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

```asm
<puts>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%rdx
  mov    $0x19,%esi
  mov    $0x2100,%edi
  call   <__send_to_ppe@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int puts(char const *s) {
  ;
  return __send_to_ppe(0x2100, 25, &s);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<puts>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x19,%esi
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%rdx
  mov    $0x2100,%edi
  call   <__send_to_ppe@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int puts(char const *s) {
  ;
  return __send_to_ppe(0x2100, 25, &s);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<puts>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x19,%esi
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%rdx
  mov    $0x2100,%edi
  call   <__send_to_ppe@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int puts(char const *s) {
  ;
  return __send_to_ppe(0x2100, 25, &s);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ggetline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L27
L48:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L27:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L1
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L1
  cmpl   $0xa,-0x8(%rbp)
  jne    L48
L1:
  cmpl   $0xa,-0x8(%rbp)
  jne    L0
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int ggetline(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ggetline>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L22
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L11:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D4(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L49
  cmp    $0xa,%eax
  je     L49
  mov    %al,0x0(%rbp)
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L11
  mov    %r12d,%ebx
  jmp    L10
L22:
  mov    $0x0,%ebx
L49:
  cmp    $0xa,%eax
  je     L2
L10:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  add    $0x1,%ebx
  jmp    L10
```

assistant: ```c
int ggetline(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ggetline>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L5
  mov    D8(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L31:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L17
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L34
  mov    %edx,%r14d
L46:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L31
  cmp    $0xa,%eax
  je     L0
L5:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  lea    0x1(%r14),%edx
L17:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L34:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int ggetline(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ggetline>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L40
  mov    D6(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L10:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L46
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L59
  mov    %edx,%r14d
L21:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L10
  cmp    $0xa,%eax
  je     L4
L40:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  lea    0x1(%r14),%edx
L46:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L59:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int ggetline(char s[], int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<unescape>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L40
L169:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  je     L168
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  jmp    L172
L168:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x76,%eax
  jg     L20
  cmp    $0x3f,%eax
  jge    L152
  cmp    $0x22,%eax
  je     L166
  cmp    $0x27,%eax
  je     L25
  jmp    L20
L152:
  sub    $0x3f,%eax
  cmp    $0x37,%eax
  ja     L20
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D12(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D12(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x7,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x8,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0xc,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0xa,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0xd,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x9,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0xb,(%rax)
  jmp    L172
L25:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x27,(%rax)
  jmp    L172
L166:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x22,(%rax)
  jmp    L172
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x3f,(%rax)
  jmp    L172
L20:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L172:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x8(%rbp)
L40:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L169
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void unescape(char *source, char *target) {
  while (*source != '\0') {
    if (*source != '\\')
      *target++ = *source;
    else {
      switch (*(++source)) {
      case 'a':
        *target++ = '\a';
        break;
      case 'b':
        *target++ = '\b';
        break;
      case 'f':
        *target++ = '\f';
        break;
      case 'n':
        *target++ = '\n';
        break;
      case 'r':
        *target++ = '\r';
        break;
      case 't':
        *target++ = '\t';
        break;
      case 'v':
        *target++ = '\v';
        break;
      case '\'':
        *target++ = '\'';
        break;
      case '"':
        *target++ = '\"';
        break;
      case '?':
        *target++ = '\?';
        break;
      default:
        *target++ = *source;
      }
    }
    *target = '\0';
    source++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<unescape>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L121
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rdx
  lea    D7(%rip),%r8
  mov    $0xb,%r13d
  mov    $0x9,%r12d
  mov    $0xd,%ebp
  mov    $0xa,%ebx
  mov    $0xc,%r11d
  mov    $0x8,%r10d
  mov    $0x7,%r9d
  jmp    L37
L163:
  lea    0x1(%rdi),%rsi
  movzbl 0x1(%rdi),%eax
  cmp    $0x76,%al
  jg     L97
  cmp    $0x3e,%al
  jle    L21
  lea    -0x3f(%rax),%ecx
  cmp    $0x37,%cl
  ja     L97
  movzbl %cl,%ecx
  movslq (%r8,%rcx,4),%rcx
  add    %r8,%rcx
  notrack jmp *%rcx
L21:
  lea    0x1(%rdx),%rcx
  cmp    $0x22,%al
  je     L126
  cmp    $0x27,%al
  jne    L97
  lea    0x1(%rdx),%rcx
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %r9d,%eax
L126:
  mov    %al,(%rdx)
  movb   $0x0,(%rcx)
  lea    0x1(%rsi),%rdi
  movzbl 0x1(%rsi),%eax
  test   %al,%al
  je     L136
  mov    %rcx,%rdx
L37:
  cmp    $0x5c,%al
  je     L163
  lea    0x1(%rdx),%rcx
  mov    %rdi,%rsi
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %r10d,%eax
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %r11d,%eax
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %ebx,%eax
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %ebp,%eax
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %r12d,%eax
  jmp    L126
  lea    0x1(%rdx),%rcx
  mov    %r13d,%eax
  jmp    L126
  lea    0x1(%rdx),%rcx
  jmp    L126
L97:
  lea    0x1(%rdx),%rcx
  jmp    L126
L136:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L121:
  ret
```

assistant: ```c
void unescape(char *source, char *target) {
  while (*source != '\0') {
    if (*source != '\\')
      *target++ = *source;
    else {
      switch (*(++source)) {
      case 'a':
        *target++ = '\a';
        break;
      case 'b':
        *target++ = '\b';
        break;
      case 'f':
        *target++ = '\f';
        break;
      case 'n':
        *target++ = '\n';
        break;
      case 'r':
        *target++ = '\r';
        break;
      case 't':
        *target++ = '\t';
        break;
      case 'v':
        *target++ = '\v';
        break;
      case '\'':
        *target++ = '\'';
        break;
      case '"':
        *target++ = '\"';
        break;
      case '?':
        *target++ = '\?';
        break;
      default:
        *target++ = *source;
      }
    }
    *target = '\0';
    source++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<unescape>:
  endbr64
  movzbl (%rdi),%eax
  lea    D9(%rip),%r8
  test   %al,%al
  jne    L17
  jmp    L14
  nopl   0x0(%rax)
L27:
  mov    %rdi,%rdx
L54:
  mov    %al,-0x1(%rsi)
  lea    0x1(%rdx),%rdi
  movb   $0x0,(%rsi)
  movzbl 0x1(%rdx),%eax
  test   %al,%al
  je     L14
L17:
  lea    0x1(%rdi),%rdx
  add    $0x1,%rsi
  cmp    $0x5c,%al
  jne    L27
  movzbl 0x1(%rdi),%eax
  cmp    $0x76,%al
  jg     L54
  cmp    $0x3e,%al
  jle    L54
  lea    -0x3f(%rax),%ecx
  cmp    $0x37,%cl
  ja     L54
  movzbl %cl,%ecx
  movslq (%r8,%rcx,4),%rcx
  add    %r8,%rcx
  notrack jmp *%rcx
  nopw   0x0(%rax,%rax,1)
L14:
  ret
  mov    $0x9,%eax
  jmp    L54
  mov    $0xd,%eax
  jmp    L54
  mov    $0xa,%eax
  jmp    L54
  mov    $0xc,%eax
  jmp    L54
  mov    $0xb,%eax
  jmp    L54
  mov    $0x7,%eax
  jmp    L54
  mov    $0x8,%eax
  jmp    L54
```

assistant: ```c
void unescape(char *source, char *target) {
  while (*source != '\0') {
    if (*source != '\\')
      *target++ = *source;
    else {
      switch (*(++source)) {
      case 'a':
        *target++ = '\a';
        break;
      case 'b':
        *target++ = '\b';
        break;
      case 'f':
        *target++ = '\f';
        break;
      case 'n':
        *target++ = '\n';
        break;
      case 'r':
        *target++ = '\r';
        break;
      case 't':
        *target++ = '\t';
        break;
      case 'v':
        *target++ = '\v';
        break;
      case '\'':
        *target++ = '\'';
        break;
      case '"':
        *target++ = '\"';
        break;
      case '?':
        *target++ = '\?';
        break;
      default:
        *target++ = *source;
      }
    }
    *target = '\0';
    source++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unescape>:
  endbr64
  movzbl (%rdi),%eax
  lea    D8(%rip),%r8
  test   %al,%al
  jne    L109
  jmp    L9
  nopl   0x0(%rax)
L16:
  mov    %rdi,%rdx
L30:
  mov    %al,-0x1(%rsi)
  lea    0x1(%rdx),%rdi
  movb   $0x0,(%rsi)
  movzbl 0x1(%rdx),%eax
  test   %al,%al
  je     L9
L109:
  lea    0x1(%rdi),%rdx
  add    $0x1,%rsi
  cmp    $0x5c,%al
  jne    L16
  movzbl 0x1(%rdi),%eax
  cmp    $0x76,%al
  jg     L30
  cmp    $0x3e,%al
  jle    L30
  lea    -0x3f(%rax),%ecx
  cmp    $0x37,%cl
  ja     L30
  movzbl %cl,%ecx
  movslq (%r8,%rcx,4),%rcx
  add    %r8,%rcx
  notrack jmp *%rcx
  nopw   0x0(%rax,%rax,1)
L9:
  ret
  mov    $0x9,%eax
  jmp    L30
  mov    $0xd,%eax
  jmp    L30
  mov    $0xa,%eax
  jmp    L30
  mov    $0xc,%eax
  jmp    L30
  mov    $0xb,%eax
  jmp    L30
  mov    $0x7,%eax
  jmp    L30
  mov    $0x8,%eax
  jmp    L30
```

assistant: ```c
void unescape(char *source, char *target) {
  while (*source != '\0') {
    if (*source != '\\')
      *target++ = *source;
    else {
      switch (*(++source)) {
      case 'a':
        *target++ = '\a';
        break;
      case 'b':
        *target++ = '\b';
        break;
      case 'f':
        *target++ = '\f';
        break;
      case 'n':
        *target++ = '\n';
        break;
      case 'r':
        *target++ = '\r';
        break;
      case 't':
        *target++ = '\t';
        break;
      case 'v':
        *target++ = '\v';
        break;
      case '\'':
        *target++ = '\'';
        break;
      case '"':
        *target++ = '\"';
        break;
      case '?':
        *target++ = '\?';
        break;
      default:
        *target++ = *source;
      }
    }
    *target = '\0';
    source++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L19
L16:
  addq   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *my_basename(char *path) {
  char *p;
  p = strrchr(path, '/');
  if (!p)
    p = path;
  else
    p++;
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<my_basename>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2f,%esi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  cmovne %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *my_basename(char *path) {
  char *p;
  p = strrchr(path, '/');
  if (!p)
    p = path;
  else
    p++;
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<my_basename>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *my_basename(char *path) {
  char *p;
  p = strrchr(path, '/');
  if (!p)
    p = path;
  else
    p++;
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<my_basename>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *my_basename(char *path) {
  char *p;
  p = strrchr(path, '/');
  if (!p)
    p = path;
  else
    p++;
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x6f77206f6c6c6548,%rax
  mov    %rax,-0x16(%rbp)
  movl   $0x2e646c72,-0xe(%rbp)
  movw   $0xa,-0xa(%rbp)
  lea    -0x16(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
type: string

tool: D3: %s

assistant: ```c
int main() {
  char str[] = "Hello world.\n";
  printf("%s", str);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x6f77206f6c6c6548,%rax
  mov    %rax,0xa(%rsp)
  movl   $0x2e646c72,0x12(%rsp)
  movw   $0xa,0x16(%rsp)
  lea    0xa(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type_guessed: string

tool: D6: %s

assistant: ```c
int main() {
  char str[] = "Hello world.\n";
  printf("%s", str);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D1(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x6f77206f6c6c6548,%rax
  lea    0xa(%rsp),%rsi
  movl   $0x2e646c72,0x12(%rsp)
  mov    %rax,0xa(%rsp)
  mov    $0xa,%eax
  mov    %ax,0x16(%rsp)
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type: string

tool: D1: %s

assistant: ```c
int main() {
  char str[] = "Hello world.\n";
  printf("%s", str);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x6f77206f6c6c6548,%rax
  lea    0xa(%rsp),%rsi
  movl   $0x2e646c72,0x12(%rsp)
  mov    %rax,0xa(%rsp)
  mov    $0xa,%eax
  mov    %ax,0x16(%rsp)
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_guessed: string

tool: D7: %s

assistant: ```c
int main() {
  char str[] = "Hello world.\n";
  printf("%s", str);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x61616161,-0x20(%rbp)
  movl   $0x61616161,-0x1c(%rbp)
  movl   $0x61616161,-0x18(%rbp)
  movl   $0x61616161,-0x14(%rbp)
  movl   $0x61616161,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <Zero@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L47
L11:
  movl   $0x0,-0x24(%rbp)
  jmp    L24
L46:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rbp,%rax
  add    %rdx,%rax
  sub    $0x20,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L24:
  cmpl   $0x3,-0x24(%rbp)
  jle    L46
  addl   $0x1,-0x28(%rbp)
L47:
  cmpl   $0x4,-0x28(%rbp)
  jle    L11
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_guessed: string

tool: D2: us[%d][%d] is %c


assistant: ```c
main() {
  int i, j;
  char us[5][4] = {'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a'};
  Zero((void *)us[4], 4);
  for (i = 0; i < 5; i++)
    for (j = 0; j < 4; j++)
      printf("us[%d][%d] is %c\n", i, j, us[i][j]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x61616161,(%rsp)
  movl   $0x61616161,0x4(%rsp)
  movl   $0x61616161,0x8(%rsp)
  movl   $0x61616161,0xc(%rsp)
  movl   $0x61616161,0x10(%rsp)
  mov    %rsp,%r12
  lea    0x10(%rsp),%rdi
  mov    $0x4,%esi
  call   <Zero@plt>
  mov    $0x0,%ebp
  lea    D7(%rip),%r13
  jmp    L30
L17:
  add    $0x1,%ebp
  add    $0x4,%r12
  cmp    $0x5,%ebp
  je     L9
L30:
  mov    $0x0,%ebx
L6:
  movsbl (%r12,%rbx,1),%ecx
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x4,%rbx
  jne    L6
  jmp    L17
L9:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D7: us[%d][%d] is %c


assistant: ```c
main() {
  int i, j;
  char us[5][4] = {'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a'};
  Zero((void *)us[4], 4);
  for (i = 0; i < 5; i++)
    for (j = 0; j < 4; j++)
      printf("us[%d][%d] is %c\n", i, j, us[i][j]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x4,%esi
  lea    D15(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x28,%rsp
  movdqa D8(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    %rsp,%r12
  movl   $0x61616161,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <Zero@plt>
L9:
  xor    %ebx,%ebx
L26:
  movsbl (%r12,%rbx,1),%ecx
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L26
  add    $0x1,%ebp
  add    $0x4,%r12
  cmp    $0x5,%ebp
  jne    L9
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D15
datatype: string
name: disassemble_data
arguments: label: D8
datatype: byte[16]

tool: D15: us[%d][%d] is %c


tool: {"D8": [0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61]}

assistant: ```c
main() {
  int i, j;
  char us[5][4] = {'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a'};
  Zero((void *)us[4], 4);
  for (i = 0; i < 5; i++)
    for (j = 0; j < 4; j++)
      printf("us[%d][%d] is %c\n", i, j, us[i][j]);
}
```