ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_snprintf_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x1,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L52
L2:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L52:
  cmpl   $0x9,-0xc(%rbp)
  jg     L2
  cmpl   $0x0,-0x14(%rbp)
  jns    L0
  addl   $0x1,-0x10(%rbp)
L0:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L57:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x8(%rbp)
  jae    L17
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L17:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L57
  cmpl   $0x0,-0x14(%rbp)
  jns    L37
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
L37:
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int _snprintf_int(int num, char *buf, int buf_size) {
  int len = 1;
  char *p;
  int i = num < 0 ? -num : num;
  for (; i >= 10; i /= 10, len++)
    ;
  if (num < 0)
    len++;
  i = num;
  p = buf + len - 1;
  do {
    if (p < buf + buf_size)
      *p-- = '0' + i % 10;
    i /= 10;
  } while (i != 0);
  if (num < 0)
    *p = '-';
  return len < buf_size ? len : buf_size;
}
```
user: What is the c source code of the assembly code below?

```asm
<_snprintf_int>:
  endbr64
  mov    %rsi,%r9
  mov    %edx,%esi
  mov    %edi,%ecx
  neg    %ecx
  cmovs  %edi,%ecx
  cmp    $0x9,%ecx
  jle    L32
  mov    $0x1,%eax
L19:
  mov    %ecx,%r8d
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sar    $0x1f,%ecx
  mov    %ecx,%r10d
  mov    %edx,%ecx
  sub    %r10d,%ecx
  add    $0x1,%eax
  cmp    $0x63,%r8d
  jg     L19
L23:
  cmp    $0x80000000,%edi
  sbb    $0xffffffff,%eax
  movslq %eax,%rdx
  lea    -0x1(%r9,%rdx,1),%r8
  mov    %edi,%ecx
  movslq %esi,%rdx
  add    %rdx,%r9
  jmp    L45
L32:
  mov    $0x1,%eax
  jmp    L23
L33:
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%edx
  lea    (%rdx,%rdx,4),%r10d
  add    %r10d,%r10d
  mov    %ecx,%edx
  sub    %r10d,%edx
  add    $0x30,%edx
  mov    %dl,(%r8)
  lea    -0x1(%r8),%r8
L41:
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  je     L3
L45:
  cmp    %r9,%r8
  jae    L41
  jmp    L33
L3:
  test   %edi,%edi
  js     L53
L26:
  cmp    %esi,%eax
  cmovg  %esi,%eax
  ret
L53:
  movb   $0x2d,(%r8)
  jmp    L26
```

assistant: ```c
int _snprintf_int(int num, char *buf, int buf_size) {
  int len = 1;
  char *p;
  int i = num < 0 ? -num : num;
  for (; i >= 10; i /= 10, len++)
    ;
  if (num < 0)
    len++;
  i = num;
  p = buf + len - 1;
  do {
    if (p < buf + buf_size)
      *p-- = '0' + i % 10;
    i /= 10;
  } while (i != 0);
  if (num < 0)
    *p = '-';
  return len < buf_size ? len : buf_size;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_snprintf_int>:
  endbr64
  mov    %edx,%r8d
  mov    %edi,%edx
  mov    $0x1,%eax
  neg    %edx
  cmovs  %edi,%edx
  cmp    $0x9,%edx
  jle    L16
  mov    $0xcccccccd,%r9d
  xchg   %ax,%ax
L23:
  mov    %edx,%edx
  add    $0x1,%eax
  mov    %rdx,%rcx
  imul   %r9,%rdx
  shr    $0x23,%rdx
  cmp    $0x63,%ecx
  jg     L23
L16:
  cmp    $0x80000000,%edi
  mov    %edi,%ecx
  sbb    $0xffffffff,%eax
  movslq %eax,%rdx
  lea    -0x1(%rsi,%rdx,1),%r9
  movslq %r8d,%rdx
  lea    (%rsi,%rdx,1),%r10
  nop
L28:
  movslq %ecx,%rdx
  cmp    %r10,%r9
  jae    L35
  imul   $0x66666667,%rdx,%rsi
  mov    %ecx,%r11d
  sub    $0x1,%r9
  sar    $0x1f,%r11d
  sar    $0x22,%rsi
  sub    %r11d,%esi
  lea    (%rsi,%rsi,4),%r11d
  mov    %ecx,%esi
  add    %r11d,%r11d
  sub    %r11d,%esi
  add    $0x30,%esi
  mov    %sil,0x1(%r9)
L35:
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  mov    %ecx,%esi
  sar    $0x22,%rdx
  mov    %edx,%ecx
  sub    %esi,%ecx
  jne    L28
  test   %edi,%edi
  jns    L15
  movb   $0x2d,(%r9)
L15:
  cmp    %r8d,%eax
  cmovg  %r8d,%eax
  ret
```

assistant: ```c
int _snprintf_int(int num, char *buf, int buf_size) {
  int len = 1;
  char *p;
  int i = num < 0 ? -num : num;
  for (; i >= 10; i /= 10, len++)
    ;
  if (num < 0)
    len++;
  i = num;
  p = buf + len - 1;
  do {
    if (p < buf + buf_size)
      *p-- = '0' + i % 10;
    i /= 10;
  } while (i != 0);
  if (num < 0)
    *p = '-';
  return len < buf_size ? len : buf_size;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_snprintf_int>:
  endbr64
  mov    %edx,%r8d
  mov    %edi,%edx
  mov    $0x1,%eax
  neg    %edx
  cmovs  %edi,%edx
  cmp    $0x9,%edx
  jle    L23
  mov    $0xcccccccd,%r9d
  xchg   %ax,%ax
L12:
  mov    %edx,%edx
  add    $0x1,%eax
  mov    %rdx,%rcx
  imul   %r9,%rdx
  shr    $0x23,%rdx
  cmp    $0x63,%ecx
  jg     L12
L23:
  cmp    $0x80000000,%edi
  sbb    $0xffffffff,%eax
  movslq %eax,%rdx
  lea    -0x1(%rsi,%rdx,1),%r9
  movslq %r8d,%rdx
  lea    (%rsi,%rdx,1),%r10
  mov    %edi,%edx
  nop
L2:
  cmp    %r10,%r9
  jae    L6
  movslq %edx,%rcx
  mov    %edx,%esi
  lea    -0x1(%r9),%r11
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%esi
  sar    $0x22,%rcx
  sub    %esi,%ecx
  lea    (%rcx,%rcx,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%r9)
  mov    %ecx,%edx
  mov    %r11,%r9
  test   %ecx,%ecx
  jne    L2
L49:
  test   %edi,%edi
  jns    L59
  movb   $0x2d,(%r9)
L59:
  cmp    %r8d,%eax
  cmovg  %r8d,%eax
  ret
  nopl   0x0(%rax)
L6:
  movslq %edx,%rcx
  mov    %edx,%esi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%esi
  sar    $0x22,%rcx
  sub    %esi,%ecx
  mov    %ecx,%edx
  je     L49
L47:
  movslq %edx,%rcx
  mov    %edx,%esi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%esi
  sar    $0x22,%rcx
  mov    %ecx,%edx
  sub    %esi,%edx
  jne    L47
  jmp    L49
```

assistant: ```c
int _snprintf_int(int num, char *buf, int buf_size) {
  int len = 1;
  char *p;
  int i = num < 0 ? -num : num;
  for (; i >= 10; i /= 10, len++)
    ;
  if (num < 0)
    len++;
  i = num;
  p = buf + len - 1;
  do {
    if (p < buf + buf_size)
      *p-- = '0' + i % 10;
    i /= 10;
  } while (i != 0);
  if (num < 0)
    *p = '-';
  return len < buf_size ? len : buf_size;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<F5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movb   $0x0,-0x1(%rbp)
  movzbl -0x14(%rbp),%eax
  xor    $0x1,%eax
  test   %al,%al
  jne    L9
  movzbl -0x18(%rbp),%eax
  xor    $0x1,%eax
  test   %al,%al
  je     L5
L9:
  movb   $0x1,-0x1(%rbp)
L5:
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
_Bool F5(_Bool B1, _Bool B2) {
  _Bool R = 0;
  if (!(B1 && B2))
    R = 1;
  return R;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<F5>:
  endbr64
  and    %edi,%esi
  mov    %esi,%eax
  xor    $0x1,%eax
  ret
```

assistant: ```c
_Bool F5(_Bool B1, _Bool B2) {
  _Bool R = 0;
  if (!(B1 && B2))
    R = 1;
  return R;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<F5>:
  endbr64
  and    %edi,%esi
  mov    %esi,%eax
  xor    $0x1,%eax
  ret
```

assistant: ```c
_Bool F5(_Bool B1, _Bool B2) {
  _Bool R = 0;
  if (!(B1 && B2))
    R = 1;
  return R;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<F5>:
  endbr64
  and    %edi,%esi
  mov    %esi,%eax
  xor    $0x1,%eax
  ret
```

assistant: ```c
_Bool F5(_Bool B1, _Bool B2) {
  _Bool R = 0;
  if (!(B1 && B2))
    R = 1;
  return R;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getSigFreq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x18,-0x4(%rbp)
  ja     L49
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D16(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D16(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x5de70060,%eax
  jmp    L53
  mov    $0x5de70060,%eax
  jmp    L53
  mov    $0x492bb080,%eax
  jmp    L53
  mov    $0x492bb080,%eax
  jmp    L53
  mov    $0x461f33d0,%eax
  jmp    L53
  mov    $0x5de70060,%eax
  jmp    L53
  mov    $0x5de70060,%eax
  jmp    L53
  mov    $0x492bb080,%eax
  jmp    L53
  mov    $0x492bb080,%eax
  jmp    L53
  mov    $0x5de70060,%eax
  jmp    L53
  mov    $0x5de70060,%eax
  jmp    L53
  mov    $0x4c382d30,%eax
  jmp    L53
  mov    $0x4c382d30,%eax
  jmp    L53
  mov    $0x461f33d0,%eax
  jmp    L53
  mov    $0x461f33d0,%eax
  jmp    L53
  mov    $0x461f33d0,%eax
  jmp    L53
  mov    $0x5de70060,%eax
  jmp    L53
L49:
  mov    $0x5de70060,%eax
L53:
  pop    %rbp
  ret
```

assistant: ```c
int getSigFreq(int _signType) {
  switch (_signType) {
  case 0:
    return 1.57542E9;
  case 1:
    return 1.57542E9;
  case 2:
    return 1.22760E9;
  case 3:
    return 1.22760E9;
  case 4:
    return 1.17645E9;
  case 8:
    return 1.57542E9;
  case 9:
    return 1.57542E9;
  case 10:
    return 1.22760E9;
  case 11:
    return 1.22760E9;
  case 16:
    return 1.57542E9;
  case 17:
    return 1.57542E9;
  case 18:
    return 1.27875E9;
  case 19:
    return 1.27875E9;
  case 20:
    return 1.17645E9;
  case 21:
    return 1.17645E9;
  case 22:
    return 1.17645E9;
  case 24:
    return 1.57542E9;
  }
  return 1.57542E9;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getSigFreq>:
  endbr64
  cmp    $0x16,%edi
  ja     L6
  mov    %edi,%edi
  lea    D1(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x492bb080,%eax
  ret
  mov    $0x461f33d0,%eax
  ret
  mov    $0x4c382d30,%eax
  ret
  mov    $0x4c382d30,%eax
  ret
  mov    $0x461f33d0,%eax
  ret
  mov    $0x461f33d0,%eax
  ret
  mov    $0x461f33d0,%eax
  ret
L6:
  mov    $0x5de70060,%eax
  ret
```

assistant: ```c
int getSigFreq(int _signType) {
  switch (_signType) {
  case 0:
    return 1.57542E9;
  case 1:
    return 1.57542E9;
  case 2:
    return 1.22760E9;
  case 3:
    return 1.22760E9;
  case 4:
    return 1.17645E9;
  case 8:
    return 1.57542E9;
  case 9:
    return 1.57542E9;
  case 10:
    return 1.22760E9;
  case 11:
    return 1.22760E9;
  case 16:
    return 1.57542E9;
  case 17:
    return 1.57542E9;
  case 18:
    return 1.27875E9;
  case 19:
    return 1.27875E9;
  case 20:
    return 1.17645E9;
  case 21:
    return 1.17645E9;
  case 22:
    return 1.17645E9;
  case 24:
    return 1.57542E9;
  }
  return 1.57542E9;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<digest_fold>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L17
L15:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  xor    %al,-0x5(%rbp)
  addl   $0x1,-0x4(%rbp)
L17:
  cmpl   $0xf,-0x4(%rbp)
  jle    L15
  movzbl -0x5(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char digest_fold(const unsigned char *mdigest) {
  int i;
  unsigned char rval;
  rval = mdigest[0];
  for (i = 1; i < 16; i++) {
    rval = rval ^ mdigest[i];
  }
  return (rval);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<digest_fold>:
  endbr64
  movzbl (%rdi),%edx
  lea    0x1(%rdi),%rax
  lea    0x10(%rdi),%rcx
L3:
  xor    (%rax),%dl
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L3
  mov    %edx,%eax
  ret
```

assistant: ```c
unsigned char digest_fold(const unsigned char *mdigest) {
  int i;
  unsigned char rval;
  rval = mdigest[0];
  for (i = 1; i < 16; i++) {
    rval = rval ^ mdigest[i];
  }
  return (rval);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<digest_fold>:
  endbr64
  movzbl (%rdi),%r8d
  lea    0x1(%rdi),%rax
  lea    0x10(%rdi),%rdx
L1:
  xor    (%rax),%r8b
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L1
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned char digest_fold(const unsigned char *mdigest) {
  int i;
  unsigned char rval;
  rval = mdigest[0];
  for (i = 1; i < 16; i++) {
    rval = rval ^ mdigest[i];
  }
  return (rval);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<digest_fold>:
  endbr64
  mov    0x1(%rdi),%rax
  movzbl 0x9(%rdi),%edx
  xor    0xa(%rdi),%dl
  xor    (%rdi),%dl
  mov    %eax,%ecx
  xor    0xb(%rdi),%dl
  xor    0xc(%rdi),%dl
  xor    0xd(%rdi),%dl
  xor    0xe(%rdi),%dl
  xor    0xf(%rdi),%dl
  xor    %ecx,%edx
  movzbl %ah,%ecx
  xor    %ecx,%edx
  mov    %rax,%rcx
  shr    $0x10,%rcx
  xor    %ecx,%edx
  mov    %rax,%rcx
  shr    $0x18,%rcx
  xor    %ecx,%edx
  mov    %rax,%rcx
  shr    $0x20,%rcx
  xor    %ecx,%edx
  mov    %rax,%rcx
  shr    $0x28,%rcx
  xor    %ecx,%edx
  mov    %rax,%rcx
  shr    $0x38,%rax
  shr    $0x30,%rcx
  xor    %ecx,%edx
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned char digest_fold(const unsigned char *mdigest) {
  int i;
  unsigned char rval;
  rval = mdigest[0];
  for (i = 1; i < 16; i++) {
    rval = rval ^ mdigest[i];
  }
  return (rval);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_collatz_conjecture>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L12
  mov    $0x1,%eax
  jmp    L24
L12:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L29
  mov    -0x4(%rbp),%eax
  shr    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L1
L29:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <ft_collatz_conjecture>
L24:
  leave
  ret
```

assistant: ```c
unsigned int ft_collatz_conjecture(unsigned int base) {
  if (base == 1)
    return (1);
  if ((base % 2) == 0)
    base = base / 2;
  else
    base = (base * 3) + 1;
  return (ft_collatz_conjecture(base));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_collatz_conjecture>:
  endbr64
  mov    %edi,%eax
  cmp    $0x1,%edi
  je     L25
  sub    $0x8,%rsp
  test   $0x1,%dil
  jne    L6
  shr    %edi
L19:
  call   <ft_collatz_conjecture>
  add    $0x8,%rsp
  ret
L6:
  lea    0x1(%rdi,%rdi,2),%edi
  jmp    L19
L25:
  ret
```

assistant: ```c
unsigned int ft_collatz_conjecture(unsigned int base) {
  if (base == 1)
    return (1);
  if ((base % 2) == 0)
    base = base / 2;
  else
    base = (base * 3) + 1;
  return (ft_collatz_conjecture(base));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_collatz_conjecture>:
  endbr64
  nopl   0x0(%rax)
L22:
  cmp    $0x1,%edi
  je     L10
L25:
  test   $0x1,%dil
  jne    L39
  shr    %edi
  cmp    $0x1,%edi
  jne    L25
L10:
  mov    $0x1,%eax
  ret
L39:
  lea    0x1(%rdi,%rdi,2),%edi
  jmp    L22
```

assistant: ```c
unsigned int ft_collatz_conjecture(unsigned int base) {
  if (base == 1)
    return (1);
  if ((base % 2) == 0)
    base = base / 2;
  else
    base = (base * 3) + 1;
  return (ft_collatz_conjecture(base));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_collatz_conjecture>:
  endbr64
  nopl   0x0(%rax)
L35:
  cmp    $0x1,%edi
  je     L11
L30:
  test   $0x1,%dil
  jne    L2
  shr    %edi
  cmp    $0x1,%edi
  jne    L30
L11:
  mov    $0x1,%eax
  ret
L2:
  lea    0x1(%rdi,%rdi,2),%edi
  jmp    L35
```

assistant: ```c
unsigned int ft_collatz_conjecture(unsigned int base) {
  if (base == 1)
    return (1);
  if ((base % 2) == 0)
    base = base / 2;
  else
    base = (base * 3) + 1;
  return (ft_collatz_conjecture(base));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<h2net>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  shr    $0x8,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  cmpl   $0x2,-0x18(%rbp)
  je     L10
  mov    -0x14(%rbp),%eax
  shr    $0x10,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L15
L10:
  nop
L15:
  pop    %rbp
  ret
```

assistant: ```c
void h2net(unsigned int h, unsigned char *net, int n) {
  int i = 0;
  net[i++] = h & 0xff;
  net[i++] = (h >> 8) & 0xff;
  if (n == 2)
    return;
  net[i++] = (h >> 16) & 0xff;
  net[i++] = (h >> 24) & 0xff;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<h2net>:
  endbr64
  mov    %edi,%eax
  mov    %dil,(%rsi)
  movzbl %ah,%ecx
  mov    %cl,0x1(%rsi)
  cmp    $0x2,%edx
  je     L2
  mov    %edi,%edx
  shr    $0x10,%edx
  mov    %dl,0x2(%rsi)
  shr    $0x18,%eax
  mov    %al,0x3(%rsi)
L2:
  ret
```

assistant: ```c
void h2net(unsigned int h, unsigned char *net, int n) {
  int i = 0;
  net[i++] = h & 0xff;
  net[i++] = (h >> 8) & 0xff;
  if (n == 2)
    return;
  net[i++] = (h >> 16) & 0xff;
  net[i++] = (h >> 24) & 0xff;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<h2net>:
  endbr64
  mov    %di,(%rsi)
  cmp    $0x2,%edx
  je     L4
  mov    %edi,%eax
  shr    $0x18,%edi
  shr    $0x10,%eax
  mov    %dil,0x3(%rsi)
  mov    %al,0x2(%rsi)
L4:
  ret
```

assistant: ```c
void h2net(unsigned int h, unsigned char *net, int n) {
  int i = 0;
  net[i++] = h & 0xff;
  net[i++] = (h >> 8) & 0xff;
  if (n == 2)
    return;
  net[i++] = (h >> 16) & 0xff;
  net[i++] = (h >> 24) & 0xff;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<h2net>:
  endbr64
  mov    %di,(%rsi)
  cmp    $0x2,%edx
  je     L4
  mov    %edi,%eax
  shr    $0x18,%edi
  shr    $0x10,%eax
  mov    %dil,0x3(%rsi)
  mov    %al,0x2(%rsi)
L4:
  ret
```

assistant: ```c
void h2net(unsigned int h, unsigned char *net, int n) {
  int i = 0;
  net[i++] = h & 0xff;
  net[i++] = (h >> 8) & 0xff;
  if (n == 2)
    return;
  net[i++] = (h >> 16) & 0xff;
  net[i++] = (h >> 24) & 0xff;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L107
L89:
  lea    -0x28(%rbp),%rdx
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  movl   $0x1,-0x20(%rbp)
  jmp    L62
L42:
  movl   $0x1,-0x1c(%rbp)
  jmp    L21
L35:
  cmpl   $0x1,-0x1c(%rbp)
  jne    L70
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L22
L70:
  subl   $0x1,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L22:
  mov    -0x28(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jne    L101
  mov    -0x2c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L101
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L119
L101:
  addl   $0x1,-0x10(%rbp)
  addl   $0x1,-0x1c(%rbp)
L21:
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L35
L119:
  addl   $0x1,-0x20(%rbp)
L62:
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  cmp    %eax,-0x20(%rbp)
  jle    L42
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L107:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jg     L89
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D21
type_infered: string
name: unpack_data
arguments: data_tag: D17
type_infered: string
name: unpack_data
arguments: data_tag: D20
type_infered: string

tool: D21: %d

tool: D17: %d %d

tool: D20: %d


assistant: ```c
int main() {
  int i, j, a, b, rank, u, v, T, max;
  rank = 1;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d %d", &u, &v);
    rank = 1;
    for (i = 1; i <= u + v + 1; i++) {
      for (j = 1; j <= i; j++) {
        if (j == 1) {
          a = i - 1;
          b = 0;
        } else {
          a--;
          b++;
        }
        if (a == v && b == u) {
          max = rank;
          break;
        }
        rank++;
      }
    }
    printf("%d\n", max);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x14(%rsp)
  test   %eax,%eax
  jle    L103
  lea    0x10(%rsp),%r15
  lea    0xc(%rsp),%r14
  lea    D3(%rip),%r13
  jmp    L125
L24:
  lea    -0x1(%rcx),%ebp
  mov    $0x0,%ebx
  jmp    L111
L121:
  add    $0x1,%edx
  add    $0x1,%eax
  cmp    %ecx,%eax
  jg     L118
L83:
  sub    $0x1,%ebp
  add    $0x1,%ebx
  cmp    $0x1,%eax
  je     L24
L111:
  cmp    %esi,%ebx
  jne    L121
  cmp    %edi,%ebp
  jne    L121
  mov    %edx,%r12d
L118:
  add    $0x1,%ecx
  cmp    %ecx,%r8d
  je     L17
L47:
  mov    $0x1,%eax
  test   %ecx,%ecx
  jg     L83
  jmp    L118
L17:
  mov    %r12d,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x14(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x14(%rsp)
  test   %eax,%eax
  jle    L103
L125:
  mov    %r15,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x10(%rsp),%edi
  mov    %esi,%r8d
  add    %edi,%r8d
  js     L17
  add    $0x2,%r8d
  mov    $0x1,%edx
  mov    $0x1,%ecx
  jmp    L47
L103:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D23
data_type: string
name: parse_data
arguments: data_label: D3
data_type: string
name: parse_data
arguments: data_label: D16
data_type: string

tool: D23: %d

tool: D3: %d %d

tool: D16: %d


assistant: ```c
int main() {
  int i, j, a, b, rank, u, v, T, max;
  rank = 1;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d %d", &u, &v);
    rank = 1;
    for (i = 1; i <= u + v + 1; i++) {
      for (j = 1; j <= i; j++) {
        if (j == 1) {
          a = i - 1;
          b = 0;
        } else {
          a--;
          b++;
        }
        if (a == v && b == u) {
          max = rank;
          break;
        }
        rank++;
      }
    }
    printf("%d\n", max);
  }
  return 0;
}
```