ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<Clean_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L29
  jmp    L34
L42:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L26
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <safestrchr@plt>
  test   %rax,%rax
  jne    L26
  mov    -0x18(%rbp),%rax
  movb   $0x5f,(%rax)
L26:
  addq   $0x1,-0x18(%rbp)
L34:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L42
L29:
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type_infered: string

tool: D7: -_.

assistant: ```c
void Clean_name(char *s) {
  int c;
  if (s) {
    for (; (c = (int)(*(unsigned const char *)(s))); ++s) {
      if (!(isalnum(c) || safestrchr("-_.", c)))
        *s = '_';
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Clean_name>:
  endbr64
  test   %rdi,%rdi
  je     L29
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%ebp
  test   %ebp,%ebp
  je     L41
  lea    D0(%rip),%r12
  jmp    L37
L52:
  add    $0x1,%rbx
  movzbl (%rbx),%ebp
  test   %ebp,%ebp
  je     L41
L37:
  mov    %ebp,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L52
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <safestrchr@plt>
  test   %rax,%rax
  jne    L52
  movb   $0x5f,(%rbx)
  jmp    L52
L41:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  ret
```

tool_calls: name: read_data
arguments: tag: D0
datatype: string

tool: D0: -_.

assistant: ```c
void Clean_name(char *s) {
  int c;
  if (s) {
    for (; (c = (int)(*(unsigned const char *)(s))); ++s) {
      if (!(isalnum(c) || safestrchr("-_.", c)))
        *s = '_';
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Clean_name>:
  endbr64
  test   %rdi,%rdi
  je     L62
  push   %r12
  push   %rbp
  push   %rbx
  movzbl (%rdi),%ebp
  mov    %rdi,%rbx
  test   %ebp,%ebp
  je     L12
  lea    D8(%rip),%r12
  jmp    L22
L47:
  movzbl 0x1(%rbx),%ebp
  add    $0x1,%rbx
  test   %ebp,%ebp
  je     L12
L22:
  mov    %ebp,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L47
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <safestrchr@plt>
  test   %rax,%rax
  jne    L47
  movb   $0x5f,(%rbx)
  movzbl 0x1(%rbx),%ebp
  add    $0x1,%rbx
  test   %ebp,%ebp
  jne    L22
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L62:
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type_guessed: string

tool: D8: -_.

assistant: ```c
void Clean_name(char *s) {
  int c;
  if (s) {
    for (; (c = (int)(*(unsigned const char *)(s))); ++s) {
      if (!(isalnum(c) || safestrchr("-_.", c)))
        *s = '_';
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Clean_name>:
  endbr64
  test   %rdi,%rdi
  je     L9
  push   %r12
  push   %rbp
  push   %rbx
  movzbl (%rdi),%ebp
  mov    %rdi,%rbx
  test   %ebp,%ebp
  je     L66
  lea    D7(%rip),%r12
  jmp    L23
L15:
  movzbl 0x1(%rbx),%ebp
  add    $0x1,%rbx
  test   %ebp,%ebp
  je     L66
L23:
  mov    %ebp,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L15
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <safestrchr@plt>
  test   %rax,%rax
  jne    L15
  movb   $0x5f,(%rbx)
  movzbl 0x1(%rbx),%ebp
  add    $0x1,%rbx
  test   %ebp,%ebp
  jne    L23
L66:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type_infered: string

tool: D7: -_.

assistant: ```c
void Clean_name(char *s) {
  int c;
  if (s) {
    for (; (c = (int)(*(unsigned const char *)(s))); ++s) {
      if (!(isalnum(c) || safestrchr("-_.", c)))
        *s = '_';
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<busquedaBin1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movsd  %xmm0,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L43
  mov    $0xffffffff,%eax
  jmp    L60
L43:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x28(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L89
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L26
L89:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L26:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  ucomisd -0x28(%rbp),%xmm0
  jp     L48
  ucomisd -0x28(%rbp),%xmm0
  je     L51
L48:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L43
L51:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  ucomisd -0x28(%rbp),%xmm0
  jp     L6
  ucomisd -0x28(%rbp),%xmm0
  jne    L6
  mov    -0x4(%rbp),%eax
  jmp    L60
L6:
  mov    $0xffffffff,%eax
L60:
  pop    %rbp
  ret
```

assistant: ```c
int busquedaBin1(double m[], int n, double v) {
  int mitad, inf = 0, sup = n - 1;
  if (n == 0)
    return -1;
  do {
    mitad = (inf + sup) / 2;
    if (v > m[mitad])
      inf = mitad + 1;
    else
      sup = mitad - 1;
  } while (m[mitad] != v && inf <= sup);
  if (m[mitad] == v)
    return mitad;
  else
    return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<busquedaBin1>:
  endbr64
  lea    -0x1(%rsi),%ecx
  test   %esi,%esi
  je     L64
  mov    $0x0,%esi
  mov    $0x1,%r9d
  jmp    L55
L99:
  lea    0x1(%rax),%esi
L89:
  cmp    %ecx,%esi
  jg     L26
  ucomisd %xmm0,%xmm1
  setp   %al
  cmovne %r9d,%eax
  test   %al,%al
  je     L26
L55:
  lea    (%rsi,%rcx,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%r8d
  movslq %eax,%rdx
  movsd  (%rdi,%rdx,8),%xmm1
  comisd %xmm1,%xmm0
  ja     L99
  lea    -0x1(%rax),%ecx
  jmp    L89
L26:
  ucomisd %xmm0,%xmm1
  jp     L3
  jne    L3
L8:
  mov    %r8d,%eax
  ret
L3:
  mov    $0xffffffff,%r8d
  jmp    L8
L64:
  mov    $0xffffffff,%r8d
  jmp    L8
```

assistant: ```c
int busquedaBin1(double m[], int n, double v) {
  int mitad, inf = 0, sup = n - 1;
  if (n == 0)
    return -1;
  do {
    mitad = (inf + sup) / 2;
    if (v > m[mitad])
      inf = mitad + 1;
    else
      sup = mitad - 1;
  } while (m[mitad] != v && inf <= sup);
  if (m[mitad] == v)
    return mitad;
  else
    return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<busquedaBin1>:
  endbr64
  lea    -0x1(%rsi),%ecx
  test   %esi,%esi
  je     L79
  xor    %esi,%esi
  mov    $0x1,%r8d
  jmp    L39
  nopl   (%rax)
L60:
  lea    0x1(%rax),%esi
  cmp    %ecx,%esi
  jg     L30
L74:
  ucomisd %xmm0,%xmm1
  setp   %dl
  cmovne %r8d,%edx
  test   %dl,%dl
  je     L30
L39:
  lea    (%rsi,%rcx,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  movslq %eax,%rdx
  movsd  (%rdi,%rdx,8),%xmm1
  comisd %xmm1,%xmm0
  ja     L60
  lea    -0x1(%rax),%ecx
  cmp    %ecx,%esi
  jle    L74
L30:
  ucomisd %xmm0,%xmm1
  jp     L79
  jne    L79
  ret
L79:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int busquedaBin1(double m[], int n, double v) {
  int mitad, inf = 0, sup = n - 1;
  if (n == 0)
    return -1;
  do {
    mitad = (inf + sup) / 2;
    if (v > m[mitad])
      inf = mitad + 1;
    else
      sup = mitad - 1;
  } while (m[mitad] != v && inf <= sup);
  if (m[mitad] == v)
    return mitad;
  else
    return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<busquedaBin1>:
  endbr64
  lea    -0x1(%rsi),%ecx
  test   %esi,%esi
  je     L35
  xor    %esi,%esi
  mov    $0x1,%r8d
  jmp    L51
  nopl   (%rax)
L70:
  ucomisd %xmm1,%xmm0
  lea    0x1(%rax),%esi
  setp   %dl
  cmovne %r8d,%edx
  test   %dl,%dl
  je     L78
L49:
  cmp    %esi,%ecx
  jl     L78
L51:
  lea    (%rcx,%rsi,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  movslq %eax,%rdx
  movsd  (%rdi,%rdx,8),%xmm1
  comisd %xmm1,%xmm0
  ja     L70
  ucomisd %xmm1,%xmm0
  lea    -0x1(%rax),%ecx
  setp   %dl
  cmovne %r8d,%edx
  test   %dl,%dl
  jne    L49
L78:
  ucomisd %xmm1,%xmm0
  jp     L35
  jne    L35
  ret
L35:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int busquedaBin1(double m[], int n, double v) {
  int mitad, inf = 0, sup = n - 1;
  if (n == 0)
    return -1;
  do {
    mitad = (inf + sup) / 2;
    if (v > m[mitad])
      inf = mitad + 1;
    else
      sup = mitad - 1;
  } while (m[mitad] != v && inf <= sup);
  if (m[mitad] == v)
    return mitad;
  else
    return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<fmax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  movsd  -0x8(%rbp),%xmm0
  comisd -0x10(%rbp),%xmm0
  jbe    L15
  movsd  -0x8(%rbp),%xmm0
  jmp    L19
L15:
  movsd  -0x10(%rbp),%xmm0
L19:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double fmax(double a, double b) {
  if (a > b) {
    return (a);
  } else {
    return (b);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fmax>:
  endbr64
  maxsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double fmax(double a, double b) {
  if (a > b) {
    return (a);
  } else {
    return (b);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fmax>:
  endbr64
  maxsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double fmax(double a, double b) {
  if (a > b) {
    return (a);
  } else {
    return (b);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fmax>:
  endbr64
  maxsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double fmax(double a, double b) {
  if (a > b) {
    return (a);
  } else {
    return (b);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<unpack_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L20
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jge    L16
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  movzbl -0x11(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x10(%rbp)
  addl   $0x8,-0x8(%rbp)
L16:
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%eax
  and    %eax,%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    %dx,(%rax)
  mov    -0x24(%rbp),%eax
  mov    %eax,%ecx
  shrl   %cl,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  sub    %eax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L20:
  mov    -0x4(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jg     L38
  cmpl   $0x77,-0xc(%rbp)
  jle    L19
L38:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int unpack_bytes(int bits, int blocksize, const unsigned char *block, short *samples) {
  unsigned int in_buffer = 0;
  unsigned char in_byte;
  int k, in_bits = 0, bindex = 0;
  for (k = 0; bindex <= blocksize && k < (3 * 5 * 8); k++) {
    if (in_bits < bits) {
      in_byte = block[bindex++];
      in_buffer |= (in_byte << in_bits);
      in_bits += 8;
    }
    samples[k] = in_buffer & ((1 << bits) - 1);
    in_buffer >>= bits;
    in_bits -= bits;
  };
  return k;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<unpack_bytes>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  js     L22
  mov    %esi,%r11d
  mov    %rdx,%rbp
  mov    %rcx,%rbx
  mov    $0x1,%r10d
  mov    %edi,%ecx
  shl    %cl,%r10d
  sub    $0x1,%r10d
  mov    $0x1,%eax
  mov    $0x0,%r8d
  mov    $0x0,%edx
  mov    $0x0,%esi
  jmp    L46
L0:
  mov    %r10d,%ecx
  and    %esi,%ecx
  mov    %cx,-0x2(%rbx,%rax,2)
  mov    %edi,%ecx
  shr    %cl,%esi
  sub    %edi,%edx
  mov    %eax,%r12d
  cmp    %r11d,%r8d
  setle  %r9b
  cmp    $0x77,%eax
  setle  %cl
  add    $0x1,%rax
  test   %cl,%r9b
  je     L48
L46:
  cmp    %edx,%edi
  jle    L0
  movslq %r8d,%rcx
  movzbl 0x0(%rbp,%rcx,1),%r9d
  mov    %edx,%ecx
  shl    %cl,%r9d
  or     %r9d,%esi
  add    $0x8,%edx
  lea    0x1(%r8),%r8d
  jmp    L0
L22:
  mov    $0x0,%r12d
L48:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int unpack_bytes(int bits, int blocksize, const unsigned char *block, short *samples) {
  unsigned int in_buffer = 0;
  unsigned char in_byte;
  int k, in_bits = 0, bindex = 0;
  for (k = 0; bindex <= blocksize && k < (3 * 5 * 8); k++) {
    if (in_bits < bits) {
      in_byte = block[bindex++];
      in_buffer |= (in_byte << in_bits);
      in_bits += 8;
    }
    samples[k] = in_buffer & ((1 << bits) - 1);
    in_buffer >>= bits;
    in_bits -= bits;
  };
  return k;
}
```
user: Translate the below assembly code into C programming language:

```asm
<unpack_bytes>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  js     L25
  mov    %rcx,%rbx
  mov    $0x1,%r10d
  mov    %edi,%ecx
  mov    %esi,%r11d
  shl    %cl,%r10d
  mov    %rdx,%rbp
  mov    $0x1,%eax
  xor    %r8d,%r8d
  sub    $0x1,%r10d
  xor    %edx,%edx
  xor    %esi,%esi
L22:
  cmp    %edx,%edi
  jle    L18
  movslq %r8d,%rcx
  add    $0x1,%r8d
  movzbl 0x0(%rbp,%rcx,1),%r9d
  mov    %edx,%ecx
  add    $0x8,%edx
  shl    %cl,%r9d
  or     %r9d,%esi
L18:
  mov    %r10d,%ecx
  sub    %edi,%edx
  mov    %eax,%r12d
  and    %esi,%ecx
  mov    %cx,-0x2(%rbx,%rax,2)
  mov    %edi,%ecx
  shr    %cl,%esi
  cmp    %r11d,%r8d
  setle  %r9b
  cmp    $0x77,%eax
  setle  %cl
  add    $0x1,%rax
  test   %cl,%r9b
  jne    L22
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L25:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int unpack_bytes(int bits, int blocksize, const unsigned char *block, short *samples) {
  unsigned int in_buffer = 0;
  unsigned char in_byte;
  int k, in_bits = 0, bindex = 0;
  for (k = 0; bindex <= blocksize && k < (3 * 5 * 8); k++) {
    if (in_bits < bits) {
      in_byte = block[bindex++];
      in_buffer |= (in_byte << in_bits);
      in_bits += 8;
    }
    samples[k] = in_buffer & ((1 << bits) - 1);
    in_buffer >>= bits;
    in_bits -= bits;
  };
  return k;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unpack_bytes>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  js     L24
  mov    %rcx,%rbx
  mov    $0x1,%r9d
  mov    %edi,%ecx
  mov    %esi,%r10d
  shl    %cl,%r9d
  mov    %rdx,%rbp
  mov    $0x1,%eax
  xor    %esi,%esi
  sub    $0x1,%r9d
  xor    %edx,%edx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L46:
  cmp    %edx,%edi
  jle    L18
L25:
  lea    0x1(%rsi),%r11d
  movzbl 0x0(%rbp,%rsi,1),%esi
  mov    %edx,%ecx
  add    $0x8,%edx
  sub    %edi,%edx
  mov    %eax,%r12d
  shl    %cl,%esi
  mov    %r9d,%ecx
  or     %r8d,%esi
  and    %esi,%ecx
  mov    %cx,-0x2(%rbx,%rax,2)
  mov    %edi,%ecx
  shr    %cl,%esi
  cmp    $0x77,%eax
  mov    %esi,%r8d
  setle  %sil
  cmp    %r11d,%r10d
  setge  %cl
  add    $0x1,%rax
  test   %cl,%sil
  je     L10
  movslq %r11d,%rsi
  cmp    %edx,%edi
  jg     L25
L18:
  mov    %r9d,%ecx
  sub    %edi,%edx
  mov    %eax,%r12d
  and    %r8d,%ecx
  mov    %cx,-0x2(%rbx,%rax,2)
  mov    %edi,%ecx
  shr    %cl,%r8d
  cmp    %esi,%r10d
  setge  %r11b
  cmp    $0x77,%eax
  setle  %cl
  add    $0x1,%rax
  test   %cl,%r11b
  jne    L46
L10:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L24:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int unpack_bytes(int bits, int blocksize, const unsigned char *block, short *samples) {
  unsigned int in_buffer = 0;
  unsigned char in_byte;
  int k, in_bits = 0, bindex = 0;
  for (k = 0; bindex <= blocksize && k < (3 * 5 * 8); k++) {
    if (in_bits < bits) {
      in_byte = block[bindex++];
      in_buffer |= (in_byte << in_bits);
      in_bits += 8;
    }
    samples[k] = in_buffer & ((1 << bits) - 1);
    in_buffer >>= bits;
    in_bits -= bits;
  };
  return k;
}
```
user: Translate the below assembly code into C programming language:

```asm
<float_to_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    -0x14(%rbp),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
uint32_t float_to_bits(const float f) {
  uint32_t bits = 0;
  memcpy(&bits, &f, sizeof(float));
  return bits;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<float_to_bits>:
  endbr64
  sub    $0x28,%rsp
  movss  %xmm0,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t float_to_bits(const float f) {
  uint32_t bits = 0;
  memcpy(&bits, &f, sizeof(float));
  return bits;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<float_to_bits>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x4,%edx
  movss  %xmm0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  call   <memcpy@plt>
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t float_to_bits(const float f) {
  uint32_t bits = 0;
  memcpy(&bits, &f, sizeof(float));
  return bits;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<float_to_bits>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x4,%edx
  movss  %xmm0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  call   <memcpy@plt>
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t float_to_bits(const float f) {
  uint32_t bits = 0;
  memcpy(&bits, &f, sizeof(float));
  return bits;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_start>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <main@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  mov    $0x0,%eax
  int    $0x80
L6:
  jmp    L6
```

assistant: ```c
void _start() {
  int result = main(0, 0x0);
  __asm__ volatile("movl $0, %%eax\r\n"
                   "int $0x80\r\n"
                   :
                   : "c"(result));
  while (1) {
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_start>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <main@plt>
  mov    %eax,%ecx
  mov    $0x0,%eax
  int    $0x80
L4:
  jmp    L4
```

assistant: ```c
void _start() {
  int result = main(0, 0x0);
  __asm__ volatile("movl $0, %%eax\r\n"
                   "int $0x80\r\n"
                   :
                   : "c"(result));
  while (1) {
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_start>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  xor    %eax,%eax
  call   <main@plt>
  mov    %eax,%ecx
  mov    $0x0,%eax
  int    $0x80
L8:
  jmp    L8
```

assistant: ```c
void _start() {
  int result = main(0, 0x0);
  __asm__ volatile("movl $0, %%eax\r\n"
                   "int $0x80\r\n"
                   :
                   : "c"(result));
  while (1) {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<_start>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  xor    %eax,%eax
  call   <main@plt>
  mov    %eax,%ecx
  mov    $0x0,%eax
  int    $0x80
L2:
  jmp    L2
```

assistant: ```c
void _start() {
  int result = main(0, 0x0);
  __asm__ volatile("movl $0, %%eax\r\n"
                   "int $0x80\r\n"
                   :
                   : "c"(result));
  while (1) {
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<copyout>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int copyout(const void *src, void *dst, size_t nbytes) {
  memcpy(dst, src, nbytes);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<copyout>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rdi
  mov    %rax,%rsi
  call   <memcpy@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copyout(const void *src, void *dst, size_t nbytes) {
  memcpy(dst, src, nbytes);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<copyout>:
  endbr64
  mov    %rdi,%r8
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  mov    %r8,%rsi
  call   <memcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copyout(const void *src, void *dst, size_t nbytes) {
  memcpy(dst, src, nbytes);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copyout>:
  endbr64
  mov    %rdi,%r8
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  mov    %r8,%rsi
  call   <memcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copyout(const void *src, void *dst, size_t nbytes) {
  memcpy(dst, src, nbytes);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<harbol_unilist_new_from_hashmap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L19
  mov    $0x0,%eax
  jmp    L16
L19:
  mov    $0x0,%eax
  call   <harbol_unilist_new@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <harbol_unilist_from_hashmap@plt>
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
struct HarbolUniList *harbol_unilist_new_from_hashmap(const struct HarbolHashmap *const map) {
  if (!map)
    return NULL;
  struct HarbolUniList *const unilist = harbol_unilist_new();
  harbol_unilist_from_hashmap(unilist, map);
  return unilist;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<harbol_unilist_new_from_hashmap>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L16
  mov    $0x0,%eax
  call   <harbol_unilist_new@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <harbol_unilist_from_hashmap@plt>
L14:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    %rdi,%rbp
  jmp    L14
```

assistant: ```c
struct HarbolUniList *harbol_unilist_new_from_hashmap(const struct HarbolHashmap *const map) {
  if (!map)
    return NULL;
  struct HarbolUniList *const unilist = harbol_unilist_new();
  harbol_unilist_from_hashmap(unilist, map);
  return unilist;
}
```