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

```asm
<atoi>:
  endbr64
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L22
  nopl   0x0(%rax)
L14:
  add    $0x1,%r12d
L22:
  movsbl 0x0(%rbp),%edi
  mov    %rbp,%r13
  add    $0x1,%rbp
  call   <is_space@plt>
  test   %rax,%rax
  jne    L14
  movzbl 0x0(%r13),%eax
  cmp    $0x2d,%al
  je     L41
  mov    $0x1,%r13d
  cmp    $0x2b,%al
  je     L46
L48:
  movslq %r12d,%r12
  xor    %ebp,%ebp
  add    %r12,%rbx
  jmp    L40
  cs nopw 0x0(%rax,%rax,1)
L43:
  movsbl (%rbx),%eax
  lea    0x0(%rbp,%rbp,4),%edx
  add    $0x1,%rbx
  lea    -0x30(%rax,%rdx,2),%ebp
L40:
  movsbl (%rbx),%edi
  call   <is_digit@plt>
  test   %rax,%rax
  jne    L43
  add    $0x8,%rsp
  mov    %ebp,%eax
  pop    %rbx
  imul   %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L41:
  mov    $0xffffffff,%r13d
L46:
  add    $0x1,%r12d
  jmp    L48
```

assistant: ```c
int atoi(char s[]) {
  int i, n, sign;
  for (i = 0; is_space(s[i]); i++)
    ;
  sign = (s[i] == '-') ? -1 : 1;
  if (s[i] == '+' || s[i] == '-')
    i++;
  for (n = 0; is_digit(s[i]); i++)
    n = 10 * n + (s[i] - '0');
  return sign * n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fcitx_utils_strcmp0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L76
  cmpq   $0x0,-0x10(%rbp)
  jne    L76
  mov    $0x0,%eax
  jmp    L71
L76:
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  cmpq   $0x0,-0x10(%rbp)
  je     L0
  mov    $0xffffffff,%eax
  jmp    L71
L0:
  cmpq   $0x0,-0x8(%rbp)
  je     L70
  cmpq   $0x0,-0x10(%rbp)
  jne    L70
  mov    $0x1,%eax
  jmp    L71
L70:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
L71:
  leave
  ret
```

assistant: ```c
__attribute__((always_)) int fcitx_utils_strcmp0(const char *a, const char *b) {
  if (a == NULL && b == NULL)
    return 0;
  if (a == NULL && b)
    return -1;
  if (a && b == NULL)
    return 1;
  return strcmp(a, b);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fcitx_utils_strcmp0>:
  endbr64
  mov    $0x0,%eax
  mov    %rdi,%rdx
  or     %rsi,%rdx
  je     L46
  test   %rdi,%rdi
  jne    L41
  test   %rsi,%rsi
  jne    L10
L41:
  test   %rsi,%rsi
  jne    L38
  mov    $0x1,%eax
  test   %rdi,%rdi
  je     L38
L46:
  ret
L38:
  sub    $0x8,%rsp
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
L10:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
__attribute__((always_)) int fcitx_utils_strcmp0(const char *a, const char *b) {
  if (a == NULL && b == NULL)
    return 0;
  if (a == NULL && b)
    return -1;
  if (a && b == NULL)
    return 1;
  return strcmp(a, b);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fcitx_utils_strcmp0>:
  endbr64
  mov    %rdi,%rax
  or     %rsi,%rax
  je     L41
  test   %rdi,%rdi
  jne    L27
  test   %rsi,%rsi
  jne    L20
L22:
  mov    $0x1,%eax
  test   %rdi,%rdi
  jne    L30
  jmp    1050 <strcmp@plt>
  nopl   (%rax)
L27:
  test   %rsi,%rsi
  je     L22
  jmp    1050 <strcmp@plt>
L20:
  mov    $0xffffffff,%eax
L30:
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  xor    %eax,%eax
  ret
```

assistant: ```c
__attribute__((always_)) int fcitx_utils_strcmp0(const char *a, const char *b) {
  if (a == NULL && b == NULL)
    return 0;
  if (a == NULL && b)
    return -1;
  if (a && b == NULL)
    return 1;
  return strcmp(a, b);
}
```
user: What is the c source code of the assembly code below:

```asm
<fcitx_utils_strcmp0>:
  endbr64
  mov    %rdi,%rax
  or     %rsi,%rax
  je     L16
  test   %rdi,%rdi
  jne    L24
  test   %rsi,%rsi
  jne    L11
L14:
  mov    $0x1,%eax
  test   %rdi,%rdi
  jne    L0
  jmp    1050 <strcmp@plt>
  nopl   (%rax)
L24:
  test   %rsi,%rsi
  je     L14
  jmp    1050 <strcmp@plt>
L11:
  mov    $0xffffffff,%eax
L0:
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
__attribute__((always_)) int fcitx_utils_strcmp0(const char *a, const char *b) {
  if (a == NULL && b == NULL)
    return 0;
  if (a == NULL && b)
    return -1;
  if (a && b == NULL)
    return 1;
  return strcmp(a, b);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %edi,-0x1c(%rbp)
  mov    %rsi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    $0x3,%eax
  mov    -0x1c(%rbp),%esi
  mov    -0x28(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    %esi,%ebx
  int    $0x85
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long read(int fd, void *buf, unsigned long count) {
  long res;
  asm volatile("int $0x85" : "=a"(res) : "a"(3), "b"(fd), "c"(buf), "d"(count) : "memory");
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read>:
  endbr64
  push   %rbx
  mov    $0x3,%eax
  mov    %edi,%ebx
  mov    %rsi,%rcx
  int    $0x85
  pop    %rbx
  ret
```

assistant: ```c
long read(int fd, void *buf, unsigned long count) {
  long res;
  asm volatile("int $0x85" : "=a"(res) : "a"(3), "b"(fd), "c"(buf), "d"(count) : "memory");
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read>:
  endbr64
  push   %rbx
  mov    $0x3,%eax
  mov    %edi,%ebx
  mov    %rsi,%rcx
  int    $0x85
  pop    %rbx
  ret
```

assistant: ```c
long read(int fd, void *buf, unsigned long count) {
  long res;
  asm volatile("int $0x85" : "=a"(res) : "a"(3), "b"(fd), "c"(buf), "d"(count) : "memory");
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read>:
  endbr64
  push   %rbx
  mov    $0x3,%eax
  mov    %edi,%ebx
  mov    %rsi,%rcx
  int    $0x85
  pop    %rbx
  ret
```

assistant: ```c
long read(int fd, void *buf, unsigned long count) {
  long res;
  asm volatile("int $0x85" : "=a"(res) : "a"(3), "b"(fd), "c"(buf), "d"(count) : "memory");
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<areAnagrams>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x410(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x80,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    -0x428(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x414(%rbp)
  mov    -0x430(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x414(%rbp)
  je     L88
  mov    $0x0,%eax
  jmp    L19
L88:
  movl   $0x0,-0x418(%rbp)
  jmp    L43
L57:
  mov    -0x418(%rbp),%edx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movslq %eax,%rdx
  mov    -0x410(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x410(%rbp,%rax,4)
  mov    -0x418(%rbp),%edx
  mov    -0x430(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movslq %eax,%rdx
  mov    -0x410(%rbp,%rdx,4),%edx
  sub    $0x1,%edx
  cltq
  mov    %edx,-0x410(%rbp,%rax,4)
  addl   $0x1,-0x418(%rbp)
L43:
  mov    -0x418(%rbp),%eax
  cmp    -0x414(%rbp),%eax
  jb     L57
  movl   $0x0,-0x418(%rbp)
  jmp    L14
L76:
  mov    -0x418(%rbp),%eax
  mov    -0x410(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L83
  mov    $0x0,%eax
  jmp    L19
L83:
  addl   $0x1,-0x418(%rbp)
L14:
  mov    -0x418(%rbp),%eax
  cmp    -0x414(%rbp),%eax
  jb     L76
  mov    $0x1,%eax
L19:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  leave
  ret
```

assistant: ```c
bool areAnagrams(const char *str1, const char *str2) {
  int lookup[256] = {0};
  const unsigned int len = strlen(str1);
  if (strlen(str2) != len)
    return false;
  unsigned int i;
  for (i = 0; i < len; ++i) {
    ++lookup[str1[i]];
    --lookup[str2[i]];
  }
  for (i = 0; i < len; ++i)
    if (lookup[i] != 0)
      return false;
  return true;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<areAnagrams>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x80,%ecx
  rep stos %rax,%es:(%rdi)
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    %r12d,%edx
  jne    L25
  test   %r12d,%r12d
  je     L56
  lea    -0x1(%r12),%ecx
  mov    $0x0,%eax
L51:
  movsbq (%rbx,%rax,1),%rdx
  addl   $0x1,(%rsp,%rdx,4)
  movsbq 0x0(%rbp,%rax,1),%rdx
  subl   $0x1,(%rsp,%rdx,4)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L51
  mov    %rsp,%rax
  lea    0x4(%rsp,%rcx,4),%rdx
L42:
  cmpl   $0x0,(%rax)
  jne    L3
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L42
  mov    $0x1,%eax
  jmp    L25
L56:
  mov    $0x1,%eax
  jmp    L25
L3:
  mov    $0x0,%eax
L25:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L73
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L73:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool areAnagrams(const char *str1, const char *str2) {
  int lookup[256] = {0};
  const unsigned int len = strlen(str1);
  if (strlen(str2) != len)
    return false;
  unsigned int i;
  for (i = 0; i < len; ++i) {
    ++lookup[str1[i]];
    --lookup[str2[i]];
  }
  for (i = 0; i < len; ++i)
    if (lookup[i] != 0)
      return false;
  return true;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<areAnagrams>:
  endbr64
  push   %r13
  mov    $0x80,%ecx
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  cmp    %r13d,%eax
  jne    L50
  lea    -0x1(%rax),%esi
  mov    %eax,%ecx
  xor    %edx,%edx
  test   %eax,%eax
  je     L28
  nopl   0x0(%rax,%rax,1)
L64:
  movsbq 0x0(%rbp,%rdx,1),%rax
  addl   $0x1,(%rsp,%rax,4)
  movsbq (%r12,%rdx,1),%rax
  add    $0x1,%rdx
  subl   $0x1,(%rsp,%rax,4)
  cmp    %rcx,%rdx
  jne    L64
  mov    %rbx,%rdi
  lea    0x4(%rsp,%rsi,4),%rax
  jmp    L51
  cs nopw 0x0(%rax,%rax,1)
L67:
  add    $0x4,%rdi
  cmp    %rdi,%rax
  je     L28
L51:
  mov    (%rdi),%edx
  test   %edx,%edx
  je     L67
L50:
  xor    %eax,%eax
L27:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L53
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  mov    $0x1,%eax
  jmp    L27
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool areAnagrams(const char *str1, const char *str2) {
  int lookup[256] = {0};
  const unsigned int len = strlen(str1);
  if (strlen(str2) != len)
    return false;
  unsigned int i;
  for (i = 0; i < len; ++i) {
    ++lookup[str1[i]];
    --lookup[str2[i]];
  }
  for (i = 0; i < len; ++i)
    if (lookup[i] != 0)
      return false;
  return true;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<areAnagrams>:
  endbr64
  push   %r13
  mov    $0x80,%ecx
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  cmp    %r13d,%eax
  jne    L35
  lea    -0x1(%rax),%esi
  mov    %eax,%ecx
  xor    %edx,%edx
  test   %eax,%eax
  je     L79
  nopl   0x0(%rax,%rax,1)
L78:
  movsbq 0x0(%rbp,%rdx,1),%rax
  addl   $0x1,(%rsp,%rax,4)
  movsbq (%r12,%rdx,1),%rax
  add    $0x1,%rdx
  subl   $0x1,(%rsp,%rax,4)
  cmp    %rdx,%rcx
  jne    L78
  mov    %rbx,%rdi
  lea    0x4(%rsp,%rsi,4),%rax
  jmp    L57
  cs nopw 0x0(%rax,%rax,1)
L36:
  add    $0x4,%rdi
  cmp    %rdi,%rax
  je     L79
L57:
  mov    (%rdi),%edx
  test   %edx,%edx
  je     L36
L35:
  xor    %eax,%eax
L16:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L51
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L79:
  mov    $0x1,%eax
  jmp    L16
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool areAnagrams(const char *str1, const char *str2) {
  int lookup[256] = {0};
  const unsigned int len = strlen(str1);
  if (strlen(str2) != len)
    return false;
  unsigned int i;
  for (i = 0; i < len; ++i) {
    ++lookup[str1[i]];
    --lookup[str2[i]];
  }
  for (i = 0; i < len; ++i)
    if (lookup[i] != 0)
      return false;
  return true;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<COM_StripFilename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <Q_strncpyz@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <COM_SkipPath@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void COM_StripFilename(char *in, char *out) {
  char *end;
  Q_strncpyz(out, in, strlen(in));
  end = COM_SkipPath(out);
  *end = 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<COM_StripFilename>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <Q_strncpyz@plt>
  mov    %rbx,%rdi
  call   <COM_SkipPath@plt>
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void COM_StripFilename(char *in, char *out) {
  char *end;
  Q_strncpyz(out, in, strlen(in));
  end = COM_SkipPath(out);
  *end = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<COM_StripFilename>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %eax,%edx
  call   <Q_strncpyz@plt>
  mov    %rbp,%rdi
  call   <COM_SkipPath@plt>
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void COM_StripFilename(char *in, char *out) {
  char *end;
  Q_strncpyz(out, in, strlen(in));
  end = COM_SkipPath(out);
  *end = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<COM_StripFilename>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %eax,%edx
  call   <Q_strncpyz@plt>
  mov    %rbp,%rdi
  call   <COM_SkipPath@plt>
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void COM_StripFilename(char *in, char *out) {
  char *end;
  Q_strncpyz(out, in, strlen(in));
  end = COM_SkipPath(out);
  *end = 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dead_result1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x2,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dead_result1(int *data) {
  int j;
  j = data[2];
  data[j] = 2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dead_result1>:
  endbr64
  movslq 0x8(%rdi),%rax
  movl   $0x2,(%rdi,%rax,4)
  ret
```

assistant: ```c
void dead_result1(int *data) {
  int j;
  j = data[2];
  data[j] = 2;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dead_result1>:
  endbr64
  movslq 0x8(%rdi),%rax
  movl   $0x2,(%rdi,%rax,4)
  ret
```

assistant: ```c
void dead_result1(int *data) {
  int j;
  j = data[2];
  data[j] = 2;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dead_result1>:
  endbr64
  movslq 0x8(%rdi),%rax
  movl   $0x2,(%rdi,%rax,4)
  ret
```

assistant: ```c
void dead_result1(int *data) {
  int j;
  j = data[2];
  data[j] = 2;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<levelSetCurvatureCoefficientsEvaluate>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %r9,0x8(%rsp)
  test   %edi,%edi
  jle    L50
  mov    %esi,%r12d
  mov    %rdx,%rbp
  mov    %rcx,%r15
  mov    %r8,%r14
  mov    %edi,%eax
  mov    %rax,0x10(%rsp)
  test   %esi,%esi
  mov    $0x0,%eax
  cmovns %esi,%eax
  mov    %eax,0x1c(%rsp)
  mov    $0x0,%r13d
  mov    $0x0,%ebx
  jmp    L123
L98:
  pxor   %xmm0,%xmm0
  jmp    L86
L122:
  call   <sqrt@plt>
  mov    %r12d,%eax
  comisd D3(%rip),%xmm0
  ja     L106
L62:
  add    %r13d,%eax
  cltq
  movq   $0x0,(%r14,%rax,8)
L8:
  add    $0x1,%rbx
  add    %r12d,%r13d
  cmp    0x10(%rsp),%rbx
  je     L50
L123:
  movsd  (%r15,%rbx,8),%xmm0
  mov    0x8(%rsp),%rax
  movsd  %xmm0,(%rax,%rbx,8)
  mov    0x60(%rsp),%rax
  mov    D19(%rip),%rsi
  mov    %rsi,(%rax,%rbx,8)
  test   %r12d,%r12d
  jle    L98
  movslq %r13d,%rcx
  lea    0x0(%rbp,%rcx,8),%rax
  lea    -0x1(%r12),%edx
  add    %rcx,%rdx
  lea    0x8(%rbp,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L33:
  movsd  (%rax),%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L33
  pxor   %xmm2,%xmm2
  ucomisd %xmm0,%xmm2
  ja     L122
L86:
  sqrtsd %xmm0,%xmm0
  comisd D3(%rip),%xmm0
  ja     L64
  mov    0x1c(%rsp),%eax
  jmp    L62
L64:
  test   %r12d,%r12d
  jle    L8
L106:
  movslq %r13d,%rcx
  lea    0x0(,%rcx,8),%rax
  mov    %r12d,%edx
  add    %rcx,%rdx
  shl    $0x3,%rdx
L118:
  movsd  0x0(%rbp,%rax,1),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%r14,%rax,1)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L118
  jmp    L8
L50:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: label: D3
type_infered: f64[2]
name: read_data
arguments: label: D19
type_infered: f64

tool: {"D3": [0.0,1.0]}

tool: D19: 1

assistant: ```c
void levelSetCurvatureCoefficientsEvaluate(int nPoints, int nSpace, double *grad_phi, double *u, double *f, double *r, double *dr) {
  int i, I;
  double norm_grad_phi;
  for (i = 0; i < nPoints; i++) {
    r[i] = u[i];
    dr[i] = 1.0;
    norm_grad_phi = 0.0;
    for (I = 0; I < nSpace; I++)
      norm_grad_phi += grad_phi[i * nSpace + I] * grad_phi[i * nSpace + I];
    norm_grad_phi = sqrt(norm_grad_phi);
    if (norm_grad_phi > 0.0) {
      for (I = 0; I < nSpace; I++)
        f[i * nSpace + I] = grad_phi[i * nSpace + I] / norm_grad_phi;
    } else
      f[i * nSpace + I] = 0.0;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<levelSetCurvatureCoefficientsEvaluate>:
  endbr64
  push   %r15
  movslq %edi,%r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  mov    %rcx,%rdx
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    0x70(%rsp),%rcx
  test   %r15d,%r15d
  jle    L90
  xor    %edi,%edi
  test   %esi,%esi
  pxor   %xmm2,%xmm2
  mov    %esi,%r14d
  cmovns %esi,%edi
  movsd  D9(%rip),%xmm3
  lea    -0x1(%rsi),%esi
  xor    %ebp,%ebp
  xor    %r13d,%r13d
  lea    0x1(%rsi),%r10
  nopl   0x0(%rax,%rax,1)
L3:
  movsd  (%rdx,%rbp,8),%xmm0
  movsd  %xmm0,(%r9,%rbp,8)
  movsd  %xmm3,(%rcx,%rbp,8)
  test   %r14d,%r14d
  jle    L93
  movslq %r13d,%rbx
  movapd %xmm2,%xmm0
  lea    (%rsi,%rbx,1),%r11
  lea    (%r12,%rbx,8),%rax
  lea    0x8(%r12,%r11,8),%r11
  nopl   0x0(%rax)
L59:
  movsd  (%rax),%xmm1
  add    $0x8,%rax
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %r11,%rax
  jne    L59
  ucomisd %xmm0,%xmm2
  ja     L94
  sqrtsd %xmm0,%xmm0
  comisd %xmm2,%xmm0
  ja     L85
L93:
  mov    %edi,%eax
L98:
  add    %r13d,%eax
  cltq
  movq   $0x0,(%r8,%rax,8)
L28:
  add    $0x1,%rbp
  add    %r14d,%r13d
  cmp    %r15,%rbp
  jne    L3
L90:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L85:
  lea    0x0(,%rbx,8),%rax
  add    %r10,%rbx
  shl    $0x3,%rbx
  nop
L75:
  movsd  (%r12,%rax,1),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%r8,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rbx
  jne    L75
  jmp    L28
L94:
  mov    %r10,0x28(%rsp)
  mov    %rsi,0x20(%rsp)
  mov    %rcx,0x70(%rsp)
  mov    %r9,0x18(%rsp)
  mov    %r8,0x10(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    %edi,0x4(%rsp)
  call   <sqrt@plt>
  pxor   %xmm2,%xmm2
  mov    0x4(%rsp),%edi
  mov    D9(%rip),%rax
  comisd %xmm2,%xmm0
  mov    0x8(%rsp),%rdx
  mov    0x10(%rsp),%r8
  mov    0x18(%rsp),%r9
  mov    0x70(%rsp),%rcx
  movq   %rax,%xmm3
  mov    0x20(%rsp),%rsi
  mov    0x28(%rsp),%r10
  ja     L85
  mov    %r14d,%eax
  jmp    L98
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: f64

tool: D9: 1

assistant: ```c
void levelSetCurvatureCoefficientsEvaluate(int nPoints, int nSpace, double *grad_phi, double *u, double *f, double *r, double *dr) {
  int i, I;
  double norm_grad_phi;
  for (i = 0; i < nPoints; i++) {
    r[i] = u[i];
    dr[i] = 1.0;
    norm_grad_phi = 0.0;
    for (I = 0; I < nSpace; I++)
      norm_grad_phi += grad_phi[i * nSpace + I] * grad_phi[i * nSpace + I];
    norm_grad_phi = sqrt(norm_grad_phi);
    if (norm_grad_phi > 0.0) {
      for (I = 0; I < nSpace; I++)
        f[i * nSpace + I] = grad_phi[i * nSpace + I] / norm_grad_phi;
    } else
      f[i * nSpace + I] = 0.0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<argv_argc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L14
  jmp    L30
L28:
  addl   $0x1,-0x4(%rbp)
L30:
  cmpl   $0x7fff,-0x4(%rbp)
  jg     L14
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L28
L14:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int argv_argc(char **vector) {
  int argc = 0;
  if (vector != 0) {
    for (; argc < 32768 && vector[argc] != 0; argc++)
      ;
  }
  return argc;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<argv_argc>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L0
L21:
  cmpq   $0x0,(%rdi)
  je     L0
  add    $0x1,%eax
  add    $0x8,%rdi
  cmp    $0x8000,%eax
  jne    L21
L0:
  ret
```

assistant: ```c
int argv_argc(char **vector) {
  int argc = 0;
  if (vector != 0) {
    for (; argc < 32768 && vector[argc] != 0; argc++)
      ;
  }
  return argc;
}
```
user: What is the c source code of the assembly code below:

```asm
<argv_argc>:
  endbr64
  xor    %eax,%eax
  xor    %r8d,%r8d
  test   %rdi,%rdi
  jne    L14
  jmp    L5
L13:
  add    $0x1,%rax
  cmp    $0x8000,%rax
  je     L18
L14:
  cmpq   $0x0,(%rdi,%rax,8)
  mov    %eax,%r8d
  jne    L13
L5:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    $0x8000,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int argv_argc(char **vector) {
  int argc = 0;
  if (vector != 0) {
    for (; argc < 32768 && vector[argc] != 0; argc++)
      ;
  }
  return argc;
}
```
user: Translate the below assembly code into C programming language:

```asm
<argv_argc>:
  endbr64
  xor    %eax,%eax
  xor    %r8d,%r8d
  test   %rdi,%rdi
  jne    L33
  jmp    L17
L38:
  add    $0x1,%rax
  cmp    $0x8000,%rax
  je     L25
L33:
  cmpq   $0x0,(%rdi,%rax,8)
  mov    %eax,%r8d
  jne    L38
L17:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  mov    $0x8000,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int argv_argc(char **vector) {
  int argc = 0;
  if (vector != 0) {
    for (; argc < 32768 && vector[argc] != 0; argc++)
      ;
  }
  return argc;
}
```