ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<wots_gen_pk>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r14
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %rdi,%rbx
  mov    $0x0,%ebp
L6:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <set_chain_addr@plt>
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <wots_gen_sk@plt>
  mov    %r12,%r9
  mov    %r13,%r8
  mov    $0xf,%ecx
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <gen_chain@plt>
  add    $0x1,%ebp
  add    $0x10,%rbx
  cmp    $0x23,%ebp
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void wots_gen_pk(unsigned char *pk, const unsigned char *sk_seed, const unsigned char *pub_seed, uint32_t addr[8]) {
  uint32_t i;
  for (i = 0; i < ((8 * 16 / 4) + 3); i++) {
    set_chain_addr(addr, i);
    wots_gen_sk(pk + i * 16, sk_seed, addr);
    gen_chain(pk + i * 16, pk + i * 16, 0, 16 - 1, pub_seed, addr);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<wots_gen_pk>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %ebp,%esi
  mov    %r12,%rdi
  add    $0x1,%ebp
  call   <set_chain_addr@plt>
  mov    %rbx,%rdi
  mov    %r12,%rdx
  mov    %r14,%rsi
  call   <wots_gen_sk@plt>
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  mov    %r12,%r9
  mov    %r13,%r8
  mov    $0xf,%ecx
  xor    %edx,%edx
  add    $0x10,%rbx
  call   <gen_chain@plt>
  cmp    $0x23,%ebp
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void wots_gen_pk(unsigned char *pk, const unsigned char *sk_seed, const unsigned char *pub_seed, uint32_t addr[8]) {
  uint32_t i;
  for (i = 0; i < ((8 * 16 / 4) + 3); i++) {
    set_chain_addr(addr, i);
    wots_gen_sk(pk + i * 16, sk_seed, addr);
    gen_chain(pk + i * 16, pk + i * 16, 0, 16 - 1, pub_seed, addr);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<wots_gen_pk>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %ebp,%esi
  mov    %r12,%rdi
  add    $0x1,%ebp
  call   <set_chain_addr@plt>
  mov    %rbx,%rdi
  mov    %r12,%rdx
  mov    %r14,%rsi
  call   <wots_gen_sk@plt>
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  mov    %r12,%r9
  mov    %r13,%r8
  mov    $0xf,%ecx
  xor    %edx,%edx
  add    $0x10,%rbx
  call   <gen_chain@plt>
  cmp    $0x23,%ebp
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void wots_gen_pk(unsigned char *pk, const unsigned char *sk_seed, const unsigned char *pub_seed, uint32_t addr[8]) {
  uint32_t i;
  for (i = 0; i < ((8 * 16 / 4) + 3); i++) {
    set_chain_addr(addr, i);
    wots_gen_sk(pk + i * 16, sk_seed, addr);
    gen_chain(pk + i * 16, pk + i * 16, 0, 16 - 1, pub_seed, addr);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<htab_djb_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L11
L7:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  subq   $0x1,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  xor    %rdx,%rax
  mov    %rax,-0x18(%rbp)
L11:
  cmpq   $0x0,-0x28(%rbp)
  jne    L7
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long htab_djb_hash(unsigned long hash, const void *_buf, size_t buflen) {
  const unsigned char *buf = _buf;
  int c;
  while (buflen > 0) {
    c = *buf++;
    buflen--;
    hash = ((hash << 5) + hash) ^ c;
  }
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<htab_djb_hash>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L9
  add    %rsi,%rdx
L17:
  add    $0x1,%rsi
  mov    %rax,%rcx
  shl    $0x5,%rcx
  add    %rcx,%rax
  movzbl -0x1(%rsi),%ecx
  xor    %rcx,%rax
  cmp    %rsi,%rdx
  jne    L17
L9:
  ret
```

assistant: ```c
unsigned long htab_djb_hash(unsigned long hash, const void *_buf, size_t buflen) {
  const unsigned char *buf = _buf;
  int c;
  while (buflen > 0) {
    c = *buf++;
    buflen--;
    hash = ((hash << 5) + hash) ^ c;
  }
  return hash;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<htab_djb_hash>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L13
  add    %rsi,%rdx
  nop
L0:
  mov    %rax,%rcx
  add    $0x1,%rsi
  shl    $0x5,%rcx
  add    %rcx,%rax
  movzbl -0x1(%rsi),%ecx
  xor    %rcx,%rax
  cmp    %rdx,%rsi
  jne    L0
L13:
  ret
```

assistant: ```c
unsigned long htab_djb_hash(unsigned long hash, const void *_buf, size_t buflen) {
  const unsigned char *buf = _buf;
  int c;
  while (buflen > 0) {
    c = *buf++;
    buflen--;
    hash = ((hash << 5) + hash) ^ c;
  }
  return hash;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<htab_djb_hash>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L8
  add    %rsi,%rdx
  nop
L19:
  mov    %rax,%rcx
  add    $0x1,%rsi
  shl    $0x5,%rcx
  add    %rcx,%rax
  movzbl -0x1(%rsi),%ecx
  xor    %rcx,%rax
  cmp    %rdx,%rsi
  jne    L19
L8:
  ret
```

assistant: ```c
unsigned long htab_djb_hash(unsigned long hash, const void *_buf, size_t buflen) {
  const unsigned char *buf = _buf;
  int c;
  while (buflen > 0) {
    c = *buf++;
    buflen--;
    hash = ((hash << 5) + hash) ^ c;
  }
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<titleToNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L3
L5:
  subl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x40,%eax
  imul   -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  imul   $0x1a,%eax,%eax
  mov    %eax,-0x4(%rbp)
L3:
  cmpl   $0x0,-0xc(%rbp)
  jne    L5
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int titleToNumber(char *s) {
  int i = strlen(s);
  int ret = 0;
  int n = 1;
  while (i) {
    i--;
    ret += (*(s + i) - 'A' + 1) * n;
    n *= 26;
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<titleToNumber>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L7
  movslq %eax,%rdx
  lea    -0x1(%rbx,%rdx,1),%rcx
  lea    -0x2(%rbx,%rdx,1),%rdi
  lea    -0x1(%rax),%eax
  sub    %rax,%rdi
  mov    $0x1,%esi
  mov    $0x0,%eax
L1:
  movsbl (%rcx),%edx
  sub    $0x40,%edx
  imul   %esi,%edx
  add    %edx,%eax
  imul   $0x1a,%esi,%esi
  sub    $0x1,%rcx
  cmp    %rdi,%rcx
  jne    L1
L7:
  pop    %rbx
  ret
```

assistant: ```c
int titleToNumber(char *s) {
  int i = strlen(s);
  int ret = 0;
  int n = 1;
  while (i) {
    i--;
    ret += (*(s + i) - 'A' + 1) * n;
    n *= 26;
  }
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<titleToNumber>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L5
  movslq %eax,%rdx
  sub    $0x1,%eax
  mov    $0x1,%esi
  lea    -0x2(%rbx,%rdx,1),%rdi
  lea    -0x1(%rbx,%rdx,1),%rcx
  sub    %rax,%rdi
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L10:
  movsbl (%rcx),%edx
  sub    $0x1,%rcx
  sub    $0x40,%edx
  imul   %esi,%edx
  imul   $0x1a,%esi,%esi
  add    %edx,%eax
  cmp    %rdi,%rcx
  jne    L10
L5:
  pop    %rbx
  ret
```

assistant: ```c
int titleToNumber(char *s) {
  int i = strlen(s);
  int ret = 0;
  int n = 1;
  while (i) {
    i--;
    ret += (*(s + i) - 'A' + 1) * n;
    n *= 26;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<titleToNumber>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L1
  movslq %eax,%rdx
  sub    $0x1,%eax
  mov    $0x1,%esi
  lea    -0x2(%rbx,%rdx,1),%rdi
  lea    -0x1(%rbx,%rdx,1),%rcx
  sub    %rax,%rdi
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L9:
  movsbl (%rcx),%edx
  sub    $0x1,%rcx
  sub    $0x40,%edx
  imul   %esi,%edx
  imul   $0x1a,%esi,%esi
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L9
L1:
  pop    %rbx
  ret
```

assistant: ```c
int titleToNumber(char *s) {
  int i = strlen(s);
  int ret = 0;
  int n = 1;
  while (i) {
    i--;
    ret += (*(s + i) - 'A' + 1) * n;
    n *= 26;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<MergeSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L18
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x4(%rbp),%eax
  lea    (%rdx,%rax,1),%edi
  mov    -0x24(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <MergeSort>
  mov    -0x1c(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  lea    0x1(%rax),%esi
  mov    -0x24(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <MergeSort>
  mov    -0x1c(%rbp),%edx
  mov    -0x4(%rbp),%eax
  lea    (%rdx,%rax,1),%edi
  mov    -0x24(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <Merge@plt>
  jmp    L13
L18:
  nop
L13:
  leave
  ret
```

assistant: ```c
void MergeSort(double **data, int first, int last, int col) {
  if (first >= last)
    return;
  int mid = (last - first) / 2;
  MergeSort(data, first, first + mid, col);
  MergeSort(data, first + mid + 1, last, col);
  Merge(data, first, first + mid, last, col);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MergeSort>:
  endbr64
  cmp    %edx,%esi
  jl     L7
  ret
L7:
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%ebx
  mov    %edx,%ebp
  mov    %ecx,%r14d
  mov    %edx,%eax
  sub    %esi,%eax
  mov    %eax,%r12d
  shr    $0x1f,%r12d
  add    %eax,%r12d
  sar    %r12d
  add    %esi,%r12d
  mov    %r12d,%edx
  call   <MergeSort>
  lea    0x1(%r12),%esi
  mov    %r14d,%ecx
  mov    %ebp,%edx
  mov    %r13,%rdi
  call   <MergeSort>
  mov    %r14d,%r8d
  mov    %ebp,%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <Merge@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void MergeSort(double **data, int first, int last, int col) {
  if (first >= last)
    return;
  int mid = (last - first) / 2;
  MergeSort(data, first, first + mid, col);
  MergeSort(data, first + mid + 1, last, col);
  Merge(data, first, first + mid, last, col);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<MergeSort>:
  endbr64
  cmp    %edx,%esi
  jl     L9
  ret
  nopl   0x0(%rax)
L9:
  push   %r14
  mov    %edx,%r14d
  sub    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  sar    %r14d
  mov    %edx,%r12d
  push   %rbp
  add    %esi,%r14d
  mov    %esi,%ebp
  push   %rbx
  mov    %r14d,%edx
  mov    %ecx,%ebx
  call   <MergeSort>
  lea    0x1(%r14),%esi
  mov    %ebx,%ecx
  mov    %r12d,%edx
  mov    %r13,%rdi
  call   <MergeSort>
  mov    %ebx,%r8d
  mov    %r12d,%ecx
  pop    %rbx
  mov    %r14d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <Merge@plt>
```

assistant: ```c
void MergeSort(double **data, int first, int last, int col) {
  if (first >= last)
    return;
  int mid = (last - first) / 2;
  MergeSort(data, first, first + mid, col);
  MergeSort(data, first + mid + 1, last, col);
  Merge(data, first, first + mid, last, col);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<str2int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L48
L39:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L8
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L4
L8:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  jmp    L15
L4:
  addq   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
L15:
  leave
  ret
```

assistant: ```c
int str2int(char *s, int *value) {
  char *start;
  for (start = s; *start; start++) {
    if (*start < '0' || *start > '9')
      return *value;
  }
  *value = atoi(s);
  return *value;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<str2int>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L31
  mov    %rdi,%rdx
L24:
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L27
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L24
L31:
  call   <atoi@plt>
  mov    %eax,(%rbx)
L22:
  pop    %rbx
  ret
L27:
  mov    (%rbx),%eax
  jmp    L22
```

assistant: ```c
int str2int(char *s, int *value) {
  char *start;
  for (start = s; *start; start++) {
    if (*start < '0' || *start > '9')
      return *value;
  }
  *value = atoi(s);
  return *value;
}
```
user: What is the c source code of the assembly code below:

```asm
<str2int>:
  endbr64
  movzbl (%rdi),%eax
  push   %rbx
  mov    %rsi,%rbx
  test   %al,%al
  je     L6
  mov    %rdi,%rdx
  jmp    L11
  nopl   0x0(%rax)
L12:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L6
L11:
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L12
  mov    (%rbx),%eax
  pop    %rbx
  ret
  nop
L6:
  call   <atoi@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int str2int(char *s, int *value) {
  char *start;
  for (start = s; *start; start++) {
    if (*start < '0' || *start > '9')
      return *value;
  }
  *value = atoi(s);
  return *value;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<str2int>:
  endbr64
  movzbl (%rdi),%eax
  push   %rbx
  mov    %rsi,%rbx
  test   %al,%al
  je     L17
  mov    %rdi,%rdx
  jmp    L26
  nopl   0x0(%rax)
L32:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L17
L26:
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L32
  mov    (%rbx),%eax
  pop    %rbx
  ret
  nop
L17:
  call   <atoi@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int str2int(char *s, int *value) {
  char *start;
  for (start = s; *start; start++) {
    if (*start < '0' || *start > '9')
      return *value;
  }
  *value = atoi(s);
  return *value;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<SetBackLight>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x60,%edi
  call   <SendChar@plt>
  movzbl -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  nop
  leave
  ret
```

assistant: ```c
void SetBackLight(unsigned char light_level) {
  SendChar((0XEE) & 0xFF);
  SendChar((0x60) & 0xFF);
  SendChar((light_level)&0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: What is the c source code of the assembly code below?

```asm
<SetBackLight>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x60,%edi
  call   <SendChar@plt>
  movzbl %bl,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  pop    %rbx
  ret
```

assistant: ```c
void SetBackLight(unsigned char light_level) {
  SendChar((0XEE) & 0xFF);
  SendChar((0x60) & 0xFF);
  SendChar((light_level)&0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: Translate the below assembly code into C programming language:

```asm
<SetBackLight>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x60,%edi
  call   <SendChar@plt>
  movzbl %bl,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  pop    %rbx
  jmp    1050 <SendChar@plt>
```

assistant: ```c
void SetBackLight(unsigned char light_level) {
  SendChar((0XEE) & 0xFF);
  SendChar((0x60) & 0xFF);
  SendChar((light_level)&0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<SetBackLight>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xee,%edi
  call   <SendChar@plt>
  mov    $0x60,%edi
  call   <SendChar@plt>
  movzbl %bl,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xfc,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  call   <SendChar@plt>
  mov    $0xff,%edi
  pop    %rbx
  jmp    1050 <SendChar@plt>
```

assistant: ```c
void SetBackLight(unsigned char light_level) {
  SendChar((0XEE) & 0xFF);
  SendChar((0x60) & 0xFF);
  SendChar((light_level)&0xFF);
  SendChar((((0XFFFCFFFF) >> 16) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) >> 16) & 0xFF);
  SendChar((((0XFFFCFFFF) & 0xFFFF) >> 8) & 0xFF);
  SendChar(((0XFFFCFFFF) & 0xFFFF) & 0xFF);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readlines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x410,%rsp
  mov    %rdi,-0x408(%rbp)
  mov    %esi,-0x40c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x400(%rbp)
  jmp    L53
L1:
  mov    -0x400(%rbp),%eax
  cmp    -0x40c(%rbp),%eax
  jg     L21
  mov    -0x3fc(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x3f8(%rbp)
  cmpq   $0x0,-0x3f8(%rbp)
  jne    L50
L21:
  mov    $0xffffffff,%eax
  jmp    L34
L50:
  mov    -0x3fc(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movb   $0x0,-0x3f0(%rbp,%rax,1)
  lea    -0x3f0(%rbp),%rdx
  mov    -0x3f8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x400(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x400(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x408(%rbp),%rax
  add    %rax,%rdx
  mov    -0x3f8(%rbp),%rax
  mov    %rax,(%rdx)
L53:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <_getline@plt>
  mov    %eax,-0x3fc(%rbp)
  cmpl   $0x0,-0x3fc(%rbp)
  jg     L1
  mov    -0x400(%rbp),%eax
L34:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
int readlines(char *lineptr[], int maxlines) {
  char line[1000], *p;
  int len, nlines = 0;
  while ((len = _getline(line, 1000)) > 0) {
    if (nlines > maxlines || (p = malloc(len)) == NULL) {
      return -1;
    } else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
    }
  }
  return nlines;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readlines>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %rdi,%r15
  mov    %esi,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r12d
  mov    %rsp,%r13
  jmp    L35
L44:
  sub    $0x1,%ebx
  movslq %ebx,%rbx
  movb   $0x0,(%rsp,%rbx,1)
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbp,(%r15,%r12,8)
  add    $0x1,%r12
L35:
  mov    %r12d,%ebp
  mov    $0x3e8,%esi
  mov    %r13,%rdi
  call   <_getline@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L7
  cmp    %r12d,%r14d
  jl     L45
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L44
  mov    $0xffffffff,%ebp
  jmp    L7
L45:
  mov    $0xffffffff,%ebp
L7:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    %ebp,%eax
  add    $0x3f8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readlines(char *lineptr[], int maxlines) {
  char line[1000], *p;
  int len, nlines = 0;
  while ((len = _getline(line, 1000)) > 0) {
    if (nlines > maxlines || (p = malloc(len)) == NULL) {
      return -1;
    } else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
    }
  }
  return nlines;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<readlines>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  jmp    L23
  nopl   (%rax)
L0:
  cmp    %ebp,%r13d
  jl     L37
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L37
  sub    $0x1,%ebx
  mov    %r12,%rsi
  mov    %rax,%rdi
  movslq %ebx,%rbx
  movb   $0x0,(%rsp,%rbx,1)
  call   <strcpy@plt>
  mov    %r15,(%r14,%rbp,8)
  add    $0x1,%rbp
L23:
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  mov    %ebp,%r15d
  call   <_getline@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L0
L51:
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x3f8,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  mov    $0xffffffff,%r15d
  jmp    L51
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readlines(char *lineptr[], int maxlines) {
  char line[1000], *p;
  int len, nlines = 0;
  while ((len = _getline(line, 1000)) > 0) {
    if (nlines > maxlines || (p = malloc(len)) == NULL) {
      return -1;
    } else {
      line[len - 1] = '\0';
      strcpy(p, line);
      lineptr[nlines++] = p;
    }
  }
  return nlines;
}
```