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

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <expecti@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <expecti@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <expecti@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <expecti@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <expecti@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <expecti@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <expecti@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <expecti@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1050 <expecti@plt>
```

assistant: ```c
void test() {
  int a = 0;
  if (0)
    a = 1;
  expecti(a, 0);
  if (1)
    a = 1;
  expecti(a, 1);
  if (0) {
    a = 0;
  } else
    expecti(a, 1);
  if (1) {
    a = 0;
  }
  expecti(a, 0);
  if (0)
    expecti(0, 1);
  else {
    expecti(1, 1);
  }
  if (1) {
    expecti(1, 1);
  } else
    a = 1;
  expecti(a, 0);
  if (1) {
    if (0) {
      expecti(1, 0);
    } else
      expecti(1, 1);
    if (1) {
      if (1)
        expecti(1, 1);
      else {
        expecti(1, 0);
      }
    } else
      expecti(0, 1);
  } else {
    expecti(1, 1);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<read_until_delim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
L63:
  lea    -0xd(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  test   %rax,%rax
  jg     L7
  mov    $0xffffffff,%eax
  jmp    L65
L7:
  movzbl -0xd(%rbp),%eax
  cmp    %al,-0x24(%rbp)
  je     L11
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jb     L57
  mov    $0xffffffff,%eax
  jmp    L65
L57:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  jmp    L63
L11:
  nop
  mov    -0xc(%rbp),%eax
L65:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

assistant: ```c
int read_until_delim(int fd, char *buf, unsigned int size, char endchar) {
  char ch;
  unsigned int total = 0;
  while (1) {
    if (read(fd, &ch, 1) <= 0) {
      return -1;
    }
    if (ch == endchar)
      break;
    if (total >= size)
      return -1;
    buf[total++] = ch;
  }
  return (int)total;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_until_delim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r14
  mov    %ecx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %edx,%r13d
  mov    $0x0,%ebx
  lea    0x7(%rsp),%r15
  jmp    L36
L15:
  mov    %al,(%r14,%rbx,1)
  add    $0x1,%rbx
L36:
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  test   %rax,%rax
  jle    L17
  movzbl 0x7(%rsp),%eax
  cmp    %r12b,%al
  je     L66
  cmp    %r13,%rbx
  jne    L15
  mov    $0xffffffff,%eax
  jmp    L31
L66:
  mov    %ebx,%eax
  jmp    L31
L17:
  mov    $0xffffffff,%eax
L31:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_until_delim(int fd, char *buf, unsigned int size, char endchar) {
  char ch;
  unsigned int total = 0;
  while (1) {
    if (read(fd, &ch, 1) <= 0) {
      return -1;
    }
    if (ch == endchar)
      break;
    if (total >= size)
      return -1;
    buf[total++] = ch;
  }
  return (int)total;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_until_delim>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r15
  jmp    L56
  cs nopw 0x0(%rax,%rax,1)
L42:
  movzbl 0x7(%rsp),%eax
  cmp    %r12b,%al
  je     L31
  cmp    %r13,%rbx
  je     L26
  mov    %al,(%r14,%rbx,1)
  add    $0x1,%rbx
L56:
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  test   %rax,%rax
  jg     L42
L26:
  mov    $0xffffffff,%eax
L37:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L41
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L31:
  mov    %ebx,%eax
  jmp    L37
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_until_delim(int fd, char *buf, unsigned int size, char endchar) {
  char ch;
  unsigned int total = 0;
  while (1) {
    if (read(fd, &ch, 1) <= 0) {
      return -1;
    }
    if (ch == endchar)
      break;
    if (total >= size)
      return -1;
    buf[total++] = ch;
  }
  return (int)total;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_until_delim>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r15
  jmp    L2
  cs nopw 0x0(%rax,%rax,1)
L5:
  movzbl 0x7(%rsp),%eax
  cmp    %r12b,%al
  je     L21
  cmp    %r13,%rbx
  je     L8
  mov    %al,(%r14,%rbx,1)
  add    $0x1,%rbx
L2:
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  test   %rax,%rax
  jg     L5
L8:
  mov    $0xffffffff,%eax
L27:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L21:
  mov    %ebx,%eax
  jmp    L27
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_until_delim(int fd, char *buf, unsigned int size, char endchar) {
  char ch;
  unsigned int total = 0;
  while (1) {
    if (read(fd, &ch, 1) <= 0) {
      return -1;
    }
    if (ch == endchar)
      break;
    if (total >= size)
      return -1;
    buf[total++] = ch;
  }
  return (int)total;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<wycc_type_subsume_lists>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <wycc_type_down@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <wycc_type_down@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <wycc_type_subsume_tokens@plt>
  leave
  ret
```

assistant: ```c
int wycc_type_subsume_lists(int itok, int atok) {
  int istp;
  int astp;
  istp = wycc_type_down(itok);
  astp = wycc_type_down(atok);
  return wycc_type_subsume_tokens(istp, astp);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<wycc_type_subsume_lists>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  call   <wycc_type_down@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <wycc_type_down@plt>
  mov    %eax,%esi
  mov    %ebx,%edi
  call   <wycc_type_subsume_tokens@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int wycc_type_subsume_lists(int itok, int atok) {
  int istp;
  int astp;
  istp = wycc_type_down(itok);
  astp = wycc_type_down(atok);
  return wycc_type_subsume_tokens(istp, astp);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<wycc_type_subsume_lists>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  sub    $0x8,%rsp
  call   <wycc_type_down@plt>
  mov    %r12d,%edi
  mov    %eax,%ebp
  call   <wycc_type_down@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  mov    %eax,%esi
  pop    %rbp
  pop    %r12
  jmp    1060 <wycc_type_subsume_tokens@plt>
```

assistant: ```c
int wycc_type_subsume_lists(int itok, int atok) {
  int istp;
  int astp;
  istp = wycc_type_down(itok);
  astp = wycc_type_down(atok);
  return wycc_type_subsume_tokens(istp, astp);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<wycc_type_subsume_lists>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  sub    $0x8,%rsp
  call   <wycc_type_down@plt>
  mov    %r12d,%edi
  mov    %eax,%ebp
  call   <wycc_type_down@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  mov    %eax,%esi
  pop    %rbp
  pop    %r12
  jmp    1060 <wycc_type_subsume_tokens@plt>
```

assistant: ```c
int wycc_type_subsume_lists(int itok, int atok) {
  int istp;
  int astp;
  istp = wycc_type_down(itok);
  astp = wycc_type_down(atok);
  return wycc_type_subsume_tokens(istp, astp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<kernel1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
L29:
  movl   $0x0,-0x4(%rbp)
  jmp    L27
L59:
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L5:
  movl   $0x0,-0x8(%rbp)
  jmp    L16
L54:
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <lin_solve_computation@plt>
  addl   $0x1,-0x8(%rbp)
L16:
  cmpl   $0x96,-0x8(%rbp)
  jle    L54
  addl   $0x1,-0xc(%rbp)
L33:
  cmpl   $0x96,-0xc(%rbp)
  jle    L5
  addl   $0x1,-0x4(%rbp)
L27:
  cmpl   $0xf,-0x4(%rbp)
  jle    L59
  jmp    L29
```

assistant: ```c
int kernel1() {
  int i, j, k;
  while (1) {
#pragma acr grid(5)
#pragma acr monitor(temporary_array[l][m], max)
#pragma acr alternative low(parameter, N = 1)
#pragma acr alternative medium(parameter, N = 2)
#pragma acr alternative high(parameter, N = 3)
#pragma acr strategy direct(1, low)
#pragma acr strategy direct(2, medium)
#pragma acr strategy direct(3, high)
#pragma scop
    for (k = 0; k <= 15; ++k)
      for (i = 0; i <= 150; ++i)
        for (j = 0; j <= 150; ++j)
          lin_solve_computation(k, i, j);
#pragma endscop
  }
#pragma acr destroy
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<kernel1>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
L24:
  mov    $0x0,%r12d
L2:
  mov    $0x0,%ebp
L23:
  mov    $0x0,%ebx
L35:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <lin_solve_computation@plt>
  add    $0x1,%ebx
  cmp    $0x97,%ebx
  jne    L35
  add    $0x1,%ebp
  cmp    $0x97,%ebp
  jne    L23
  add    $0x1,%r12d
  cmp    $0x10,%r12d
  jne    L2
  jmp    L24
```

assistant: ```c
int kernel1() {
  int i, j, k;
  while (1) {
#pragma acr grid(5)
#pragma acr monitor(temporary_array[l][m], max)
#pragma acr alternative low(parameter, N = 1)
#pragma acr alternative medium(parameter, N = 2)
#pragma acr alternative high(parameter, N = 3)
#pragma acr strategy direct(1, low)
#pragma acr strategy direct(2, medium)
#pragma acr strategy direct(3, high)
#pragma scop
    for (k = 0; k <= 15; ++k)
      for (i = 0; i <= 150; ++i)
        for (j = 0; j <= 150; ++j)
          lin_solve_computation(k, i, j);
#pragma endscop
  }
#pragma acr destroy
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<kernel1>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
L8:
  xor    %r12d,%r12d
  xor    %ebp,%ebp
L28:
  xor    %ebx,%ebx
  nop
L18:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12d,%edi
  add    $0x1,%ebx
  call   <lin_solve_computation@plt>
  cmp    $0x97,%ebx
  jne    L18
  add    $0x1,%ebp
  cmp    $0x97,%ebp
  jne    L28
  add    $0x1,%r12d
  cmp    $0x10,%r12d
  je     L8
  xor    %ebp,%ebp
  jmp    L28
```

assistant: ```c
int kernel1() {
  int i, j, k;
  while (1) {
#pragma acr grid(5)
#pragma acr monitor(temporary_array[l][m], max)
#pragma acr alternative low(parameter, N = 1)
#pragma acr alternative medium(parameter, N = 2)
#pragma acr alternative high(parameter, N = 3)
#pragma acr strategy direct(1, low)
#pragma acr strategy direct(2, medium)
#pragma acr strategy direct(3, high)
#pragma scop
    for (k = 0; k <= 15; ++k)
      for (i = 0; i <= 150; ++i)
        for (j = 0; j <= 150; ++j)
          lin_solve_computation(k, i, j);
#pragma endscop
  }
#pragma acr destroy
}
```
user: Translate the below assembly code into C programming language:

```asm
<kernel1>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
L19:
  xor    %r12d,%r12d
  xor    %ebp,%ebp
L9:
  xor    %ebx,%ebx
  nop
L25:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12d,%edi
  add    $0x1,%ebx
  call   <lin_solve_computation@plt>
  cmp    $0x97,%ebx
  jne    L25
  add    $0x1,%ebp
  cmp    $0x97,%ebp
  jne    L9
  add    $0x1,%r12d
  cmp    $0x10,%r12d
  je     L19
  xor    %ebp,%ebp
  jmp    L9
```

assistant: ```c
int kernel1() {
  int i, j, k;
  while (1) {
#pragma acr grid(5)
#pragma acr monitor(temporary_array[l][m], max)
#pragma acr alternative low(parameter, N = 1)
#pragma acr alternative medium(parameter, N = 2)
#pragma acr alternative high(parameter, N = 3)
#pragma acr strategy direct(1, low)
#pragma acr strategy direct(2, medium)
#pragma acr strategy direct(3, high)
#pragma scop
    for (k = 0; k <= 15; ++k)
      for (i = 0; i <= 150; ++i)
        for (j = 0; j <= 150; ++j)
          lin_solve_computation(k, i, j);
#pragma endscop
  }
#pragma acr destroy
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<add_CRC>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1c(%rbp),%eax
  lea    -0x2(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <CRC_16@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x2,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void add_CRC(uint8_t *buf, int size, uint16_t start_mask) {
  uint16_t crc;
  crc = CRC_16(buf, size - 2, start_mask);
  memcpy(&buf[size - 2], &crc, 2);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<add_CRC>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    -0x2(%rsi),%esi
  call   <CRC_16@plt>
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  movslq %ebx,%rbx
  lea    -0x8(%rbp,%rbx,4),%rdi
  mov    $0x2,%edx
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void add_CRC(uint8_t *buf, int size, uint16_t start_mask) {
  uint16_t crc;
  crc = CRC_16(buf, size - 2, start_mask);
  memcpy(&buf[size - 2], &crc, 2);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<add_CRC>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  lea    -0x2(%rbx),%esi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <CRC_16@plt>
  lea    0x4(%rsp),%rsi
  lea    -0x8(%rbp,%rbx,4),%rdi
  mov    $0x2,%edx
  mov    %eax,0x4(%rsp)
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void add_CRC(uint8_t *buf, int size, uint16_t start_mask) {
  uint16_t crc;
  crc = CRC_16(buf, size - 2, start_mask);
  memcpy(&buf[size - 2], &crc, 2);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<add_CRC>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  lea    -0x2(%rbx),%esi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <CRC_16@plt>
  lea    0x4(%rsp),%rsi
  lea    -0x8(%rbp,%rbx,4),%rdi
  mov    $0x2,%edx
  mov    %eax,0x4(%rsp)
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void add_CRC(uint8_t *buf, int size, uint16_t start_mask) {
  uint16_t crc;
  crc = CRC_16(buf, size - 2, start_mask);
  memcpy(&buf[size - 2], &crc, 2);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calcul_mu_moyen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L67
L72:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L67:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L72
  movl   $0x0,-0x8(%rbp)
  jmp    L78
L79:
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L52:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L52
  addl   $0x1,-0x8(%rbp)
L78:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L79
  movl   $0x0,-0x4(%rbp)
  jmp    L56
L75:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L56:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L75
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void calcul_mu_moyen(int nb_points, double **coordonnees, int nb_dimensions, double *mu) {
  int i, j;
  double norme;
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] = 0;
  }
  for (i = 0; i < nb_points; i++) {
    for (j = 0; j < nb_dimensions; j++) {
      mu[j] += coordonnees[i][j];
    }
  }
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] /= (double)nb_points;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calcul_mu_moyen>:
  endbr64
  mov    %edi,%r9d
  mov    %edx,%r10d
  mov    %rcx,%rdx
  test   %r10d,%r10d
  jle    L22
  mov    %rcx,%rax
  lea    -0x1(%r10),%ecx
  lea    0x8(%rdx,%rcx,8),%rcx
L71:
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L71
  test   %r9d,%r9d
  jle    L80
L50:
  lea    -0x1(%r10),%eax
  lea    0x8(,%rax,8),%rdi
  mov    $0x0,%r8d
  jmp    L72
L62:
  mov    (%rsi),%rcx
  movsd  (%rdx,%rax,1),%xmm0
  addsd  (%rcx,%rax,1),%xmm0
  movsd  %xmm0,(%rdx,%rax,1)
  add    $0x8,%rax
  cmp    %rdi,%rax
  jne    L62
L83:
  add    $0x1,%r8d
  add    $0x8,%rsi
  cmp    %r8d,%r9d
  jle    L68
L72:
  mov    $0x0,%eax
  test   %r10d,%r10d
  jg     L62
  jmp    L83
L68:
  test   %r10d,%r10d
  jle    L66
L80:
  mov    %rdx,%rax
  lea    -0x1(%r10),%ecx
  lea    0x8(%rdx,%rcx,8),%rdx
  pxor   %xmm1,%xmm1
  cvtsi2sd %r9d,%xmm1
L96:
  movsd  (%rax),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rax,%rdx
  jne    L96
  ret
L22:
  test   %edi,%edi
  jg     L50
L66:
  ret
```

assistant: ```c
void calcul_mu_moyen(int nb_points, double **coordonnees, int nb_dimensions, double *mu) {
  int i, j;
  double norme;
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] = 0;
  }
  for (i = 0; i < nb_points; i++) {
    for (j = 0; j < nb_dimensions; j++) {
      mu[j] += coordonnees[i][j];
    }
  }
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] /= (double)nb_points;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<calcul_mu_moyen>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  test   %edx,%edx
  jle    L86
  movslq %edx,%rdx
  mov    %rcx,%rdi
  xor    %esi,%esi
  shl    $0x3,%rdx
  call   <memset@plt>
  mov    %rax,%rcx
  test   %ebp,%ebp
  jle    L89
L53:
  xor    %edi,%edi
  movslq %ebx,%rsi
  nopl   0x0(%rax)
L76:
  test   %ebx,%ebx
  jle    L24
  mov    (%r12,%rdi,8),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L29:
  movsd  (%rcx,%rax,8),%xmm0
  addsd  (%rdx,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L29
L24:
  add    $0x1,%rdi
  cmp    %edi,%ebp
  jg     L76
  test   %ebx,%ebx
  jle    L9
L89:
  pxor   %xmm1,%xmm1
  lea    -0x1(%rbx),%eax
  cvtsi2sd %ebp,%xmm1
  lea    0x8(%rcx,%rax,8),%rax
  nopl   0x0(%rax)
L27:
  movsd  (%rcx),%xmm0
  add    $0x8,%rcx
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rcx)
  cmp    %rcx,%rax
  jne    L27
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L86:
  test   %edi,%edi
  jg     L53
  jmp    L9
```

assistant: ```c
void calcul_mu_moyen(int nb_points, double **coordonnees, int nb_dimensions, double *mu) {
  int i, j;
  double norme;
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] = 0;
  }
  for (i = 0; i < nb_points; i++) {
    for (j = 0; j < nb_dimensions; j++) {
      mu[j] += coordonnees[i][j];
    }
  }
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] /= (double)nb_points;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<calcul_mu_moyen>:
  endbr64
  test   %edx,%edx
  jle    L26
  push   %r13
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  mov    %rcx,%rdi
  push   %rbx
  mov    %edx,%ebx
  movslq %edx,%rdx
  shl    $0x3,%rdx
  sub    $0x8,%rsp
  call   <memset@plt>
  mov    %rax,%rcx
  test   %ebp,%ebp
  jle    L24
  mov    %ebx,%eax
  mov    %ebx,%r13d
  xor    %r9d,%r9d
  mov    %ebx,%r10d
  shr    %eax
  and    $0xfffffffe,%r13d
  lea    -0x1(%rbx),%r11d
  lea    -0x1(%rax),%edi
  mov    %r13d,%r8d
  add    $0x1,%rdi
  shl    $0x3,%r8
  shl    $0x4,%rdi
  lea    (%rcx,%r8,1),%rsi
  nopl   0x0(%rax)
L59:
  mov    (%r12,%r9,8),%rdx
  lea    0x8(%rdx),%rax
  cmp    %rax,%rcx
  je     L116
  xor    %eax,%eax
  cmp    $0x1,%r11d
  jbe    L116
  nopl   (%rax)
L85:
  movupd (%rcx,%rax,1),%xmm0
  movupd (%rdx,%rax,1),%xmm3
  addpd  %xmm3,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L85
  cmp    %r13d,%ebx
  je     L107
  movsd  (%rsi),%xmm0
  addsd  (%rdx,%r8,1),%xmm0
  movsd  %xmm0,(%rsi)
L107:
  add    $0x1,%r9
  cmp    %r9d,%ebp
  jg     L59
L24:
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebp,%xmm1
  cmp    $0x1,%ebx
  je     L125
  mov    %ebx,%edx
  movapd %xmm1,%xmm2
  mov    %rcx,%rax
  shr    %edx
  unpcklpd %xmm2,%xmm2
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rcx,%rdx,1),%rdx
  nopl   0x0(%rax,%rax,1)
L124:
  movupd (%rax),%xmm0
  add    $0x10,%rax
  divpd  %xmm2,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L124
  mov    %ebx,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%ebx
  je     L118
L49:
  cltq
  lea    (%rcx,%rax,8),%rax
  movsd  (%rax),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
L118:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L116:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L62:
  movsd  (%rcx,%rax,8),%xmm0
  addsd  (%rdx,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%r10
  jne    L62
  jmp    L107
L26:
  ret
L125:
  xor    %eax,%eax
  jmp    L49
```

assistant: ```c
void calcul_mu_moyen(int nb_points, double **coordonnees, int nb_dimensions, double *mu) {
  int i, j;
  double norme;
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] = 0;
  }
  for (i = 0; i < nb_points; i++) {
    for (j = 0; j < nb_dimensions; j++) {
      mu[j] += coordonnees[i][j];
    }
  }
  for (j = 0; j < nb_dimensions; j++) {
    mu[j] /= (double)nb_points;
  }
}
```