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

```asm
<atoi_eol>:
  endbr64
  movzbl (%rdi),%r9d
  mov    %esi,%ecx
  cmp    $0x2d,%r9b
  je     L21
  movsbl %r9b,%eax
  cmp    %sil,%r9b
  je     L48
L38:
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L14:
  lea    (%r8,%r8,4),%edx
  add    $0x1,%rdi
  lea    (%rax,%rdx,2),%edx
  movsbl (%rdi),%eax
  lea    -0x30(%rdx),%r8d
  cmp    %cl,%al
  jne    L14
  mov    $0x30,%eax
  sub    %edx,%eax
  cmp    $0x2d,%r9b
  cmove  %eax,%r8d
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L21:
  movsbl 0x1(%rdi),%eax
  lea    0x1(%rdi),%rdx
  cmp    %sil,%al
  je     L48
  mov    %rdx,%rdi
  jmp    L38
  nopw   0x0(%rax,%rax,1)
L48:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int atoi_eol(const char *str, char eol) {
  const char *s = str;
  int value = 0;
  if (*s == '-')
    s++;
  while (*s != eol)
    value = 10 * value + *s++ - '0';
  if (*str == '-')
    value = -value;
  return value;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x2,-0x18(%rbp)
  je     L70
  cmpq   $0x3,-0x18(%rbp)
  jne    L114
L70:
  mov    $0x1,%eax
  jmp    L129
L114:
  cmpq   $0x1,-0x18(%rbp)
  jbe    L25
  mov    -0x18(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  jne    L128
L25:
  mov    $0x0,%eax
  jmp    L129
L128:
  mov    -0x18(%rbp),%rcx
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  mov    %rax,%rdx
  add    %rdx,%rdx
  add    %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jne    L8
  mov    $0x0,%eax
  jmp    L129
L8:
  cmpq   $0x8,-0x18(%rbp)
  ja     L95
  mov    $0x1,%eax
  jmp    L129
L95:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <sqrt@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x5,-0x10(%rbp)
  jmp    L5
L110:
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  je     L145
  mov    -0x10(%rbp),%rax
  lea    0x2(%rax),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L28
L145:
  mov    $0x0,%eax
  jmp    L129
L28:
  addq   $0x6,-0x10(%rbp)
L5:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L110
  mov    $0x1,%eax
L129:
  leave
  ret
```

assistant: ```c
int isPrime(unsigned long n) {
  if (n == 2 || n == 3)
    return 1;
  if (n < 2 || n % 2 == 0)
    return 0;
  if (n % 3 == 0)
    return 0;
  if (n < 9)
    return 1;
  unsigned long r = (unsigned long)sqrt(n);
  unsigned long f = 5;
  while (f <= r) {
    if (n % f == 0 || n % (f + 2) == 0)
      return 0;
    f += 6;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isPrime>:
  endbr64
  lea    -0x2(%rdi),%rdx
  mov    $0x1,%eax
  cmp    $0x1,%rdx
  jbe    L19
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x0,%eax
  cmp    $0x1,%rdi
  jbe    L26
  test   $0x1,%dil
  je     L26
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  and    $0xfffffffffffffffe,%rdx
  add    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%rdi
  je     L26
  mov    $0x1,%eax
  cmp    $0x8,%rdi
  ja     L90
L26:
  pop    %rbx
  ret
L90:
  call   <sqrt@plt>
  mov    %rax,%rdi
  cmp    $0x4,%rax
  jbe    L50
  movabs $0xcccccccccccccccd,%rdx
  mov    %rbx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%rdx
  add    %rax,%rdx
  cmp    %rdx,%rbx
  je     L62
  movabs $0x2492492492492493,%rdx
  mov    %rbx,%rax
  mul    %rdx
  mov    %rbx,%rax
  sub    %rdx,%rax
  shr    %rax
  add    %rax,%rdx
  shr    $0x2,%rdx
  lea    0x0(,%rdx,8),%rax
  sub    %rdx,%rax
  cmp    %rax,%rbx
  je     L135
  mov    $0x5,%ecx
L24:
  mov    %rcx,%rsi
  add    $0x6,%rcx
  cmp    %rcx,%rdi
  jb     L85
  mov    %rbx,%rax
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L110
  add    $0x8,%rsi
  mov    %rbx,%rax
  mov    $0x0,%edx
  div    %rsi
  test   %rdx,%rdx
  jne    L24
  mov    $0x0,%eax
  jmp    L26
L85:
  mov    $0x1,%eax
  jmp    L26
L50:
  mov    $0x1,%eax
  jmp    L26
L62:
  mov    $0x0,%eax
  jmp    L26
L135:
  mov    $0x0,%eax
  jmp    L26
L110:
  mov    $0x0,%eax
  jmp    L26
L19:
  ret
```

assistant: ```c
int isPrime(unsigned long n) {
  if (n == 2 || n == 3)
    return 1;
  if (n < 2 || n % 2 == 0)
    return 0;
  if (n % 3 == 0)
    return 0;
  if (n < 9)
    return 1;
  unsigned long r = (unsigned long)sqrt(n);
  unsigned long f = 5;
  while (f <= r) {
    if (n % f == 0 || n % (f + 2) == 0)
      return 0;
    f += 6;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<isPrime>:
  endbr64
  lea    -0x2(%rdi),%rax
  cmp    $0x1,%rax
  jbe    L99
  push   %r12
  mov    %rdi,%r12
  cmp    $0x1,%rdi
  jbe    L128
  test   $0x1,%dil
  je     L128
  movabs $0xaaaaaaaaaaaaaaab,%rax
  movabs $0x5555555555555555,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L128
  cmp    $0x8,%rdi
  ja     L106
L44:
  mov    $0x1,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L128:
  xor    %eax,%eax
  pop    %r12
  ret
  nopl   (%rax)
L106:
  call   <sqrt@plt>
  mov    %rax,%rdi
  cmp    $0x4,%rax
  jbe    L44
  movabs $0xcccccccccccccccd,%rax
  movabs $0x3333333333333333,%rdx
  imul   %r12,%rax
  cmp    %rdx,%rax
  jbe    L128
  movabs $0x6db6db6db6db6db7,%rax
  movabs $0x2492492492492492,%rdx
  imul   %r12,%rax
  cmp    %rdx,%rax
  jbe    L128
  mov    $0x5,%ecx
  jmp    L121
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L128
  add    $0x8,%rsi
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rsi
  test   %rdx,%rdx
  je     L128
L121:
  mov    %rcx,%rsi
  add    $0x6,%rcx
  cmp    %rcx,%rdi
  jae    L64
  jmp    L44
  nop
L99:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isPrime(unsigned long n) {
  if (n == 2 || n == 3)
    return 1;
  if (n < 2 || n % 2 == 0)
    return 0;
  if (n % 3 == 0)
    return 0;
  if (n < 9)
    return 1;
  unsigned long r = (unsigned long)sqrt(n);
  unsigned long f = 5;
  while (f <= r) {
    if (n % f == 0 || n % (f + 2) == 0)
      return 0;
    f += 6;
  }
  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
<isPrime>:
  endbr64
  lea    -0x2(%rdi),%rax
  cmp    $0x1,%rax
  jbe    L107
  push   %r12
  mov    %rdi,%r12
  cmp    $0x1,%rdi
  jbe    L116
  test   $0x1,%dil
  je     L116
  movabs $0xaaaaaaaaaaaaaaab,%rax
  movabs $0x5555555555555555,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  jbe    L116
  cmp    $0x8,%rdi
  ja     L87
L129:
  mov    $0x1,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L116:
  xor    %eax,%eax
  pop    %r12
  ret
  nopl   (%rax)
L87:
  call   <sqrt@plt>
  mov    %rax,%rdi
  cmp    $0x4,%rax
  jbe    L129
  movabs $0xcccccccccccccccd,%rax
  movabs $0x3333333333333333,%rdx
  imul   %r12,%rax
  cmp    %rdx,%rax
  jbe    L116
  movabs $0x6db6db6db6db6db7,%rax
  movabs $0x2492492492492492,%rdx
  imul   %r12,%rax
  cmp    %rdx,%rax
  jbe    L116
  mov    $0x5,%ecx
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L69:
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L116
  add    $0x8,%rsi
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rsi
  test   %rdx,%rdx
  je     L116
L6:
  mov    %rcx,%rsi
  add    $0x6,%rcx
  cmp    %rcx,%rdi
  jae    L69
  jmp    L129
  nop
L107:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int isPrime(unsigned long n) {
  if (n == 2 || n == 3)
    return 1;
  if (n < 2 || n % 2 == 0)
    return 0;
  if (n % 3 == 0)
    return 0;
  if (n < 9)
    return 1;
  unsigned long r = (unsigned long)sqrt(n);
  unsigned long f = 5;
  while (f <= r) {
    if (n % f == 0 || n % (f + 2) == 0)
      return 0;
    f += 6;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<all_ff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L37
L24:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xff,%al
  je     L25
  mov    $0x0,%eax
  jmp    L23
L25:
  addl   $0x1,-0x4(%rbp)
L37:
  cmpl   $0x17,-0x4(%rbp)
  jle    L24
  mov    $0x1,%eax
L23:
  pop    %rbp
  ret
```

assistant: ```c
int all_ff(const char *guid) {
  int i;
  for (i = 0; i < 24; i++)
    if (guid[i] != (char)0xff)
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<all_ff>:
  endbr64
  lea    0x18(%rdi),%rax
L10:
  cmpb   $0xff,(%rdi)
  jne    L18
  add    $0x1,%rdi
  cmp    %rax,%rdi
  jne    L10
  mov    $0x1,%eax
  ret
L18:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int all_ff(const char *guid) {
  int i;
  for (i = 0; i < 24; i++)
    if (guid[i] != (char)0xff)
      return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<all_ff>:
  endbr64
  lea    0x18(%rdi),%rax
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L19:
  add    $0x1,%rdi
  cmp    %rax,%rdi
  je     L26
L15:
  cmpb   $0xff,(%rdi)
  je     L19
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L26:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int all_ff(const char *guid) {
  int i;
  for (i = 0; i < 24; i++)
    if (guid[i] != (char)0xff)
      return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<all_ff>:
  endbr64
  lea    0x18(%rdi),%rax
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L26:
  add    $0x1,%rdi
  cmp    %rax,%rdi
  je     L23
L15:
  cmpb   $0xff,(%rdi)
  je     L26
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L23:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int all_ff(const char *guid) {
  int i;
  for (i = 0; i < 24; i++)
    if (guid[i] != (char)0xff)
      return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L28
L11:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L28:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L26
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
L26:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int my_strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++)
    ;
  return *(unsigned char *)l - *(unsigned char *)r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_strcmp>:
  endbr64
  movzbl (%rdi),%eax
  movzbl (%rsi),%ecx
  test   %al,%al
  je     L34
  cmp    %cl,%al
  jne    L34
  mov    $0x1,%edx
L12:
  movzbl (%rdi,%rdx,1),%eax
  movzbl (%rsi,%rdx,1),%ecx
  add    $0x1,%rdx
  test   %al,%al
  je     L34
  cmp    %cl,%al
  je     L12
L34:
  movzbl %al,%eax
  movzbl %cl,%ecx
  sub    %ecx,%eax
  ret
```

assistant: ```c
int my_strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++)
    ;
  return *(unsigned char *)l - *(unsigned char *)r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<my_strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  mov    $0x1,%eax
  cmp    %dl,%cl
  je     L1
  jmp    L13
  nopl   (%rax)
L20:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L13
L1:
  test   %dl,%dl
  jne    L20
  xor    %eax,%eax
  sub    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int my_strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++)
    ;
  return *(unsigned char *)l - *(unsigned char *)r;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<my_strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  mov    $0x1,%eax
  cmp    %dl,%cl
  je     L11
  jmp    L38
  nopl   (%rax)
L20:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L38
L11:
  test   %dl,%dl
  jne    L20
  xor    %eax,%eax
  sub    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  movzbl %dl,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int my_strcmp(const char *l, const char *r) {
  for (; *l == *r && *l; l++, r++)
    ;
  return *(unsigned char *)l - *(unsigned char *)r;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_add_epi32@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <_mm_cvtsi128_si32@plt>
  mov    -0x30(%rbp),%rdx
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void func(__m128i *foo, size_t a, size_t b, int *dst) {
  __m128i x = foo[a];
  __m128i y = foo[b];
  __m128i sum = _mm_add_epi32(x, y);
  *dst = _mm_cvtsi128_si32(sum);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<func>:
  endbr64
  push   %rbx
  mov    %rsi,%rax
  mov    %rcx,%rbx
  mov    (%rdi,%rdx,4),%esi
  mov    (%rdi,%rax,4),%edi
  call   <_mm_add_epi32@plt>
  mov    %eax,%edi
  call   <_mm_cvtsi128_si32@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void func(__m128i *foo, size_t a, size_t b, int *dst) {
  __m128i x = foo[a];
  __m128i y = foo[b];
  __m128i sum = _mm_add_epi32(x, y);
  *dst = _mm_cvtsi128_si32(sum);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<func>:
  endbr64
  push   %rbx
  mov    %rsi,%r8
  mov    (%rdi,%rdx,4),%esi
  mov    %rcx,%rbx
  mov    (%rdi,%r8,4),%edi
  call   <_mm_add_epi32@plt>
  mov    %eax,%edi
  call   <_mm_cvtsi128_si32@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void func(__m128i *foo, size_t a, size_t b, int *dst) {
  __m128i x = foo[a];
  __m128i y = foo[b];
  __m128i sum = _mm_add_epi32(x, y);
  *dst = _mm_cvtsi128_si32(sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inv_fifo_protocol_helper_decode_pressure_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <FifoProtocol_memcpy@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void inv_fifo_protocol_helper_decode_pressure_data(const void *in, uint16_t len, uint32_t *pressure) {
  FifoProtocol_memcpy((pressure), (in), (sizeof(uint32_t)));
  *pressure = *pressure;
  (void)len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inv_fifo_protocol_helper_decode_pressure_data>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    %rdx,%rdi
  mov    $0x4,%edx
  call   <FifoProtocol_memcpy@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void inv_fifo_protocol_helper_decode_pressure_data(const void *in, uint16_t len, uint32_t *pressure) {
  FifoProtocol_memcpy((pressure), (in), (sizeof(uint32_t)));
  *pressure = *pressure;
  (void)len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<inv_fifo_protocol_helper_decode_pressure_data>:
  endbr64
  mov    %rdi,%rsi
  mov    %rdx,%rdi
  mov    $0x4,%edx
  jmp    1050 <FifoProtocol_memcpy@plt>
```

assistant: ```c
void inv_fifo_protocol_helper_decode_pressure_data(const void *in, uint16_t len, uint32_t *pressure) {
  FifoProtocol_memcpy((pressure), (in), (sizeof(uint32_t)));
  *pressure = *pressure;
  (void)len;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<inv_fifo_protocol_helper_decode_pressure_data>:
  endbr64
  mov    %rdi,%rsi
  mov    %rdx,%rdi
  mov    $0x4,%edx
  jmp    1050 <FifoProtocol_memcpy@plt>
```

assistant: ```c
void inv_fifo_protocol_helper_decode_pressure_data(const void *in, uint16_t len, uint32_t *pressure) {
  FifoProtocol_memcpy((pressure), (in), (sizeof(uint32_t)));
  *pressure = *pressure;
  (void)len;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rul__initialize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <rul__mol_init@plt>
  mov    $0x0,%eax
  call   <rul__ios_init@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rul__initialize(void) {
  rul__mol_init();
  rul__ios_init();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rul__initialize>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <rul__mol_init@plt>
  mov    $0x0,%eax
  call   <rul__ios_init@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rul__initialize(void) {
  rul__mol_init();
  rul__ios_init();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rul__initialize>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <rul__mol_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <rul__ios_init@plt>
```

assistant: ```c
void rul__initialize(void) {
  rul__mol_init();
  rul__ios_init();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rul__initialize>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <rul__mol_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <rul__ios_init@plt>
```

assistant: ```c
void rul__initialize(void) {
  rul__mol_init();
  rul__ios_init();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<MFRC522_Anticoll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0xd(%rbp)
  mov    $0x0,%esi
  mov    $0xd,%edi
  call   <Write_MFRC522@plt>
  mov    -0x18(%rbp),%rax
  movb   $0x93,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movb   $0x20,(%rax)
  lea    -0xc(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%r8
  mov    %rdx,%rcx
  mov    $0x2,%edx
  mov    %rax,%rsi
  mov    $0xc,%edi
  call   <MFRC522_ToCard@plt>
  mov    %al,-0xf(%rbp)
  cmpb   $0x0,-0xf(%rbp)
  jne    L29
  movb   $0x0,-0xe(%rbp)
  jmp    L12
L22:
  movzbl -0xe(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  xor    %al,-0xd(%rbp)
  movzbl -0xe(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0xe(%rbp)
L12:
  cmpb   $0x3,-0xe(%rbp)
  jbe    L22
  movzbl -0xe(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xd(%rbp)
  je     L29
  movb   $0x2,-0xf(%rbp)
L29:
  movzbl -0xf(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
unsigned char MFRC522_Anticoll(unsigned char *serNum) {
  unsigned char status;
  unsigned char i;
  unsigned char serNumCheck = 0;
  unsigned int unLen;
  Write_MFRC522(0x0D, 0x00);
  serNum[0] = 0x93;
  serNum[1] = 0x20;
  status = MFRC522_ToCard(0x0C, serNum, 2, serNum, &unLen);
  if (status == 0) {
    for (i = 0; i < 4; i++) {
      serNumCheck ^= serNum[i];
    }
    if (serNumCheck != serNum[i]) {
      status = 2;
    }
  }
  return status;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<MFRC522_Anticoll>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%esi
  mov    $0xd,%edi
  call   <Write_MFRC522@plt>
  movb   $0x93,(%rbx)
  movb   $0x20,0x1(%rbx)
  lea    0x4(%rsp),%r8
  mov    %rbx,%rcx
  mov    $0x2,%edx
  mov    %rbx,%rsi
  mov    $0xc,%edi
  call   <MFRC522_ToCard@plt>
  test   %al,%al
  jne    L0
  mov    %rbx,%rdx
  lea    0x4(%rbx),%rcx
L2:
  xor    (%rdx),%al
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L2
  cmp    %al,0x4(%rbx)
  setne  %al
  add    %eax,%eax
L0:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x10,%rsp
  pop    %rbx
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char MFRC522_Anticoll(unsigned char *serNum) {
  unsigned char status;
  unsigned char i;
  unsigned char serNumCheck = 0;
  unsigned int unLen;
  Write_MFRC522(0x0D, 0x00);
  serNum[0] = 0x93;
  serNum[1] = 0x20;
  status = MFRC522_ToCard(0x0C, serNum, 2, serNum, &unLen);
  if (status == 0) {
    for (i = 0; i < 4; i++) {
      serNumCheck ^= serNum[i];
    }
    if (serNumCheck != serNum[i]) {
      status = 2;
    }
  }
  return status;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<MFRC522_Anticoll>:
  endbr64
  push   %rbx
  xor    %esi,%esi
  mov    %rdi,%rbx
  mov    $0xd,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <Write_MFRC522@plt>
  mov    $0x2093,%eax
  mov    %rbx,%rcx
  mov    %rbx,%rsi
  mov    %ax,(%rbx)
  lea    0x4(%rsp),%r8
  mov    $0x2,%edx
  mov    $0xc,%edi
  call   <MFRC522_ToCard@plt>
  test   %al,%al
  jne    L26
  mov    %rbx,%rdx
  lea    0x4(%rbx),%rcx
L12:
  xor    (%rdx),%al
  add    $0x1,%rdx
  cmp    %rdx,%rcx
  jne    L12
  cmp    %al,0x4(%rbx)
  setne  %al
  add    %eax,%eax
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char MFRC522_Anticoll(unsigned char *serNum) {
  unsigned char status;
  unsigned char i;
  unsigned char serNumCheck = 0;
  unsigned int unLen;
  Write_MFRC522(0x0D, 0x00);
  serNum[0] = 0x93;
  serNum[1] = 0x20;
  status = MFRC522_ToCard(0x0C, serNum, 2, serNum, &unLen);
  if (status == 0) {
    for (i = 0; i < 4; i++) {
      serNumCheck ^= serNum[i];
    }
    if (serNumCheck != serNum[i]) {
      status = 2;
    }
  }
  return status;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MFRC522_Anticoll>:
  endbr64
  push   %rbx
  xor    %esi,%esi
  mov    %rdi,%rbx
  mov    $0xd,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <Write_MFRC522@plt>
  mov    $0x2093,%eax
  mov    %rbx,%rcx
  mov    %rbx,%rsi
  mov    %ax,(%rbx)
  lea    0x4(%rsp),%r8
  mov    $0x2,%edx
  mov    $0xc,%edi
  call   <MFRC522_ToCard@plt>
  test   %al,%al
  jne    L2
  movzbl 0x1(%rbx),%eax
  xor    0x2(%rbx),%al
  xor    (%rbx),%al
  xor    0x3(%rbx),%al
  cmp    0x4(%rbx),%al
  setne  %al
  add    %eax,%eax
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x10,%rsp
  pop    %rbx
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char MFRC522_Anticoll(unsigned char *serNum) {
  unsigned char status;
  unsigned char i;
  unsigned char serNumCheck = 0;
  unsigned int unLen;
  Write_MFRC522(0x0D, 0x00);
  serNum[0] = 0x93;
  serNum[1] = 0x20;
  status = MFRC522_ToCard(0x0C, serNum, 2, serNum, &unLen);
  if (status == 0) {
    for (i = 0; i < 4; i++) {
      serNumCheck ^= serNum[i];
    }
    if (serNumCheck != serNum[i]) {
      status = 2;
    }
  }
  return status;
}
```