
  je     L40
  movb   $0x0,0x8(%rcx)
L40:
  pop    %rbx
  mov    $0x8,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
uint8_t snprint_x8(char *buffer, size_t buf_size, uint8_t num) {
  char str[sizeof(num) * 2];
  uint8_t i = sizeof(num) * 2;
  uint8_t tmp;
  while (i) {
    i--;
    tmp = num & 0x0F;
    if (tmp < 10) {
      str[i] = '0' + tmp;
    } else {
      str[i] = 'A' - 10 + tmp;
    }
    num >>= 4;
  }
  while (i <= sizeof(num) * 2 - 1) {
    buffer[i] = str[i];
    i++;
    if (i == buf_size)
      return (sizeof(num) * 2);
  }
  buffer[i] = 0;
  return (sizeof(num) * 2);
}
```