ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inIntervalo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    %r8d,-0x34(%rbp)
  mov    %r9d,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <time@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x38(%rbp),%r8d
  mov    -0x34(%rbp),%edi
  mov    -0x30(%rbp),%ecx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  jne    L41
  mov    -0x14(%rbp),%r8d
  mov    -0x18(%rbp),%edi
  mov    -0x1c(%rbp),%ecx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  jne    L67
  mov    -0x38(%rbp),%r8d
  mov    -0x34(%rbp),%edi
  mov    -0x30(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x1c(%rbp),%eax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  jne    L67
  mov    $0x1,%eax
  jmp    L59
L67:
  mov    $0x0,%eax
  jmp    L59
L41:
  mov    -0x2c(%rbp),%r8d
  mov    -0x28(%rbp),%edi
  mov    -0x24(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x1c(%rbp),%eax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <timeCmp@plt>
  cmp    $0x1,%eax
  je     L36
  mov    -0x38(%rbp),%r8d
  mov    -0x34(%rbp),%edi
  mov    -0x30(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x1c(%rbp),%eax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  jne    L53
L36:
  mov    $0x1,%eax
  jmp    L59
L53:
  mov    $0x0,%eax
L59:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

assistant: ```c
int inIntervalo(int h1, int m1, int s1, int h2, int m2, int s2) {
  struct tm *taim;
  time_t tiempo;
  time(&tiempo);
  taim = localtime(&tiempo);
  int ha = taim->tm_hour, ma = taim->tm_min, sa = taim->tm_sec;
  if (timeCmp(h1, m1, s1, h2, m2, s2) == -1) {
    if ((timeCmp(h1, m1, s1, ha, ma, sa) == -1) && (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
      return 1;
    else
      return 0;
  } else if ((timeCmp(ha, ma, sa, h1, m1, s1) == 1) || (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
    return 1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<inIntervalo>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebx
  mov    %edx,%ebp
  mov    %ecx,0x4(%rsp)
  mov    %r8d,0x8(%rsp)
  mov    %r9d,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r13
  mov    %r13,%rdi
  call   <time@plt>
  mov    %r13,%rdi
  call   <localtime@plt>
  mov    (%rax),%r13d
  mov    0x4(%rax),%r14d
  mov    0x8(%rax),%r15d
  mov    0xc(%rsp),%r9d
  mov    0x8(%rsp),%r8d
  mov    0x4(%rsp),%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12d,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  je     L24
  mov    %ebp,%r9d
  mov    %ebx,%r8d
  mov    %r12d,%ecx
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0x1,%eax
  je     L37
  mov    0xc(%rsp),%r9d
  mov    0x8(%rsp),%r8d
  mov    0x4(%rsp),%ecx
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
L37:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  mov    %r15d,%r9d
  mov    %r14d,%r8d
  mov    %r13d,%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12d,%edi
  call   <timeCmp@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0xffffffff,%edx
  jne    L37
  mov    0xc(%rsp),%r9d
  mov    0x8(%rsp),%r8d
  mov    0x4(%rsp),%ecx
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L37
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int inIntervalo(int h1, int m1, int s1, int h2, int m2, int s2) {
  struct tm *taim;
  time_t tiempo;
  time(&tiempo);
  taim = localtime(&tiempo);
  int ha = taim->tm_hour, ma = taim->tm_min, sa = taim->tm_sec;
  if (timeCmp(h1, m1, s1, h2, m2, s2) == -1) {
    if ((timeCmp(h1, m1, s1, ha, ma, sa) == -1) && (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
      return 1;
    else
      return 0;
  } else if ((timeCmp(ha, ma, sa, h1, m1, s1) == 1) || (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
    return 1;
  else
    return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<inIntervalo>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %r9d,%ebp
  push   %rbx
  mov    %r8d,%ebx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  lea    0x14(%rsp),%r13
  mov    %edx,0x4(%rsp)
  mov    %r13,%rdi
  mov    %ecx,(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  mov    %r13,%rdi
  call   <localtime@plt>
  mov    0x4(%rsp),%edx
  mov    0xc(%rsp),%esi
  mov    %ebp,%r9d
  mov    (%rsp),%ecx
  mov    %ebx,%r8d
  mov    %r12d,%edi
  mov    (%rax),%r13d
  mov    0x4(%rax),%r14d
  mov    0x8(%rax),%r15d
  mov    %edx,0x8(%rsp)
  mov    %esi,0x4(%rsp)
  call   <timeCmp@plt>
  mov    0x4(%rsp),%esi
  mov    0x8(%rsp),%edx
  cmp    $0xffffffff,%eax
  je     L33
  mov    %edx,%r9d
  mov    %esi,%r8d
  mov    %r12d,%ecx
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0x1,%eax
  je     L9
L14:
  mov    (%rsp),%ecx
  mov    %ebp,%r9d
  mov    %ebx,%r8d
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
L9:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L33:
  mov    %r14d,%r8d
  mov    %r15d,%r9d
  mov    %r13d,%ecx
  mov    %r12d,%edi
  call   <timeCmp@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0xffffffff,%r8d
  jne    L9
  jmp    L14
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int inIntervalo(int h1, int m1, int s1, int h2, int m2, int s2) {
  struct tm *taim;
  time_t tiempo;
  time(&tiempo);
  taim = localtime(&tiempo);
  int ha = taim->tm_hour, ma = taim->tm_min, sa = taim->tm_sec;
  if (timeCmp(h1, m1, s1, h2, m2, s2) == -1) {
    if ((timeCmp(h1, m1, s1, ha, ma, sa) == -1) && (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
      return 1;
    else
      return 0;
  } else if ((timeCmp(ha, ma, sa, h1, m1, s1) == 1) || (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
    return 1;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<inIntervalo>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %r9d,%ebp
  push   %rbx
  mov    %r8d,%ebx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  lea    0x14(%rsp),%r13
  mov    %edx,0x4(%rsp)
  mov    %r13,%rdi
  mov    %ecx,(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  mov    %r13,%rdi
  call   <localtime@plt>
  mov    0x4(%rsp),%edx
  mov    0xc(%rsp),%esi
  mov    %ebp,%r9d
  mov    (%rsp),%ecx
  mov    %ebx,%r8d
  mov    %r12d,%edi
  mov    (%rax),%r13d
  mov    0x4(%rax),%r14d
  mov    0x8(%rax),%r15d
  mov    %edx,0x8(%rsp)
  mov    %esi,0x4(%rsp)
  call   <timeCmp@plt>
  mov    0x4(%rsp),%esi
  mov    0x8(%rsp),%edx
  cmp    $0xffffffff,%eax
  je     L30
  mov    %edx,%r9d
  mov    %esi,%r8d
  mov    %r12d,%ecx
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0x1,%eax
  je     L22
L24:
  mov    (%rsp),%ecx
  mov    %ebp,%r9d
  mov    %ebx,%r8d
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13d,%edi
  call   <timeCmp@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
L22:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L30:
  mov    %r14d,%r8d
  mov    %r15d,%r9d
  mov    %r13d,%ecx
  mov    %r12d,%edi
  call   <timeCmp@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0xffffffff,%r8d
  jne    L22
  jmp    L24
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int inIntervalo(int h1, int m1, int s1, int h2, int m2, int s2) {
  struct tm *taim;
  time_t tiempo;
  time(&tiempo);
  taim = localtime(&tiempo);
  int ha = taim->tm_hour, ma = taim->tm_min, sa = taim->tm_sec;
  if (timeCmp(h1, m1, s1, h2, m2, s2) == -1) {
    if ((timeCmp(h1, m1, s1, ha, ma, sa) == -1) && (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
      return 1;
    else
      return 0;
  } else if ((timeCmp(ha, ma, sa, h1, m1, s1) == 1) || (timeCmp(ha, ma, sa, h2, m2, s2) == -1))
    return 1;
  else
    return 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
<slow_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L104
L3:
  cmpl   $0x1,-0x14(%rbp)
  jne    L92
  mov    $0x0,%eax
  jmp    L104
L92:
  cmpl   $0x2,-0x14(%rbp)
  jne    L22
  mov    $0x1,%eax
  jmp    L104
L22:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L37
  mov    $0x0,%eax
  jmp    L104
L37:
  movl   $0x2,-0x4(%rbp)
  jmp    L42
L85:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L94
  mov    $0x0,%eax
  jmp    L104
L94:
  addl   $0x1,-0x4(%rbp)
L42:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  cmp    %eax,-0x4(%rbp)
  jle    L85
  mov    $0x1,%eax
L104:
  leave
  ret
```

assistant: ```c
int slow_prime(int n) {
  int i;
  if (n == 0) {
    return 0;
  }
  if (n == 1) {
    return 0;
  }
  if (n == 2) {
    return 1;
  }
  if (n % 2 == 0) {
    return 0;
  }
  for (i = 2; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<slow_prime>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%edi
  jbe    L37
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x1,%eax
  cmp    $0x2,%edi
  je     L52
  mov    $0x0,%eax
  mov    $0x2,%ebp
  test   $0x1,%dil
  je     L52
L11:
  mov    %ebx,%edi
  call   <sqrt@plt>
  cmp    %ebp,%eax
  jl     L57
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %edx,%eax
  test   %edx,%edx
  je     L52
  add    $0x1,%ebp
  jmp    L11
L57:
  mov    $0x1,%eax
L52:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L37:
  ret
```

assistant: ```c
int slow_prime(int n) {
  int i;
  if (n == 0) {
    return 0;
  }
  if (n == 1) {
    return 0;
  }
  if (n == 2) {
    return 1;
  }
  if (n % 2 == 0) {
    return 0;
  }
  for (i = 2; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<slow_prime>:
  endbr64
  cmp    $0x1,%edi
  jbe    L33
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L59
  mov    $0x2,%ebx
  test   $0x1,%dil
  jne    L14
  jmp    L11
  nopl   (%rax)
L4:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L11
  add    $0x1,%ebx
L14:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L4
L59:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %r12
  ret
  xchg   %ax,%ax
L11:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  xor    %eax,%eax
  ret
```

assistant: ```c
int slow_prime(int n) {
  int i;
  if (n == 0) {
    return 0;
  }
  if (n == 1) {
    return 0;
  }
  if (n == 2) {
    return 1;
  }
  if (n % 2 == 0) {
    return 0;
  }
  for (i = 2; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<slow_prime>:
  endbr64
  cmp    $0x1,%edi
  jbe    L47
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L55
  mov    $0x2,%ebx
  test   $0x1,%dil
  jne    L52
  jmp    L22
  nopl   (%rax)
L28:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L22
  add    $0x1,%ebx
L52:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L28
L55:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %r12
  ret
  xchg   %ax,%ax
L22:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  xor    %eax,%eax
  ret
```

assistant: ```c
int slow_prime(int n) {
  int i;
  if (n == 0) {
    return 0;
  }
  if (n == 1) {
    return 0;
  }
  if (n == 2) {
    return 1;
  }
  if (n % 2 == 0) {
    return 0;
  }
  for (i = 2; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_citrus_bcs_skip_ws>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L18
L22:
  addq   $0x1,-0x8(%rbp)
L18:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L19
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <_citrus_bcs_isspace@plt>
  test   %rax,%rax
  jne    L22
L19:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
const char *_citrus_bcs_skip_ws(const char *p) {
  while (*p && _citrus_bcs_isspace(*p))
    p++;
  return (p);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_citrus_bcs_skip_ws>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  mov    %rbx,%rax
  test   %dil,%dil
  je     L33
L18:
  movsbl %dil,%edi
  call   <_citrus_bcs_isspace@plt>
  test   %rax,%rax
  je     L22
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L18
  mov    %rbx,%rax
  jmp    L33
L22:
  mov    %rbx,%rax
L33:
  pop    %rbx
  ret
```

assistant: ```c
const char *_citrus_bcs_skip_ws(const char *p) {
  while (*p && _citrus_bcs_isspace(*p))
    p++;
  return (p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_citrus_bcs_skip_ws>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L26
  jmp    L22
  nopl   0x0(%rax,%rax,1)
L11:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L22
L26:
  call   <_citrus_bcs_isspace@plt>
  test   %rax,%rax
  jne    L11
L22:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *_citrus_bcs_skip_ws(const char *p) {
  while (*p && _citrus_bcs_isspace(*p))
    p++;
  return (p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_citrus_bcs_skip_ws>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L4
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L17:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L26
L4:
  call   <_citrus_bcs_isspace@plt>
  test   %rax,%rax
  jne    L17
L26:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *_citrus_bcs_skip_ws(const char *p) {
  while (*p && _citrus_bcs_isspace(*p))
    p++;
  return (p);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sg_put_unaligned_le32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    -0x4(%rbp),%eax
  sar    $0x10,%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <sg_put_unaligned_le16@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <sg_put_unaligned_le16@plt>
  nop
  leave
  ret
```

assistant: ```c
void sg_put_unaligned_le32(uint32_t val, void *p) {
  sg_put_unaligned_le16(val >> 16, (uint8_t *)p + 2);
  sg_put_unaligned_le16(val, p);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sg_put_unaligned_le32>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  lea    0x2(%rsi),%rsi
  sar    $0x10,%edi
  call   <sg_put_unaligned_le16@plt>
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <sg_put_unaligned_le16@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sg_put_unaligned_le32(uint32_t val, void *p) {
  sg_put_unaligned_le16(val >> 16, (uint8_t *)p + 2);
  sg_put_unaligned_le16(val, p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sg_put_unaligned_le32>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    0x2(%rsi),%rsi
  push   %rbp
  mov    %edi,%ebp
  sar    $0x10,%edi
  sub    $0x8,%rsp
  call   <sg_put_unaligned_le16@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <sg_put_unaligned_le16@plt>
```

assistant: ```c
void sg_put_unaligned_le32(uint32_t val, void *p) {
  sg_put_unaligned_le16(val >> 16, (uint8_t *)p + 2);
  sg_put_unaligned_le16(val, p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sg_put_unaligned_le32>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    0x2(%rsi),%rsi
  push   %rbp
  mov    %edi,%ebp
  sar    $0x10,%edi
  sub    $0x8,%rsp
  call   <sg_put_unaligned_le16@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <sg_put_unaligned_le16@plt>
```

assistant: ```c
void sg_put_unaligned_le32(uint32_t val, void *p) {
  sg_put_unaligned_le16(val >> 16, (uint8_t *)p + 2);
  sg_put_unaligned_le16(val, p);
}
```
user: What is the c source code of the assembly code below?

```asm
<s_pread>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L41
L26:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,%rdi
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  mov    %rdi,%rdx
  mov    %eax,%edi
  call   <pread@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L78
  mov    D28(%rip),%rax
  mov    (%rax),%rdx
  mov    D3(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L41
  mov    D22(%rip),%rax
  mov    (%rax),%rdx
  mov    D3(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L81
  jmp    L41
L81:
  mov    $0xffffffff,%eax
  jmp    L13
L78:
  cmpl   $0x0,-0x4(%rbp)
  jne    L21
  mov    -0xc(%rbp),%eax
  jmp    L13
L21:
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x8(%rbp)
L41:
  mov    -0xc(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L26
  mov    -0xc(%rbp),%eax
L13:
  leave
  ret
```

assistant: ```c
int s_pread(int fd, void *buf, size_t len, off_t off) {
  int total;
  int thistime;
  off_t thisoffset;
  thisoffset = off;
  for (total = 0; total < len;) {
    thistime = pread(fd, buf + total, len - total, thisoffset);
    if (thistime < 0) {
      if (EINTR == errno || EAGAIN == errno)
        continue;
      return -1;
    } else if (thistime == 0) {
      return total;
    }
    total += thistime;
    thisoffset += total;
  }
  return total;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<s_pread>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdx,%rdx
  je     L16
  mov    %edi,%r13d
  mov    %rsi,%r14
  mov    %rdx,%r12
  mov    %ecx,%ebp
  mov    $0x0,%ebx
  mov    $0x0,%esi
  jmp    L52
L50:
  mov    D19(%rip),%rax
  mov    (%rax),%rax
  mov    D22(%rip),%rdx
  cmp    %rax,(%rdx)
  je     L79
  mov    D18(%rip),%rdx
  cmp    (%rdx),%rax
  jne    L20
L79:
  movslq %ebx,%rsi
  cmp    %r12,%rsi
  jae    L63
L52:
  mov    %r12,%rdx
  sub    %rsi,%rdx
  add    %r14,%rsi
  mov    %ebp,%ecx
  mov    %r13d,%edi
  call   <pread@plt>
  test   %eax,%eax
  js     L50
  je     L63
  add    %eax,%ebx
  add    %ebx,%ebp
  jmp    L79
L20:
  mov    $0xffffffff,%ebx
L63:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  mov    $0x0,%ebx
  jmp    L63
```

assistant: ```c
int s_pread(int fd, void *buf, size_t len, off_t off) {
  int total;
  int thistime;
  off_t thisoffset;
  thisoffset = off;
  for (total = 0; total < len;) {
    thistime = pread(fd, buf + total, len - total, thisoffset);
    if (thistime < 0) {
      if (EINTR == errno || EAGAIN == errno)
        continue;
      return -1;
    } else if (thistime == 0) {
      return total;
    }
    total += thistime;
    thisoffset += total;
  }
  return total;
}
```
user: What is the c source code of the assembly code below:

```asm
<s_pread>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdx,%rdx
  je     L73
  mov    %rsi,%r12
  mov    %edi,%ebp
  mov    %rdx,%rbx
  mov    %ecx,%r14d
  xor    %r13d,%r13d
  xor    %esi,%esi
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L19:
  je     L57
  add    %eax,%r13d
  add    %r13d,%r14d
L52:
  movslq %r13d,%rsi
  cmp    %rbx,%rsi
  jae    L57
L11:
  mov    %rbx,%rdx
  mov    %r14d,%ecx
  mov    %ebp,%edi
  sub    %rsi,%rdx
  add    %r12,%rsi
  call   <pread@plt>
  test   %eax,%eax
  jns    L19
  mov    D7(%rip),%rax
  mov    D12(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L52
  mov    D14(%rip),%rdx
  cmp    (%rdx),%rax
  je     L52
  mov    $0xffffffff,%r13d
L57:
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L73:
  xor    %r13d,%r13d
  jmp    L57
```

assistant: ```c
int s_pread(int fd, void *buf, size_t len, off_t off) {
  int total;
  int thistime;
  off_t thisoffset;
  thisoffset = off;
  for (total = 0; total < len;) {
    thistime = pread(fd, buf + total, len - total, thisoffset);
    if (thistime < 0) {
      if (EINTR == errno || EAGAIN == errno)
        continue;
      return -1;
    } else if (thistime == 0) {
      return total;
    }
    total += thistime;
    thisoffset += total;
  }
  return total;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<s_pread>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdx,%rdx
  je     L38
  mov    %rsi,%r12
  mov    %edi,%ebp
  mov    %rdx,%rbx
  mov    %ecx,%r14d
  xor    %r13d,%r13d
  xor    %esi,%esi
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L16:
  je     L71
  add    %eax,%r13d
  add    %r13d,%r14d
L9:
  movslq %r13d,%rsi
  cmp    %rbx,%rsi
  jae    L71
L37:
  mov    %rbx,%rdx
  mov    %r14d,%ecx
  mov    %ebp,%edi
  sub    %rsi,%rdx
  add    %r12,%rsi
  call   <pread@plt>
  test   %eax,%eax
  jns    L16
  mov    D19(%rip),%rax
  mov    D11(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L9
  mov    D21(%rip),%rdx
  cmp    (%rdx),%rax
  je     L9
  mov    $0xffffffff,%r13d
L71:
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L38:
  xor    %r13d,%r13d
  jmp    L71
```

assistant: ```c
int s_pread(int fd, void *buf, size_t len, off_t off) {
  int total;
  int thistime;
  off_t thisoffset;
  thisoffset = off;
  for (total = 0; total < len;) {
    thistime = pread(fd, buf + total, len - total, thisoffset);
    if (thistime < 0) {
      if (EINTR == errno || EAGAIN == errno)
        continue;
      return -1;
    } else if (thistime == 0) {
      return total;
    }
    total += thistime;
    thisoffset += total;
  }
  return total;
}
```