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
<rtems_shell_make_args>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  lea    -0x1(%rcx),%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L127:
  cmpb   $0x0,(%rbx)
  jne    L124
  jmp    L47
  nop
L162:
  add    $0x1,%rbx
L124:
  movzbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L162
  movzbl (%rbx),%eax
  test   %al,%al
  je     L47
  cmp    $0x22,%al
  je     L109
  mov    %rbx,(%r12)
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L40
  jmp    L161
  nopw   0x0(%rax,%rax,1)
L141:
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L161
L40:
  call   <isspace@plt>
  test   %rax,%rax
  je     L141
  movzbl (%rbx),%edx
  mov    %rbx,%rax
L140:
  add    $0x1,%ebp
  test   %dl,%dl
  je     L47
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rbx
  add    $0x8,%r12
  cmp    %ebp,%r13d
  jne    L127
  mov    $0xffffffff,%eax
  jmp    L102
  nopl   0x0(%rax,%rax,1)
L161:
  add    $0x1,%ebp
L47:
  xor    %eax,%eax
L102:
  movslq %ebp,%rdx
  movq   $0x0,(%r14,%rdx,8)
  mov    %ebp,(%r15)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L109:
  lea    0x1(%rbx),%rax
  mov    %rax,(%r12)
  movzbl 0x1(%rbx),%edx
  test   %dl,%dl
  jne    L155
  jmp    L140
  nopw   0x0(%rax,%rax,1)
L62:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L140
L155:
  cmp    $0x22,%dl
  jne    L62
  jmp    L140
```

assistant: ```c
int rtems_shell_make_args(char *commandLine, int *argc_p, char **argv_p, int max_args) {
  int argc;
  char *ch;
  int status = 0;
  argc = 0;
  ch = commandLine;
  while (*ch) {
    while (isspace((unsigned char)*ch))
      ch++;
    if (*ch == '\0')
      break;
    if (*ch == '"') {
      argv_p[argc] = ++ch;
      while ((*ch != '\0') && (*ch != '"'))
        ch++;
    } else {
      argv_p[argc] = ch;
      while ((*ch != '\0') && (!isspace((unsigned char)*ch)))
        ch++;
    }
    argc++;
    if (*ch == '\0')
      break;
    *ch++ = '\0';
    if (argc == (max_args - 1)) {
      status = -1;
      break;
    }
  }
  argv_p[argc] = NULL;
  *argc_p = argc;
  return status;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L51
  cmpl   $0x9,-0x4(%rbp)
  jg     L51
  mov    -0x4(%rbp),%eax
  add    $0x30,%eax
  jmp    L29
L51:
  cmpl   $0x9,-0x4(%rbp)
  jle    L34
  cmpl   $0xf,-0x4(%rbp)
  jg     L34
  mov    -0x4(%rbp),%eax
  add    $0x37,%eax
  jmp    L29
L34:
  mov    $0x30,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
char itoa(int i) {
  if (i >= 0 && i <= 9)
    return i + '0';
  else if (i >= 0xA && i <= 0xF)
    return i - 0xA + 'A';
  else
    return '0';
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<itoa>:
  endbr64
  cmp    $0x9,%edi
  jbe    L8
  lea    -0xa(%rdi),%eax
  add    $0x37,%edi
  cmp    $0x6,%eax
  mov    $0x30,%eax
  cmovb  %edi,%eax
  ret
L8:
  lea    0x30(%rdi),%eax
  ret
```

assistant: ```c
char itoa(int i) {
  if (i >= 0 && i <= 9)
    return i + '0';
  else if (i >= 0xA && i <= 0xF)
    return i - 0xA + 'A';
  else
    return '0';
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<itoa>:
  endbr64
  cmp    $0x9,%edi
  jbe    L1
  lea    -0xa(%rdi),%eax
  add    $0x37,%edi
  cmp    $0x6,%eax
  mov    $0x30,%eax
  cmovb  %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  lea    0x30(%rdi),%eax
  ret
```

assistant: ```c
char itoa(int i) {
  if (i >= 0 && i <= 9)
    return i + '0';
  else if (i >= 0xA && i <= 0xF)
    return i - 0xA + 'A';
  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
<itoa>:
  endbr64
  cmp    $0x9,%edi
  jbe    L0
  lea    -0xa(%rdi),%eax
  add    $0x37,%edi
  cmp    $0x6,%eax
  mov    $0x30,%eax
  cmovb  %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  lea    0x30(%rdi),%eax
  ret
```

assistant: ```c
char itoa(int i) {
  if (i >= 0 && i <= 9)
    return i + '0';
  else if (i >= 0xA && i <= 0xF)
    return i - 0xA + 'A';
  else
    return '0';
}
```
user: What is the c source code of the assembly code below:

```asm
<read_block>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x0,-0xd(%rbp)
  mov    $0x0,%eax
  call   <get_word@plt>
  mov    %eax,%eax
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <get_word@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L16
L0:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <put_char@plt>
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  add    %al,-0xd(%rbp)
L16:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jne    L0
  movzbl -0xd(%rbp),%eax
  mov    %eax,%edi
  call   <put_char@plt>
  nop
  leave
  ret
```

assistant: ```c
void read_block(void) {
  unsigned char checksum = 0;
  unsigned char *p = (unsigned char *)get_word();
  unsigned length = get_word();
  while (length--) {
    put_char(*p);
    checksum += *p++;
  }
  put_char(checksum);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<read_block>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <get_word@plt>
  mov    %eax,%ebp
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <get_word@plt>
  test   %eax,%eax
  je     L27
  lea    -0x1(%rax),%edx
  mov    %ebp,%ebp
  mov    %edx,%edx
  lea    0x1(%rbp,%rdx,1),%r12
  mov    $0x0,%ebp
L0:
  movzbl (%rbx),%edi
  call   <put_char@plt>
  mov    %rbx,%rax
  add    $0x1,%rbx
  add    (%rax),%bpl
  cmp    %r12,%rbx
  jne    L0
L23:
  movzbl %bpl,%edi
  call   <put_char@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  mov    $0x0,%ebp
  jmp    L23
```

assistant: ```c
void read_block(void) {
  unsigned char checksum = 0;
  unsigned char *p = (unsigned char *)get_word();
  unsigned length = get_word();
  while (length--) {
    put_char(*p);
    checksum += *p++;
  }
  put_char(checksum);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_block>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <get_word@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <get_word@plt>
  test   %eax,%eax
  je     L23
  lea    -0x1(%rax),%r13d
  lea    0x1(%rbp),%rbx
  xor    %r12d,%r12d
  add    %rbx,%r13
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L17:
  add    $0x1,%rbx
L24:
  movzbl 0x0(%rbp),%edi
  call   <put_char@plt>
  mov    %rbp,%rax
  mov    %rbx,%rbp
  add    (%rax),%r12b
  cmp    %r13,%rbx
  jne    L17
  add    $0x8,%rsp
  movzbl %r12b,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <put_char@plt>
  cs nopw 0x0(%rax,%rax,1)
L23:
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <put_char@plt>
```

assistant: ```c
void read_block(void) {
  unsigned char checksum = 0;
  unsigned char *p = (unsigned char *)get_word();
  unsigned length = get_word();
  while (length--) {
    put_char(*p);
    checksum += *p++;
  }
  put_char(checksum);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<read_block>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <get_word@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <get_word@plt>
  test   %eax,%eax
  je     L19
  lea    -0x1(%rax),%r13d
  lea    0x1(%rbp),%rbx
  xor    %r12d,%r12d
  add    %rbx,%r13
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L5:
  add    $0x1,%rbx
L14:
  movzbl 0x0(%rbp),%edi
  call   <put_char@plt>
  mov    %rbp,%rax
  mov    %rbx,%rbp
  add    (%rax),%r12b
  cmp    %r13,%rbx
  jne    L5
  add    $0x8,%rsp
  movzbl %r12b,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <put_char@plt>
  cs nopw 0x0(%rax,%rax,1)
L19:
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <put_char@plt>
```

assistant: ```c
void read_block(void) {
  unsigned char checksum = 0;
  unsigned char *p = (unsigned char *)get_word();
  unsigned length = get_word();
  while (length--) {
    put_char(*p);
    checksum += *p++;
  }
  put_char(checksum);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  shl    $0x7,%eax
  and    $0x4334b500,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L58
  mov    -0xc(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L41
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shr    $0x2,%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L8
L41:
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L8
L58:
  mov    -0xc(%rbp),%eax
  and    $0x10,%eax
  test   %eax,%eax
  jne    L8
  lea    -0xc(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  lea    -0xc(%rbp),%rax
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
L8:
  mov    -0xc(%rbp),%eax
  xor    $0x2c68860e,%eax
  and    $0x3cbeecbf,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    if ((state[0UL] >> 1U) & 1U) {
      if ((state[0UL] >> 1U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 2U) & 7U) | 1UL;
      } else {
        state[0UL] <<= (state[0UL] & 7U) | 1UL;
      }
    } else if (!((state[0UL] >> 4U) & 1U)) {
      copy12 = *((char *)(&state[0UL]) + 3);
      *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
    }
    output[0UL] = (state[0UL] ^ 745047566UL) & 1019145407U;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x7,%eax
  mov    %eax,%edx
  and    $0x4334b500,%edx
  and    $0x34b500,%eax
  sar    $0x18,%edx
  mov    %dl,%al
  xor    $0x2c68860e,%eax
  and    $0x3cbeecbf,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    if ((state[0UL] >> 1U) & 1U) {
      if ((state[0UL] >> 1U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 2U) & 7U) | 1UL;
      } else {
        state[0UL] <<= (state[0UL] & 7U) | 1UL;
      }
    } else if (!((state[0UL] >> 4U) & 1U)) {
      copy12 = *((char *)(&state[0UL]) + 3);
      *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
    }
    output[0UL] = (state[0UL] ^ 745047566UL) & 1019145407U;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x7,%eax
  mov    %eax,%edx
  and    $0x34b500,%eax
  and    $0x4334b500,%edx
  sar    $0x18,%edx
  mov    %dl,%al
  xor    $0x2c68860e,%eax
  and    $0x3cbeecbf,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    if ((state[0UL] >> 1U) & 1U) {
      if ((state[0UL] >> 1U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 2U) & 7U) | 1UL;
      } else {
        state[0UL] <<= (state[0UL] & 7U) | 1UL;
      }
    } else if (!((state[0UL] >> 4U) & 1U)) {
      copy12 = *((char *)(&state[0UL]) + 3);
      *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
    }
    output[0UL] = (state[0UL] ^ 745047566UL) & 1019145407U;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x7,%eax
  mov    %eax,%edx
  and    $0x34b500,%eax
  and    $0x4334b500,%edx
  sar    $0x18,%edx
  mov    %dl,%al
  xor    $0x2c68860e,%eax
  and    $0x3cbeecbf,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    if ((state[0UL] >> 1U) & 1U) {
      if ((state[0UL] >> 1U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 2U) & 7U) | 1UL;
      } else {
        state[0UL] <<= (state[0UL] & 7U) | 1UL;
      }
    } else if (!((state[0UL] >> 4U) & 1U)) {
      copy12 = *((char *)(&state[0UL]) + 3);
      *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
    }
    output[0UL] = (state[0UL] ^ 745047566UL) & 1019145407U;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<calu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %esi,-0x58(%rbp)
  mov    %edx,-0x5c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x40(%rbp)
  movl   $0x1f,-0x3c(%rbp)
  movl   $0x3b,-0x38(%rbp)
  movl   $0x5a,-0x34(%rbp)
  movl   $0x78,-0x30(%rbp)
  movl   $0x97,-0x2c(%rbp)
  movl   $0xb5,-0x28(%rbp)
  movl   $0xd4,-0x24(%rbp)
  movl   $0xf3,-0x20(%rbp)
  movl   $0x111,-0x1c(%rbp)
  movl   $0x130,-0x18(%rbp)
  movl   $0x14e,-0x14(%rbp)
  mov    -0x54(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L58
  mov    -0x54(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L23
L58:
  mov    -0x54(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L12
L23:
  mov    $0x1,%eax
  jmp    L49
L12:
  mov    $0x0,%eax
L49:
  mov    %eax,-0x4c(%rbp)
  mov    -0x58(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%edx
  mov    -0x5c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x48(%rbp)
  cmpl   $0x0,-0x4c(%rbp)
  je     L62
  cmpl   $0x2,-0x58(%rbp)
  jle    L62
  mov    -0x48(%rbp),%eax
  add    $0x1,%eax
  jmp    L67
L62:
  mov    -0x48(%rbp),%eax
L67:
  mov    %eax,-0x48(%rbp)
  mov    -0x54(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x54(%rbp),%eax
  sub    $0x1,%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  add    %eax,%ecx
  mov    -0x54(%rbp),%eax
  sub    $0x1,%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  sub    %edx,%eax
  add    %eax,%ecx
  mov    -0x54(%rbp),%eax
  sub    $0x1,%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x7,%edx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x48(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x3,%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

assistant: ```c
int calu(int y, int m, int d) {
  int days[12] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
  int leap_year = (((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0)) ? 1 : 0;
  int D = days[m - 1] + d;
  D = ((leap_year && m > 2) ? D + 1 : D);
  int W = y - 1 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + D;
  return W % 7;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<calu>:
  endbr64
  sub    $0x48,%rsp
  mov    %edi,%eax
  mov    %esi,%ecx
  mov    %edx,%esi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x38(%rsp)
  xor    %edx,%edx
  movl   $0x0,(%rsp)
  movl   $0x1f,0x4(%rsp)
  movl   $0x3b,0x8(%rsp)
  movl   $0x5a,0xc(%rsp)
  movl   $0x78,0x10(%rsp)
  movl   $0x97,0x14(%rsp)
  movl   $0xb5,0x18(%rsp)
  movl   $0xd4,0x1c(%rsp)
  movl   $0xf3,0x20(%rsp)
  movl   $0x111,0x24(%rsp)
  movl   $0x130,0x28(%rsp)
  movl   $0x14e,0x2c(%rsp)
  test   $0x3,%al
  jne    L28
  movslq %edi,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sar    $0x1f,%edi
  sub    %edi,%edx
  imul   $0x64,%edx,%edx
  mov    $0x1,%edi
  cmp    %edx,%eax
  jne    L31
L28:
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x27,%rdx
  mov    %eax,%edi
  sar    $0x1f,%edi
  sub    %edi,%edx
  imul   $0x190,%edx,%edx
  cmp    %edx,%eax
  sete   %dil
  movzbl %dil,%edi
L31:
  lea    -0x1(%rcx),%edx
  movslq %edx,%rdx
  add    (%rsp,%rdx,4),%esi
  test   %edi,%edi
  je     L23
  cmp    $0x2,%ecx
  setg   %dl
  cmp    $0x1,%dl
  sbb    $0xffffffff,%esi
L23:
  lea    0x2(%rax),%edx
  sub    $0x1,%eax
  cmovns %eax,%edx
  sar    $0x2,%edx
  add    %eax,%edx
  movslq %eax,%rcx
  imul   $0x51eb851f,%rcx,%rcx
  mov    %rcx,%rdi
  sar    $0x25,%rdi
  mov    %rdi,%r8
  sar    $0x1f,%eax
  mov    %eax,%edi
  sub    %r8d,%edi
  add    %edi,%edx
  sar    $0x27,%rcx
  sub    %eax,%ecx
  add    %ecx,%edx
  add    %esi,%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    0x0(,%rax,8),%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x48,%rsp
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int calu(int y, int m, int d) {
  int days[12] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
  int leap_year = (((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0)) ? 1 : 0;
  int D = days[m - 1] + d;
  D = ((leap_year && m > 2) ? D + 1 : D);
  int W = y - 1 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + D;
  return W % 7;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<calu>:
  endbr64
  movabs $0x1f00000000,%rcx
  sub    $0x48,%rsp
  mov    %edi,%eax
  mov    %edx,%edi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x38(%rsp)
  xor    %edx,%edx
  mov    %rcx,(%rsp)
  movabs $0x5a0000003b,%rcx
  movabs $0x111000000f3,%rdx
  mov    %rcx,0x8(%rsp)
  movabs $0x9700000078,%rcx
  mov    %rcx,0x10(%rsp)
  movabs $0xd4000000b5,%rcx
  mov    %rcx,0x18(%rsp)
  movabs $0x14e00000130,%rcx
  mov    %rdx,0x20(%rsp)
  mov    %rcx,0x28(%rsp)
  test   $0x3,%al
  jne    L13
  imul   $0xc28f5c29,%eax,%edx
  add    $0x51eb850,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L13
L32:
  lea    -0x1(%rsi),%edx
  movslq %edx,%rdx
  add    (%rsp,%rdx,4),%edi
  xor    %edx,%edx
  cmp    $0x2,%esi
  setg   %dl
  add    %edx,%edi
L39:
  lea    0x2(%rax),%edx
  sub    $0x1,%eax
  movslq %eax,%rcx
  cmovns %eax,%edx
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x2,%edx
  add    %eax,%edx
  sar    $0x1f,%eax
  mov    %rcx,%rsi
  sar    $0x27,%rcx
  sar    $0x25,%rsi
  sub    %eax,%ecx
  mov    %rsi,%r8
  mov    %eax,%esi
  sub    %r8d,%esi
  add    %esi,%edx
  add    %ecx,%edx
  add    %edi,%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%ecx
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  sub    %ecx,%eax
  lea    0x0(,%rax,8),%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L34
  add    $0x48,%rsp
  ret
  nopl   (%rax)
L13:
  imul   $0xc28f5c29,%eax,%ecx
  add    $0x51eb850,%ecx
  ror    $0x4,%ecx
  cmp    $0xa3d70a,%ecx
  jbe    L32
  sub    $0x1,%esi
  movslq %esi,%rsi
  add    (%rsp,%rsi,4),%edi
  jmp    L39
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int calu(int y, int m, int d) {
  int days[12] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
  int leap_year = (((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0)) ? 1 : 0;
  int D = days[m - 1] + d;
  D = ((leap_year && m > 2) ? D + 1 : D);
  int W = y - 1 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + D;
  return W % 7;
}
```
user: What is the c source code of the assembly code below?

```asm
<calu>:
  endbr64
  sub    $0x48,%rsp
  movdqa D16(%rip),%xmm0
  mov    %edi,%eax
  mov    %edx,%edi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x38(%rsp)
  xor    %edx,%edx
  movaps %xmm0,(%rsp)
  movdqa D8(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movdqa D19(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  test   $0x3,%al
  jne    L0
  imul   $0xc28f5c29,%eax,%edx
  add    $0x51eb850,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L0
L39:
  lea    -0x1(%rsi),%edx
  movslq %edx,%rdx
  add    (%rsp,%rdx,4),%edi
  xor    %edx,%edx
  cmp    $0x2,%esi
  setg   %dl
  add    %edx,%edi
L34:
  lea    0x2(%rax),%edx
  sub    $0x1,%eax
  movslq %eax,%rcx
  cmovns %eax,%edx
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x2,%edx
  add    %eax,%edx
  sar    $0x1f,%eax
  mov    %rcx,%rsi
  sar    $0x27,%rcx
  sar    $0x25,%rsi
  sub    %eax,%ecx
  mov    %rsi,%r8
  mov    %eax,%esi
  sub    %r8d,%esi
  add    %esi,%edx
  add    %ecx,%edx
  add    %edi,%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%ecx
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  sub    %ecx,%eax
  lea    0x0(,%rax,8),%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x48,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  imul   $0xc28f5c29,%eax,%ecx
  add    $0x51eb850,%ecx
  ror    $0x4,%ecx
  cmp    $0xa3d70a,%ecx
  jbe    L39
  sub    $0x1,%esi
  movslq %esi,%rsi
  add    (%rsp,%rsi,4),%edi
  jmp    L34
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D16
type_guessed: byte[16]
name: disassemble_data
arguments: data_label: D8
type_guessed: byte[16]
name: disassemble_data
arguments: data_label: D19
type_guessed: byte[16]

tool: {"D16": [0x00,0x00,0x00,0x00,0x1f,0x00,0x00,0x00,0x3b,0x00,0x00,0x00,0x5a,0x00,0x00,0x00]}

tool: {"D8": [0x78,0x00,0x00,0x00,0x97,0x00,0x00,0x00,0xb5,0x00,0x00,0x00,0xd4,0x00,0x00,0x00]}

tool: {"D19": [0xf3,0x00,0x00,0x00,0x11,0x01,0x00,0x00,0x30,0x01,0x00,0x00,0x4e,0x01,0x00,0x00]}

assistant: ```c
int calu(int y, int m, int d) {
  int days[12] = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
  int leap_year = (((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0)) ? 1 : 0;
  int D = days[m - 1] + d;
  D = ((leap_year && m > 2) ? D + 1 : D);
  int W = y - 1 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + D;
  return W % 7;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sfill>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  mov    %rax,%r12
  mov    -0x18(%rbp),%ebx
L9:
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movb   $0x20,(%rax)
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jne    L9
  nop
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
sfill(str, cnt) {
  register char *cp;
  register int i;
  cp = str;
  i = cnt;
  do
    *cp++ = ' ';
  while (--i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sfill>:
  endbr64
  movslq %edi,%rdi
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rdx
L5:
  mov    %rdi,%rax
  add    $0x1,%rdi
  movb   $0x20,(%rax)
  cmp    %rdx,%rdi
  jne    L5
  ret
```

assistant: ```c
sfill(str, cnt) {
  register char *cp;
  register int i;
  cp = str;
  i = cnt;
  do
    *cp++ = ' ';
  while (--i);
}
```