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
<segment_to_int>:
  endbr64
  cmp    $0x1993,%edi
  je     L41
  jg     L11
  cmp    $0x1500,%edi
  je     L24
  jg     L77
  mov    $0x2,%eax
  cmp    $0xd92,%edi
  je     L65
  cmp    $0x1400,%edi
  jne    L60
  mov    $0x1,%eax
L65:
  ret
  nopl   0x0(%rax,%rax,1)
L77:
  xor    %eax,%eax
  cmp    $0x1717,%edi
  je     L65
  cmp    $0x1991,%edi
  mov    $0x5,%edx
  mov    $0xffffffff,%eax
  cmove  %edx,%eax
  ret
  xchg   %ax,%ax
L11:
  mov    $0x9,%eax
  cmp    $0x1d91,%edi
  je     L65
  jle    L62
  xor    %eax,%eax
  cmp    $0x1d93,%edi
  sete   %al
  lea    -0x1(%rax,%rax,8),%eax
  ret
L60:
  neg    %edi
  sbb    %eax,%eax
  ret
  nopl   0x0(%rax)
L62:
  mov    $0x4,%eax
  cmp    $0x1c81,%edi
  je     L65
  xor    %eax,%eax
  cmp    $0x1d90,%edi
  sete   %al
  lea    -0x1(,%rax,4),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  mov    $0x7,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L41:
  mov    $0x6,%eax
  ret
```

assistant: ```c
int segment_to_int(int segment_bitfield) {
  switch (segment_bitfield) {
  case 0:
  case 0x1717:
    return 0;
  case 0x1400:
    return 1;
  case 0x0d92:
    return 2;
  case 0x1d90:
    return 3;
  case 0x1c81:
    return 4;
  case 0x1991:
    return 5;
  case 0x1993:
    return 6;
  case 0x1500:
    return 7;
  case 0x1d93:
    return 8;
  case 0x1d91:
    return 9;
  default:
    return -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
<ReverseNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L9:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0x4(%rbp)
  mov    -0x14(%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,-0x14(%rbp)
L7:
  cmpl   $0x0,-0x14(%rbp)
  jne    L9
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ReverseNumber(int number) {
  int reversedNumber = 0;
  while (number) {
    reversedNumber *= 10;
    reversedNumber += number % 10;
    number /= 10;
  }
  return reversedNumber;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ReverseNumber>:
  endbr64
  test   %edi,%edi
  je     L13
  mov    $0x0,%edx
L27:
  lea    (%rdx,%rdx,4),%ecx
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    (%rdi,%rcx,2),%edx
  mov    %eax,%edi
  test   %eax,%eax
  jne    L27
L3:
  mov    %edx,%eax
  ret
L13:
  mov    %edi,%edx
  jmp    L3
```

assistant: ```c
int ReverseNumber(int number) {
  int reversedNumber = 0;
  while (number) {
    reversedNumber *= 10;
    reversedNumber += number % 10;
    number /= 10;
  }
  return reversedNumber;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ReverseNumber>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L14
  nopl   0x0(%rax,%rax,1)
L9:
  movslq %edi,%rax
  mov    %edi,%edx
  lea    (%r8,%r8,4),%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    (%rdi,%rcx,2),%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L9
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ReverseNumber(int number) {
  int reversedNumber = 0;
  while (number) {
    reversedNumber *= 10;
    reversedNumber += number % 10;
    number /= 10;
  }
  return reversedNumber;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ReverseNumber>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L11
  nopl   0x0(%rax,%rax,1)
L19:
  movslq %edi,%rax
  mov    %edi,%edx
  lea    (%r8,%r8,4),%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    (%rdi,%rcx,2),%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L19
L11:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ReverseNumber(int number) {
  int reversedNumber = 0;
  while (number) {
    reversedNumber *= 10;
    reversedNumber += number % 10;
    number /= 10;
  }
  return reversedNumber;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bitcount1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L19:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  and    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x0,-0x14(%rbp)
  jne    L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int bitcount1(int n) {
  int c;
  for (c = 0; n; c++)
    n = n & n - 1;
  return c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitcount1>:
  endbr64
  test   %edi,%edi
  je     L19
  mov    $0x0,%edx
L6:
  lea    -0x1(%rdi),%eax
  add    $0x1,%edx
  and    %eax,%edi
  jne    L6
L25:
  mov    %edx,%eax
  ret
L19:
  mov    %edi,%edx
  jmp    L25
```

assistant: ```c
int bitcount1(int n) {
  int c;
  for (c = 0; n; c++)
    n = n & n - 1;
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bitcount1>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L16
  nopl   0x0(%rax,%rax,1)
L18:
  lea    -0x1(%rdi),%eax
  add    $0x1,%r8d
  and    %eax,%edi
  jne    L18
L16:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitcount1(int n) {
  int c;
  for (c = 0; n; c++)
    n = n & n - 1;
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bitcount1>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L16
  nopl   0x0(%rax,%rax,1)
L9:
  lea    -0x1(%rdi),%eax
  add    $0x1,%r8d
  and    %eax,%edi
  jne    L9
L16:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitcount1(int n) {
  int c;
  for (c = 0; n; c++)
    n = n & n - 1;
  return c;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ENGINE_register_all_pkey_meths>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <ENGINE_get_first@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L0
L2:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <ENGINE_register_pkey_meths@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,-0x8(%rbp)
L0:
  cmpq   $0x0,-0x8(%rbp)
  jne    L2
  nop
  nop
  leave
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_meths(e);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ENGINE_register_all_pkey_meths>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <ENGINE_get_first@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L5
L9:
  mov    %rbx,%rdi
  call   <ENGINE_register_pkey_meths@plt>
  mov    %rbx,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L9
L5:
  pop    %rbx
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_meths(e);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ENGINE_register_all_pkey_meths>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <ENGINE_get_first@plt>
  test   %rax,%rax
  je     L10
  mov    %rax,%rbp
  nopl   0x0(%rax)
L18:
  mov    %rbp,%rdi
  call   <ENGINE_register_pkey_meths@plt>
  mov    %rbp,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L18
L10:
  pop    %rbp
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_meths(e);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ENGINE_register_all_pkey_meths>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <ENGINE_get_first@plt>
  test   %rax,%rax
  je     L18
  mov    %rax,%rbp
  nopl   0x0(%rax)
L1:
  mov    %rbp,%rdi
  call   <ENGINE_register_pkey_meths@plt>
  mov    %rbp,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L1
L18:
  pop    %rbp
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_meths(e);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Hex_To_Int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L4
L57:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  cmpb   $0x2f,-0xd(%rbp)
  jle    L82
  cmpb   $0x39,-0xd(%rbp)
  jg     L82
  movzbl -0xd(%rbp),%eax
  sub    $0x30,%eax
  mov    %al,-0xd(%rbp)
  jmp    L48
L82:
  cmpb   $0x60,-0xd(%rbp)
  jle    L87
  cmpb   $0x66,-0xd(%rbp)
  jg     L87
  movzbl -0xd(%rbp),%eax
  sub    $0x57,%eax
  mov    %al,-0xd(%rbp)
  jmp    L48
L87:
  cmpb   $0x40,-0xd(%rbp)
  jle    L48
  cmpb   $0x46,-0xd(%rbp)
  jg     L48
  movzbl -0xd(%rbp),%eax
  sub    $0x37,%eax
  mov    %al,-0xd(%rbp)
L48:
  mov    -0xc(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movsbl -0xd(%rbp),%eax
  and    $0xf,%eax
  or     %edx,%eax
  mov    %eax,-0xc(%rbp)
L4:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L57
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t Hex_To_Int(char *hax) {
  char *hex = NULL;
  hex = hax;
  uint64_t val = 0;
  while (*hex) {
    uint8_t byte = *hex++;
    if (byte >= '0' && byte <= '9')
      byte = byte - '0';
    else if (byte >= 'a' && byte <= 'f')
      byte = byte - 'a' + 10;
    else if (byte >= 'A' && byte <= 'F')
      byte = byte - 'A' + 10;
    val = (val << 4) | (byte & 0xF);
  }
  return val;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Hex_To_Int>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L68
  mov    $0x0,%eax
  jmp    L45
L24:
  lea    -0x61(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L29
  lea    -0x57(%rsi),%edx
L50:
  shl    $0x4,%eax
  and    $0xf,%edx
  or     %edx,%eax
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L44
L45:
  add    $0x1,%rdi
  mov    %edx,%esi
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  ja     L24
  mov    %ecx,%edx
  jmp    L50
L29:
  lea    -0x41(%rsi),%ecx
  sub    $0x37,%esi
  cmp    $0x6,%cl
  cmovb  %esi,%edx
  jmp    L50
L44:
  ret
L68:
  mov    $0x0,%eax
  ret
```

assistant: ```c
uint64_t Hex_To_Int(char *hax) {
  char *hex = NULL;
  hex = hax;
  uint64_t val = 0;
  while (*hex) {
    uint8_t byte = *hex++;
    if (byte >= '0' && byte <= '9')
      byte = byte - '0';
    else if (byte >= 'a' && byte <= 'f')
      byte = byte - 'a' + 10;
    else if (byte >= 'A' && byte <= 'F')
      byte = byte - 'A' + 10;
    val = (val << 4) | (byte & 0xF);
  }
  return val;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Hex_To_Int>:
  endbr64
  movzbl (%rdi),%ecx
  xor    %eax,%eax
  test   %cl,%cl
  jne    L15
  jmp    L32
  nop
L25:
  lea    -0x57(%rcx),%edx
L31:
  movzbl (%rdi),%ecx
  shl    $0x4,%eax
  movsbl %dl,%edx
  or     %edx,%eax
  test   %cl,%cl
  je     L40
L15:
  lea    -0x30(%rcx),%edx
  add    $0x1,%rdi
  cmp    $0x9,%dl
  jbe    L31
  lea    -0x61(%rcx),%edx
  cmp    $0x5,%dl
  jbe    L25
  lea    -0x41(%rcx),%esi
  mov    %ecx,%edx
  sub    $0x37,%ecx
  and    $0xf,%edx
  cmp    $0x6,%sil
  cmovb  %ecx,%edx
  movzbl (%rdi),%ecx
  shl    $0x4,%eax
  movsbl %dl,%edx
  or     %edx,%eax
  test   %cl,%cl
  jne    L15
L40:
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  ret
```

assistant: ```c
uint64_t Hex_To_Int(char *hax) {
  char *hex = NULL;
  hex = hax;
  uint64_t val = 0;
  while (*hex) {
    uint8_t byte = *hex++;
    if (byte >= '0' && byte <= '9')
      byte = byte - '0';
    else if (byte >= 'a' && byte <= 'f')
      byte = byte - 'a' + 10;
    else if (byte >= 'A' && byte <= 'F')
      byte = byte - 'A' + 10;
    val = (val << 4) | (byte & 0xF);
  }
  return val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Hex_To_Int>:
  endbr64
  movzbl (%rdi),%ecx
  xor    %eax,%eax
  test   %cl,%cl
  jne    L3
  jmp    L20
  nop
L14:
  lea    -0x57(%rcx),%edx
L8:
  movzbl (%rdi),%ecx
  shl    $0x4,%eax
  movsbl %dl,%edx
  or     %edx,%eax
  test   %cl,%cl
  je     L26
L3:
  lea    -0x30(%rcx),%edx
  add    $0x1,%rdi
  cmp    $0x9,%dl
  jbe    L8
  lea    -0x61(%rcx),%edx
  cmp    $0x5,%dl
  jbe    L14
  lea    -0x41(%rcx),%esi
  lea    -0x37(%rcx),%edx
  and    $0xf,%ecx
  cmp    $0x5,%sil
  cmova  %ecx,%edx
  movzbl (%rdi),%ecx
  shl    $0x4,%eax
  movsbl %dl,%edx
  or     %edx,%eax
  test   %cl,%cl
  jne    L3
L26:
  ret
  cs nopw 0x0(%rax,%rax,1)
L20:
  ret
```

assistant: ```c
uint64_t Hex_To_Int(char *hax) {
  char *hex = NULL;
  hex = hax;
  uint64_t val = 0;
  while (*hex) {
    uint8_t byte = *hex++;
    if (byte >= '0' && byte <= '9')
      byte = byte - '0';
    else if (byte >= 'a' && byte <= 'f')
      byte = byte - 'a' + 10;
    else if (byte >= 'A' && byte <= 'F')
      byte = byte - 'A' + 10;
    val = (val << 4) | (byte & 0xF);
  }
  return val;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int compareTags(const void *const one, const void *const two) {
  const char *const line1 = *(const char *const *)one;
  const char *const line2 = *(const char *const *)two;
  return strcmp(line1, line2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compareTags>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int compareTags(const void *const one, const void *const two) {
  const char *const line1 = *(const char *const *)one;
  const char *const line2 = *(const char *const *)two;
  return strcmp(line1, line2);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<compareTags>:
  endbr64
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  jmp    1050 <strcmp@plt>
```

assistant: ```c
int compareTags(const void *const one, const void *const two) {
  const char *const line1 = *(const char *const *)one;
  const char *const line2 = *(const char *const *)two;
  return strcmp(line1, line2);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<compareTags>:
  endbr64
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  jmp    1050 <strcmp@plt>
```

assistant: ```c
int compareTags(const void *const one, const void *const two) {
  const char *const line1 = *(const char *const *)one;
  const char *const line2 = *(const char *const *)two;
  return strcmp(line1, line2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<decrypt_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L7
L1:
  mov    -0x8(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rdx,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    -0x8(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  xor    -0x9(%rbp),%al
  mov    %al,(%rdx)
  addq   $0x1,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jl     L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void decrypt_buffer(long klen, const unsigned char *key, long blen, unsigned char *buf) {
  long n = 0;
  while (n < blen) {
    unsigned char c = buf[n];
    buf[n] = c ^ key[n % klen];
    n++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<decrypt_buffer>:
  endbr64
  mov    %rdx,%r9
  test   %rdx,%rdx
  jle    L17
  mov    $0x0,%r8d
L14:
  mov    %r8,%rax
  cqto
  idiv   %rdi
  movzbl (%rsi,%rdx,1),%eax
  xor    %al,(%rcx,%r8,1)
  add    $0x1,%r8
  cmp    %r8,%r9
  jne    L14
L17:
  ret
```

assistant: ```c
void decrypt_buffer(long klen, const unsigned char *key, long blen, unsigned char *buf) {
  long n = 0;
  while (n < blen) {
    unsigned char c = buf[n];
    buf[n] = c ^ key[n % klen];
    n++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<decrypt_buffer>:
  endbr64
  mov    %rdx,%r9
  test   %rdx,%rdx
  jle    L8
  xor    %r8d,%r8d
  nop
L0:
  mov    %r8,%rax
  cqto
  idiv   %rdi
  movzbl (%rsi,%rdx,1),%eax
  xor    %al,(%rcx,%r8,1)
  add    $0x1,%r8
  cmp    %r8,%r9
  jne    L0
L8:
  ret
```

assistant: ```c
void decrypt_buffer(long klen, const unsigned char *key, long blen, unsigned char *buf) {
  long n = 0;
  while (n < blen) {
    unsigned char c = buf[n];
    buf[n] = c ^ key[n % klen];
    n++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<decrypt_buffer>:
  endbr64
  mov    %rdx,%r9
  test   %rdx,%rdx
  jle    L9
  xor    %r8d,%r8d
  nop
L16:
  mov    %r8,%rax
  cqto
  idiv   %rdi
  movzbl (%rsi,%rdx,1),%eax
  xor    %al,(%rcx,%r8,1)
  add    $0x1,%r8
  cmp    %r8,%r9
  jne    L16
L9:
  ret
```

assistant: ```c
void decrypt_buffer(long klen, const unsigned char *key, long blen, unsigned char *buf) {
  long n = 0;
  while (n < blen) {
    unsigned char c = buf[n];
    buf[n] = c ^ key[n % klen];
    n++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<compute_occ_list>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %rax,-0x18(%rbp)
  jmp    L36
L33:
  mov    -0x40(%rbp),%eax
  sub    -0x28(%rbp),%eax
  mov    $0x40,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    %eax,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L10
L40:
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  mov    %edx,%ecx
  shr    %cl,%rax
  and    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L11
  mov    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%eax
  add    %eax,%edx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  addl   $0x1,-0x24(%rbp)
L11:
  addq   $0x1,-0x10(%rbp)
L10:
  mov    -0x20(%rbp),%eax
  cltq
  cmp    %rax,-0x10(%rbp)
  jb     L40
  addl   $0x40,-0x28(%rbp)
  subq   $0x1,-0x18(%rbp)
L36:
  cmpq   $0x0,-0x18(%rbp)
  jne    L33
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *compute_occ_list(uint64_t *string, int nset, int norb, int nelec) {
  size_t k, i;
  int *occ = malloc(sizeof(int) * nelec);
  int off = 0;
  int occ_ind = 0;
  for (k = nset; k > 0; --k) {
    int i_max = ((norb - off) < 64 ? (norb - off) : 64);
    for (i = 0; i < i_max; ++i) {
      int i_occ = (string[k - 1] >> i) & 1;
      if (i_occ) {
        occ[occ_ind] = i + off;
        occ_ind++;
      }
    }
    off += 64;
  }
  return occ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<compute_occ_list>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %edx,%ebx
  lea    0x0(,%rcx,4),%edi
  call   <malloc@plt>
  mov    %rax,%r11
  movslq %ebp,%rax
  test   %rax,%rax
  je     L67
  lea    (%r12,%rax,8),%rdi
  shl    $0x6,%ebp
  mov    $0x0,%esi
  mov    $0x0,%r9d
  jmp    L22
L60:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L7
L24:
  mov    -0x8(%rdi),%rdx
  bt     %rax,%rdx
  jae    L60
  movslq %esi,%rdx
  lea    (%r10,%rax,1),%r8d
  mov    %r8d,(%r11,%rdx,4)
  add    $0x1,%esi
  jmp    L60
L7:
  add    $0x40,%r9d
  sub    $0x8,%rdi
  cmp    %ebp,%r9d
  je     L67
L22:
  mov    %r9d,%r10d
  mov    %ebx,%ecx
  sub    %r9d,%ecx
  mov    $0x40,%eax
  cmp    %eax,%ecx
  cmovg  %eax,%ecx
  movslq %ecx,%rcx
  test   %rcx,%rcx
  je     L7
  mov    $0x0,%eax
  jmp    L24
L67:
  mov    %r11,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *compute_occ_list(uint64_t *string, int nset, int norb, int nelec) {
  size_t k, i;
  int *occ = malloc(sizeof(int) * nelec);
  int off = 0;
  int occ_ind = 0;
  for (k = nset; k > 0; --k) {
    int i_max = ((norb - off) < 64 ? (norb - off) : 64);
    for (i = 0; i < i_max; ++i) {
      int i_occ = (string[k - 1] >> i) & 1;
      if (i_occ) {
        occ[occ_ind] = i + off;
        occ_ind++;
      }
    }
    off += 64;
  }
  return occ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<compute_occ_list>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rcx,4),%edi
  call   <malloc@plt>
  movslq %r12d,%r11
  test   %r11,%r11
  je     L40
  xor    %esi,%esi
  xor    %r10d,%r10d
  nopl   0x0(%rax)
L42:
  mov    %ebp,%ecx
  mov    $0x40,%edx
  sub    %r10d,%ecx
  cmp    %edx,%ecx
  cmovg  %edx,%ecx
  movslq %ecx,%rcx
  test   %rcx,%rcx
  je     L19
  mov    -0x8(%rbx,%r11,8),%rdi
  xor    %edx,%edx
  xchg   %ax,%ax
L25:
  bt     %rdx,%rdi
  jae    L26
  movslq %esi,%r8
  lea    (%r10,%rdx,1),%r9d
  add    $0x1,%esi
  mov    %r9d,(%rax,%r8,4)
L26:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L25
L19:
  add    $0x40,%r10d
  sub    $0x1,%r11
  jne    L42
L40:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *compute_occ_list(uint64_t *string, int nset, int norb, int nelec) {
  size_t k, i;
  int *occ = malloc(sizeof(int) * nelec);
  int off = 0;
  int occ_ind = 0;
  for (k = nset; k > 0; --k) {
    int i_max = ((norb - off) < 64 ? (norb - off) : 64);
    for (i = 0; i < i_max; ++i) {
      int i_occ = (string[k - 1] >> i) & 1;
      if (i_occ) {
        occ[occ_ind] = i + off;
        occ_ind++;
      }
    }
    off += 64;
  }
  return occ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<compute_occ_list>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rcx,4),%edi
  call   <malloc@plt>
  movslq %r12d,%r11
  test   %r11,%r11
  je     L26
  xor    %esi,%esi
  xor    %r10d,%r10d
  nopl   0x0(%rax)
L10:
  mov    %ebp,%ecx
  mov    $0x40,%edx
  sub    %r10d,%ecx
  cmp    %edx,%ecx
  cmovg  %edx,%ecx
  movslq %ecx,%rcx
  test   %rcx,%rcx
  je     L7
  mov    -0x8(%rbx,%r11,8),%rdi
  xor    %edx,%edx
  xchg   %ax,%ax
L15:
  bt     %rdx,%rdi
  jae    L42
  movslq %esi,%r8
  lea    (%r10,%rdx,1),%r9d
  add    $0x1,%esi
  mov    %r9d,(%rax,%r8,4)
L42:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L15
L7:
  add    $0x40,%r10d
  sub    $0x1,%r11
  jne    L10
L26:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *compute_occ_list(uint64_t *string, int nset, int norb, int nelec) {
  size_t k, i;
  int *occ = malloc(sizeof(int) * nelec);
  int off = 0;
  int occ_ind = 0;
  for (k = nset; k > 0; --k) {
    int i_max = ((norb - off) < 64 ? (norb - off) : 64);
    for (i = 0; i < i_max; ++i) {
      int i_occ = (string[k - 1] >> i) & 1;
      if (i_occ) {
        occ[occ_ind] = i + off;
        occ_ind++;
      }
    }
    off += 64;
  }
  return occ;
}
```