ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<getpwnam_error_translate>:
  endbr64
  mov    D15(%rip),%rdx
  mov    %edi,%eax
  cmp    %edi,(%rdx)
  je     L35
  mov    D1(%rip),%rdx
  cmp    %edi,(%rdx)
  je     L35
  mov    D2(%rip),%rdx
  cmp    %edi,(%rdx)
  je     L35
  mov    D48(%rip),%rdx
  cmp    %edi,(%rdx)
  je     L35
  mov    D18(%rip),%rdx
  cmp    %edi,(%rdx)
  je     L35
  mov    D33(%rip),%rax
  mov    (%rax),%eax
L35:
  ret
```

assistant: ```c
int getpwnam_error_translate(int err) {
  if ((err == EIO) || (err == EMFILE) || (err == ENFILE) || (err == ENOMEM) || (err == ERANGE)) {
    return err;
  }
  return ENOENT;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readpw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  jmp    L52
L75:
  mov    -0x8(%rbp),%eax
  cltq
  lea    D49(%rip),%rdx
  add    %rdx,%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  je     L62
  mov    $0x0,%eax
  jmp    L26
L62:
  mov    -0x8(%rbp),%eax
  cltq
  lea    D49(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  cmp    $0xd,%al
  jne    L76
  mov    -0x8(%rbp),%eax
  cltq
  lea    D49(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  jmp    L65
L76:
  mov    -0x8(%rbp),%eax
  cltq
  lea    D49(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  cmp    $0xa,%al
  jne    L65
  mov    -0x8(%rbp),%eax
  cltq
  lea    D49(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  lea    D49(%rip),%rax
  jmp    L26
L65:
  addl   $0x1,-0x8(%rbp)
L52:
  mov    -0x8(%rbp),%eax
  cmp    $0x3f,%eax
  jbe    L75
  mov    $0x0,%eax
L26:
  leave
  ret
```

assistant: ```c
const char *readpw(void) {
  static char pw[64];
  for (int i = 0; i < sizeof(pw); i++) {
    int r = read(0, &pw[i], 1);
    if (r != 1)
      return 0;
    if (pw[i] == '\r') {
      pw[i] = 0;
    } else if (pw[i] == '\n') {
      pw[i] = 0;
      return pw;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readpw>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D14(%rip),%rbx
  mov    $0x0,%ebp
  jmp    L18
L42:
  movb   $0x0,(%rbx)
L22:
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    $0x40,%ebp
  je     L24
L18:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L29
  movzbl (%rbx),%eax
  cmp    $0xd,%al
  je     L42
  cmp    $0xa,%al
  jne    L22
  lea    D14(%rip),%rax
  movslq %ebp,%rbp
  movb   $0x0,(%rax,%rbp,1)
  jmp    L64
L24:
  mov    $0x0,%eax
  jmp    L64
L29:
  mov    $0x0,%eax
L64:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *readpw(void) {
  static char pw[64];
  for (int i = 0; i < sizeof(pw); i++) {
    int r = read(0, &pw[i], 1);
    if (r != 1)
      return 0;
    if (pw[i] == '\r') {
      pw[i] = 0;
    } else if (pw[i] == '\n') {
      pw[i] = 0;
      return pw;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<readpw>:
  endbr64
  push   %r12
  lea    D17(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %r12,%rbx
  jmp    L27
  cs nopw 0x0(%rax,%rax,1)
L6:
  cmp    $0xa,%al
  je     L7
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    $0x40,%ebp
  je     L40
L27:
  xor    %edi,%edi
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L40
  movzbl (%rbx),%eax
  cmp    $0xd,%al
  jne    L6
  add    $0x1,%ebp
  movb   $0x0,(%rbx)
  add    $0x1,%rbx
  cmp    $0x40,%ebp
  jne    L27
L40:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  movslq %ebp,%rbp
  pop    %rbx
  lea    D17(%rip),%rax
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *readpw(void) {
  static char pw[64];
  for (int i = 0; i < sizeof(pw); i++) {
    int r = read(0, &pw[i], 1);
    if (r != 1)
      return 0;
    if (pw[i] == '\r') {
      pw[i] = 0;
    } else if (pw[i] == '\n') {
      pw[i] = 0;
      return pw;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readpw>:
  endbr64
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %r12,%rbx
  jmp    L12
  cs nopw 0x0(%rax,%rax,1)
L17:
  cmp    $0xa,%al
  je     L51
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    $0x40,%ebp
  je     L0
L12:
  xor    %edi,%edi
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L0
  movzbl (%rbx),%eax
  cmp    $0xd,%al
  jne    L17
  add    $0x1,%ebp
  movb   $0x0,(%rbx)
  add    $0x1,%rbx
  cmp    $0x40,%ebp
  jne    L12
L0:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L51:
  movslq %ebp,%rbp
  pop    %rbx
  lea    D6(%rip),%rax
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *readpw(void) {
  static char pw[64];
  for (int i = 0; i < sizeof(pw); i++) {
    int r = read(0, &pw[i], 1);
    if (r != 1)
      return 0;
    if (pw[i] == '\r') {
      pw[i] = 0;
    } else if (pw[i] == '\n') {
      pw[i] = 0;
      return pw;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<chrIsnam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L31
  mov    $0x0,%eax
  jmp    L12
L31:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L48
L28:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <chrIsalm@plt>
  mov    %eax,-0xc(%rbp)
L48:
  cmpl   $0x0,-0xc(%rbp)
  je     L22
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
L22:
  mov    -0xc(%rbp),%eax
L12:
  leave
  ret
```

assistant: ```c
int chrIsnam(const char *string) {
  const char *pr;
  int result;
  if (!string)
    return 0;
  result = isalpha(string[0]);
  pr = string;
  while (result && *(++pr)) {
    result = chrIsalm(*pr);
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<chrIsnam>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L17
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  call   <isalpha@plt>
  test   %eax,%eax
  je     L31
L32:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L31
  movsbl %dil,%edi
  call   <chrIsalm@plt>
  test   %eax,%eax
  jne    L32
L31:
  pop    %rbx
  ret
L17:
  ret
```

assistant: ```c
int chrIsnam(const char *string) {
  const char *pr;
  int result;
  if (!string)
    return 0;
  result = isalpha(string[0]);
  pr = string;
  while (result && *(++pr)) {
    result = chrIsalm(*pr);
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<chrIsnam>:
  endbr64
  test   %rdi,%rdi
  je     L40
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  call   <isalpha@plt>
  test   %eax,%eax
  jne    L38
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L16:
  call   <chrIsalm@plt>
  test   %eax,%eax
  je     L26
L38:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L16
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L40:
  xor    %eax,%eax
  ret
```

assistant: ```c
int chrIsnam(const char *string) {
  const char *pr;
  int result;
  if (!string)
    return 0;
  result = isalpha(string[0]);
  pr = string;
  while (result && *(++pr)) {
    result = chrIsalm(*pr);
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<chrIsnam>:
  endbr64
  test   %rdi,%rdi
  je     L42
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  call   <isalpha@plt>
  test   %eax,%eax
  jne    L38
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L47:
  call   <chrIsalm@plt>
  test   %eax,%eax
  je     L0
L38:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L47
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L42:
  xor    %eax,%eax
  ret
```

assistant: ```c
int chrIsnam(const char *string) {
  const char *pr;
  int result;
  if (!string)
    return 0;
  result = isalpha(string[0]);
  pr = string;
  while (result && *(++pr)) {
    result = chrIsalm(*pr);
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bubble_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L20
L0:
  movl   $0x0,-0x8(%rbp)
  jmp    L22
L13:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L40
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L40:
  addl   $0x1,-0x8(%rbp)
L22:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L13
  addl   $0x1,-0xc(%rbp)
L20:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bubble_sort(int arr[], int len) {
  int i, j, temp;
  for (i = 0; i < len - 1; i++) {
    for (j = 0; j < len - 1 - i; j++) {
      if (arr[j] > arr[j + 1]) {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<bubble_sort>:
  endbr64
  lea    -0x1(%rsi),%r8d
  test   %r8d,%r8d
  jle    L17
  lea    0x4(%rdi),%r9
  jmp    L4
L18:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L13
L33:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L18
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  jmp    L18
L13:
  sub    $0x1,%r8d
  je     L17
L4:
  test   %r8d,%r8d
  jle    L13
  mov    %rdi,%rax
  lea    -0x1(%r8),%edx
  lea    (%r9,%rdx,4),%rsi
  jmp    L33
L17:
  ret
```

assistant: ```c
void bubble_sort(int arr[], int len) {
  int i, j, temp;
  for (i = 0; i < len - 1; i++) {
    for (j = 0; j < len - 1 - i; j++) {
      if (arr[j] > arr[j + 1]) {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bubble_sort>:
  endbr64
  lea    -0x1(%rsi),%edx
  test   %edx,%edx
  jle    L33
  lea    0x4(%rdi),%r9
  nop
L31:
  sub    $0x1,%edx
  mov    %rdi,%rax
  mov    %rdx,%r8
  lea    (%r9,%rdx,4),%rsi
  nopl   (%rax)
L6:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L16
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
L16:
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L6
  mov    %r8d,%edx
  test   %r8d,%r8d
  jne    L31
L33:
  ret
```

assistant: ```c
void bubble_sort(int arr[], int len) {
  int i, j, temp;
  for (i = 0; i < len - 1; i++) {
    for (j = 0; j < len - 1 - i; j++) {
      if (arr[j] > arr[j + 1]) {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bubble_sort>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %eax,%eax
  jle    L21
  lea    0x4(%rdi),%r9
  nop
L24:
  sub    $0x1,%eax
  mov    %rdi,%rdx
  mov    %rax,%r8
  lea    (%r9,%rax,4),%rsi
  nopl   (%rax)
L0:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%ecx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L15
  movq   %xmm1,(%rdx)
L15:
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L0
  mov    %r8d,%eax
  test   %r8d,%r8d
  jne    L24
L21:
  ret
```

assistant: ```c
void bubble_sort(int arr[], int len) {
  int i, j, temp;
  for (i = 0; i < len - 1; i++) {
    for (j = 0; j < len - 1 - i; j++) {
      if (arr[j] > arr[j + 1]) {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<u8shiftcc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L35
  movl   $0x0,-0x8(%rbp)
  jmp    L109
L28:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L93
  mov    -0x18(%rbp),%rax
  jmp    L26
L93:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jns    L66
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  jmp    L104
L52:
  addq   $0x1,-0x18(%rbp)
L104:
  shlb   -0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  test   %al,%al
  js     L52
L66:
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0x8(%rbp)
L109:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L28
  jmp    L19
L35:
  cmpl   $0x0,-0x1c(%rbp)
  jns    L19
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L84
L20:
  subq   $0x1,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L84:
  cmpl   $0x0,-0x4(%rbp)
  js     L20
L19:
  mov    -0x18(%rbp),%rax
L26:
  pop    %rbp
  ret
```

assistant: ```c
const char *u8shiftcc(const char *str, int len) {
  if (len > 0)
    for (int i = 0; i < len; ++i) {
      if (*str == 0)
        return str;
      if (*str < 0) {
        unsigned char c = *str;
        while ((c <<= 1) & 128)
          ++str;
      }
      ++str;
    }
  else if (len < 0)
    for (int i = len; i < 0; ++i)
      do {
        --str;
      } while (*str && 192 == 128);
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<u8shiftcc>:
  endbr64
  test   %esi,%esi
  jle    L73
  mov    $0x0,%edx
  jmp    L35
L28:
  add    $0x1,%rdi
  add    $0x1,%edx
  cmp    %edx,%esi
  je     L14
L35:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L78
  jns    L28
  add    %al,%al
  jns    L28
L96:
  add    $0x1,%rdi
  add    %al,%al
  js     L96
  jmp    L28
L14:
  mov    %rdi,%rax
  ret
L73:
  mov    %rdi,%rax
  jns    L15
  mov    %esi,%eax
L71:
  add    $0x1,%eax
  jne    L71
  not    %esi
  not    %rsi
  lea    (%rdi,%rsi,1),%rax
  ret
L78:
  mov    %rdi,%rax
L15:
  ret
```

assistant: ```c
const char *u8shiftcc(const char *str, int len) {
  if (len > 0)
    for (int i = 0; i < len; ++i) {
      if (*str == 0)
        return str;
      if (*str < 0) {
        unsigned char c = *str;
        while ((c <<= 1) & 128)
          ++str;
      }
      ++str;
    }
  else if (len < 0)
    for (int i = len; i < 0; ++i)
      do {
        --str;
      } while (*str && 192 == 128);
  return str;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<u8shiftcc>:
  endbr64
  mov    %rdi,%rax
  test   %esi,%esi
  jle    L1
  xor    %ecx,%ecx
  jmp    L64
  nop
L44:
  add    $0x1,%ecx
  add    $0x1,%rax
  cmp    %ecx,%esi
  je     L16
  nopl   0x0(%rax,%rax,1)
L64:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L45
  js     L74
  add    $0x1,%ecx
  add    $0x1,%rax
  cmp    %ecx,%esi
  jne    L64
L16:
  ret
  nopl   (%rax)
L62:
  add    $0x1,%rax
L74:
  add    %dl,%dl
  js     L62
  jmp    L44
  nopw   0x0(%rax,%rax,1)
L1:
  je     L45
  not    %esi
  not    %rsi
  add    %rsi,%rax
L45:
  ret
```

assistant: ```c
const char *u8shiftcc(const char *str, int len) {
  if (len > 0)
    for (int i = 0; i < len; ++i) {
      if (*str == 0)
        return str;
      if (*str < 0) {
        unsigned char c = *str;
        while ((c <<= 1) & 128)
          ++str;
      }
      ++str;
    }
  else if (len < 0)
    for (int i = len; i < 0; ++i)
      do {
        --str;
      } while (*str && 192 == 128);
  return str;
}
```
user: What is the c source code of the assembly code below?

```asm
<u8shiftcc>:
  endbr64
  mov    %rdi,%rax
  test   %esi,%esi
  jle    L60
  xor    %ecx,%ecx
  jmp    L2
  nop
L32:
  add    $0x1,%ecx
  add    $0x1,%rax
  cmp    %ecx,%esi
  je     L67
  nopl   0x0(%rax,%rax,1)
L2:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L6
  js     L4
  add    $0x1,%ecx
  add    $0x1,%rax
  cmp    %ecx,%esi
  jne    L2
L67:
  ret
  nopl   (%rax)
L23:
  add    $0x1,%rax
L4:
  add    %dl,%dl
  js     L23
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L60:
  je     L6
  not    %esi
  not    %rsi
  add    %rsi,%rax
L6:
  ret
```

assistant: ```c
const char *u8shiftcc(const char *str, int len) {
  if (len > 0)
    for (int i = 0; i < len; ++i) {
      if (*str == 0)
        return str;
      if (*str < 0) {
        unsigned char c = *str;
        while ((c <<= 1) & 128)
          ++str;
      }
      ++str;
    }
  else if (len < 0)
    for (int i = len; i < 0; ++i)
      do {
        --str;
      } while (*str && 192 == 128);
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<SendI2CStop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <I2CAL_output_data@plt>
  mov    $0x1,%edi
  call   <I2CAL_output_clock@plt>
  mov    $0x1,%edi
  call   <I2CAL_output_data@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void SendI2CStop(void) {
  I2CAL_output_data(0);
  I2CAL_output_clock(1);
  I2CAL_output_data(1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<SendI2CStop>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <I2CAL_output_data@plt>
  mov    $0x1,%edi
  call   <I2CAL_output_clock@plt>
  mov    $0x1,%edi
  call   <I2CAL_output_data@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void SendI2CStop(void) {
  I2CAL_output_data(0);
  I2CAL_output_clock(1);
  I2CAL_output_data(1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<SendI2CStop>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <I2CAL_output_data@plt>
  mov    $0x1,%edi
  call   <I2CAL_output_clock@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <I2CAL_output_data@plt>
```

assistant: ```c
void SendI2CStop(void) {
  I2CAL_output_data(0);
  I2CAL_output_clock(1);
  I2CAL_output_data(1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<SendI2CStop>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <I2CAL_output_data@plt>
  mov    $0x1,%edi
  call   <I2CAL_output_clock@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <I2CAL_output_data@plt>
```

assistant: ```c
void SendI2CStop(void) {
  I2CAL_output_data(0);
  I2CAL_output_clock(1);
  I2CAL_output_data(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiply264>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L27
L31:
  movl   $0x0,-0x3c(%rbp)
  jmp    L12
L3:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L12:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L3
  addl   $0x1,-0x40(%rbp)
L27:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L31
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply264(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<multiply264>:
  endbr64
  test   %edi,%edi
  jle    L9
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L10:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L19:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L19
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  ret
```

assistant: ```c
void multiply264(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<multiply264>:
  endbr64
  test   %edi,%edi
  jle    L14
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L24:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L24
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L6
  pop    %rbx
  ret
L14:
  ret
```

assistant: ```c
void multiply264(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```