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
<GetMcGainReg>:
  endbr64
  xor    %eax,%eax
  cmp    $0x117f,%di
  jle    L10
  movswl %di,%edi
  mov    $0x1,%eax
  sub    $0xffffff80,%edi
  mov    %edi,%edx
  sar    $0x8,%edx
  cmp    $0x16ff,%edi
  jg     L8
L10:
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  cmp    $0x1b,%dx
  setg   %al
  add    $0x2,%eax
  ret
```

assistant: ```c
signed short GetMcGainReg(signed short swVol) {
  signed short swGain = (swVol + 128) / 256;
  signed short swRet = 3;
  if (swGain < 18) {
    swRet = 0;
  } else if (swGain < 23) {
    swRet = 1;
  } else if (swGain < 28) {
    swRet = 2;
  } else {
  }
  return swRet;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<GetMcGainReg>:
  endbr64
  xor    %eax,%eax
  cmp    $0x117f,%di
  jle    L19
  movswl %di,%edi
  mov    $0x1,%eax
  sub    $0xffffff80,%edi
  mov    %edi,%edx
  sar    $0x8,%edx
  cmp    $0x16ff,%edi
  jg     L2
L19:
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  cmp    $0x1b,%dx
  setg   %al
  add    $0x2,%eax
  ret
```

assistant: ```c
signed short GetMcGainReg(signed short swVol) {
  signed short swGain = (swVol + 128) / 256;
  signed short swRet = 3;
  if (swGain < 18) {
    swRet = 0;
  } else if (swGain < 23) {
    swRet = 1;
  } else if (swGain < 28) {
    swRet = 2;
  } else {
  }
  return swRet;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xfff,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  call   <Mem_Dump@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  assert(Mem_Alloc(4095) == NULL);
  Mem_Dump();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0xfff,%edi
  call   <Mem_Alloc@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  call   <Mem_Dump@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  assert(Mem_Alloc(4095) == NULL);
  Mem_Dump();
  exit(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0xfff,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <Mem_Dump@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  assert(Mem_Alloc(4095) == NULL);
  Mem_Dump();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0xfff,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <Mem_Dump@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  assert(Mem_Alloc(4095) == NULL);
  Mem_Dump();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

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

assistant: ```c
void CRTFree(void *mm_context, void *ptr) {
  (void)(1 ? (void)0 : ((void)mm_context));
  free(ptr);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<CRTFree>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void CRTFree(void *mm_context, void *ptr) {
  (void)(1 ? (void)0 : ((void)mm_context));
  free(ptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CRTFree>:
  endbr64
  mov    %rsi,%rdi
  jmp    1050 <free@plt>
```

assistant: ```c
void CRTFree(void *mm_context, void *ptr) {
  (void)(1 ? (void)0 : ((void)mm_context));
  free(ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<CRTFree>:
  endbr64
  mov    %rsi,%rdi
  jmp    1050 <free@plt>
```

assistant: ```c
void CRTFree(void *mm_context, void *ptr) {
  (void)(1 ? (void)0 : ((void)mm_context));
  free(ptr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <left@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <left@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <left@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <left@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <right@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <right@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <right@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <right@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <leftp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <leftp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <leftp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <leftp@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <rightp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <rightp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <rightp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <rightp@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main() {
  left(2, 1);
  left(23, 4);
  left(5, 10);
  left(237, 7);
  right(2, 1);
  right(23, 4);
  right(5, 10);
  right(237, 7);
  leftp(2, 1);
  leftp(23, 4);
  leftp(5, 10);
  leftp(237, 7);
  rightp(2, 1);
  rightp(23, 4);
  rightp(5, 10);
  rightp(237, 7);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <left@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <left@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <left@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <left@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <right@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <right@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <right@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <right@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <leftp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <leftp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <leftp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <leftp@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <rightp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <rightp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <rightp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <rightp@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void main() {
  left(2, 1);
  left(23, 4);
  left(5, 10);
  left(237, 7);
  right(2, 1);
  right(23, 4);
  right(5, 10);
  right(237, 7);
  leftp(2, 1);
  leftp(23, 4);
  leftp(5, 10);
  leftp(237, 7);
  rightp(2, 1);
  rightp(23, 4);
  rightp(5, 10);
  rightp(237, 7);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <left@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <left@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <left@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <left@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <right@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <right@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <right@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <right@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <leftp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <leftp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <leftp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <leftp@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <rightp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <rightp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <rightp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  add    $0x8,%rsp
  jmp    10b0 <rightp@plt>
```

assistant: ```c
void main() {
  left(2, 1);
  left(23, 4);
  left(5, 10);
  left(237, 7);
  right(2, 1);
  right(23, 4);
  right(5, 10);
  right(237, 7);
  leftp(2, 1);
  leftp(23, 4);
  leftp(5, 10);
  leftp(237, 7);
  rightp(2, 1);
  rightp(23, 4);
  rightp(5, 10);
  rightp(237, 7);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <left@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <left@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <left@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <left@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <right@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <right@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <right@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <right@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <leftp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <leftp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <leftp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  call   <leftp@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <rightp@plt>
  mov    $0x4,%esi
  mov    $0x17,%edi
  call   <rightp@plt>
  mov    $0xa,%esi
  mov    $0x5,%edi
  call   <rightp@plt>
  mov    $0x7,%esi
  mov    $0xed,%edi
  add    $0x8,%rsp
  jmp    10b0 <rightp@plt>
```

assistant: ```c
void main() {
  left(2, 1);
  left(23, 4);
  left(5, 10);
  left(237, 7);
  right(2, 1);
  right(23, 4);
  right(5, 10);
  right(237, 7);
  leftp(2, 1);
  leftp(23, 4);
  leftp(5, 10);
  leftp(237, 7);
  rightp(2, 1);
  rightp(23, 4);
  rightp(5, 10);
  rightp(237, 7);
}
```
user: What is the c source code of the assembly code below:

```asm
<msclock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  cmp    $0x1,%eax
  jne    L10
  mov    $0xffffffff,%eax
  jmp    L16
L10:
  mov    -0x18(%rbp),%rax
  imul   $0xf4240,%rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
L16:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
clock_t msclock(void) {
  long long int usecs;
  struct timeval tv;
  if (gettimeofday(&tv, NULL) == 1)
    return (-1);
  usecs = ((long long int)tv.tv_sec) * ((long long int)1000000) + tv.tv_usec;
  return ((clock_t)(usecs / (1000000 / 1000)));
}
```
user: What is the c source code of the assembly code below:

```asm
<msclock>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  cmp    $0x1,%eax
  je     L23
  imul   $0xf4240,0x8(%rsp),%rcx
  add    (%rsp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  sar    $0x3f,%rcx
  sub    %ecx,%eax
L21:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L14
  add    $0x28,%rsp
  ret
L23:
  mov    $0xffffffff,%eax
  jmp    L21
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
clock_t msclock(void) {
  long long int usecs;
  struct timeval tv;
  if (gettimeofday(&tv, NULL) == 1)
    return (-1);
  usecs = ((long long int)tv.tv_sec) * ((long long int)1000000) + tv.tv_usec;
  return ((clock_t)(usecs / (1000000 / 1000)));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<msclock>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  cmp    $0x1,%eax
  je     L25
  imul   $0xf4240,0x8(%rsp),%rcx
  add    (%rsp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rcx
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  sub    %ecx,%eax
L2:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x28,%rsp
  ret
L25:
  mov    $0xffffffff,%eax
  jmp    L2
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
clock_t msclock(void) {
  long long int usecs;
  struct timeval tv;
  if (gettimeofday(&tv, NULL) == 1)
    return (-1);
  usecs = ((long long int)tv.tv_sec) * ((long long int)1000000) + tv.tv_usec;
  return ((clock_t)(usecs / (1000000 / 1000)));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<msclock>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  cmp    $0x1,%eax
  je     L16
  imul   $0xf4240,0x8(%rsp),%rcx
  add    (%rsp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rcx
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  sub    %ecx,%eax
L27:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L13
  add    $0x28,%rsp
  ret
L16:
  mov    $0xffffffff,%eax
  jmp    L27
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
clock_t msclock(void) {
  long long int usecs;
  struct timeval tv;
  if (gettimeofday(&tv, NULL) == 1)
    return (-1);
  usecs = ((long long int)tv.tv_sec) * ((long long int)1000000) + tv.tv_usec;
  return ((clock_t)(usecs / (1000000 / 1000)));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  nop
L0:
  mov    -0x20(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strcpy(char *dst, const char *src) {
  char *ret = dst;
  while ((*dst++ = *src++))
    ;
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  mov    $0x0,%edx
L6:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L6
  ret
```

assistant: ```c
char *strcpy(char *dst, const char *src) {
  char *ret = dst;
  while ((*dst++ = *src++))
    ;
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L2:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L2
  ret
```

assistant: ```c
char *strcpy(char *dst, const char *src) {
  char *ret = dst;
  while ((*dst++ = *src++))
    ;
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L1:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L1
  ret
```

assistant: ```c
char *strcpy(char *dst, const char *src) {
  char *ret = dst;
  while ((*dst++ = *src++))
    ;
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<variance>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L19
L13:
  movl   $0x0,-0x24(%rbp)
  jmp    L21
L8:
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x38(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  add    %eax,-0x18(%rbp)
  addl   $0x1,-0x24(%rbp)
L21:
  cmpl   $0xf,-0x24(%rbp)
  jle    L8
  mov    -0x3c(%rbp),%eax
  cltq
  sub    $0x10,%rax
  add    %rax,-0x38(%rbp)
  addl   $0x1,-0x20(%rbp)
L19:
  cmpl   $0xf,-0x20(%rbp)
  jle    L13
  mov    -0x1c(%rbp),%eax
  imul   %eax,%eax
  shr    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  sub    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int variance(unsigned char *p, int lx) {
  int i;
  int j;
  unsigned int v;
  unsigned int s;
  unsigned int s2;
  unsigned int sa;
  unsigned int s2a;
  unsigned char *tmp;
  {
    s2a = 0U;
    sa = s2a;
    s2 = 0U;
    s = s2;
    j = 0;
    while (j < 16) {
      i = 0;
      while (i < 16) {
        tmp = p;
        p++;
        v = (unsigned int)*tmp;
        s += v;
        s2 += v * v;
        i++;
      }
      p += lx - 16;
      j++;
    }
    return ((int)(s2 - (s * s) / 256U));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<variance>:
  endbr64
  movslq %esi,%r9
  add    $0x10,%rdi
  mov    $0x10,%r8d
  mov    $0x0,%eax
  mov    $0x0,%esi
L2:
  lea    -0x10(%rdi),%rcx
L14:
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%edx
  add    %edx,%esi
  imul   %edx,%edx
  add    %edx,%eax
  cmp    %rdi,%rcx
  jne    L14
  add    %r9,%rdi
  sub    $0x1,%r8d
  jne    L2
  imul   %esi,%esi
  shr    $0x8,%esi
  sub    %esi,%eax
  ret
```

assistant: ```c
int variance(unsigned char *p, int lx) {
  int i;
  int j;
  unsigned int v;
  unsigned int s;
  unsigned int s2;
  unsigned int sa;
  unsigned int s2a;
  unsigned char *tmp;
  {
    s2a = 0U;
    sa = s2a;
    s2 = 0U;
    s = s2;
    j = 0;
    while (j < 16) {
      i = 0;
      while (i < 16) {
        tmp = p;
        p++;
        v = (unsigned int)*tmp;
        s += v;
        s2 += v * v;
        i++;
      }
      p += lx - 16;
      j++;
    }
    return ((int)(s2 - (s * s) / 256U));
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<variance>:
  endbr64
  movslq %esi,%r9
  add    $0x10,%rdi
  mov    $0x10,%r8d
  xor    %eax,%eax
  xor    %esi,%esi
  nopl   (%rax)
L19:
  lea    -0x10(%rdi),%rcx
  nopl   0x0(%rax)
L13:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  add    %edx,%esi
  imul   %edx,%edx
  add    %edx,%eax
  cmp    %rdi,%rcx
  jne    L13
  lea    (%rcx,%r9,1),%rdi
  sub    $0x1,%r8d
  jne    L19
  imul   %esi,%esi
  shr    $0x8,%esi
  sub    %esi,%eax
  ret
```

assistant: ```c
int variance(unsigned char *p, int lx) {
  int i;
  int j;
  unsigned int v;
  unsigned int s;
  unsigned int s2;
  unsigned int sa;
  unsigned int s2a;
  unsigned char *tmp;
  {
    s2a = 0U;
    sa = s2a;
    s2 = 0U;
    s = s2;
    j = 0;
    while (j < 16) {
      i = 0;
      while (i < 16) {
        tmp = p;
        p++;
        v = (unsigned int)*tmp;
        s += v;
        s2 += v * v;
        i++;
      }
      p += lx - 16;
      j++;
    }
    return ((int)(s2 - (s * s) / 256U));
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<variance>:
  endbr64
  push   %r15
  movslq %esi,%rax
  mov    %rdi,%r9
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rax,-0x10(%rsp)
  movl   $0x10,-0x4(%rsp)
  movl   $0x0,-0x8(%rsp)
  movl   $0x0,-0x14(%rsp)
  nopl   0x0(%rax)
L6:
  movzbl (%r9),%r10d
  movzbl 0x1(%r9),%r15d
  movzbl 0x8(%r9),%eax
  movzbl 0x2(%r9),%r8d
  movzbl 0x3(%r9),%edi
  movzbl 0x4(%r9),%esi
  movzbl 0x5(%r9),%ecx
  movzbl 0x6(%r9),%edx
  mov    %eax,-0x20(%rsp)
  lea    (%r10,%r15,1),%eax
  add    %r8d,%eax
  imul   %r15d,%r15d
  movzbl 0x7(%r9),%r14d
  movzbl 0x9(%r9),%ebx
  add    %edi,%eax
  imul   %r10d,%r10d
  movzbl 0xa(%r9),%r11d
  movzbl 0xb(%r9),%r13d
  add    %esi,%eax
  imul   %r8d,%r8d
  mov    %ebx,-0x1c(%rsp)
  movzbl 0xc(%r9),%r12d
  add    %ecx,%eax
  imul   %edi,%edi
  mov    %r11d,-0x18(%rsp)
  movzbl 0xd(%r9),%ebp
  add    %edx,%eax
  imul   %esi,%esi
  add    %r15d,%r10d
  movzbl 0xe(%r9),%ebx
  add    %r14d,%eax
  add    -0x20(%rsp),%eax
  add    -0x1c(%rsp),%eax
  add    %r10d,%r8d
  add    -0x18(%rsp),%eax
  imul   %ecx,%ecx
  add    %r8d,%edi
  movzbl 0xf(%r9),%r11d
  add    %r13d,%eax
  imul   %edx,%edx
  add    %edi,%esi
  add    %r12d,%eax
  imul   %r14d,%r14d
  add    %ebp,%eax
  add    %esi,%ecx
  add    %ebx,%eax
  add    %r11d,%eax
  add    %eax,-0x14(%rsp)
  lea    (%rcx,%rdx,1),%eax
  mov    -0x20(%rsp),%edx
  add    %r14d,%eax
  imul   %edx,%edx
  add    %edx,%eax
  mov    -0x1c(%rsp),%edx
  imul   %edx,%edx
  add    %edx,%eax
  mov    -0x18(%rsp),%edx
  imul   %r13d,%r13d
  add    -0x10(%rsp),%r9
  imul   %r12d,%r12d
  imul   %edx,%edx
  imul   %ebp,%ebp
  imul   %ebx,%ebx
  add    %edx,%eax
  imul   %r11d,%r11d
  add    %eax,%r13d
  add    %r13d,%r12d
  add    %r12d,%ebp
  add    %ebp,%ebx
  add    %ebx,%r11d
  add    %r11d,-0x8(%rsp)
  subl   $0x1,-0x4(%rsp)
  jne    L6
  mov    -0x14(%rsp),%edx
  mov    -0x8(%rsp),%eax
  pop    %rbx
  pop    %rbp
  imul   %edx,%edx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  shr    $0x8,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int variance(unsigned char *p, int lx) {
  int i;
  int j;
  unsigned int v;
  unsigned int s;
  unsigned int s2;
  unsigned int sa;
  unsigned int s2a;
  unsigned char *tmp;
  {
    s2a = 0U;
    sa = s2a;
    s2 = 0U;
    s = s2;
    j = 0;
    while (j < 16) {
      i = 0;
      while (i < 16) {
        tmp = p;
        p++;
        v = (unsigned int)*tmp;
        s += v;
        s2 += v * v;
        i++;
      }
      p += lx - 16;
      j++;
    }
    return ((int)(s2 - (s * s) / 256U));
  }
}
```