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
<strcat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    %r12d,%eax
  cmp    %eax,%r12d
  jge    L19
  movslq %r12d,%r12
  mov    %r12,%rdx
  mov    %rbp,%rsi
  sub    %r12,%rsi
L28:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L28
L22:
  cltq
  movb   $0x0,(%rbx,%rax,1)
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  mov    %r12d,%eax
  jmp    L22
```

assistant: ```c
int strcat(char *str1, char *str2) {
  int i = 0;
  int j = 0;
  int len1 = 0, len2 = 0;
  len1 = strlen(str1);
  len2 = strlen(str2);
  for (i = len1; i < len1 + len2; i++) {
    str1[i] = str2[j];
    j++;
  }
  str1[i] = '\0';
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  add    %ebx,%eax
  cmp    %eax,%ebx
  jge    L13
  movslq %ebx,%rdx
  sub    %rdx,%rbp
  nopl   0x0(%rax)
L16:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r12,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L16
  cltq
  pop    %rbx
  pop    %rbp
  movb   $0x0,(%r12,%rax,1)
  xor    %eax,%eax
  pop    %r12
  ret
  nop
L13:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  cltq
  movb   $0x0,(%r12,%rax,1)
  xor    %eax,%eax
  pop    %r12
  ret
```

assistant: ```c
int strcat(char *str1, char *str2) {
  int i = 0;
  int j = 0;
  int len1 = 0, len2 = 0;
  len1 = strlen(str1);
  len2 = strlen(str2);
  for (i = len1; i < len1 + len2; i++) {
    str1[i] = str2[j];
    j++;
  }
  str1[i] = '\0';
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<strcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  lea    0x0(%rbp,%rax,1),%edx
  cmp    %edx,%ebp
  jge    L40
  mov    %eax,%ecx
  lea    -0x1(%rax),%esi
  lea    0x1(%rbx),%r10
  mov    %eax,%r8d
  movslq %ebp,%rax
  lea    (%r12,%rax,1),%rdi
  mov    %rdi,%r9
  sub    %r10,%r9
  cmp    $0xe,%r9
  jbe    L68
  cmp    $0x6,%esi
  jbe    L68
  cmp    $0xe,%esi
  jbe    L15
  mov    %ecx,%esi
  xor    %eax,%eax
  shr    $0x4,%esi
  shl    $0x4,%rsi
  nopl   0x0(%rax)
L139:
  movdqu (%rbx,%rax,1),%xmm0
  movups %xmm0,(%rdi,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L139
  mov    %ecx,%eax
  and    $0xfffffff0,%eax
  mov    %eax,%esi
  add    %eax,%ebp
  cmp    %eax,%ecx
  je     L4
  sub    %eax,%ecx
  mov    %ecx,%r8d
  lea    -0x1(%rcx),%ecx
  cmp    $0x6,%ecx
  jbe    L59
L94:
  mov    (%rbx,%rax,1),%rcx
  mov    %rcx,(%rdi,%rax,1)
  mov    %r8d,%eax
  and    $0xfffffff8,%eax
  add    %eax,%ebp
  add    %eax,%esi
  cmp    %eax,%r8d
  je     L4
L59:
  movslq %esi,%rax
  movzbl (%rbx,%rax,1),%ecx
  movslq %ebp,%rax
  mov    %cl,(%r12,%rax,1)
  lea    0x1(%rbp),%eax
  lea    0x1(%rsi),%ecx
  cmp    %eax,%edx
  jle    L4
  movslq %ecx,%rcx
  cltq
  movzbl (%rbx,%rcx,1),%ecx
  mov    %cl,(%r12,%rax,1)
  lea    0x2(%rbp),%eax
  lea    0x2(%rsi),%ecx
  cmp    %eax,%edx
  jle    L4
  movslq %ecx,%rcx
  cltq
  movzbl (%rbx,%rcx,1),%ecx
  mov    %cl,(%r12,%rax,1)
  lea    0x3(%rbp),%eax
  lea    0x3(%rsi),%ecx
  cmp    %eax,%edx
  jle    L4
  movslq %ecx,%rcx
  cltq
  movzbl (%rbx,%rcx,1),%ecx
  mov    %cl,(%r12,%rax,1)
  lea    0x4(%rbp),%eax
  lea    0x4(%rsi),%ecx
  cmp    %eax,%edx
  jle    L4
  movslq %ecx,%rcx
  cltq
  movzbl (%rbx,%rcx,1),%ecx
  mov    %cl,(%r12,%rax,1)
  lea    0x5(%rbp),%eax
  lea    0x5(%rsi),%ecx
  cmp    %eax,%edx
  jle    L4
  movslq %ecx,%rcx
  cltq
  add    $0x6,%ebp
  add    $0x6,%esi
  movzbl (%rbx,%rcx,1),%ecx
  mov    %cl,(%r12,%rax,1)
  cmp    %ebp,%edx
  jle    L4
  movslq %esi,%rsi
  movslq %ebp,%rbp
  movzbl (%rbx,%rsi,1),%eax
  mov    %al,(%r12,%rbp,1)
L4:
  movslq %edx,%rdx
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  movb   $0x0,(%r12,%rdx,1)
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L68:
  sub    %rax,%rbx
  nopl   0x0(%rax,%rax,1)
L73:
  movzbl (%rbx,%rax,1),%ecx
  mov    %cl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L73
  movslq %edx,%rdx
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  movb   $0x0,(%r12,%rdx,1)
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L40:
  mov    %ebp,%edx
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  movslq %edx,%rdx
  movb   $0x0,(%r12,%rdx,1)
  pop    %r12
  ret
L15:
  xor    %eax,%eax
  xor    %esi,%esi
  jmp    L94
```

assistant: ```c
int strcat(char *str1, char *str2) {
  int i = 0;
  int j = 0;
  int len1 = 0, len2 = 0;
  len1 = strlen(str1);
  len2 = strlen(str2);
  for (i = len1; i < len1 + len2; i++) {
    str1[i] = str2[j];
    j++;
  }
  str1[i] = '\0';
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strtok_r>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L24
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  jmp    L87
L24:
  mov    -0x18(%rbp),%rax
L87:
  mov    %rax,-0x10(%rbp)
  jmp    L68
L21:
  addq   $0x1,-0x10(%rbp)
L68:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L12
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L21
L12:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L77
  mov    -0x28(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    $0x0,%eax
  jmp    L65
L77:
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L96
L2:
  addq   $0x1,-0x8(%rbp)
L96:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L89
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L2
L89:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L28
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L45
L28:
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
L45:
  mov    -0x10(%rbp),%rax
L65:
  leave
  ret
```

assistant: ```c
char *strtok_r(char *str, const char *sep, char **state) {
  char *start, *end;
  start = str ? str : *state;
  while (*start && strchr(sep, *start)) {
    start++;
  }
  if (*start == '\0') {
    *state = start;
    return NULL;
  }
  end = start;
  while (*end && !strchr(sep, *end)) {
    end++;
  }
  if (*end != '\0') {
    *end = '\0';
    *state = end + 1;
  } else {
    *state = end;
  }
  return start;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strtok_r>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  test   %rdi,%rdi
  je     L12
L106:
  movzbl (%rbx),%esi
  test   %sil,%sil
  je     L70
L33:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L66
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L33
L70:
  mov    %rbx,%r13
  mov    $0x0,%eax
  jmp    L92
L12:
  mov    (%rdx),%rbx
  jmp    L106
L66:
  movzbl (%rbx),%esi
  test   %sil,%sil
  je     L70
  mov    %rbx,%r13
L51:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L42
  add    $0x1,%r13
  movzbl 0x0(%r13),%esi
  test   %sil,%sil
  jne    L51
L29:
  mov    %rbx,%rax
  jmp    L92
L42:
  cmpb   $0x0,0x0(%r13)
  je     L29
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
  mov    %rbx,%rax
L92:
  mov    %r13,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strtok_r(char *str, const char *sep, char **state) {
  char *start, *end;
  start = str ? str : *state;
  while (*start && strchr(sep, *start)) {
    start++;
  }
  if (*start == '\0') {
    *state = start;
    return NULL;
  }
  end = start;
  while (*end && !strchr(sep, *end)) {
    end++;
  }
  if (*end != '\0') {
    *end = '\0';
    *state = end + 1;
  } else {
    *state = end;
  }
  return start;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strtok_r>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  jne    L75
  jmp    L48
  nopl   0x0(%rax)
L2:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L78
  add    $0x1,%rbx
L75:
  movsbl (%rbx),%esi
  test   %sil,%sil
  jne    L2
L50:
  mov    %rbx,%r13
  xor    %eax,%eax
  mov    %r13,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L78:
  movsbl (%rbx),%esi
  test   %sil,%sil
  je     L50
  mov    %rbx,%r13
  jmp    L79
  nopl   (%rax)
L11:
  movsbl 0x1(%r13),%esi
  add    $0x1,%r13
  test   %sil,%sil
  je     L53
L79:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L11
  cmpb   $0x0,0x0(%r13)
  je     L53
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
  mov    %rbx,%rax
  mov    %r13,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L53:
  mov    %r13,(%r12)
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L48:
  mov    (%rdx),%rbx
  jmp    L75
```

assistant: ```c
char *strtok_r(char *str, const char *sep, char **state) {
  char *start, *end;
  start = str ? str : *state;
  while (*start && strchr(sep, *start)) {
    start++;
  }
  if (*start == '\0') {
    *state = start;
    return NULL;
  }
  end = start;
  while (*end && !strchr(sep, *end)) {
    end++;
  }
  if (*end != '\0') {
    *end = '\0';
    *state = end + 1;
  } else {
    *state = end;
  }
  return start;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strtok_r>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  jne    L68
  jmp    L18
  nopl   0x0(%rax)
L91:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L58
  add    $0x1,%rbx
L68:
  movsbl (%rbx),%esi
  test   %sil,%sil
  jne    L91
L62:
  mov    %rbx,%r13
  xor    %eax,%eax
  mov    %r13,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L58:
  movsbl (%rbx),%esi
  test   %sil,%sil
  je     L62
  mov    %rbx,%r13
  jmp    L25
  nopl   (%rax)
L37:
  movsbl 0x1(%r13),%esi
  add    $0x1,%r13
  test   %sil,%sil
  je     L92
L25:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L37
  cmpb   $0x0,0x0(%r13)
  je     L92
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
  mov    %rbx,%rax
  mov    %r13,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L92:
  mov    %r13,(%r12)
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    (%rdx),%rbx
  jmp    L68
```

assistant: ```c
char *strtok_r(char *str, const char *sep, char **state) {
  char *start, *end;
  start = str ? str : *state;
  while (*start && strchr(sep, *start)) {
    start++;
  }
  if (*start == '\0') {
    *state = start;
    return NULL;
  }
  end = start;
  while (*end && !strchr(sep, *end)) {
    end++;
  }
  if (*end != '\0') {
    *end = '\0';
    *state = end + 1;
  } else {
    *state = end;
  }
  return start;
}
```
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
  mov    $0x5f,%esi
  mov    $0x0,%edi
  call   <motor@plt>
  mov    $0x64,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xe10,%edi
  call   <msleep@plt>
  mov    $0x0,%eax
  call   <ao@plt>
  mov    $0x3e8,%edi
  call   <msleep@plt>
  mov    $0xffffff9c,%esi
  mov    $0x0,%edi
  call   <motor@plt>
  mov    $0xffffffa1,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xed8,%edi
  call   <msleep@plt>
  mov    $0x0,%eax
  call   <ao@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  motor(0, 95);
  motor(2, 100);
  msleep(3600);
  ao();
  msleep(1000);
  motor(0, -100);
  motor(2, -95);
  msleep(3800);
  ao();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5f,%esi
  mov    $0x0,%edi
  call   <motor@plt>
  mov    $0x64,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xe10,%edi
  call   <msleep@plt>
  mov    $0x0,%eax
  call   <ao@plt>
  mov    $0x3e8,%edi
  call   <msleep@plt>
  mov    $0xffffff9c,%esi
  mov    $0x0,%edi
  call   <motor@plt>
  mov    $0xffffffa1,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xed8,%edi
  call   <msleep@plt>
  mov    $0x0,%eax
  call   <ao@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  motor(0, 95);
  motor(2, 100);
  msleep(3600);
  ao();
  msleep(1000);
  motor(0, -100);
  motor(2, -95);
  msleep(3800);
  ao();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5f,%esi
  xor    %edi,%edi
  call   <motor@plt>
  mov    $0x64,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xe10,%edi
  call   <msleep@plt>
  xor    %eax,%eax
  call   <ao@plt>
  mov    $0x3e8,%edi
  call   <msleep@plt>
  mov    $0xffffff9c,%esi
  xor    %edi,%edi
  call   <motor@plt>
  mov    $0xffffffa1,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xed8,%edi
  call   <msleep@plt>
  xor    %eax,%eax
  call   <ao@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  motor(0, 95);
  motor(2, 100);
  msleep(3600);
  ao();
  msleep(1000);
  motor(0, -100);
  motor(2, -95);
  msleep(3800);
  ao();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5f,%esi
  xor    %edi,%edi
  call   <motor@plt>
  mov    $0x64,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xe10,%edi
  call   <msleep@plt>
  xor    %eax,%eax
  call   <ao@plt>
  mov    $0x3e8,%edi
  call   <msleep@plt>
  mov    $0xffffff9c,%esi
  xor    %edi,%edi
  call   <motor@plt>
  mov    $0xffffffa1,%esi
  mov    $0x2,%edi
  call   <motor@plt>
  mov    $0xed8,%edi
  call   <msleep@plt>
  xor    %eax,%eax
  call   <ao@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  motor(0, 95);
  motor(2, 100);
  msleep(3600);
  ao();
  msleep(1000);
  motor(0, -100);
  motor(2, -95);
  msleep(3800);
  ao();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L20
L12:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
L20:
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jg     L12
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D11
data_type: string
name: read_data
arguments: tag: D18
data_type: string
name: read_data
arguments: tag: D19
data_type: string
name: read_data
arguments: tag: D44
data_type: string

tool: D11: Enter number to be reversed: 

tool: D18: %d

tool: D19: Your number reversed is: 

tool: D44: .

assistant: ```c
int main(void) {
  int number;
  printf("Enter number to be reversed: ");
  scanf("%d", &number);
  printf("Your number reversed is: ");
  for (; number > 0; number /= 10)
    printf("%d", number % 10);
  printf(".\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L14
  lea    D13(%rip),%rbx
L3:
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  movslq %eax,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  cltd
  sub    %edx,%esi
  mov    %esi,0x4(%rsp)
  cmp    $0x9,%eax
  jg     L3
L14:
  lea    D9(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D40
data_type: string
name: disassemble_data
arguments: data_tag: D13
data_type: string
name: disassemble_data
arguments: data_tag: D28
data_type: string
name: disassemble_data
arguments: data_tag: D9
data_type: string

tool: D40: Enter number to be reversed: 

tool: D13: %d

tool: D28: Your number reversed is: 

tool: D9: .

assistant: ```c
int main(void) {
  int number;
  printf("Enter number to be reversed: ");
  scanf("%d", &number);
  printf("Your number reversed is: ");
  for (; number > 0; number /= 10)
    printf("%d", number % 10);
  printf(".\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D23(%rip),%rdi
  lea    D22(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L14
  nop
L8:
  movslq %esi,%rax
  mov    %esi,%edx
  mov    %rbx,%rdi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0x4(%rsp),%rsi
  mov    %rsi,%rax
  imul   $0x66666667,%rsi,%rsi
  cltd
  sar    $0x22,%rsi
  sub    %edx,%esi
  mov    %esi,0x4(%rsp)
  cmp    $0x9,%eax
  jg     L8
L14:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D23
type_infered: string
name: unpack_data
arguments: data_label: D22
type_infered: string
name: unpack_data
arguments: data_label: D39
type_infered: string
name: unpack_data
arguments: data_label: D34
type_infered: string

tool: D23: Enter number to be reversed: 

tool: D22: %d

tool: D39: Your number reversed is: 

tool: D34: .

assistant: ```c
int main(void) {
  int number;
  printf("Enter number to be reversed: ");
  scanf("%d", &number);
  printf("Your number reversed is: ");
  for (; number > 0; number /= 10)
    printf("%d", number % 10);
  printf(".\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D19(%rip),%rdi
  lea    D11(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L28
  nop
L14:
  movslq %esi,%rax
  mov    %esi,%edx
  mov    %rbx,%rdi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0x4(%rsp),%rsi
  mov    %rsi,%rax
  imul   $0x66666667,%rsi,%rsi
  cltd
  sar    $0x22,%rsi
  sub    %edx,%esi
  mov    %esi,0x4(%rsp)
  cmp    $0x9,%eax
  jg     L14
L28:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
datatype: string
name: read_data
arguments: data_label: D11
datatype: string
name: read_data
arguments: data_label: D31
datatype: string
name: read_data
arguments: data_label: D35
datatype: string

tool: D19: Enter number to be reversed: 

tool: D11: %d

tool: D31: Your number reversed is: 

tool: D35: .

assistant: ```c
int main(void) {
  int number;
  printf("Enter number to be reversed: ");
  scanf("%d", &number);
  printf("Your number reversed is: ");
  for (; number > 0; number /= 10)
    printf("%d", number % 10);
  printf(".\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<format_fdd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <format_log@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jne    L27
  mov    $0xc,%edi
  call   <textattr@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <cprintf@plt>
  mov    $0x7,%edi
  call   <textattr@plt>
  jmp    L11
L27:
  cmpl   $0x1,-0x4(%rbp)
  jne    L7
  mov    $0xc,%edi
  call   <textattr@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <cprintf@plt>
  mov    $0x7,%edi
  call   <textattr@plt>
  jmp    L11
L7:
  mov    $0x3,%edi
  call   <textattr@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <cprintf@plt>
  mov    $0x7,%edi
  call   <textattr@plt>
L11:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D22
data_type: string
name: retrieve_data
arguments: data_label: D29
data_type: string
name: retrieve_data
arguments: data_label: D23
data_type: string

tool: D22: 
Format incomplete. Data error. STATUS: %d

tool: D29: 
Format incomplete. Failed. STATUS: %d

tool: D23: 
Format complete. STATUS: %d

assistant: ```c
int format_fdd() {
  int status;
  status = format_log();
  if (status == 2) {
    textattr(0x0C);
    cprintf("\n\rFormat incomplete. Data error. STATUS: %d", status);
    textattr(0x07);
  } else if (status == 1) {
    textattr(0x0C);
    cprintf("\n\rFormat incomplete. Failed. STATUS: %d", status);
    textattr(0x07);
  } else {
    textattr(0x03);
    cprintf("\n\rFormat complete. STATUS: %d", status);
    textattr(0x07);
  }
  return status;
}
```