ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<echoLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <move@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  mov    $0x0,%eax
  call   <clrtoeol@plt>
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printw@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  nop
  leave
  ret
```

assistant: ```c
void echoLine(fmt, arg1, arg2, arg3) char *fmt;
int arg1;
int arg2;
int arg3;
{
  move(0, 0);
  refresh();
  clrtoeol();
  printw(fmt, arg1, arg2, arg3);
  refresh();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<echoLine>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %edx,%r12d
  mov    %ecx,%r13d
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <move@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  mov    $0x0,%eax
  call   <clrtoeol@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printw@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void echoLine(fmt, arg1, arg2, arg3) char *fmt;
int arg1;
int arg2;
int arg3;
{
  move(0, 0);
  refresh();
  clrtoeol();
  printw(fmt, arg1, arg2, arg3);
  refresh();
}
```
user: Translate the below assembly code into C programming language:

```asm
<echoLine>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %ecx,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  sub    $0x8,%rsp
  call   <move@plt>
  xor    %eax,%eax
  call   <refresh@plt>
  xor    %eax,%eax
  call   <clrtoeol@plt>
  mov    %r14d,%ecx
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printw@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10b0 <refresh@plt>
```

assistant: ```c
void echoLine(fmt, arg1, arg2, arg3) char *fmt;
int arg1;
int arg2;
int arg3;
{
  move(0, 0);
  refresh();
  clrtoeol();
  printw(fmt, arg1, arg2, arg3);
  refresh();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<echoLine>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %ecx,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  sub    $0x8,%rsp
  call   <move@plt>
  xor    %eax,%eax
  call   <refresh@plt>
  xor    %eax,%eax
  call   <clrtoeol@plt>
  mov    %r14d,%ecx
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printw@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10b0 <refresh@plt>
```

assistant: ```c
void echoLine(fmt, arg1, arg2, arg3) char *fmt;
int arg1;
int arg2;
int arg3;
{
  move(0, 0);
  refresh();
  clrtoeol();
  printw(fmt, arg1, arg2, arg3);
  refresh();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dump_matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L6
L29:
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L10:
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L10
  addl   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L29
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dump_matrix(int **vals, int rows, int cols) {
  int i, j;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      ;
    };
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dump_matrix>:
  endbr64
  mov    $0x0,%ecx
  test   %esi,%esi
  jg     L33
  ret
L23:
  add    $0x1,%eax
  cmp    %eax,%edx
  jne    L23
L17:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  je     L40
L33:
  mov    $0x0,%eax
  test   %edx,%edx
  jg     L23
  jmp    L17
L40:
  ret
```

assistant: ```c
void dump_matrix(int **vals, int rows, int cols) {
  int i, j;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      ;
    };
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dump_matrix>:
  endbr64
  ret
```

assistant: ```c
void dump_matrix(int **vals, int rows, int cols) {
  int i, j;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      ;
    };
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dump_matrix>:
  endbr64
  ret
```

assistant: ```c
void dump_matrix(int **vals, int rows, int cols) {
  int i, j;
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      ;
    };
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getIntNum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpb   $0x30,-0x14(%rbp)
  je     L121
  movzbl -0x14(%rbp),%eax
  mov    %al,-0x9(%rbp)
L103:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movsbl -0x9(%rbp),%eax
  sub    $0x30,%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x9(%rbp)
  movsbl -0x9(%rbp),%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L103
  jmp    L84
L121:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x9(%rbp)
  cmpb   $0x2f,-0x9(%rbp)
  jle    L38
  cmpb   $0x37,-0x9(%rbp)
  jg     L38
L21:
  mov    -0x8(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  movsbl -0x9(%rbp),%eax
  sub    $0x30,%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x9(%rbp)
  cmpb   $0x2f,-0x9(%rbp)
  jg     L21
  cmpb   $0x37,-0x9(%rbp)
  jle    L21
  jmp    L84
L38:
  cmpb   $0x58,-0x9(%rbp)
  je     L61
  cmpb   $0x78,-0x9(%rbp)
  jne    L11
  jmp    L61
L63:
  mov    -0x8(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
L61:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x9(%rbp)
  movsbl -0x9(%rbp),%eax
  mov    %eax,%edi
  call   <hexValue@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L63
  jmp    L84
L11:
  movl   $0x0,-0x8(%rbp)
L84:
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  movsbl -0x9(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <ungetc@plt>
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getIntNum(char firstCharacter) {
  int num = 0;
  int value;
  char ch;
  if (firstCharacter != '0') {
    ch = firstCharacter;
    do {
      num = 10 * num + (int)(ch - '0');
      ch = getchar();
    } while (isdigit(ch));
  } else {
    ch = getchar();
    if ((ch >= '0') && (ch <= '7'))
      do {
        num = 8 * num + (int)(ch - '0');
        ch = getchar();
      } while ((ch >= '0') || (ch <= '7'));
    else if ((ch == 'X') || (ch == 'x')) {
      while ((value = hexValue(ch = getchar())) != -1)
        num = 16 * num + value;
    } else
      num = 0;
  }
  ungetc(ch, stdin);
  return num;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getIntNum>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x30,%dil
  je     L56
  mov    %edi,%ebx
  mov    $0x0,%ebp
L34:
  lea    0x0(%rbp,%rbp,4),%eax
  movsbl %bl,%ebx
  lea    -0x30(%rbx,%rax,2),%ebp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  movsbl %al,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L34
L3:
  movsbl %bl,%edi
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  call   <ungetc@plt>
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L56:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  lea    -0x30(%rbx),%eax
  cmp    $0x7,%al
  ja     L22
L1:
  mov    $0x0,%eax
  call   <getchar@plt>
  jmp    L1
L22:
  mov    %ebx,%eax
  and    $0xffffffdf,%eax
  mov    $0x0,%ebp
  cmp    $0x58,%al
  jne    L3
  mov    $0x0,%ebp
  jmp    L45
L28:
  shl    $0x4,%ebp
  add    %eax,%ebp
L45:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  movsbl %al,%edi
  call   <hexValue@plt>
  cmp    $0xffffffff,%eax
  jne    L28
  jmp    L3
```

assistant: ```c
int getIntNum(char firstCharacter) {
  int num = 0;
  int value;
  char ch;
  if (firstCharacter != '0') {
    ch = firstCharacter;
    do {
      num = 10 * num + (int)(ch - '0');
      ch = getchar();
    } while (isdigit(ch));
  } else {
    ch = getchar();
    if ((ch >= '0') && (ch <= '7'))
      do {
        num = 8 * num + (int)(ch - '0');
        ch = getchar();
      } while ((ch >= '0') || (ch <= '7'));
    else if ((ch == 'X') || (ch == 'x')) {
      while ((value = hexValue(ch = getchar())) != -1)
        num = 16 * num + value;
    } else
      num = 0;
  }
  ungetc(ch, stdin);
  return num;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getIntNum>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x30,%dil
  je     L61
  movsbl %dil,%ebx
  xor    %r12d,%r12d
  nopl   (%rax)
L27:
  lea    (%r12,%r12,4),%eax
  lea    -0x30(%rbx,%rax,2),%r12d
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%ebx
  mov    %ebx,%edi
  mov    %ebx,%ebp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L27
L57:
  mov    D4(%rip),%rax
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <ungetc@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L61:
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x30(%rax),%edx
  cmp    $0x7,%dl
  jbe    L53
  mov    %eax,%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  je     L1
  movsbl %al,%ebp
  xor    %r12d,%r12d
  jmp    L57
L53:
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  jmp    L53
  nopl   0x0(%rax)
L1:
  xor    %r12d,%r12d
  jmp    L40
  nopl   (%rax)
L43:
  shl    $0x4,%r12d
  add    %eax,%r12d
L40:
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%ebp
  mov    %ebp,%edi
  call   <hexValue@plt>
  cmp    $0xffffffff,%eax
  jne    L43
  jmp    L57
```

assistant: ```c
int getIntNum(char firstCharacter) {
  int num = 0;
  int value;
  char ch;
  if (firstCharacter != '0') {
    ch = firstCharacter;
    do {
      num = 10 * num + (int)(ch - '0');
      ch = getchar();
    } while (isdigit(ch));
  } else {
    ch = getchar();
    if ((ch >= '0') && (ch <= '7'))
      do {
        num = 8 * num + (int)(ch - '0');
        ch = getchar();
      } while ((ch >= '0') || (ch <= '7'));
    else if ((ch == 'X') || (ch == 'x')) {
      while ((value = hexValue(ch = getchar())) != -1)
        num = 16 * num + value;
    } else
      num = 0;
  }
  ungetc(ch, stdin);
  return num;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getIntNum>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x30,%dil
  je     L59
  movsbl %dil,%ebx
  xor    %r12d,%r12d
  nopl   (%rax)
L68:
  lea    (%r12,%r12,4),%eax
  lea    -0x30(%rbx,%rax,2),%r12d
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%ebx
  mov    %ebx,%edi
  mov    %ebx,%ebp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L68
L3:
  mov    D7(%rip),%rax
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <ungetc@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L59:
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x30(%rax),%edx
  cmp    $0x7,%dl
  jbe    L72
  mov    %eax,%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  je     L7
  movsbl %al,%ebp
  xor    %r12d,%r12d
  jmp    L3
L72:
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  jmp    L72
  nopl   0x0(%rax)
L7:
  xor    %r12d,%r12d
  jmp    L4
  nopl   (%rax)
L15:
  shl    $0x4,%r12d
  add    %eax,%r12d
L4:
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%ebp
  mov    %ebp,%edi
  call   <hexValue@plt>
  cmp    $0xffffffff,%eax
  jne    L15
  jmp    L3
```

assistant: ```c
int getIntNum(char firstCharacter) {
  int num = 0;
  int value;
  char ch;
  if (firstCharacter != '0') {
    ch = firstCharacter;
    do {
      num = 10 * num + (int)(ch - '0');
      ch = getchar();
    } while (isdigit(ch));
  } else {
    ch = getchar();
    if ((ch >= '0') && (ch <= '7'))
      do {
        num = 8 * num + (int)(ch - '0');
        ch = getchar();
      } while ((ch >= '0') || (ch <= '7'));
    else if ((ch == 'X') || (ch == 'x')) {
      while ((value = hexValue(ch = getchar())) != -1)
        num = 16 * num + value;
    } else
      num = 0;
  }
  ungetc(ch, stdin);
  return num;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<autohelperowl_attackpat287>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    $0xffffffff,%edi
  call   <offset@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x2,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%edi
  mov    -0xc(%rbp),%esi
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%eax
  sub    $0x8,%rsp
  mov    -0x4(%rbp),%ecx
  push   %rcx
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    %edx,%ecx
  mov    $0x3,%edx
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <play_attack_defend_n@plt>
  add    $0x10,%rsp
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int autohelperowl_attackpat287(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, C;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, 1, move, transformation);
  C = offset(0, 2, move, transformation);
  return !play_attack_defend_n(color, 0, 3, move, a, b, C);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<autohelperowl_attackpat287>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    %ecx,%r12d
  mov    %esi,%ecx
  mov    $0x0,%esi
  mov    $0xffffffff,%edi
  call   <offset@plt>
  mov    %eax,%r13d
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    %eax,%r14d
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    $0x2,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  sub    $0x8,%rsp
  push   %rax
  mov    %r14d,%r9d
  mov    %r13d,%r8d
  mov    %ebx,%ecx
  mov    $0x3,%edx
  mov    $0x0,%esi
  mov    %r12d,%edi
  call   <play_attack_defend_n@plt>
  add    $0x10,%rsp
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int autohelperowl_attackpat287(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, C;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, 1, move, transformation);
  C = offset(0, 2, move, transformation);
  return !play_attack_defend_n(color, 0, 3, move, a, b, C);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<autohelperowl_attackpat287>:
  endbr64
  push   %r14
  mov    $0xffffffff,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %ecx,%ebp
  mov    %esi,%ecx
  xor    %esi,%esi
  push   %rbx
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0x1,%esi
  mov    %eax,%ebx
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0x2,%esi
  mov    %eax,%r14d
  call   <offset@plt>
  sub    $0x8,%rsp
  mov    %r12d,%ecx
  mov    %r14d,%r9d
  push   %rax
  mov    %ebx,%r8d
  mov    $0x3,%edx
  xor    %esi,%esi
  mov    %ebp,%edi
  call   <play_attack_defend_n@plt>
  pop    %rdx
  pop    %rcx
  test   %eax,%eax
  pop    %rbx
  pop    %rbp
  sete   %al
  pop    %r12
  pop    %r13
  movzbl %al,%eax
  pop    %r14
  ret
```

assistant: ```c
int autohelperowl_attackpat287(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, C;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, 1, move, transformation);
  C = offset(0, 2, move, transformation);
  return !play_attack_defend_n(color, 0, 3, move, a, b, C);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<autohelperowl_attackpat287>:
  endbr64
  push   %r14
  mov    $0xffffffff,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %ecx,%ebp
  mov    %esi,%ecx
  xor    %esi,%esi
  push   %rbx
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0x1,%esi
  mov    %eax,%ebx
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0x2,%esi
  mov    %eax,%r14d
  call   <offset@plt>
  sub    $0x8,%rsp
  mov    %r12d,%ecx
  mov    %r14d,%r9d
  push   %rax
  mov    %ebx,%r8d
  mov    $0x3,%edx
  xor    %esi,%esi
  mov    %ebp,%edi
  call   <play_attack_defend_n@plt>
  pop    %rdx
  pop    %rcx
  test   %eax,%eax
  pop    %rbx
  pop    %rbp
  sete   %al
  pop    %r12
  pop    %r13
  movzbl %al,%eax
  pop    %r14
  ret
```

assistant: ```c
int autohelperowl_attackpat287(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, C;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, 1, move, transformation);
  C = offset(0, 2, move, transformation);
  return !play_attack_defend_n(color, 0, 3, move, a, b, C);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_fileext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    $0x0,%esi
  mov    $0x10,%edi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L15
  addq   $0x1,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L12
L10:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x14(%rbp)
L12:
  cmpl   $0xf,-0x14(%rbp)
  jle    L10
L15:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *get_fileext(char *file) {
  int i;
  char *ext = calloc(sizeof(char) * 16, 0);
  char *v = strrchr(file, '.');
  if (v) {
    v++;
    for (i = 0; i < 16; i++) {
      ext[i] = v[i];
    }
  }
  return ext;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_fileext>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%esi
  mov    $0x10,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L6
  mov    $0x1,%edx
L4:
  movzbl (%rax,%rdx,1),%ecx
  mov    %cl,-0x1(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    $0x11,%rdx
  jne    L4
L6:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *get_fileext(char *file) {
  int i;
  char *ext = calloc(sizeof(char) * 16, 0);
  char *v = strrchr(file, '.');
  if (v) {
    v++;
    for (i = 0; i < 16; i++) {
      ext[i] = v[i];
    }
  }
  return ext;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_fileext>:
  endbr64
  push   %r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x10,%edi
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strrchr@plt>
  test   %rax,%rax
  je     L6
  mov    $0x1,%edx
  nopl   0x0(%rax)
L16:
  movzbl (%rax,%rdx,1),%ecx
  mov    %cl,-0x1(%r12,%rdx,1)
  add    $0x1,%rdx
  cmp    $0x11,%rdx
  jne    L16
L6:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *get_fileext(char *file) {
  int i;
  char *ext = calloc(sizeof(char) * 16, 0);
  char *v = strrchr(file, '.');
  if (v) {
    v++;
    for (i = 0; i < 16; i++) {
      ext[i] = v[i];
    }
  }
  return ext;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_fileext>:
  endbr64
  push   %r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x10,%edi
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strrchr@plt>
  test   %rax,%rax
  je     L2
  lea    0x2(%rax),%rcx
  mov    %r12,%rdx
  sub    %rcx,%rdx
  cmp    $0xe,%rdx
  jbe    L0
  movdqu 0x1(%rax),%xmm0
  movups %xmm0,(%r12)
L2:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  movzbl 0x1(%rax),%edx
  mov    %dl,(%r12)
  movzbl 0x2(%rax),%edx
  mov    %dl,0x1(%r12)
  movzbl 0x3(%rax),%edx
  mov    %dl,0x2(%r12)
  movzbl 0x4(%rax),%edx
  mov    %dl,0x3(%r12)
  movzbl 0x5(%rax),%edx
  mov    %dl,0x4(%r12)
  movzbl 0x6(%rax),%edx
  mov    %dl,0x5(%r12)
  movzbl 0x7(%rax),%edx
  mov    %dl,0x6(%r12)
  movzbl 0x8(%rax),%edx
  mov    %dl,0x7(%r12)
  movzbl 0x9(%rax),%edx
  mov    %dl,0x8(%r12)
  movzbl 0xa(%rax),%edx
  mov    %dl,0x9(%r12)
  movzbl 0xb(%rax),%edx
  mov    %dl,0xa(%r12)
  movzbl 0xc(%rax),%edx
  mov    %dl,0xb(%r12)
  movzbl 0xd(%rax),%edx
  mov    %dl,0xc(%r12)
  movzbl 0xe(%rax),%edx
  mov    %dl,0xd(%r12)
  movzbl 0xf(%rax),%edx
  mov    %dl,0xe(%r12)
  movzbl 0x10(%rax),%eax
  mov    %al,0xf(%r12)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *get_fileext(char *file) {
  int i;
  char *ext = calloc(sizeof(char) * 16, 0);
  char *v = strrchr(file, '.');
  if (v) {
    v++;
    for (i = 0; i < 16; i++) {
      ext[i] = v[i];
    }
  }
  return ext;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<USBDevMode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x41c,%rax
  movb   $0x3,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void USBDevMode(unsigned long ulBase) {
  ;
  (*((volatile unsigned char *)(ulBase + 0x0000041C))) = 0x00000002 | 0x00000001;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<USBDevMode>:
  endbr64
  movb   $0x3,0x41c(%rdi)
  ret
```

assistant: ```c
void USBDevMode(unsigned long ulBase) {
  ;
  (*((volatile unsigned char *)(ulBase + 0x0000041C))) = 0x00000002 | 0x00000001;
}
```