ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<read_exactly>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L11
L33:
  lea    -0x10(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <receive@plt>
  test   %rax,%rax
  je     L48
  mov    $0xffffffff,%eax
  jmp    L12
L48:
  mov    -0x10(%rbp),%rax
  sub    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  shl    $0x2,%rax
  add    %rax,-0x20(%rbp)
L11:
  cmpq   $0x0,-0x28(%rbp)
  jne    L33
  mov    $0x0,%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
int read_exactly(int fd, uint8_t *buf, size_t n) {
  size_t rx;
  while (n > 0) {
    if (receive(fd, buf, n, &rx) != 0)
      return -1;
    n -= rx;
    buf += n;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_exactly>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L16
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rsp,%r13
L10:
  mov    %r13,%rcx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <receive@plt>
  test   %rax,%rax
  jne    L26
  sub    (%rsp),%rbx
  lea    0x0(%rbp,%rbx,4),%rbp
  jne    L10
  jmp    L28
L16:
  mov    $0x0,%eax
  jmp    L28
L26:
  mov    $0xffffffff,%eax
L28:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_exactly(int fd, uint8_t *buf, size_t n) {
  size_t rx;
  while (n > 0) {
    if (receive(fd, buf, n, &rx) != 0)
      return -1;
    n -= rx;
    buf += n;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_exactly>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L10
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rsp,%r13
  jmp    L49
  nopl   0x0(%rax)
L14:
  sub    (%rsp),%rbx
  lea    0x0(%rbp,%rbx,4),%rbp
  je     L10
L49:
  mov    %r13,%rcx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <receive@plt>
  test   %rax,%rax
  je     L14
  mov    $0xffffffff,%eax
  jmp    L1
L10:
  xor    %eax,%eax
L1:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L48
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_exactly(int fd, uint8_t *buf, size_t n) {
  size_t rx;
  while (n > 0) {
    if (receive(fd, buf, n, &rx) != 0)
      return -1;
    n -= rx;
    buf += n;
  }
  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
<read_exactly>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L50
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rsp,%r13
  jmp    L22
  nopl   0x0(%rax)
L37:
  sub    (%rsp),%rbx
  lea    0x0(%rbp,%rbx,4),%rbp
  je     L50
L22:
  mov    %r13,%rcx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <receive@plt>
  test   %rax,%rax
  je     L37
  mov    $0xffffffff,%eax
  jmp    L43
L50:
  xor    %eax,%eax
L43:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_exactly(int fd, uint8_t *buf, size_t n) {
  size_t rx;
  while (n > 0) {
    if (receive(fd, buf, n, &rx) != 0)
      return -1;
    n -= rx;
    buf += n;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  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)
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x58(%rbp),%rax
  lea    0x8(%rax),%rcx
  lea    -0x58(%rbp),%rax
  lea    0x4(%rax),%rdx
  lea    -0x58(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x54(%rbp),%edx
  mov    -0x58(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  je     L6
  mov    -0x54(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x48(%rbp)
  mov    -0x58(%rbp),%eax
  mov    %eax,-0x4c(%rbp)
  mov    -0x50(%rbp),%eax
  mov    %eax,-0x44(%rbp)
  jmp    L13
L6:
  mov    -0x58(%rbp),%eax
  cmp    $0xc,%eax
  jne    L39
  movl   $0x1,-0x48(%rbp)
  movl   $0x1,-0x4c(%rbp)
  mov    -0x50(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x44(%rbp)
  jmp    L13
L39:
  movl   $0x1,-0x48(%rbp)
  mov    -0x58(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4c(%rbp)
  mov    -0x50(%rbp),%eax
  mov    %eax,-0x44(%rbp)
L13:
  mov    -0x44(%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
  mov    -0x48(%rbp),%edx
  mov    -0x4c(%rbp),%esi
  mov    %eax,%ecx
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
datatype: string
name: disassemble_data
arguments: data_label: D17
datatype: string
name: disassemble_data
arguments: data_label: D25
datatype: string

tool: D6: Enter today's date (mm dd yyyy): 

tool: D17: %i%i%i

tool: D25: Tomorrow's date is %i/%i/%.2i.


assistant: ```c
int main(void) {
  struct date {
    int month;
    int day;
    int year;
  };
  struct date today, tomorrow;
  const int daysPerMonth[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  printf("Enter today's date (mm dd yyyy): ");
  scanf("%i%i%i", &today.month, &today.day, &today.year);
  if (today.day != daysPerMonth[today.month - 1]) {
    tomorrow.day = today.day + 1;
    tomorrow.month = today.month;
    tomorrow.year = today.year;
  } else if (today.month == 12) {
    tomorrow.day = 1;
    tomorrow.month = 1;
    tomorrow.year = today.year + 1;
  } else {
    tomorrow.day = 1;
    tomorrow.month = today.month + 1;
    tomorrow.year = today.year;
  }
  printf("Tomorrow's date is %i/%i/%.2i.\n", tomorrow.month, tomorrow.day, tomorrow.year % 100);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x1f,0x10(%rsp)
  movl   $0x1c,0x14(%rsp)
  movl   $0x1f,0x18(%rsp)
  movl   $0x1e,0x1c(%rsp)
  movl   $0x1f,0x20(%rsp)
  movl   $0x1e,0x24(%rsp)
  movl   $0x1f,0x28(%rsp)
  movl   $0x1f,0x2c(%rsp)
  movl   $0x1e,0x30(%rsp)
  movl   $0x1f,0x34(%rsp)
  movl   $0x1e,0x38(%rsp)
  movl   $0x1f,0x3c(%rsp)
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    0x4(%rsp),%esi
  lea    -0x1(%rsi),%eax
  cltq
  cmp    0x10(%rsp,%rax,4),%edx
  je     L38
  add    $0x1,%edx
  mov    0xc(%rsp),%ecx
L6:
  movslq %ecx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  imul   $0x64,%eax,%eax
  sub    %eax,%ecx
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L38:
  cmp    $0xc,%esi
  je     L27
  add    $0x1,%esi
  mov    0xc(%rsp),%ecx
  mov    $0x1,%edx
  jmp    L6
L27:
  mov    0xc(%rsp),%eax
  lea    0x1(%rax),%ecx
  mov    $0x1,%edx
  mov    $0x1,%esi
  jmp    L6
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
type_infered: string
name: retrieve_data
arguments: label: D1
type_infered: string
name: retrieve_data
arguments: label: D22
type_infered: string

tool: D10: Enter today's date (mm dd yyyy): 

tool: D1: %i%i%i

tool: D22: Tomorrow's date is %i/%i/%.2i.


assistant: ```c
int main(void) {
  struct date {
    int month;
    int day;
    int year;
  };
  struct date today, tomorrow;
  const int daysPerMonth[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  printf("Enter today's date (mm dd yyyy): ");
  scanf("%i%i%i", &today.month, &today.day, &today.year);
  if (today.day != daysPerMonth[today.month - 1]) {
    tomorrow.day = today.day + 1;
    tomorrow.month = today.month;
    tomorrow.year = today.year;
  } else if (today.month == 12) {
    tomorrow.day = 1;
    tomorrow.month = 1;
    tomorrow.year = today.year + 1;
  } else {
    tomorrow.day = 1;
    tomorrow.month = today.month + 1;
    tomorrow.year = today.year;
  }
  printf("Tomorrow's date is %i/%i/%.2i.\n", tomorrow.month, tomorrow.day, tomorrow.year % 100);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  lea    D28(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x1c0000001f,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x1e0000001f,%rax
  mov    %rax,0x18(%rsp)
  mov    %rax,0x20(%rsp)
  movabs $0x1f0000001f,%rax
  mov    %rax,0x28(%rsp)
  sub    $0x1,%rax
  mov    %rax,0x30(%rsp)
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    0x8(%rsp),%rdx
  lea    0xc(%rsp),%rcx
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    0x8(%rsp),%edx
  lea    -0x1(%rsi),%eax
  cltq
  cmp    0x10(%rsp,%rax,4),%edx
  je     L48
  mov    0xc(%rsp),%ecx
  add    $0x1,%edx
L28:
  movslq %ecx,%rax
  mov    %ecx,%edi
  imul   $0x51eb851f,%rax,%rax
  sar    $0x1f,%edi
  sar    $0x25,%rax
  sub    %edi,%eax
  lea    D1(%rip),%rdi
  imul   $0x64,%eax,%eax
  sub    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L48:
  mov    0xc(%rsp),%ecx
  cmp    $0xc,%esi
  je     L35
  add    $0x1,%esi
  mov    $0x1,%edx
  jmp    L28
L35:
  add    $0x1,%ecx
  mov    $0x1,%edx
  mov    $0x1,%esi
  jmp    L28
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D28
data_type: string
name: read_data
arguments: data_tag: D25
data_type: string
name: read_data
arguments: data_tag: D1
data_type: string

tool: D28: Enter today's date (mm dd yyyy): 

tool: D25: %i%i%i

tool: D1: Tomorrow's date is %i/%i/%.2i.


assistant: ```c
int main(void) {
  struct date {
    int month;
    int day;
    int year;
  };
  struct date today, tomorrow;
  const int daysPerMonth[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  printf("Enter today's date (mm dd yyyy): ");
  scanf("%i%i%i", &today.month, &today.day, &today.year);
  if (today.day != daysPerMonth[today.month - 1]) {
    tomorrow.day = today.day + 1;
    tomorrow.month = today.month;
    tomorrow.year = today.year;
  } else if (today.month == 12) {
    tomorrow.day = 1;
    tomorrow.month = 1;
    tomorrow.year = today.year + 1;
  } else {
    tomorrow.day = 1;
    tomorrow.month = today.month + 1;
    tomorrow.year = today.year;
  }
  printf("Tomorrow's date is %i/%i/%.2i.\n", tomorrow.month, tomorrow.day, tomorrow.year % 100);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  movdqa D44(%rip),%xmm0
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  movdqa D8(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movdqa D58(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    0x8(%rsp),%rdx
  lea    0xc(%rsp),%rcx
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    0x8(%rsp),%edx
  lea    -0x1(%rsi),%eax
  cltq
  cmp    0x10(%rsp,%rax,4),%edx
  je     L40
  mov    0xc(%rsp),%ecx
  add    $0x1,%edx
L5:
  movslq %ecx,%rax
  mov    %ecx,%edi
  imul   $0x51eb851f,%rax,%rax
  sar    $0x1f,%edi
  sar    $0x25,%rax
  sub    %edi,%eax
  lea    D43(%rip),%rdi
  imul   $0x64,%eax,%eax
  sub    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L40:
  mov    0xc(%rsp),%ecx
  cmp    $0xc,%esi
  je     L33
  add    $0x1,%esi
  mov    $0x1,%edx
  jmp    L5
L33:
  add    $0x1,%ecx
  mov    $0x1,%edx
  mov    $0x1,%esi
  jmp    L5
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D44
type: byte[16]
name: disassemble_data
arguments: tag: D2
type: string
name: disassemble_data
arguments: tag: D8
type: byte[16]
name: disassemble_data
arguments: tag: D58
type: byte[16]
name: disassemble_data
arguments: tag: D13
type: string
name: disassemble_data
arguments: tag: D43
type: string

tool: {"D44": [0x1f,0x00,0x00,0x00,0x1c,0x00,0x00,0x00,0x1f,0x00,0x00,0x00,0x1e,0x00,0x00,0x00]}

tool: D2: Enter today's date (mm dd yyyy): 

tool: {"D8": [0x1f,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x1f,0x00,0x00,0x00,0x1f,0x00,0x00,0x00]}

tool: {"D58": [0x1e,0x00,0x00,0x00,0x1f,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x1f,0x00,0x00,0x00]}

tool: D13: %i%i%i

tool: D43: Tomorrow's date is %i/%i/%.2i.


assistant: ```c
int main(void) {
  struct date {
    int month;
    int day;
    int year;
  };
  struct date today, tomorrow;
  const int daysPerMonth[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  printf("Enter today's date (mm dd yyyy): ");
  scanf("%i%i%i", &today.month, &today.day, &today.year);
  if (today.day != daysPerMonth[today.month - 1]) {
    tomorrow.day = today.day + 1;
    tomorrow.month = today.month;
    tomorrow.year = today.year;
  } else if (today.month == 12) {
    tomorrow.day = 1;
    tomorrow.month = 1;
    tomorrow.year = today.year + 1;
  } else {
    tomorrow.day = 1;
    tomorrow.month = today.month + 1;
    tomorrow.year = today.year;
  }
  printf("Tomorrow's date is %i/%i/%.2i.\n", tomorrow.month, tomorrow.day, tomorrow.year % 100);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memrchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L27
L16:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    -0x1c(%rbp),%edx
  cmp    %dl,%al
  jne    L5
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  jmp    L35
L5:
  subl   $0x1,-0xc(%rbp)
L27:
  cmpl   $0x0,-0xc(%rbp)
  jns    L16
  mov    $0x0,%eax
L35:
  pop    %rbp
  ret
```

assistant: ```c
void *memrchr(const void *s, int c, size_t n) {
  unsigned char *src = (unsigned char *)s;
  int i;
  for (i = n - 1; i >= 0; i--) {
    if (src[i] == (unsigned char)c)
      return (void *)(src + i);
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<memrchr>:
  endbr64
  sub    $0x1,%edx
  js     L39
  movslq %edx,%rax
  add    %rax,%rdi
  mov    %rdi,%rax
  sub    %edi,%edx
L33:
  mov    %rax,%rcx
  cmp    %sil,(%rax)
  je     L3
  sub    $0x1,%rax
  mov    %edx,%edi
  add    %eax,%edi
  jns    L33
  mov    $0x0,%ecx
L3:
  mov    %rcx,%rax
  ret
L39:
  mov    $0x0,%ecx
  jmp    L3
```

assistant: ```c
void *memrchr(const void *s, int c, size_t n) {
  unsigned char *src = (unsigned char *)s;
  int i;
  for (i = n - 1; i >= 0; i--) {
    if (src[i] == (unsigned char)c)
      return (void *)(src + i);
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memrchr>:
  endbr64
  sub    $0x1,%edx
  js     L24
  movslq %edx,%rax
  add    %rdi,%rax
  sub    %eax,%edx
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L17:
  sub    $0x1,%rax
  mov    %edx,%ecx
  add    %eax,%ecx
  js     L24
L33:
  cmp    %sil,(%rax)
  jne    L17
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memrchr(const void *s, int c, size_t n) {
  unsigned char *src = (unsigned char *)s;
  int i;
  for (i = n - 1; i >= 0; i--) {
    if (src[i] == (unsigned char)c)
      return (void *)(src + i);
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<memrchr>:
  endbr64
  sub    $0x1,%edx
  js     L26
  movslq %edx,%rax
  add    %rdi,%rax
  sub    %eax,%edx
  jmp    L20
  nopl   0x0(%rax,%rax,1)
L21:
  sub    $0x1,%rax
  mov    %edx,%ecx
  add    %eax,%ecx
  js     L26
L20:
  cmp    %sil,(%rax)
  jne    L21
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memrchr(const void *s, int c, size_t n) {
  unsigned char *src = (unsigned char *)s;
  int i;
  for (i = n - 1; i >= 0; i--) {
    if (src[i] == (unsigned char)c)
      return (void *)(src + i);
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0xc(%rbp)
L16:
  cmpl   $0x0,-0xc(%rbp)
  je     L2
  movl   $0x5,-0x4(%rbp)
  jmp    L16
L2:
  nop
  movl   $0x0,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int i;
  int i___0;
  int a;
  {
    i = 0;
    {
      while (1) {
      while_continue:;
        if (i) {
        } else {
          goto while_break;
        }
        i___0 = 5;
      }
    while_break:;
    }
    a = 0;
    return (0);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int i;
  int i___0;
  int a;
  {
    i = 0;
    {
      while (1) {
      while_continue:;
        if (i) {
        } else {
          goto while_break;
        }
        i___0 = 5;
      }
    while_break:;
    }
    a = 0;
    return (0);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int i;
  int i___0;
  int a;
  {
    i = 0;
    {
      while (1) {
      while_continue:;
        if (i) {
        } else {
          goto while_break;
        }
        i___0 = 5;
      }
    while_break:;
    }
    a = 0;
    return (0);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int i;
  int i___0;
  int a;
  {
    i = 0;
    {
      while (1) {
      while_continue:;
        if (i) {
        } else {
          goto while_break;
        }
        i___0 = 5;
      }
    while_break:;
    }
    a = 0;
    return (0);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<kernel_dgecp_2_0_lib4>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  mov    %rcx,(%rsp)
  cmp    $0x1,%edi
  sete   %al
  movzbl %al,%eax
  lea    (%rsi,%rax,1),%ebx
  test   %ebx,%ebx
  jle    L26
  mov    %rdx,%r12
  cmp    $0x3,%ebx
  jle    L13
  lea    -0x4(%rbx),%r14d
  shr    $0x2,%r14d
  lea    0x1(%r14),%eax
  mov    %eax,%eax
  shl    $0x7,%rax
  mov    %rax,%r15
  lea    (%rdx,%rax,1),%rbp
  mov    %rcx,%r13
L68:
  mov    %r12,%rdi
  call   <_mm_loadu_pd@plt>
  mov    %eax,%esi
  mov    %r13,%rdi
  call   <_mm_storeu_pd@plt>
  lea    0x20(%r12),%rdi
  call   <_mm_loadu_pd@plt>
  mov    %eax,%esi
  lea    0x20(%r13),%rdi
  call   <_mm_storeu_pd@plt>
  lea    0x40(%r12),%rdi
  call   <_mm_loadu_pd@plt>
  mov    %eax,%esi
  lea    0x40(%r13),%rdi
  call   <_mm_storeu_pd@plt>
  lea    0x60(%r12),%rdi
  call   <_mm_loadu_pd@plt>
  mov    %eax,%esi
  lea    0x60(%r13),%rdi
  call   <_mm_storeu_pd@plt>
  sub    $0xffffffffffffff80,%r12
  sub    $0xffffffffffffff80,%r13
  cmp    %rbp,%r12
  jne    L68
  add    %r15,(%rsp)
  lea    0x4(,%r14,4),%r15d
  cmp    %r15d,%ebx
  jle    L5
L45:
  mov    %r15d,%r14d
  mov    (%rsp),%r13
  mov    %rbp,%r12
L74:
  mov    %r12,%rdi
  call   <_mm_loadu_pd@plt>
  mov    %eax,%esi
  mov    %r13,%rdi
  call   <_mm_storeu_pd@plt>
  add    $0x20,%r12
  add    $0x20,%r13
  add    $0x1,%r14d
  cmp    %r14d,%ebx
  jg     L74
  lea    -0x1(%rbx),%eax
  sub    %r15d,%eax
  add    $0x1,%rax
  shl    $0x5,%rax
  cmp    %r15d,%ebx
  mov    $0x20,%edx
  cmovle %rdx,%rax
  add    %rax,%rbp
  add    %rax,(%rsp)
L5:
  cmpl   $0x1,0xc(%rsp)
  je     L40
L26:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  mov    %rdx,%rbp
  mov    $0x0,%r15d
  jmp    L45
L40:
  lea    0x8(%rbp),%rdi
  call   <_mm_load_sd@plt>
  mov    %eax,%esi
  mov    (%rsp),%rdi
  add    $0x8,%rdi
  call   <_mm_store_sd@plt>
  jmp    L26
```

assistant: ```c
void kernel_dgecp_2_0_lib4(int tri, int kmax, double *A, double *B) {
  if (tri == 1) {
    kmax += 1;
  }
  if (kmax <= 0)
    return;
  const int bs = 4;
  __m128d a_0;
  int k;
  for (k = 0; k < kmax - 3; k += 4) {
    a_0 = _mm_loadu_pd(&A[0 + bs * 0]);
    _mm_storeu_pd(&B[0 + bs * 0], a_0);
    a_0 = _mm_loadu_pd(&A[0 + bs * 1]);
    _mm_storeu_pd(&B[0 + bs * 1], a_0);
    a_0 = _mm_loadu_pd(&A[0 + bs * 2]);
    _mm_storeu_pd(&B[0 + bs * 2], a_0);
    a_0 = _mm_loadu_pd(&A[0 + bs * 3]);
    _mm_storeu_pd(&B[0 + bs * 3], a_0);
    A += 16;
    B += 16;
  }
  for (; k < kmax; k++) {
    a_0 = _mm_loadu_pd(&A[0 + bs * 0]);
    _mm_storeu_pd(&B[0 + bs * 0], a_0);
    A += 4;
    B += 4;
  }
  if (tri == 1) {
    a_0 = _mm_load_sd(&A[1 + bs * 0]);
    _mm_store_sd(&B[1 + bs * 0], a_0);
  }
}
```