ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<bmi_test>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0xfacec0ff,%ebx
  sub    $0x8,%rsp
L10:
  mov    %ebp,%r13d
  lea    (%rbx,%rbx,2),%ebx
  and    $0x1f,%r13d
  mov    %ebx,%edi
  mov    %r13d,%esi
  call   <calc_bextr_u32@plt>
  mov    %r13d,%esi
  mov    %ebx,%edi
  mov    %eax,%r12d
  call   <__bextr_u32@plt>
  cmp    %eax,%r12d
  je     L2
  xor    %eax,%eax
  call   <abort@plt>
L2:
  sub    $0x41,%ebp
  cmp    $0xbb,%bpl
  jne    L10
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void bmi_test() {
  unsigned i;
  unsigned char start, len;
  unsigned src1 = 0xfacec0ff;
  unsigned res, res_ref, src2;
  for (i = 0; i < 5; ++i) {
    start = (i * 1983) % 32;
    len = (i + (i * 1983)) % 32;
    src1 = src1 * 3;
    src2 = start | (((unsigned)len) << 8);
    res_ref = calc_bextr_u32(src1, src2);
    res = __bextr_u32(src1, src2);
    if (res != res_ref)
      abort();
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<julian_make_y2k_year>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x4,%eax
  jne    L32
  mov    -0x28(%rbp),%rax
  jmp    L30
L32:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x2,%eax
  je     L68
  lea    D18(%rip),%rax
  jmp    L30
L68:
  lea    -0x1c(%rbp),%rdx
  lea    -0xb(%rbp),%rcx
  lea    -0xe(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <julian_get_current_century@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x1c(%rbp),%edx
  cmp    %edx,%eax
  jle    L14
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0xb(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0xa(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L41
L14:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0xe(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
L41:
  mov    -0x28(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  lea    D3(%rip),%rax
L30:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D18
data_type: string

tool: 

assistant: ```c
char *julian_make_y2k_year(char *two_digit_year_buffer) {
  static char year_buffer[5];
  char *ptr = year_buffer;
  char current_century[3];
  char prior_century[3];
  int current_two_digit_year;
  if (strlen(two_digit_year_buffer) == 4) {
    return two_digit_year_buffer;
  }
  if (strlen(two_digit_year_buffer) != 2)
    return "";
  julian_get_current_century(current_century, prior_century, &current_two_digit_year);
  if (atoi(two_digit_year_buffer) > current_two_digit_year) {
    *ptr++ = *prior_century;
    *ptr++ = *(prior_century + 1);
  } else {
    *ptr++ = *current_century;
    *ptr++ = *(current_century + 1);
  }
  *ptr++ = *two_digit_year_buffer++;
  *ptr++ = *two_digit_year_buffer++;
  *ptr = '\0';
  return year_buffer;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_ascii85_write_toneline>:
  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)
  cmpq   $0x0,-0x28(%rbp)
  je     L43
  cmpq   $0x0,-0x30(%rbp)
  je     L43
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L8:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <ascii85_encode@plt>
  mov    -0x28(%rbp),%rax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  imul   -0x30(%rbp),%rax
  cmp    %rax,%rdx
  jb     L8
  jmp    L38
L43:
  nop
L38:
  leave
  ret
```

assistant: ```c
void _ascii85_write_toneline(struct ascii85 *a, const char *buf, size_t ss, size_t count) {
  int x;
  if (ss == 0 || count == 0)
    return;
  for (x = 0; x < ss * count; x += ss)
    ascii85_encode(a, buf[x]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_ascii85_write_toneline>:
  endbr64
  test   %rdx,%rdx
  je     L34
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %rdx,%r12
  mov    %rcx,%rbp
  test   %rcx,%rcx
  je     L23
  imul   %rdx,%rbp
  test   %rbp,%rbp
  je     L23
  mov    $0x0,%ebx
  mov    $0x0,%eax
L18:
  movsbl (%r14,%rax,1),%esi
  mov    %r13,%rdi
  call   <ascii85_encode@plt>
  add    %r12d,%ebx
  movslq %ebx,%rax
  cmp    %rbp,%rax
  jb     L18
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  ret
```

assistant: ```c
void _ascii85_write_toneline(struct ascii85 *a, const char *buf, size_t ss, size_t count) {
  int x;
  if (ss == 0 || count == 0)
    return;
  for (x = 0; x < ss * count; x += ss)
    ascii85_encode(a, buf[x]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_ascii85_write_toneline>:
  endbr64
  test   %rdx,%rdx
  je     L18
  push   %r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  test   %rcx,%rcx
  je     L13
  imul   %rdx,%r12
  mov    %rdx,%rbp
  test   %r12,%r12
  je     L13
  mov    %rdi,%r13
  mov    %rsi,%r14
  xor    %ebx,%ebx
  xor    %eax,%eax
  nop
L32:
  movsbl (%r14,%rax,1),%esi
  mov    %r13,%rdi
  add    %ebp,%ebx
  call   <ascii85_encode@plt>
  movslq %ebx,%rax
  cmp    %r12,%rax
  jb     L32
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  ret
```

assistant: ```c
void _ascii85_write_toneline(struct ascii85 *a, const char *buf, size_t ss, size_t count) {
  int x;
  if (ss == 0 || count == 0)
    return;
  for (x = 0; x < ss * count; x += ss)
    ascii85_encode(a, buf[x]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_ascii85_write_toneline>:
  endbr64
  test   %rdx,%rdx
  je     L10
  push   %r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  test   %rcx,%rcx
  je     L20
  imul   %rdx,%r12
  mov    %rdx,%rbp
  test   %r12,%r12
  je     L20
  mov    %rdi,%r13
  mov    %rsi,%r14
  xor    %ebx,%ebx
  xor    %eax,%eax
  nop
L32:
  movsbl (%r14,%rax,1),%esi
  mov    %r13,%rdi
  add    %ebp,%ebx
  call   <ascii85_encode@plt>
  movslq %ebx,%rax
  cmp    %r12,%rax
  jb     L32
L20:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  ret
```

assistant: ```c
void _ascii85_write_toneline(struct ascii85 *a, const char *buf, size_t ss, size_t count) {
  int x;
  if (ss == 0 || count == 0)
    return;
  for (x = 0; x < ss * count; x += ss)
    ascii85_encode(a, buf[x]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L7
L32:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L8
L7:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L32
  jmp    L1
L8:
  nop
L1:
  mov    -0x1c(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, size_t count) {
  const unsigned char *su1 = cs, *su2 = ct, *end = su1 + count;
  int res = 0;
  while (su1 < end) {
    res = *su1++ - *su2++;
    if (res) {
      break;
    }
  }
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<memcmp>:
  endbr64
  add    %rdi,%rdx
L4:
  cmp    %rdx,%rdi
  jae    L5
  add    $0x1,%rdi
  add    $0x1,%rsi
  movzbl -0x1(%rdi),%eax
  movzbl -0x1(%rsi),%ecx
  sub    %ecx,%eax
  je     L4
  jmp    L24
L5:
  mov    $0x0,%eax
L24:
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, size_t count) {
  const unsigned char *su1 = cs, *su2 = ct, *end = su1 + count;
  int res = 0;
  while (su1 < end) {
    res = *su1++ - *su2++;
    if (res) {
      break;
    }
  }
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<memcmp>:
  endbr64
  add    %rdi,%rdx
  jmp    L4
  nopl   0x0(%rax)
L1:
  movzbl (%rdi),%eax
  movzbl (%rsi),%ecx
  add    $0x1,%rdi
  add    $0x1,%rsi
  sub    %ecx,%eax
  jne    L22
L4:
  cmp    %rdx,%rdi
  jb     L1
  xor    %eax,%eax
L22:
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, size_t count) {
  const unsigned char *su1 = cs, *su2 = ct, *end = su1 + count;
  int res = 0;
  while (su1 < end) {
    res = *su1++ - *su2++;
    if (res) {
      break;
    }
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memcmp>:
  endbr64
  add    %rdi,%rdx
  jmp    L16
  nopl   0x0(%rax)
L20:
  movzbl (%rdi),%eax
  movzbl (%rsi),%ecx
  add    $0x1,%rdi
  add    $0x1,%rsi
  sub    %ecx,%eax
  jne    L2
L16:
  cmp    %rdx,%rdi
  jb     L20
  xor    %eax,%eax
L2:
  ret
```

assistant: ```c
int memcmp(const void *cs, const void *ct, size_t count) {
  const unsigned char *su1 = cs, *su2 = ct, *end = su1 + count;
  int res = 0;
  while (su1 < end) {
    res = *su1++ - *su2++;
    if (res) {
      break;
    }
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<adfDays2Date>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %rdx,-0x68(%rbp)
  mov    %rcx,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1f,-0x40(%rbp)
  movl   $0x1c,-0x3c(%rbp)
  movl   $0x1f,-0x38(%rbp)
  movl   $0x1e,-0x34(%rbp)
  movl   $0x1f,-0x30(%rbp)
  movl   $0x1e,-0x2c(%rbp)
  movl   $0x1f,-0x28(%rbp)
  movl   $0x1f,-0x24(%rbp)
  movl   $0x1e,-0x20(%rbp)
  movl   $0x1f,-0x1c(%rbp)
  movl   $0x1e,-0x18(%rbp)
  movl   $0x1f,-0x14(%rbp)
  movl   $0x7ba,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,%edi
  call   <adfIsLeap@plt>
  test   %rax,%rax
  je     L73
  movl   $0x16e,-0x44(%rbp)
  jmp    L23
L73:
  movl   $0x16d,-0x44(%rbp)
  jmp    L23
L34:
  mov    -0x44(%rbp),%eax
  cltq
  sub    %rax,-0x58(%rbp)
  addl   $0x1,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,%edi
  call   <adfIsLeap@plt>
  test   %rax,%rax
  je     L50
  movl   $0x16e,-0x44(%rbp)
  jmp    L23
L50:
  movl   $0x16d,-0x44(%rbp)
L23:
  mov    -0x44(%rbp),%eax
  cltq
  cmp    %rax,-0x58(%rbp)
  jge    L34
  movl   $0x1,-0x48(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,%edi
  call   <adfIsLeap@plt>
  test   %rax,%rax
  je     L22
  movl   $0x1d,-0x3c(%rbp)
  jmp    L22
L36:
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  cltq
  sub    %rax,-0x58(%rbp)
  addl   $0x1,-0x48(%rbp)
L22:
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  cltq
  cmp    %rax,-0x58(%rbp)
  jge    L36
  mov    -0x60(%rbp),%rax
  mov    -0x4c(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x68(%rbp),%rax
  mov    -0x48(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x58(%rbp),%rax
  add    $0x1,%eax
  mov    %eax,%edx
  mov    -0x70(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L93
  call   <__stack_chk_fail@plt>
L93:
  leave
  ret
```

assistant: ```c
void adfDays2Date(long days, int *yy, int *mm, int *dd) {
  int y, m;
  int nd;
  int jm[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  y = 1978;
  if (adfIsLeap(y))
    nd = 366;
  else
    nd = 365;
  while (days >= nd) {
    days -= nd;
    y++;
    if (adfIsLeap(y))
      nd = 366;
    else
      nd = 365;
  }
  m = 1;
  if (adfIsLeap(y))
    jm[2 - 1] = 29;
  while (days >= jm[m - 1]) {
    days -= jm[m - 1];
    m++;
  }
  *yy = y;
  *mm = m;
  *dd = days + 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<adfDays2Date>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1f,(%rsp)
  movl   $0x1c,0x4(%rsp)
  movl   $0x1f,0x8(%rsp)
  movl   $0x1e,0xc(%rsp)
  movl   $0x1f,0x10(%rsp)
  movl   $0x1e,0x14(%rsp)
  movl   $0x1f,0x18(%rsp)
  movl   $0x1f,0x1c(%rsp)
  movl   $0x1e,0x20(%rsp)
  movl   $0x1f,0x24(%rsp)
  movl   $0x1e,0x28(%rsp)
  movl   $0x1f,0x2c(%rsp)
  mov    $0x7ba,%edi
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x16d,%eax
  sbb    $0xffffffffffffffff,%rax
  cmp    %rax,%rbx
  jl     L45
  mov    $0x7ba,%ebp
L18:
  sub    %rax,%rbx
  add    $0x1,%ebp
  mov    %ebp,%edi
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x16d,%eax
  sbb    $0xffffffffffffffff,%rax
  cmp    %rbx,%rax
  jle    L18
L47:
  mov    %ebp,%edi
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x1c,%eax
  sbb    $0xffffffff,%eax
  mov    %eax,0x4(%rsp)
  cmp    $0x1e,%rbx
  jle    L28
  mov    $0x2,%eax
  mov    $0x1f,%edx
L44:
  sub    %rdx,%rbx
  mov    %eax,%ecx
  movslq -0x4(%rsp,%rax,4),%rdx
  add    $0x1,%rax
  cmp    %rbx,%rdx
  jle    L44
L49:
  mov    %ebp,(%r14)
  mov    %ecx,0x0(%r13)
  add    $0x1,%ebx
  mov    %ebx,(%r12)
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L45:
  mov    $0x7ba,%ebp
  jmp    L47
L28:
  mov    $0x1,%ecx
  jmp    L49
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void adfDays2Date(long days, int *yy, int *mm, int *dd) {
  int y, m;
  int nd;
  int jm[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  y = 1978;
  if (adfIsLeap(y))
    nd = 366;
  else
    nd = 365;
  while (days >= nd) {
    days -= nd;
    y++;
    if (adfIsLeap(y))
      nd = 366;
    else
      nd = 365;
  }
  m = 1;
  if (adfIsLeap(y))
    jm[2 - 1] = 29;
  while (days >= jm[m - 1]) {
    days -= jm[m - 1];
    m++;
  }
  *yy = y;
  *mm = m;
  *dd = days + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<adfDays2Date>:
  endbr64
  push   %r14
  mov    $0x7ba,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x7ba,%edi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x1e0000001f,%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,0x10(%rsp)
  movabs $0x1f0000001f,%rax
  mov    %rax,0x18(%rsp)
  sub    $0x1,%rax
  mov    %rax,0x20(%rsp)
  mov    %rax,0x28(%rsp)
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x16d,%eax
  sbb    $0xffffffffffffffff,%rax
  cmp    %rax,%rbx
  jl     L8
  nopl   (%rax)
L45:
  add    $0x1,%r14d
  sub    %rax,%rbx
  mov    %r14d,%edi
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x16d,%eax
  sbb    $0xffffffffffffffff,%rax
  cmp    %rbx,%rax
  jle    L45
L8:
  mov    %r14d,%edi
  call   <adfIsLeap@plt>
  mov    $0x1c,%ecx
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ecx
  cmp    $0x1e,%rbx
  jle    L31
  mov    $0x2,%eax
  mov    $0x1f,%edx
  jmp    L47
  nopw   0x0(%rax,%rax,1)
L29:
  mov    (%rsp,%rax,4),%ecx
  add    $0x1,%rax
L47:
  sub    %rdx,%rbx
  movslq %ecx,%rdx
  mov    %eax,%esi
  cmp    %rbx,%rdx
  jle    L29
L42:
  add    $0x1,%ebx
  mov    %r14d,0x0(%r13)
  mov    %esi,(%r12)
  mov    %ebx,0x0(%rbp)
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L31:
  mov    $0x1,%esi
  jmp    L42
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void adfDays2Date(long days, int *yy, int *mm, int *dd) {
  int y, m;
  int nd;
  int jm[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  y = 1978;
  if (adfIsLeap(y))
    nd = 366;
  else
    nd = 365;
  while (days >= nd) {
    days -= nd;
    y++;
    if (adfIsLeap(y))
      nd = 366;
    else
      nd = 365;
  }
  m = 1;
  if (adfIsLeap(y))
    jm[2 - 1] = 29;
  while (days >= jm[m - 1]) {
    days -= jm[m - 1];
    m++;
  }
  *yy = y;
  *mm = m;
  *dd = days + 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<adfDays2Date>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    $0x7ba,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x7ba,%edi
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x16d,%eax
  sbb    $0xffffffffffffffff,%rax
  cmp    %rax,%rbx
  jl     L177
  nopl   0x0(%rax)
L266:
  add    $0x1,%ebp
  sub    %rax,%rbx
  mov    %ebp,%edi
  call   <adfIsLeap@plt>
  test   %rax,%rax
  je     L46
  cmp    $0x16d,%rbx
  jle    L177
  mov    $0x16e,%eax
  jmp    L266
  nopw   0x0(%rax,%rax,1)
L46:
  cmp    $0x16c,%rbx
  jle    L177
  mov    $0x16d,%eax
  jmp    L266
  nopl   0x0(%rax,%rax,1)
L177:
  mov    %ebp,%edi
  call   <adfIsLeap@plt>
  cmp    $0x1,%rax
  mov    $0x1c,%eax
  sbb    $0xffffffff,%eax
  cmp    $0x1e,%rbx
  jle    L48
  sub    $0x1f,%rbx
  cltq
  cmp    %rbx,%rax
  jg     L52
  sub    %rax,%rbx
  cmp    $0x1e,%rbx
  jle    L87
  lea    -0x1f(%rbx),%rax
  cmp    $0x1d,%rax
  jle    L120
  lea    -0x3d(%rbx),%rax
  cmp    $0x1e,%rax
  jle    L78
  lea    -0x5c(%rbx),%rax
  cmp    $0x1d,%rax
  jle    L233
  lea    -0x7a(%rbx),%rax
  cmp    $0x1e,%rax
  jle    L230
  lea    -0x99(%rbx),%rax
  cmp    $0x1e,%rax
  jle    L224
  lea    -0xb8(%rbx),%rax
  cmp    $0x1d,%rax
  jle    L39
  lea    -0xd6(%rbx),%rax
  cmp    $0x1e,%rax
  jle    L121
  lea    -0xf5(%rbx),%rax
  cmp    $0x1d,%rax
  jle    L64
  sub    $0x113,%rbx
  mov    $0xc,%eax
L26:
  add    $0x1,%ebx
  mov    %ebp,(%r14)
  mov    %eax,0x0(%r13)
  mov    %ebx,(%r12)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L233:
  mov    %rax,%rbx
  mov    $0x6,%eax
  jmp    L26
  xchg   %ax,%ax
L48:
  mov    $0x1,%eax
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L52:
  mov    $0x2,%eax
  jmp    L26
L87:
  mov    $0x3,%eax
  jmp    L26
L120:
  mov    %rax,%rbx
  mov    $0x4,%eax
  jmp    L26
L78:
  mov    %rax,%rbx
  mov    $0x5,%eax
  jmp    L26
L230:
  mov    %rax,%rbx
  mov    $0x7,%eax
  jmp    L26
L224:
  mov    %rax,%rbx
  mov    $0x8,%eax
  jmp    L26
L39:
  mov    %rax,%rbx
  mov    $0x9,%eax
  jmp    L26
L121:
  mov    %rax,%rbx
  mov    $0xa,%eax
  jmp    L26
L64:
  mov    %rax,%rbx
  mov    $0xb,%eax
  jmp    L26
```

assistant: ```c
void adfDays2Date(long days, int *yy, int *mm, int *dd) {
  int y, m;
  int nd;
  int jm[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  y = 1978;
  if (adfIsLeap(y))
    nd = 366;
  else
    nd = 365;
  while (days >= nd) {
    days -= nd;
    y++;
    if (adfIsLeap(y))
      nd = 366;
    else
      nd = 365;
  }
  m = 1;
  if (adfIsLeap(y))
    jm[2 - 1] = 29;
  while (days >= jm[m - 1]) {
    days -= jm[m - 1];
    m++;
  }
  *yy = y;
  *mm = m;
  *dd = days + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<pcm_SL16_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  jmp    L39
L4:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jns    L21
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %edx,%eax
  lea    -0x10000(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L15
L21:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
L15:
  addq   $0x2,-0x10(%rbp)
  addq   $0x4,-0x18(%rbp)
  subl   $0x1,-0x4(%rbp)
L39:
  cmpl   $0x0,-0x4(%rbp)
  jne    L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void pcm_SL16_to_int(unsigned total_samples, const unsigned char pcm_samples[], int int_samples[]) {
  for (; total_samples; total_samples--) {
    if (pcm_samples[1] & 0x80) {
      int_samples[0] = -(int)(0x10000 - ((pcm_samples[1] << 8) | pcm_samples[0]));
    } else {
      int_samples[0] = (int)(pcm_samples[1] << 8) | pcm_samples[0];
    }
    pcm_samples += 2;
    int_samples += 1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pcm_SL16_to_int>:
  endbr64
  mov    %rsi,%rcx
  test   %edi,%edi
  je     L0
  mov    %edi,%edi
  lea    (%rsi,%rdi,2),%rdi
  jmp    L2
L43:
  movzbl %al,%eax
  shl    $0x8,%eax
  movzbl (%rcx),%esi
  or     %esi,%eax
L34:
  mov    %eax,(%rdx)
  add    $0x2,%rcx
  add    $0x4,%rdx
  cmp    %rcx,%rdi
  je     L0
L2:
  movzbl 0x1(%rcx),%eax
  test   %al,%al
  jns    L43
  movzbl %al,%eax
  shl    $0x8,%eax
  movzbl (%rcx),%esi
  or     %esi,%eax
  sub    $0x10000,%eax
  jmp    L34
L0:
  ret
```

assistant: ```c
void pcm_SL16_to_int(unsigned total_samples, const unsigned char pcm_samples[], int int_samples[]) {
  for (; total_samples; total_samples--) {
    if (pcm_samples[1] & 0x80) {
      int_samples[0] = -(int)(0x10000 - ((pcm_samples[1] << 8) | pcm_samples[0]));
    } else {
      int_samples[0] = (int)(pcm_samples[1] << 8) | pcm_samples[0];
    }
    pcm_samples += 2;
    int_samples += 1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<pcm_SL16_to_int>:
  endbr64
  mov    %edi,%r8d
  xor    %eax,%eax
  test   %edi,%edi
  je     L15
  nopl   (%rax)
L7:
  cmpb   $0x0,0x1(%rsi,%rax,2)
  movzwl (%rsi,%rax,2),%ecx
  jns    L3
  sub    $0x10000,%ecx
L3:
  mov    %ecx,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r8
  jne    L7
L15:
  ret
```

assistant: ```c
void pcm_SL16_to_int(unsigned total_samples, const unsigned char pcm_samples[], int int_samples[]) {
  for (; total_samples; total_samples--) {
    if (pcm_samples[1] & 0x80) {
      int_samples[0] = -(int)(0x10000 - ((pcm_samples[1] << 8) | pcm_samples[0]));
    } else {
      int_samples[0] = (int)(pcm_samples[1] << 8) | pcm_samples[0];
    }
    pcm_samples += 2;
    int_samples += 1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<malloc_pages>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <malloc_dirty_pages@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *malloc_pages(size_t len, size_t alignment) {
  void *p;
  p = malloc_dirty_pages(len, alignment);
  memset(p, 0, len);
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<malloc_pages>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <malloc_dirty_pages@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *malloc_pages(size_t len, size_t alignment) {
  void *p;
  p = malloc_dirty_pages(len, alignment);
  memset(p, 0, len);
  return p;
}
```