ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<have_read>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <yffs_permission@plt>
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  mov    $0xa,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <getlogin_r@plt>
  test   %rax,%rax
  jne    L28
L29:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <yffs_owner@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L11
  shr    $0x3,%bl
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  and    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  shr    %bl
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  and    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L29
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_guessed: string

tool: D1: problem getting user login...


assistant: ```c
int have_read(struct yffs *fs, const char *filename) {
  char permbit = yffs_permission(fs, filename);
  char *user = (char *)malloc(sizeof(char) * 10);
  if (getlogin_r(user, 10) != 0)
    printf("problem getting user login...\n");
  char *owner = yffs_owner(fs, filename);
  if (strcmp(owner, user) == 0) {
    permbit = permbit & 8;
    if (permbit != 0)
      return 1;
    else
      return 0;
  } else {
    permbit = permbit & 2;
    if (permbit != 0)
      return 1;
    else
      return 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<have_read>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <yffs_permission@plt>
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  mov    $0xa,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <getlogin_r@plt>
  test   %rax,%rax
  jne    L12
L29:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <yffs_owner@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L9
  shr    $0x3,%bl
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  and    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  shr    %bl
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  and    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L29
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
type_guessed: string

tool: D1: problem getting user login...


assistant: ```c
int have_read(struct yffs *fs, const char *filename) {
  char permbit = yffs_permission(fs, filename);
  char *user = (char *)malloc(sizeof(char) * 10);
  if (getlogin_r(user, 10) != 0)
    printf("problem getting user login...\n");
  char *owner = yffs_owner(fs, filename);
  if (strcmp(owner, user) == 0) {
    permbit = permbit & 8;
    if (permbit != 0)
      return 1;
    else
      return 0;
  } else {
    permbit = permbit & 2;
    if (permbit != 0)
      return 1;
    else
      return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<round_up_pow2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L65
  mov    $0x1,%eax
  jmp    L73
L65:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  and    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L77
  mov    -0x14(%rbp),%eax
  jmp    L73
L77:
  mov    -0x14(%rbp),%eax
  cltq
  and    $0xffff0000,%eax
  test   %rax,%rax
  je     L61
  mov    -0x14(%rbp),%eax
  cltq
  and    $0xff000000,%eax
  test   %rax,%rax
  je     L36
  mov    $0x3,%eax
  jmp    L17
L36:
  mov    $0x2,%eax
L17:
  mov    %eax,-0x8(%rbp)
  jmp    L11
L61:
  mov    -0x14(%rbp),%eax
  cltq
  and    $0xff00,%eax
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%eax
  shl    $0x3,%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    %eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x2,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x4,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%eax
  shl    $0x3,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
L73:
  pop    %rbp
  ret
```

assistant: ```c
uint32_t round_up_pow2(uint32_t v) {
  int pos;
  if (v == 0)
    return 1;
  else if ((v & (v - 1)) == 0)
    return v;
  if (v & 0xFFFF0000UL)
    pos = (v & 0xFF000000UL) ? 3 : 2;
  else
    pos = (v & 0x0000FF00UL) ? 1 : 0;
  unsigned int byte = v >> (pos << 3);
  byte |= byte >> 1;
  byte |= byte >> 2;
  byte |= byte >> 4;
  v = (byte + 1) << (pos << 3);
  return v;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<round_up_pow2>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L5
  lea    -0x1(%rdi),%edx
  mov    %edi,%eax
  test   %edi,%edx
  je     L5
  movslq %edi,%rax
  test   $0xffff0000,%edi
  je     L32
  test   $0xff000000,%eax
  setne  %cl
  movzbl %cl,%ecx
  add    $0x2,%ecx
L34:
  shl    $0x3,%ecx
  sar    %cl,%edi
  mov    %edi,%eax
  shr    %eax
  or     %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  or     %edx,%eax
  add    $0x1,%eax
  shl    %cl,%eax
L5:
  ret
L32:
  test   $0xff00,%edi
  setne  %cl
  movzbl %cl,%ecx
  jmp    L34
```

assistant: ```c
uint32_t round_up_pow2(uint32_t v) {
  int pos;
  if (v == 0)
    return 1;
  else if ((v & (v - 1)) == 0)
    return v;
  if (v & 0xFFFF0000UL)
    pos = (v & 0xFF000000UL) ? 3 : 2;
  else
    pos = (v & 0x0000FF00UL) ? 1 : 0;
  unsigned int byte = v >> (pos << 3);
  byte |= byte >> 1;
  byte |= byte >> 2;
  byte |= byte >> 4;
  v = (byte + 1) << (pos << 3);
  return v;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<round_up_pow2>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L16
  lea    -0x1(%rdi),%edx
  mov    %edi,%eax
  test   %edi,%edx
  je     L16
  movslq %edi,%rax
  test   $0xffff0000,%edi
  je     L17
  and    $0xff000000,%eax
  cmp    $0x1,%rax
  sbb    %ecx,%ecx
  and    $0xfffffff8,%ecx
  add    $0x18,%ecx
L9:
  sar    %cl,%edi
  mov    %edi,%eax
  shr    %eax
  or     %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  or     %edx,%eax
  add    $0x1,%eax
  shl    %cl,%eax
L16:
  ret
  xchg   %ax,%ax
L17:
  xor    %ecx,%ecx
  test   $0xff00,%edi
  setne  %cl
  shl    $0x3,%ecx
  jmp    L9
```

assistant: ```c
uint32_t round_up_pow2(uint32_t v) {
  int pos;
  if (v == 0)
    return 1;
  else if ((v & (v - 1)) == 0)
    return v;
  if (v & 0xFFFF0000UL)
    pos = (v & 0xFF000000UL) ? 3 : 2;
  else
    pos = (v & 0x0000FF00UL) ? 1 : 0;
  unsigned int byte = v >> (pos << 3);
  byte |= byte >> 1;
  byte |= byte >> 2;
  byte |= byte >> 4;
  v = (byte + 1) << (pos << 3);
  return v;
}
```
user: Translate the below assembly code into C programming language:

```asm
<round_up_pow2>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L3
  lea    -0x1(%rdi),%edx
  mov    %edi,%eax
  test   %edi,%edx
  je     L3
  movslq %edi,%rax
  test   $0xffff0000,%edi
  je     L21
  and    $0xff000000,%eax
  cmp    $0x1,%rax
  sbb    %ecx,%ecx
  and    $0xfffffff8,%ecx
  add    $0x18,%ecx
L39:
  sar    %cl,%edi
  mov    %edi,%eax
  shr    %eax
  or     %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  or     %edx,%eax
  add    $0x1,%eax
  shl    %cl,%eax
L3:
  ret
  xchg   %ax,%ax
L21:
  xor    %ecx,%ecx
  test   $0xff00,%edi
  setne  %cl
  shl    $0x3,%ecx
  jmp    L39
```

assistant: ```c
uint32_t round_up_pow2(uint32_t v) {
  int pos;
  if (v == 0)
    return 1;
  else if ((v & (v - 1)) == 0)
    return v;
  if (v & 0xFFFF0000UL)
    pos = (v & 0xFF000000UL) ? 3 : 2;
  else
    pos = (v & 0x0000FF00UL) ? 1 : 0;
  unsigned int byte = v >> (pos << 3);
  byte |= byte >> 1;
  byte |= byte >> 2;
  byte |= byte >> 4;
  v = (byte + 1) << (pos << 3);
  return v;
}
```
user: What is the c source code of the assembly code below:

```asm
<append_addrinfo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L5
  mov    -0x18(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L8
L5:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L39
L26:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L26
  mov    -0x8(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
L8:
  pop    %rbp
  ret
```

assistant: ```c
void append_addrinfo(struct addrinfo **head, struct addrinfo *addr) {
  if (*head == NULL) {
    *head = addr;
    return;
  }
  struct addrinfo *ai = *head;
  while (ai->ai_next != NULL) {
    ai = ai->ai_next;
  }
  ai->ai_next = addr;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<append_addrinfo>:
  endbr64
  mov    (%rdi),%rax
  test   %rax,%rax
  je     L9
L0:
  mov    %rax,%rdx
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L0
  mov    %rsi,(%rdx)
  ret
L9:
  mov    %rsi,(%rdi)
  ret
```

assistant: ```c
void append_addrinfo(struct addrinfo **head, struct addrinfo *addr) {
  if (*head == NULL) {
    *head = addr;
    return;
  }
  struct addrinfo *ai = *head;
  while (ai->ai_next != NULL) {
    ai = ai->ai_next;
  }
  ai->ai_next = addr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<append_addrinfo>:
  endbr64
  mov    (%rdi),%rax
  test   %rax,%rax
  je     L2
  nopl   0x0(%rax)
L15:
  mov    %rax,%rdx
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L15
  mov    %rsi,(%rdx)
  ret
  nop
L2:
  mov    %rsi,(%rdi)
  ret
```

assistant: ```c
void append_addrinfo(struct addrinfo **head, struct addrinfo *addr) {
  if (*head == NULL) {
    *head = addr;
    return;
  }
  struct addrinfo *ai = *head;
  while (ai->ai_next != NULL) {
    ai = ai->ai_next;
  }
  ai->ai_next = addr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<append_addrinfo>:
  endbr64
  mov    (%rdi),%rax
  test   %rax,%rax
  je     L7
  nopl   0x0(%rax)
L11:
  mov    %rax,%rdx
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L11
  mov    %rsi,(%rdx)
  ret
  nop
L7:
  mov    %rsi,(%rdi)
  ret
```

assistant: ```c
void append_addrinfo(struct addrinfo **head, struct addrinfo *addr) {
  if (*head == NULL) {
    *head = addr;
    return;
  }
  struct addrinfo *ai = *head;
  while (ai->ai_next != NULL) {
    ai = ai->ai_next;
  }
  ai->ai_next = addr;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ExtractFileName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movb   $0x2f,-0xd(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L0
L55:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xd(%rbp)
  je     L52
  subl   $0x1,-0xc(%rbp)
L0:
  cmpl   $0x0,-0xc(%rbp)
  jns    L55
  jmp    L48
L52:
  nop
L48:
  addl   $0x1,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
L42:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L8
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L8
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
  jmp    L42
L8:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void ExtractFileName(const char *szPathName, char *szFileName) {
  int k, i, x;
  char c;
  k = strlen(szPathName);
  c = '/';
  i = k - 1;
  while (i >= 0) {
    if (szPathName[i] == c)
      break;
    i--;
  }
  i++;
  x = 0;
  for (;;) {
    szFileName[x] = szPathName[i];
    if (szFileName[x] == '\0' || szFileName[x] == '.')
      break;
    x++;
    i++;
  }
  szFileName[x] = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ExtractFileName>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L24
  movslq %eax,%rdx
  add    %rbp,%rdx
L63:
  cmpb   $0x2f,(%rdx)
  je     L24
  sub    $0x1,%eax
  sub    $0x1,%rdx
  cmp    $0xffffffff,%eax
  jne    L63
L24:
  add    $0x1,%eax
  cltq
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%rbx)
  test   %dl,%dl
  je     L37
  cmp    $0x2e,%dl
  je     L37
  lea    0x1(%rbx),%rdx
  lea    0x1(%rbp,%rax,1),%rcx
L47:
  mov    %rdx,%rbx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  add    $0x1,%rdx
  add    $0x1,%rcx
  test   %al,%al
  je     L37
  cmp    $0x2e,%al
  jne    L47
L37:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ExtractFileName(const char *szPathName, char *szFileName) {
  int k, i, x;
  char c;
  k = strlen(szPathName);
  c = '/';
  i = k - 1;
  while (i >= 0) {
    if (szPathName[i] == c)
      break;
    i--;
  }
  i++;
  x = 0;
  for (;;) {
    szFileName[x] = szPathName[i];
    if (szFileName[x] == '\0' || szFileName[x] == '.')
      break;
    x++;
    i++;
  }
  szFileName[x] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ExtractFileName>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L21
  movslq %edx,%rdx
  jmp    L79
  nopl   0x0(%rax)
L3:
  sub    $0x1,%rdx
  test   %edx,%edx
  js     L32
L79:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %edx,%eax
  cmp    $0x2f,%cl
  jne    L3
  add    $0x1,%eax
  movslq %eax,%rdx
  movzbl (%rbx,%rdx,1),%ecx
L32:
  mov    %cl,0x0(%rbp)
  test   %cl,%cl
  je     L28
  cmp    $0x2e,%cl
  je     L28
  cltq
  lea    0x1(%rbp),%rdx
  lea    0x1(%rbx,%rax,1),%rcx
  nopl   0x0(%rax)
L35:
  movzbl (%rcx),%eax
  mov    %rdx,%rbp
  add    $0x1,%rcx
  add    $0x1,%rdx
  mov    %al,-0x1(%rdx)
  test   %al,%al
  je     L28
  cmp    $0x2e,%al
  jne    L35
L28:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L21:
  movslq %eax,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  jmp    L32
```

assistant: ```c
void ExtractFileName(const char *szPathName, char *szFileName) {
  int k, i, x;
  char c;
  k = strlen(szPathName);
  c = '/';
  i = k - 1;
  while (i >= 0) {
    if (szPathName[i] == c)
      break;
    i--;
  }
  i++;
  x = 0;
  for (;;) {
    szFileName[x] = szPathName[i];
    if (szFileName[x] == '\0' || szFileName[x] == '.')
      break;
    x++;
    i++;
  }
  szFileName[x] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ExtractFileName>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L60
  movslq %edx,%rdx
  jmp    L37
  nopl   0x0(%rax)
L22:
  sub    $0x1,%rdx
  test   %edx,%edx
  js     L78
L37:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %edx,%eax
  cmp    $0x2f,%cl
  jne    L22
  add    $0x1,%eax
  movslq %eax,%rdx
  movzbl (%rbx,%rdx,1),%ecx
L78:
  mov    %cl,0x0(%rbp)
  test   %cl,%cl
  je     L6
  cmp    $0x2e,%cl
  je     L6
  cltq
  lea    0x1(%rbp),%rdx
  lea    0x1(%rbx,%rax,1),%rcx
  nopl   0x0(%rax)
L79:
  movzbl (%rcx),%eax
  mov    %rdx,%rbp
  add    $0x1,%rcx
  add    $0x1,%rdx
  mov    %al,-0x1(%rdx)
  test   %al,%al
  je     L6
  cmp    $0x2e,%al
  jne    L79
L6:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L60:
  movslq %eax,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  jmp    L78
```

assistant: ```c
void ExtractFileName(const char *szPathName, char *szFileName) {
  int k, i, x;
  char c;
  k = strlen(szPathName);
  c = '/';
  i = k - 1;
  while (i >= 0) {
    if (szPathName[i] == c)
      break;
    i--;
  }
  i++;
  x = 0;
  for (;;) {
    szFileName[x] = szPathName[i];
    if (szFileName[x] == '\0' || szFileName[x] == '.')
      break;
    x++;
    i++;
  }
  szFileName[x] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_uint32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x4,-0x10(%rbp)
  je     L15
  mov    $0xffffffff,%eax
  jmp    L16
L15:
  movzbl -0xc(%rbp),%eax
  mov    %eax,%edx
  movzbl -0xb(%rbp),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,%edx
  movzbl -0xa(%rbp),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  or     %eax,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    $0x0,%eax
L16:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int read_uint32(int fd, uint32_t *x) {
  unsigned char tmp[4];
  int n;
  n = read(fd, tmp, 4);
  if (n != 4)
    return -1;
  *x = tmp[0] | (tmp[1] << 8) | (tmp[2] << 16) | (tmp[3] << 24);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<read_uint32>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    $0x4,%edx
  call   <read@plt>
  cmp    $0x4,%eax
  jne    L29
  movzbl 0x4(%rsp),%eax
  mov    %al,(%rbx)
  mov    $0x0,%eax
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  ret
L29:
  mov    $0xffffffff,%eax
  jmp    L11
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_uint32(int fd, uint32_t *x) {
  unsigned char tmp[4];
  int n;
  n = read(fd, tmp, 4);
  if (n != 4)
    return -1;
  *x = tmp[0] | (tmp[1] << 8) | (tmp[2] << 16) | (tmp[3] << 24);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<read_uint32>:
  endbr64
  push   %rbx
  mov    $0x4,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <read@plt>
  cmp    $0x4,%eax
  jne    L24
  movzbl 0x4(%rsp),%eax
  mov    %al,(%rbx)
  xor    %eax,%eax
L22:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  ret
  nop
L24:
  mov    $0xffffffff,%eax
  jmp    L22
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_uint32(int fd, uint32_t *x) {
  unsigned char tmp[4];
  int n;
  n = read(fd, tmp, 4);
  if (n != 4)
    return -1;
  *x = tmp[0] | (tmp[1] << 8) | (tmp[2] << 16) | (tmp[3] << 24);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_uint32>:
  endbr64
  push   %rbx
  mov    $0x4,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <read@plt>
  cmp    $0x4,%eax
  jne    L19
  movzbl 0x4(%rsp),%eax
  mov    %al,(%rbx)
  xor    %eax,%eax
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  ret
  nop
L19:
  mov    $0xffffffff,%eax
  jmp    L26
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_uint32(int fd, uint32_t *x) {
  unsigned char tmp[4];
  int n;
  n = read(fd, tmp, 4);
  if (n != 4)
    return -1;
  *x = tmp[0] | (tmp[1] << 8) | (tmp[2] << 16) | (tmp[3] << 24);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L30
L22:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L21
L19:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L17
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L17:
  addl   $0x1,-0x8(%rbp)
L21:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L19
  addl   $0x1,-0xc(%rbp)
L30:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L22
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void sort(int n, int *array) {
  int i, j, temp;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (array[i] > array[j]) {
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;
      }
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sort>:
  endbr64
  test   %edi,%edi
  jle    L2
  lea    0x4(%rsi),%r8
  lea    -0x1(%rdi),%r11d
  mov    %edi,%r9d
  mov    $0x1,%edi
  lea    0x8(%rsi),%r10
  jmp    L59
L53:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L40
L62:
  mov    -0x4(%r8),%edx
  mov    (%rax),%ecx
  cmp    %ecx,%edx
  jle    L53
  mov    %ecx,-0x4(%r8)
  mov    %edx,(%rax)
  jmp    L53
L40:
  add    $0x1,%rdi
  add    $0x4,%r8
L59:
  cmp    %r9,%rdi
  je     L2
  mov    %r11d,%eax
  sub    %edi,%eax
  lea    -0x1(%rdi,%rax,1),%rax
  lea    (%r10,%rax,4),%rsi
  mov    %r8,%rax
  jmp    L62
L2:
  ret
```

assistant: ```c
void sort(int n, int *array) {
  int i, j, temp;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (array[i] > array[j]) {
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;
      }
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sort>:
  endbr64
  test   %edi,%edi
  jle    L48
  lea    -0x1(%rdi),%r11d
  mov    %edi,%r9d
  mov    $0x1,%edi
  lea    0x4(%rsi),%r8
  lea    0x8(%rsi),%r10
  cmp    %r9,%rdi
  je     L48
  nopl   0x0(%rax)
L38:
  mov    %r11d,%eax
  sub    %edi,%eax
  lea    -0x1(%rdi,%rax,1),%rax
  lea    (%r10,%rax,4),%rsi
  mov    %r8,%rax
  nopl   0x0(%rax)
L47:
  mov    -0x4(%r8),%edx
  mov    (%rax),%ecx
  cmp    %ecx,%edx
  jle    L42
  mov    %ecx,-0x4(%r8)
  mov    %edx,(%rax)
L42:
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L47
  add    $0x1,%rdi
  add    $0x4,%r8
  cmp    %r9,%rdi
  jne    L38
L48:
  ret
```

assistant: ```c
void sort(int n, int *array) {
  int i, j, temp;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (array[i] > array[j]) {
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;
      }
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sort>:
  endbr64
  cmp    $0x1,%edi
  jle    L38
  mov    %rsi,%r8
  movslq %edi,%r10
  mov    $0x1,%r9d
  nopl   (%rax)
L34:
  mov    %r9,%rax
  nopl   0x0(%rax,%rax,1)
L21:
  mov    (%r8),%edx
  mov    (%rsi,%rax,4),%ecx
  cmp    %ecx,%edx
  jle    L3
  mov    %ecx,(%r8)
  mov    %edx,(%rsi,%rax,4)
L3:
  add    $0x1,%rax
  cmp    %eax,%edi
  jg     L21
  add    $0x1,%r9
  add    $0x4,%r8
  cmp    %r10,%r9
  jne    L34
L38:
  ret
```

assistant: ```c
void sort(int n, int *array) {
  int i, j, temp;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (array[i] > array[j]) {
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;
      }
    }
  }
}
```