ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_dup>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <parse_dup_count@plt>
  cmpb   $0x2c,(%rax)
  je     L116
  mov    (%rsp),%edx
  mov    %edx,0x4(%rsp)
  mov    %edx,%esi
  mov    %edx,%ecx
L83:
  mov    D3(%rip),%rdi
  cmp    %esi,(%rdi)
  jl     L6
  test   %ecx,%ecx
  js     L6
  movzbl (%rax),%esi
  test   %r12d,%r12d
  je     L66
L119:
  cmpb   $0x7d,(%rax)
  jne    L6
  mov    %ecx,0x0(%rbp)
  add    $0x1,%rax
  mov    %edx,(%rbx)
  jmp    L68
  nopl   0x0(%rax)
L116:
  lea    0x1(%rax),%rdi
  lea    0x4(%rsp),%rsi
  call   <parse_dup_count@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%ecx
  cmp    %ecx,%edx
  jge    L50
  test   %edx,%edx
  js     L50
  nopl   (%rax)
L6:
  xor    %eax,%eax
L68:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L66:
  cmp    $0x5c,%sil
  jne    L6
  add    $0x1,%rax
  jmp    L119
  nopl   0x0(%rax)
L50:
  cmp    %ecx,%edx
  mov    %ecx,%esi
  cmovge %edx,%esi
  jmp    L83
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const char *parse_dup(const char *s, int ere, int *pmin, int *pmax) {
  int min, max;
  s = parse_dup_count(s, &min);
  if (*s == ',')
    s = parse_dup_count(s + 1, &max);
  else
    max = min;
  if ((max < min && max >= 0) || max > RE_DUP_MAX || min > RE_DUP_MAX || min < 0 || (!ere && *s++ != '\\') || *s++ != '}')
    return 0;
  *pmin = min;
  *pmax = max;
  return s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_dup>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <parse_dup_count@plt>
  cmpb   $0x2c,(%rax)
  je     L38
  mov    (%rsp),%edx
  mov    %edx,0x4(%rsp)
  mov    %edx,%esi
  mov    %edx,%ecx
L14:
  mov    D2(%rip),%rdi
  cmp    %esi,(%rdi)
  jl     L35
  test   %ecx,%ecx
  js     L35
  movzbl (%rax),%esi
  test   %r12d,%r12d
  je     L119
L47:
  cmpb   $0x7d,(%rax)
  jne    L35
  mov    %ecx,0x0(%rbp)
  add    $0x1,%rax
  mov    %edx,(%rbx)
  jmp    L3
  nopl   0x0(%rax)
L38:
  lea    0x1(%rax),%rdi
  lea    0x4(%rsp),%rsi
  call   <parse_dup_count@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%ecx
  cmp    %ecx,%edx
  jge    L115
  test   %edx,%edx
  js     L115
  nopl   (%rax)
L35:
  xor    %eax,%eax
L3:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L50
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L119:
  cmp    $0x5c,%sil
  jne    L35
  add    $0x1,%rax
  jmp    L47
  nopl   0x0(%rax)
L115:
  cmp    %ecx,%edx
  mov    %ecx,%esi
  cmovge %edx,%esi
  jmp    L14
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const char *parse_dup(const char *s, int ere, int *pmin, int *pmax) {
  int min, max;
  s = parse_dup_count(s, &min);
  if (*s == ',')
    s = parse_dup_count(s + 1, &max);
  else
    max = min;
  if ((max < min && max >= 0) || max > RE_DUP_MAX || min > RE_DUP_MAX || min < 0 || (!ere && *s++ != '\\') || *s++ != '}')
    return 0;
  *pmin = min;
  *pmax = max;
  return s;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L17:
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%ecx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jb     L17
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
void *memcpy(void *dst, const void *src, unsigned size) {
  unsigned i;
  for (i = 0; i < size; i++)
    *((char *)dst + i) = *((char *)src + i);
  return ((void *)0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<memcpy>:
  endbr64
  test   %edx,%edx
  je     L10
  mov    %edx,%edx
  mov    $0x0,%eax
L6:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L6
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
void *memcpy(void *dst, const void *src, unsigned size) {
  unsigned i;
  for (i = 0; i < size; i++)
    *((char *)dst + i) = *((char *)src + i);
  return ((void *)0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memcpy>:
  endbr64
  test   %edx,%edx
  je     L17
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L6:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L6
L17:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memcpy(void *dst, const void *src, unsigned size) {
  unsigned i;
  for (i = 0; i < size; i++)
    *((char *)dst + i) = *((char *)src + i);
  return ((void *)0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%rcx
  test   %edx,%edx
  je     L48
  lea    0x1(%rsi),%r8
  lea    -0x1(%rdx),%eax
  sub    %r8,%rdi
  cmp    $0xe,%rdi
  jbe    L16
  cmp    $0x6,%eax
  jbe    L16
  mov    %edx,%edi
  cmp    $0xe,%eax
  jbe    L31
  shr    $0x4,%edi
  xor    %eax,%eax
  shl    $0x4,%rdi
L97:
  movdqu (%rsi,%rax,1),%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L97
  mov    %edx,%eax
  and    $0xfffffff0,%eax
  test   $0xf,%dl
  je     L48
  mov    %edx,%edi
  sub    %eax,%edi
  lea    -0x1(%rdi),%r8d
  cmp    $0x6,%r8d
  jbe    L69
L63:
  mov    %eax,%r8d
  mov    (%rsi,%r8,1),%r9
  mov    %r9,(%rcx,%r8,1)
  mov    %edi,%r8d
  and    $0xfffffff8,%r8d
  add    %r8d,%eax
  cmp    %r8d,%edi
  je     L48
L69:
  mov    %eax,%edi
  movzbl (%rsi,%rdi,1),%r8d
  mov    %r8b,(%rcx,%rdi,1)
  lea    0x1(%rax),%edi
  cmp    %edi,%edx
  jbe    L48
  movzbl (%rsi,%rdi,1),%r8d
  mov    %r8b,(%rcx,%rdi,1)
  lea    0x2(%rax),%edi
  cmp    %edi,%edx
  jbe    L48
  movzbl (%rsi,%rdi,1),%r8d
  mov    %r8b,(%rcx,%rdi,1)
  lea    0x3(%rax),%edi
  cmp    %edi,%edx
  jbe    L48
  movzbl (%rsi,%rdi,1),%r8d
  mov    %r8b,(%rcx,%rdi,1)
  lea    0x4(%rax),%edi
  cmp    %edi,%edx
  jbe    L48
  movzbl (%rsi,%rdi,1),%r8d
  mov    %r8b,(%rcx,%rdi,1)
  lea    0x5(%rax),%edi
  cmp    %edi,%edx
  jbe    L48
  movzbl (%rsi,%rdi,1),%r8d
  add    $0x6,%eax
  mov    %r8b,(%rcx,%rdi,1)
  cmp    %eax,%edx
  jbe    L48
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
L48:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L16:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L131:
  movzbl (%rsi,%rax,1),%edi
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L131
  xor    %eax,%eax
  ret
L31:
  xor    %eax,%eax
  jmp    L63
```

assistant: ```c
void *memcpy(void *dst, const void *src, unsigned size) {
  unsigned i;
  for (i = 0; i < size; i++)
    *((char *)dst + i) = *((char *)src + i);
  return ((void *)0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L27
  mov    $0x0,%eax
  jmp    L2
L27:
  mov    -0x30(%rbp),%rax
  mov    -0x14(%rbp),%edx
  mov    %edx,(%rax)
  mov    $0x1,%eax
L2:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int parse_int(const char *str, int *ret) {
  char *test;
  int temp;
  temp = strtol(str, &test, 10);
  if (*test != '\0')
    return 0;
  *ret = temp;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_int>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtol@plt>
  mov    $0x0,%edx
  mov    (%rsp),%rcx
  cmpb   $0x0,(%rcx)
  jne    L11
  mov    %eax,(%rbx)
  mov    $0x1,%edx
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    %edx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_int(const char *str, int *ret) {
  char *test;
  int temp;
  temp = strtol(str, &test, 10);
  if (*test != '\0')
    return 0;
  *ret = temp;
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_int>:
  endbr64
  push   %rbx
  mov    $0xa,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    (%rsp),%rdx
  xor    %r8d,%r8d
  cmpb   $0x0,(%rdx)
  jne    L8
  mov    %eax,(%rbx)
  mov    $0x1,%r8d
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x10,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_int(const char *str, int *ret) {
  char *test;
  int temp;
  temp = strtol(str, &test, 10);
  if (*test != '\0')
    return 0;
  *ret = temp;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<parse_int>:
  endbr64
  push   %rbx
  mov    $0xa,%edx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    (%rsp),%rdx
  xor    %r8d,%r8d
  cmpb   $0x0,(%rdx)
  jne    L19
  mov    %eax,(%rbx)
  mov    $0x1,%r8d
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_int(const char *str, int *ret) {
  char *test;
  int temp;
  temp = strtol(str, &test, 10);
  if (*test != '\0')
    return 0;
  *ret = temp;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_is_sandybridge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x102,%eax
  cmp    $0x24,%eax
  seta   %dl
  test   %dl,%dl
  jne    L14
  movabs $0x1100110111,%rdx
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rax
  and    $0x1,%eax
  test   %rax,%rax
  setne  %al
  test   %al,%al
  je     L14
  movl   $0x1,-0x4(%rbp)
  jmp    L1
L14:
  nop
L1:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int _is_sandybridge(int devid) {
  int ret = 0;
  switch (devid) {
  case 0x0102:
  case 0x0112:
  case 0x0122:
  case 0x0106:
  case 0x0116:
  case 0x0126:
  case 0x010A:
    ret = 1;
    break;
  default:
    break;
  }
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_is_sandybridge>:
  endbr64
  lea    -0x102(%rdi),%ecx
  cmp    $0x24,%ecx
  ja     L1
  movabs $0x1100110111,%rax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
L1:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int _is_sandybridge(int devid) {
  int ret = 0;
  switch (devid) {
  case 0x0102:
  case 0x0112:
  case 0x0122:
  case 0x0106:
  case 0x0116:
  case 0x0126:
  case 0x010A:
    ret = 1;
    break;
  default:
    break;
  }
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_is_sandybridge>:
  endbr64
  lea    -0x102(%rdi),%ecx
  cmp    $0x24,%ecx
  ja     L4
  movabs $0x1100110111,%rax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int _is_sandybridge(int devid) {
  int ret = 0;
  switch (devid) {
  case 0x0102:
  case 0x0112:
  case 0x0122:
  case 0x0106:
  case 0x0116:
  case 0x0126:
  case 0x010A:
    ret = 1;
    break;
  default:
    break;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<_is_sandybridge>:
  endbr64
  lea    -0x102(%rdi),%ecx
  cmp    $0x24,%ecx
  ja     L9
  movabs $0x1100110111,%rax
  shr    %cl,%rax
  and    $0x1,%eax
  ret
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
int _is_sandybridge(int devid) {
  int ret = 0;
  switch (devid) {
  case 0x0102:
  case 0x0112:
  case 0x0122:
  case 0x0106:
  case 0x0116:
  case 0x0126:
  case 0x010A:
    ret = 1;
    break;
  default:
    break;
  }
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<g>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  bswap  %eax
  bswap  %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int g(int a) {
  unsigned int b = __builtin_bswap32(__builtin_bswap32(a));
  return b;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<g>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
int g(int a) {
  unsigned int b = __builtin_bswap32(__builtin_bswap32(a));
  return b;
}
```
user: What is the c source code of the assembly code below:

```asm
<g>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
int g(int a) {
  unsigned int b = __builtin_bswap32(__builtin_bswap32(a));
  return b;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<g>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
int g(int a) {
  unsigned int b = __builtin_bswap32(__builtin_bswap32(a));
  return b;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L10
L17:
  movsbl -0x9(%rbp),%eax
  mov    %eax,%edi
  call   <tpchr@plt>
L10:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x0,-0x9(%rbp)
  jne    L17
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  char *msg;
  char ch;
  while (ch = *msg++) {
    tpchr(ch);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  lea    0x1(%rax),%rbx
  movzbl (%rax),%edi
  test   %dil,%dil
  je     L14
L1:
  movsbl %dil,%edi
  call   <tpchr@plt>
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%edi
  test   %dil,%dil
  jne    L1
L14:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
main() {
  char *msg;
  char ch;
  while (ch = *msg++) {
    tpchr(ch);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  movsbl (%rax),%edi
  test   %dil,%dil
  je     L0
  push   %rbx
  lea    0x1(%rax),%rbx
  nopl   0x0(%rax,%rax,1)
L6:
  call   <tpchr@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L6
  xor    %eax,%eax
  pop    %rbx
  ret
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
main() {
  char *msg;
  char ch;
  while (ch = *msg++) {
    tpchr(ch);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  movsbl (%rax),%edi
  test   %dil,%dil
  je     L0
  push   %rbx
  lea    0x1(%rax),%rbx
  nopl   0x0(%rax,%rax,1)
L7:
  call   <tpchr@plt>
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L7
  xor    %eax,%eax
  pop    %rbx
  ret
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
main() {
  char *msg;
  char ch;
  while (ch = *msg++) {
    tpchr(ch);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hermite_interpolation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x48(%rbp)
  movsd  %xmm1,-0x50(%rbp)
  movsd  %xmm2,-0x58(%rbp)
  movsd  %xmm3,-0x60(%rbp)
  movsd  %xmm4,-0x68(%rbp)
  movsd  %xmm5,-0x70(%rbp)
  movsd  %xmm6,-0x78(%rbp)
  movsd  -0x58(%rbp),%xmm0
  subsd  -0x50(%rbp),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  -0x48(%rbp),%xmm0
  subsd  -0x50(%rbp),%xmm0
  divsd  -0x40(%rbp),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  -0x70(%rbp),%xmm0
  mulsd  -0x40(%rbp),%xmm0
  movsd  %xmm0,-0x70(%rbp)
  movsd  -0x78(%rbp),%xmm0
  mulsd  -0x40(%rbp),%xmm0
  movsd  %xmm0,-0x78(%rbp)
  movsd  -0x38(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x30(%rbp),%xmm0
  mulsd  -0x38(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x60(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x70(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x60(%rbp),%xmm1
  movsd  D16(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x70(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  subsd  %xmm0,%xmm1
  movsd  -0x68(%rbp),%xmm2
  movsd  D3(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  subsd  -0x78(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x60(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movapd %xmm0,%xmm1
  addsd  -0x70(%rbp),%xmm1
  movsd  -0x68(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  -0x78(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x28(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mulsd  -0x30(%rbp),%xmm0
  addsd  %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x38(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  addsd  -0x20(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
type: byte[8]
name: read_data
arguments: data_label: D3
type: byte[8]

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0xc0]}

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double hermite_interpolation(double x, double x0, double x1, double p0, double p1, double m0, double m1) {
  double width = x1 - x0;
  double t = (x - x0) / width;
  double t2, t3;
  double ct0, ct1, ct2, ct3;
  m0 *= width;
  m1 *= width;
  t2 = t * t;
  t3 = t2 * t;
  ct0 = p0;
  ct1 = m0;
  ct2 = -3 * p0 - 2 * m0 + 3 * p1 - m1;
  ct3 = 2 * p0 + m0 - 2 * p1 + m1;
  return ct3 * t3 + ct2 * t2 + ct1 * t + ct0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hermite_interpolation>:
  endbr64
  subsd  %xmm1,%xmm2
  subsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm5
  mulsd  %xmm6,%xmm2
  movapd %xmm1,%xmm6
  mulsd  %xmm1,%xmm6
  movapd %xmm3,%xmm0
  addsd  %xmm3,%xmm0
  addsd  %xmm5,%xmm0
  movapd %xmm4,%xmm7
  addsd  %xmm4,%xmm7
  subsd  %xmm7,%xmm0
  addsd  %xmm2,%xmm0
  movapd %xmm1,%xmm7
  mulsd  %xmm6,%xmm7
  mulsd  %xmm7,%xmm0
  movapd %xmm3,%xmm7
  mulsd  D4(%rip),%xmm7
  movapd %xmm5,%xmm8
  addsd  %xmm5,%xmm8
  subsd  %xmm8,%xmm7
  mulsd  D14(%rip),%xmm4
  addsd  %xmm7,%xmm4
  subsd  %xmm2,%xmm4
  mulsd  %xmm6,%xmm4
  addsd  %xmm4,%xmm0
  mulsd  %xmm5,%xmm1
  addsd  %xmm1,%xmm0
  addsd  %xmm3,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
type: byte[8]
name: unpack_data
arguments: label: D14
type: byte[8]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0xc0]}

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double hermite_interpolation(double x, double x0, double x1, double p0, double p1, double m0, double m1) {
  double width = x1 - x0;
  double t = (x - x0) / width;
  double t2, t3;
  double ct0, ct1, ct2, ct3;
  m0 *= width;
  m1 *= width;
  t2 = t * t;
  t3 = t2 * t;
  ct0 = p0;
  ct1 = m0;
  ct2 = -3 * p0 - 2 * m0 + 3 * p1 - m1;
  ct3 = 2 * p0 + m0 - 2 * p1 + m1;
  return ct3 * t3 + ct2 * t2 + ct1 * t + ct0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<hermite_interpolation>:
  endbr64
  subsd  %xmm1,%xmm0
  subsd  %xmm1,%xmm2
  movapd %xmm4,%xmm7
  addsd  %xmm4,%xmm7
  mulsd  D0(%rip),%xmm4
  mulsd  %xmm2,%xmm5
  movapd %xmm0,%xmm1
  movapd %xmm3,%xmm0
  divsd  %xmm2,%xmm1
  addsd  %xmm3,%xmm0
  mulsd  %xmm6,%xmm2
  addsd  %xmm5,%xmm0
  movapd %xmm5,%xmm8
  addsd  %xmm5,%xmm8
  subsd  %xmm7,%xmm0
  addsd  %xmm2,%xmm0
  movapd %xmm1,%xmm6
  movapd %xmm1,%xmm7
  mulsd  %xmm1,%xmm6
  mulsd  %xmm5,%xmm1
  mulsd  %xmm6,%xmm7
  mulsd  %xmm7,%xmm0
  movsd  D14(%rip),%xmm7
  mulsd  %xmm3,%xmm7
  subsd  %xmm8,%xmm7
  addsd  %xmm7,%xmm4
  subsd  %xmm2,%xmm4
  mulsd  %xmm6,%xmm4
  addsd  %xmm4,%xmm0
  addsd  %xmm1,%xmm0
  addsd  %xmm3,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D0
datatype: byte[8]
name: parse_data
arguments: label: D14
datatype: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0xc0]}

assistant: ```c
double hermite_interpolation(double x, double x0, double x1, double p0, double p1, double m0, double m1) {
  double width = x1 - x0;
  double t = (x - x0) / width;
  double t2, t3;
  double ct0, ct1, ct2, ct3;
  m0 *= width;
  m1 *= width;
  t2 = t * t;
  t3 = t2 * t;
  ct0 = p0;
  ct1 = m0;
  ct2 = -3 * p0 - 2 * m0 + 3 * p1 - m1;
  ct3 = 2 * p0 + m0 - 2 * p1 + m1;
  return ct3 * t3 + ct2 * t2 + ct1 * t + ct0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hermite_interpolation>:
  endbr64
  subsd  %xmm1,%xmm0
  subsd  %xmm1,%xmm2
  movapd %xmm4,%xmm7
  addsd  %xmm4,%xmm7
  mulsd  D10(%rip),%xmm4
  mulsd  %xmm2,%xmm5
  movapd %xmm0,%xmm1
  movapd %xmm3,%xmm0
  divsd  %xmm2,%xmm1
  addsd  %xmm3,%xmm0
  mulsd  %xmm6,%xmm2
  addsd  %xmm5,%xmm0
  movapd %xmm5,%xmm8
  addsd  %xmm5,%xmm8
  subsd  %xmm7,%xmm0
  addsd  %xmm2,%xmm0
  movapd %xmm1,%xmm6
  movapd %xmm1,%xmm7
  mulsd  %xmm1,%xmm6
  mulsd  %xmm5,%xmm1
  mulsd  %xmm6,%xmm7
  mulsd  %xmm7,%xmm0
  movsd  D5(%rip),%xmm7
  mulsd  %xmm3,%xmm7
  subsd  %xmm8,%xmm7
  addsd  %xmm7,%xmm4
  subsd  %xmm2,%xmm4
  mulsd  %xmm6,%xmm4
  addsd  %xmm4,%xmm0
  addsd  %xmm1,%xmm0
  addsd  %xmm3,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D10
data_type: byte[8]
name: parse_data
arguments: data_tag: D5
data_type: byte[8]

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0xc0]}

assistant: ```c
double hermite_interpolation(double x, double x0, double x1, double p0, double p1, double m0, double m1) {
  double width = x1 - x0;
  double t = (x - x0) / width;
  double t2, t3;
  double ct0, ct1, ct2, ct3;
  m0 *= width;
  m1 *= width;
  t2 = t * t;
  t3 = t2 * t;
  ct0 = p0;
  ct1 = m0;
  ct2 = -3 * p0 - 2 * m0 + 3 * p1 - m1;
  ct3 = 2 * p0 + m0 - 2 * p1 + m1;
  return ct3 * t3 + ct2 * t2 + ct1 * t + ct0;
}
```
user: What is the c source code of the assembly code below:

```asm
<timeval_subtract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%rax
  sub    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jl     L19
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  sub    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L2
L19:
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    $0xf4240,%edx
  sub    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  nop
L2:
  pop    %rbp
  ret
```

assistant: ```c
void timeval_subtract(struct timeval *result, struct timeval *start, struct timeval *end) {
  result->tv_sec = end->tv_sec - start->tv_sec;
  if (end->tv_usec >= start->tv_usec) {
    result->tv_usec = end->tv_usec - start->tv_usec;
  } else {
    result->tv_sec -= 1;
    result->tv_usec = (1000000 - start->tv_usec) + end->tv_usec;
  };
  return;
}
```