ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<scbuf_put_fd>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  test   %r12,%r12
  mov    %edx,%ebp
  push   %rbx
  setne  %dil
  mov    %rsi,%rbx
  call   <assert@plt>
  xor    %edi,%edi
  test   %rbx,%rbx
  setne  %dil
  call   <assert@plt>
  mov    (%rbx),%edi
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  xor    %edi,%edi
  test   %ebp,%ebp
  setg   %dil
  call   <assert@plt>
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L8:
  mov    D0(%rip),%rdx
  mov    D10(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L4
L25:
  mov    (%rbx),%edi
  mov    %ebp,%edx
  mov    %r12,%rsi
  call   <write@plt>
  test   %eax,%eax
  js     L8
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int scbuf_put_fd(void *srscbuf, int *pdstfd, int len) {
  int n;
  assert(srscbuf != NULL);
  assert(pdstfd != NULL);
  assert(*pdstfd >= 0);
  assert(len > 0);
  do {
    n = write(*pdstfd, srscbuf, len);
  } while ((n < 0) && (errno == EINTR));
  return (n);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scbuf_put_fd>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  test   %r12,%r12
  mov    %edx,%ebp
  push   %rbx
  setne  %dil
  mov    %rsi,%rbx
  call   <assert@plt>
  xor    %edi,%edi
  test   %rbx,%rbx
  setne  %dil
  call   <assert@plt>
  mov    (%rbx),%edi
  not    %edi
  shr    $0x1f,%edi
  call   <assert@plt>
  xor    %edi,%edi
  test   %ebp,%ebp
  setg   %dil
  call   <assert@plt>
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L18:
  mov    D18(%rip),%rdx
  mov    D6(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L29
L12:
  mov    (%rbx),%edi
  mov    %ebp,%edx
  mov    %r12,%rsi
  call   <write@plt>
  test   %eax,%eax
  js     L18
L29:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int scbuf_put_fd(void *srscbuf, int *pdstfd, int len) {
  int n;
  assert(srscbuf != NULL);
  assert(pdstfd != NULL);
  assert(*pdstfd >= 0);
  assert(len > 0);
  do {
    n = write(*pdstfd, srscbuf, len);
  } while ((n < 0) && (errno == EINTR));
  return (n);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<is_raid_level_supported>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  js     L227
  cmpl   $0x6,-0xc(%rbp)
  je     L227
  cmpl   $0x4,-0xc(%rbp)
  jne    L232
L227:
  mov    $0x0,%eax
  jmp    L182
L232:
  cmpq   $0x0,-0x8(%rbp)
  je     L41
  cmpl   $0xa,-0xc(%rbp)
  je     L213
  cmpl   $0xa,-0xc(%rbp)
  jg     L152
  cmpl   $0x5,-0xc(%rbp)
  je     L195
  cmpl   $0x5,-0xc(%rbp)
  jg     L152
  cmpl   $0x0,-0xc(%rbp)
  je     L55
  cmpl   $0x1,-0xc(%rbp)
  je     L43
  jmp    L152
L55:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <imsm_orom_has_raid0@plt>
  jmp    L182
L43:
  cmpl   $0x2,-0x10(%rbp)
  jle    L198
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <imsm_orom_has_raid1e@plt>
  jmp    L182
L198:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <imsm_orom_has_raid1@plt>
  test   %eax,%eax
  je     L2
  cmpl   $0x2,-0x10(%rbp)
  jne    L2
  mov    $0x1,%eax
  jmp    L182
L2:
  mov    $0x0,%eax
  jmp    L182
L213:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <imsm_orom_has_raid10@plt>
  test   %eax,%eax
  je     L246
  cmpl   $0x4,-0x10(%rbp)
  jne    L246
  mov    $0x1,%eax
  jmp    L182
L246:
  mov    $0x0,%eax
  jmp    L182
L195:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <imsm_orom_has_raid5@plt>
  test   %eax,%eax
  je     L127
  cmpl   $0x2,-0x10(%rbp)
  jle    L127
  mov    $0x1,%eax
  jmp    L182
L127:
  mov    $0x0,%eax
  jmp    L182
L41:
  mov    $0x1,%eax
  jmp    L182
L152:
  mov    $0x0,%eax
L182:
  leave
  ret
```

assistant: ```c
int is_raid_level_supported(const struct imsm_orom *orom, int level, int raiddisks) {
  if (level < 0 || level == 6 || level == 4)
    return 0;
  if (orom)
    switch (level) {
    case 0:
      return imsm_orom_has_raid0(orom);
    case 1:
      if (raiddisks > 2)
        return imsm_orom_has_raid1e(orom);
      return imsm_orom_has_raid1(orom) && raiddisks == 2;
    case 10:
      return imsm_orom_has_raid10(orom) && raiddisks == 4;
    case 5:
      return imsm_orom_has_raid5(orom) && raiddisks > 2;
    }
  else
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_raid_level_supported>:
  endbr64
  mov    %esi,%eax
  and    $0xfffffffd,%eax
  cmp    $0x4,%eax
  je     L104
  push   %rbx
  mov    %edx,%ebx
  mov    $0x0,%eax
  test   %esi,%esi
  js     L35
  test   %rdi,%rdi
  je     L117
  cmp    $0x5,%esi
  je     L27
  jg     L109
  test   %esi,%esi
  je     L85
  cmp    $0x1,%esi
  jne    L145
  cmp    $0x2,%edx
  jle    L148
  call   <imsm_orom_has_raid1e@plt>
  jmp    L35
L145:
  mov    $0x0,%eax
  jmp    L35
L109:
  mov    $0x0,%eax
  cmp    $0xa,%esi
  jne    L35
  call   <imsm_orom_has_raid10@plt>
  cmp    $0x4,%ebx
  sete   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  jmp    L35
L85:
  call   <imsm_orom_has_raid0@plt>
  jmp    L35
L148:
  call   <imsm_orom_has_raid1@plt>
  cmp    $0x2,%ebx
  sete   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  jmp    L35
L27:
  call   <imsm_orom_has_raid5@plt>
  cmp    $0x2,%ebx
  setg   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  jmp    L35
L117:
  mov    $0x1,%eax
L35:
  pop    %rbx
  ret
L104:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int is_raid_level_supported(const struct imsm_orom *orom, int level, int raiddisks) {
  if (level < 0 || level == 6 || level == 4)
    return 0;
  if (orom)
    switch (level) {
    case 0:
      return imsm_orom_has_raid0(orom);
    case 1:
      if (raiddisks > 2)
        return imsm_orom_has_raid1e(orom);
      return imsm_orom_has_raid1(orom) && raiddisks == 2;
    case 10:
      return imsm_orom_has_raid10(orom) && raiddisks == 4;
    case 5:
      return imsm_orom_has_raid5(orom) && raiddisks > 2;
    }
  else
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_raid_level_supported>:
  endbr64
  mov    %esi,%eax
  and    $0xfffffffd,%eax
  cmp    $0x4,%eax
  je     L73
  test   %esi,%esi
  js     L73
  test   %rdi,%rdi
  je     L32
  push   %rbx
  mov    %edx,%ebx
  cmp    $0x5,%esi
  je     L23
  jg     L63
  test   %esi,%esi
  je     L72
  cmp    $0x1,%esi
  jne    L50
  cmp    $0x2,%edx
  jle    L13
  pop    %rbx
  jmp    10c0 <imsm_orom_has_raid1e@plt>
  nopl   (%rax)
L73:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L63:
  cmp    $0xa,%esi
  jne    L50
  call   <imsm_orom_has_raid10@plt>
  cmp    $0x4,%ebx
  pop    %rbx
  sete   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
  nopl   0x0(%rax)
L32:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L50:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L72:
  pop    %rbx
  jmp    1090 <imsm_orom_has_raid0@plt>
  xchg   %ax,%ax
L23:
  call   <imsm_orom_has_raid5@plt>
  cmp    $0x2,%ebx
  pop    %rbx
  setg   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  call   <imsm_orom_has_raid1@plt>
  test   %eax,%eax
  setne  %dl
  xor    %eax,%eax
  cmp    $0x2,%ebx
  pop    %rbx
  sete   %al
  and    %edx,%eax
  ret
```

assistant: ```c
int is_raid_level_supported(const struct imsm_orom *orom, int level, int raiddisks) {
  if (level < 0 || level == 6 || level == 4)
    return 0;
  if (orom)
    switch (level) {
    case 0:
      return imsm_orom_has_raid0(orom);
    case 1:
      if (raiddisks > 2)
        return imsm_orom_has_raid1e(orom);
      return imsm_orom_has_raid1(orom) && raiddisks == 2;
    case 10:
      return imsm_orom_has_raid10(orom) && raiddisks == 4;
    case 5:
      return imsm_orom_has_raid5(orom) && raiddisks > 2;
    }
  else
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_raid_level_supported>:
  endbr64
  mov    %esi,%eax
  and    $0xfffffffd,%eax
  cmp    $0x4,%eax
  je     L68
  test   %esi,%esi
  js     L68
  test   %rdi,%rdi
  je     L85
  push   %rbx
  mov    %edx,%ebx
  cmp    $0x5,%esi
  je     L27
  jg     L81
  test   %esi,%esi
  je     L48
  cmp    $0x1,%esi
  jne    L49
  cmp    $0x2,%edx
  jle    L11
  pop    %rbx
  jmp    10c0 <imsm_orom_has_raid1e@plt>
  nopl   (%rax)
L68:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L81:
  cmp    $0xa,%esi
  jne    L49
  call   <imsm_orom_has_raid10@plt>
  cmp    $0x4,%ebx
  pop    %rbx
  sete   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
  nopl   0x0(%rax)
L85:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L49:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L48:
  pop    %rbx
  jmp    1090 <imsm_orom_has_raid0@plt>
  xchg   %ax,%ax
L27:
  call   <imsm_orom_has_raid5@plt>
  cmp    $0x2,%ebx
  pop    %rbx
  setg   %dl
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  call   <imsm_orom_has_raid1@plt>
  test   %eax,%eax
  setne  %dl
  xor    %eax,%eax
  cmp    $0x2,%ebx
  pop    %rbx
  sete   %al
  and    %edx,%eax
  ret
```

assistant: ```c
int is_raid_level_supported(const struct imsm_orom *orom, int level, int raiddisks) {
  if (level < 0 || level == 6 || level == 4)
    return 0;
  if (orom)
    switch (level) {
    case 0:
      return imsm_orom_has_raid0(orom);
    case 1:
      if (raiddisks > 2)
        return imsm_orom_has_raid1e(orom);
      return imsm_orom_has_raid1(orom) && raiddisks == 2;
    case 10:
      return imsm_orom_has_raid10(orom) && raiddisks == 4;
    case 5:
      return imsm_orom_has_raid5(orom) && raiddisks > 2;
    }
  else
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dev_to_ohci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <dev_get_drvdata@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <hcd_to_ohci@plt>
  leave
  ret
```

assistant: ```c
struct ohci_hcd *dev_to_ohci(struct device *dev) {
  struct usb_hcd *hcd = dev_get_drvdata(dev);
  return hcd_to_ohci(hcd);
}
```
user: Translate the below assembly code into C programming language:

```asm
<dev_to_ohci>:
  endbr64
  sub    $0x8,%rsp
  call   <dev_get_drvdata@plt>
  mov    %rax,%rdi
  call   <hcd_to_ohci@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
struct ohci_hcd *dev_to_ohci(struct device *dev) {
  struct usb_hcd *hcd = dev_get_drvdata(dev);
  return hcd_to_ohci(hcd);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dev_to_ohci>:
  endbr64
  sub    $0x8,%rsp
  call   <dev_get_drvdata@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <hcd_to_ohci@plt>
```

assistant: ```c
struct ohci_hcd *dev_to_ohci(struct device *dev) {
  struct usb_hcd *hcd = dev_get_drvdata(dev);
  return hcd_to_ohci(hcd);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dev_to_ohci>:
  endbr64
  sub    $0x8,%rsp
  call   <dev_get_drvdata@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <hcd_to_ohci@plt>
```

assistant: ```c
struct ohci_hcd *dev_to_ohci(struct device *dev) {
  struct usb_hcd *hcd = dev_get_drvdata(dev);
  return hcd_to_ohci(hcd);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L5:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  add    %ecx,%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L17:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  jbe    L5
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int getint(const char **p) {
  unsigned x;
  for (x = 0; **p - '0' < 10U; (*p)++)
    x = **p - '0' + 10 * x;
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getint>:
  endbr64
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  movsbl %al,%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%ecx
  ja     L23
  add    $0x1,%rdx
  mov    $0x0,%esi
L9:
  lea    (%rsi,%rsi,4),%ecx
  movsbl %al,%eax
  lea    -0x30(%rax,%rcx,2),%esi
  mov    %rdx,(%rdi)
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  movsbl %al,%ecx
  sub    $0x30,%ecx
  cmp    $0x9,%ecx
  jbe    L9
L6:
  mov    %esi,%eax
  ret
L23:
  mov    $0x0,%esi
  jmp    L6
```

assistant: ```c
int getint(const char **p) {
  unsigned x;
  for (x = 0; **p - '0' < 10U; (*p)++)
    x = **p - '0' + 10 * x;
  return x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getint>:
  endbr64
  mov    (%rdi),%rcx
  movsbl (%rcx),%edx
  lea    -0x30(%rdx),%eax
  cmp    $0x9,%eax
  ja     L16
  add    $0x1,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L10:
  lea    (%rax,%rax,4),%eax
  mov    %rcx,(%rdi)
  add    $0x1,%rcx
  lea    -0x30(%rdx,%rax,2),%eax
  movsbl -0x1(%rcx),%edx
  lea    -0x30(%rdx),%esi
  cmp    $0x9,%esi
  jbe    L10
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int getint(const char **p) {
  unsigned x;
  for (x = 0; **p - '0' < 10U; (*p)++)
    x = **p - '0' + 10 * x;
  return x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getint>:
  endbr64
  mov    (%rdi),%rcx
  movsbl (%rcx),%edx
  lea    -0x30(%rdx),%eax
  cmp    $0x9,%eax
  ja     L5
  add    $0x1,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L7:
  lea    (%rax,%rax,4),%eax
  mov    %rcx,(%rdi)
  add    $0x1,%rcx
  lea    -0x30(%rdx,%rax,2),%eax
  movsbl -0x1(%rcx),%edx
  lea    -0x30(%rdx),%esi
  cmp    $0x9,%esi
  jbe    L7
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  ret
```

assistant: ```c
int getint(const char **p) {
  unsigned x;
  for (x = 0; **p - '0' < 10U; (*p)++)
    x = **p - '0' + 10 * x;
  return x;
}
```
user: What is the c source code of the assembly code below:

```asm
<strtok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L70
  mov    -0x18(%rbp),%rax
  mov    %rax,D117(%rip)
L70:
  mov    D117(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L79
  mov    $0x0,%eax
  jmp    L42
L103:
  mov    D117(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D112(%rip)
L79:
  mov    D112(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L98
  mov    D112(%rip),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L103
L98:
  mov    D112(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    $0x0,%eax
  jmp    L42
L10:
  movq   $0x0,-0x10(%rbp)
  jmp    L40
L75:
  mov    D112(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D112(%rip)
  addq   $0x1,-0x10(%rbp)
L40:
  mov    D112(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L15
  mov    D112(%rip),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L75
L15:
  mov    D112(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L46
  mov    $0x0,%eax
  jmp    L42
L46:
  mov    D112(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    D112(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D112(%rip)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  neg    %rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
L42:
  leave
  ret
```

assistant: ```c
char *strtok(char *str, const char *delimiters) {
  static char *next_start;
  if (str != NULL)
    next_start = str;
  if (!*next_start)
    return NULL;
  while (*next_start && strchr(delimiters, *next_start) != NULL)
    ++next_start;
  if (!*next_start)
    return NULL;
  size_t length = 0;
  while (*next_start && !strchr(delimiters, *next_start)) {
    ++next_start;
    ++length;
  }
  if (!*next_start)
    return NULL;
  char *token = next_start;
  ++next_start;
  *token = '\0';
  return token - length;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strtok>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbp
  test   %rdi,%rdi
  je     L118
  mov    %rdi,D80(%rip)
L118:
  mov    D80(%rip),%rax
  movzbl (%rax),%esi
  test   %sil,%sil
  je     L28
L20:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L129
  mov    D80(%rip),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,D53(%rip)
  movzbl 0x1(%rax),%esi
  test   %sil,%sil
  jne    L20
  mov    $0x0,%eax
  jmp    L75
L129:
  mov    D53(%rip),%rax
  movzbl (%rax),%esi
  test   %sil,%sil
  je     L114
  mov    $0x0,%r12d
L58:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L49
  mov    D53(%rip),%rdx
  lea    0x1(%rdx),%rcx
  mov    %rcx,D53(%rip)
  add    $0x1,%r12
  movzbl 0x1(%rdx),%esi
  test   %sil,%sil
  jne    L58
  jmp    L75
L49:
  mov    D53(%rip),%rax
  cmpb   $0x0,(%rax)
  je     L3
  lea    0x1(%rax),%rdx
  mov    %rdx,D53(%rip)
  movb   $0x0,(%rax)
  sub    %r12,%rax
L75:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  mov    $0x0,%eax
  jmp    L75
L114:
  mov    %rbx,%rax
  jmp    L75
L3:
  mov    %rbx,%rax
  jmp    L75
```

assistant: ```c
char *strtok(char *str, const char *delimiters) {
  static char *next_start;
  if (str != NULL)
    next_start = str;
  if (!*next_start)
    return NULL;
  while (*next_start && strchr(delimiters, *next_start) != NULL)
    ++next_start;
  if (!*next_start)
    return NULL;
  size_t length = 0;
  while (*next_start && !strchr(delimiters, *next_start)) {
    ++next_start;
    ++length;
  }
  if (!*next_start)
    return NULL;
  char *token = next_start;
  ++next_start;
  *token = '\0';
  return token - length;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strtok>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L62
  mov    %rdi,D5(%rip)
L22:
  movsbl (%rdi),%esi
  test   %sil,%sil
  jne    L3
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L73:
  movsbl 0x1(%rax),%esi
  lea    0x1(%rax),%rdx
  mov    %rdx,D54(%rip)
  test   %sil,%sil
  je     L13
L3:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  mov    D54(%rip),%rax
  jne    L73
  movsbl (%rax),%esi
  test   %sil,%sil
  je     L13
  xor    %ebp,%ebp
  jmp    L56
  nopl   0x0(%rax)
L41:
  movsbl 0x1(%rax),%esi
  lea    0x1(%rax),%rdx
  add    $0x1,%rbp
  mov    %rdx,D54(%rip)
  test   %sil,%sil
  je     L13
L56:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  mov    D54(%rip),%rax
  je     L41
  cmpb   $0x0,(%rax)
  je     L13
  lea    0x1(%rax),%rdx
  movb   $0x0,(%rax)
  sub    %rbp,%rax
  mov    %rdx,D54(%rip)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L13:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L62:
  mov    D54(%rip),%rdi
  jmp    L22
```

assistant: ```c
char *strtok(char *str, const char *delimiters) {
  static char *next_start;
  if (str != NULL)
    next_start = str;
  if (!*next_start)
    return NULL;
  while (*next_start && strchr(delimiters, *next_start) != NULL)
    ++next_start;
  if (!*next_start)
    return NULL;
  size_t length = 0;
  while (*next_start && !strchr(delimiters, *next_start)) {
    ++next_start;
    ++length;
  }
  if (!*next_start)
    return NULL;
  char *token = next_start;
  ++next_start;
  *token = '\0';
  return token - length;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strtok>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L73
  mov    %rdi,D56(%rip)
L89:
  movsbl (%rdi),%esi
  test   %sil,%sil
  jne    L49
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L33:
  movsbl 0x1(%rax),%esi
  lea    0x1(%rax),%rdx
  mov    %rdx,D22(%rip)
  test   %sil,%sil
  je     L72
L49:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  mov    D22(%rip),%rax
  jne    L33
  movsbl (%rax),%esi
  test   %sil,%sil
  je     L72
  xor    %ebp,%ebp
  jmp    L109
  nopl   0x0(%rax)
L15:
  movsbl 0x1(%rax),%esi
  lea    0x1(%rax),%rdx
  add    $0x1,%rbp
  mov    %rdx,D22(%rip)
  test   %sil,%sil
  je     L72
L109:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  mov    D22(%rip),%rax
  je     L15
  cmpb   $0x0,(%rax)
  je     L72
  lea    0x1(%rax),%rdx
  movb   $0x0,(%rax)
  sub    %rbp,%rax
  mov    %rdx,D22(%rip)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L72:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L73:
  mov    D22(%rip),%rdi
  jmp    L89
```

assistant: ```c
char *strtok(char *str, const char *delimiters) {
  static char *next_start;
  if (str != NULL)
    next_start = str;
  if (!*next_start)
    return NULL;
  while (*next_start && strchr(delimiters, *next_start) != NULL)
    ++next_start;
  if (!*next_start)
    return NULL;
  size_t length = 0;
  while (*next_start && !strchr(delimiters, *next_start)) {
    ++next_start;
    ++length;
  }
  if (!*next_start)
    return NULL;
  char *token = next_start;
  ++next_start;
  *token = '\0';
  return token - length;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<user_flush_rx_tx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <user_flush_rx@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <user_flush_tx@plt>
  nop
  leave
  ret
```

assistant: ```c
void user_flush_rx_tx(struct socket *sock) {
  user_flush_rx(sock);
  user_flush_tx(sock);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<user_flush_rx_tx>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <user_flush_rx@plt>
  mov    %rbx,%rdi
  call   <user_flush_tx@plt>
  pop    %rbx
  ret
```

assistant: ```c
void user_flush_rx_tx(struct socket *sock) {
  user_flush_rx(sock);
  user_flush_tx(sock);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<user_flush_rx_tx>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <user_flush_rx@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <user_flush_tx@plt>
```

assistant: ```c
void user_flush_rx_tx(struct socket *sock) {
  user_flush_rx(sock);
  user_flush_tx(sock);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<user_flush_rx_tx>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <user_flush_rx@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <user_flush_tx@plt>
```

assistant: ```c
void user_flush_rx_tx(struct socket *sock) {
  user_flush_rx(sock);
  user_flush_tx(sock);
}
```