ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<warnx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  push   %rbx
  call   <va_start@plt>
  mov    D6(%rip),%rbx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    (%rbx),%edi
  call   <vfprintf@plt>
  mov    (%rbx),%esi
  mov    $0xa,%edi
  call   <fputc@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list varargs;
  va_start(varargs, fmt);
  vfprintf(stderr, fmt, varargs);
  fputc('\n', stderr);
  va_end(varargs);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%rax
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int ***three;
  int **two;
  ***three = **two;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main() {
  int ***three;
  int **two;
  ***three = **two;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main() {
  int ***three;
  int **two;
  ***three = **two;
}
```
user: 请从给定的汇编指令推导出C源代码：

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

assistant: ```c
int main() {
  int ***three;
  int **two;
  ***three = **two;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<TestO>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movb   $0x80,-0x2(%rbp)
  movzbl -0x2(%rbp),%eax
  add    $0x80,%al
  seto   %dl
  mov    %dl,-0x1(%rbp)
  mov    %al,-0x2(%rbp)
  cmpb   $0x0,-0x1(%rbp)
  je     L8
  movzbl -0x2(%rbp),%eax
  add    $0x1,%al
  setno  %dl
  mov    %dl,-0x1(%rbp)
  mov    %al,-0x2(%rbp)
  cmpb   $0x0,-0x1(%rbp)
  je     L8
  mov    $0x1,%eax
  jmp    L7
L8:
  mov    $0x0,%eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
char TestO() {
  char r;
  unsigned char res = 128;
  __asm__("addb $128, %1" : "=@cco"(r), "+r"(res));
  if (r) {
    __asm__("addb $1, %1" : "=@ccno"(r), "+r"(res));
    if (r)
      return 1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<TestO>:
  endbr64
  mov    $0xffffff80,%eax
  add    $0x80,%al
  seto   %dl
  jno    L3
  add    $0x1,%al
  setno  %dl
L3:
  mov    %edx,%eax
  ret
```

assistant: ```c
char TestO() {
  char r;
  unsigned char res = 128;
  __asm__("addb $128, %1" : "=@cco"(r), "+r"(res));
  if (r) {
    __asm__("addb $1, %1" : "=@ccno"(r), "+r"(res));
    if (r)
      return 1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<TestO>:
  endbr64
  mov    $0xffffff80,%eax
  add    $0x80,%al
  seto   %r8b
  jno    L7
  add    $0x1,%al
  setno  %r8b
L7:
  mov    %r8d,%eax
  ret
```

assistant: ```c
char TestO() {
  char r;
  unsigned char res = 128;
  __asm__("addb $128, %1" : "=@cco"(r), "+r"(res));
  if (r) {
    __asm__("addb $1, %1" : "=@ccno"(r), "+r"(res));
    if (r)
      return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<TestO>:
  endbr64
  mov    $0xffffff80,%eax
  add    $0x80,%al
  seto   %r8b
  jno    L1
  add    $0x1,%al
  setno  %r8b
L1:
  mov    %r8d,%eax
  ret
```

assistant: ```c
char TestO() {
  char r;
  unsigned char res = 128;
  __asm__("addb $128, %1" : "=@cco"(r), "+r"(res));
  if (r) {
    __asm__("addb $1, %1" : "=@ccno"(r), "+r"(res));
    if (r)
      return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x100,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L13
L35:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x18(%rbp)
L13:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L35
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <acc_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <acc_memcpy_to_device@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <acc_memcpy_from_device@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L3
L49:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  cmp    %eax,-0x18(%rbp)
  je     L20
  mov    $0x0,%eax
  call   <abort@plt>
L20:
  addl   $0x1,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L49
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <acc_free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const int N = 256;
  int i;
  unsigned char *h;
  void *d;
  h = (unsigned char *)malloc(N);
  for (i = 0; i < N; i++) {
    h[i] = i;
  }
  d = acc_malloc(N);
  acc_memcpy_to_device(d, h, N);
  memset(&h[0], 0, N);
  acc_memcpy_from_device(0, d, N);
  for (i = 0; i < N; i++) {
    if (h[i] != i)
      abort();
  }
  acc_free(d);
  free(h);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
L6:
  mov    %al,(%rax,%rbp,1)
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L6
  mov    $0x100,%edi
  call   <acc_malloc@plt>
  mov    %rax,%r12
  mov    $0x100,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <acc_memcpy_to_device@plt>
  mov    $0x100,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    $0x100,%edx
  mov    %r12,%rsi
  mov    $0x0,%edi
  call   <acc_memcpy_from_device@plt>
  mov    $0x0,%ebx
  jmp    L1
L20:
  add    $0x1,%rbx
  cmp    $0x100,%rbx
  je     L30
L1:
  movzbl (%rbx,%rbp,1),%eax
  cmp    %ebx,%eax
  je     L20
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L20
L30:
  mov    %r12,%rdi
  call   <acc_free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const int N = 256;
  int i;
  unsigned char *h;
  void *d;
  h = (unsigned char *)malloc(N);
  for (i = 0; i < N; i++) {
    h[i] = i;
  }
  d = acc_malloc(N);
  acc_memcpy_to_device(d, h, N);
  memset(&h[0], 0, N);
  acc_memcpy_from_device(0, d, N);
  for (i = 0; i < N; i++) {
    if (h[i] != i)
      abort();
  }
  acc_free(d);
  free(h);
  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
<main>:
  endbr64
  push   %r12
  mov    $0x100,%edi
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %al,(%rax,%rbp,1)
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L10
  mov    $0x100,%edi
  xor    %ebx,%ebx
  call   <acc_malloc@plt>
  mov    $0x100,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <acc_memcpy_to_device@plt>
  mov    $0x100,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    $0x100,%edx
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <acc_memcpy_from_device@plt>
  nopl   0x0(%rax)
L18:
  movzbl (%rbx,%rbp,1),%eax
  cmp    %ebx,%eax
  je     L19
  xor    %eax,%eax
  call   <abort@plt>
L19:
  add    $0x1,%rbx
  cmp    $0x100,%rbx
  jne    L18
  mov    %r12,%rdi
  call   <acc_free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const int N = 256;
  int i;
  unsigned char *h;
  void *d;
  h = (unsigned char *)malloc(N);
  for (i = 0; i < N; i++) {
    h[i] = i;
  }
  d = acc_malloc(N);
  acc_memcpy_to_device(d, h, N);
  memset(&h[0], 0, N);
  acc_memcpy_from_device(0, d, N);
  for (i = 0; i < N; i++) {
    if (h[i] != i)
      abort();
  }
  acc_free(d);
  free(h);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Logging_Full_Sleep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  jmp    L9
L0:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <sleep@plt>
  mov    %eax,-0x4(%rbp)
L9:
  cmpl   $0x0,-0x4(%rbp)
  jg     L0
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int Logging_Full_Sleep(int seconds) {
  while (seconds > 0)
    seconds = sleep(seconds);
  return seconds;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Logging_Full_Sleep>:
  endbr64
  test   %edi,%edi
  jle    L9
  sub    $0x8,%rsp
L1:
  call   <sleep@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jg     L1
  add    $0x8,%rsp
  ret
L9:
  mov    %edi,%eax
  ret
```

assistant: ```c
int Logging_Full_Sleep(int seconds) {
  while (seconds > 0)
    seconds = sleep(seconds);
  return seconds;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Logging_Full_Sleep>:
  endbr64
  test   %edi,%edi
  jle    L15
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L11:
  call   <sleep@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jg     L11
  add    $0x8,%rsp
  ret
L15:
  mov    %edi,%eax
  ret
```

assistant: ```c
int Logging_Full_Sleep(int seconds) {
  while (seconds > 0)
    seconds = sleep(seconds);
  return seconds;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Logging_Full_Sleep>:
  endbr64
  test   %edi,%edi
  jle    L3
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L12:
  call   <sleep@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jg     L12
  add    $0x8,%rsp
  ret
L3:
  mov    %edi,%eax
  ret
```

assistant: ```c
int Logging_Full_Sleep(int seconds) {
  while (seconds > 0)
    seconds = sleep(seconds);
  return seconds;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ipol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    -0x3c(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x40(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x3c(%rbp),%eax
  movzwl %ax,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x40(%rbp),%eax
  movzwl %ax,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x28(%rbp),%eax
  movzbl %al,%edx
  mov    -0x24(%rbp),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  add    $0x1,%eax
  movzbl %al,%edx
  mov    -0x24(%rbp),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%eax
  movzbl %al,%edx
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%eax
  add    $0x1,%eax
  movzbl %al,%edx
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    $0x10000,%eax
  sub    -0x20(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  imul   -0x14(%rbp),%eax
  add    %edx,%eax
  sar    $0x8,%eax
  mov    %eax,-0x8(%rbp)
  mov    $0x10000,%eax
  sub    -0x20(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  imul   -0xc(%rbp),%eax
  add    %edx,%eax
  sar    $0x8,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x10000,%eax
  sub    -0x1c(%rbp),%eax
  imul   -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  imul   -0x4(%rbp),%eax
  add    %edx,%eax
  sar    $0x18,%eax
  pop    %rbp
  ret
```

assistant: ```c
int ipol(uint8_t *src, int x, int y) {
  int int_x = x >> 16;
  int int_y = y >> 16;
  int frac_x = x & 0xFFFF;
  int frac_y = y & 0xFFFF;
  int s00 = src[(int_x & 255) + 256 * (int_y & 255)];
  int s01 = src[((int_x + 1) & 255) + 256 * (int_y & 255)];
  int s10 = src[(int_x & 255) + 256 * ((int_y + 1) & 255)];
  int s11 = src[((int_x + 1) & 255) + 256 * ((int_y + 1) & 255)];
  int s0 = (((1 << 16) - frac_x) * s00 + frac_x * s01) >> 8;
  int s1 = (((1 << 16) - frac_x) * s10 + frac_x * s11) >> 8;
  return (((1 << 16) - frac_y) * s0 + frac_y * s1) >> 24;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ipol>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%r10
  mov    %esi,%ecx
  sar    $0x10,%ecx
  mov    %edx,%r8d
  sar    $0x10,%r8d
  movzwl %si,%esi
  movzwl %dx,%edx
  movzbl %cl,%ebx
  movzbl %r8b,%r9d
  shl    $0x8,%r9d
  add    $0x1,%ecx
  movzbl %cl,%ecx
  add    $0x1,%r8d
  movzbl %r8b,%r8d
  shl    $0x8,%r8d
  mov    $0x10000,%ebp
  mov    %ebp,%edi
  sub    %esi,%edi
  lea    (%rbx,%r9,1),%eax
  cltq
  mov    %edi,%r11d
  imul   (%r10,%rax,4),%r11d
  mov    %r11d,%eax
  lea    (%r9,%rcx,1),%r11d
  movslq %r11d,%r11
  mov    %esi,%r9d
  imul   (%r10,%r11,4),%r9d
  add    %r9d,%eax
  sar    $0x8,%eax
  sub    %edx,%ebp
  imul   %ebp,%eax
  lea    (%rbx,%r8,1),%r9d
  movslq %r9d,%r9
  imul   (%r10,%r9,4),%edi
  add    %r8d,%ecx
  movslq %ecx,%rcx
  imul   (%r10,%rcx,4),%esi
  add    %esi,%edi
  sar    $0x8,%edi
  imul   %edx,%edi
  add    %edi,%eax
  sar    $0x18,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int ipol(uint8_t *src, int x, int y) {
  int int_x = x >> 16;
  int int_y = y >> 16;
  int frac_x = x & 0xFFFF;
  int frac_y = y & 0xFFFF;
  int s00 = src[(int_x & 255) + 256 * (int_y & 255)];
  int s01 = src[((int_x + 1) & 255) + 256 * (int_y & 255)];
  int s10 = src[(int_x & 255) + 256 * ((int_y + 1) & 255)];
  int s11 = src[((int_x + 1) & 255) + 256 * ((int_y + 1) & 255)];
  int s0 = (((1 << 16) - frac_x) * s00 + frac_x * s01) >> 8;
  int s1 = (((1 << 16) - frac_x) * s10 + frac_x * s11) >> 8;
  return (((1 << 16) - frac_y) * s0 + frac_y * s1) >> 24;
}
```
user: What is the c source code of the assembly code below:

```asm
<ipol>:
  endbr64
  mov    %edx,%r8d
  mov    %esi,%ecx
  push   %rbp
  mov    %rdi,%r10
  sar    $0x10,%r8d
  sar    $0x10,%ecx
  push   %rbx
  mov    $0x10000,%ebp
  movzbl %r8b,%r9d
  movzbl %cl,%ebx
  movzwl %si,%esi
  mov    %ebp,%edi
  shl    $0x8,%r9d
  sub    %esi,%edi
  add    $0x1,%ecx
  add    $0x1,%r8d
  lea    (%rbx,%r9,1),%eax
  movzbl %cl,%ecx
  movzbl %r8b,%r8d
  movzwl %dx,%edx
  cltq
  shl    $0x8,%r8d
  sub    %edx,%ebp
  mov    (%r10,%rax,4),%r11d
  imul   %edi,%r11d
  mov    %r11d,%eax
  lea    (%r9,%rcx,1),%r11d
  add    %r8d,%ecx
  movslq %r11d,%r11
  movslq %ecx,%rcx
  mov    (%r10,%r11,4),%r9d
  imul   %esi,%r9d
  imul   (%r10,%rcx,4),%esi
  add    %r9d,%eax
  lea    (%rbx,%r8,1),%r9d
  pop    %rbx
  movslq %r9d,%r9
  sar    $0x8,%eax
  imul   (%r10,%r9,4),%edi
  imul   %ebp,%eax
  pop    %rbp
  add    %esi,%edi
  sar    $0x8,%edi
  imul   %edx,%edi
  add    %edi,%eax
  sar    $0x18,%eax
  ret
```

assistant: ```c
int ipol(uint8_t *src, int x, int y) {
  int int_x = x >> 16;
  int int_y = y >> 16;
  int frac_x = x & 0xFFFF;
  int frac_y = y & 0xFFFF;
  int s00 = src[(int_x & 255) + 256 * (int_y & 255)];
  int s01 = src[((int_x + 1) & 255) + 256 * (int_y & 255)];
  int s10 = src[(int_x & 255) + 256 * ((int_y + 1) & 255)];
  int s11 = src[((int_x + 1) & 255) + 256 * ((int_y + 1) & 255)];
  int s0 = (((1 << 16) - frac_x) * s00 + frac_x * s01) >> 8;
  int s1 = (((1 << 16) - frac_x) * s10 + frac_x * s11) >> 8;
  return (((1 << 16) - frac_y) * s0 + frac_y * s1) >> 24;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ipol>:
  endbr64
  mov    %edx,%r8d
  mov    %esi,%ecx
  push   %rbp
  mov    %rdi,%r10
  sar    $0x10,%r8d
  sar    $0x10,%ecx
  push   %rbx
  mov    $0x10000,%ebp
  movzbl %r8b,%r9d
  movzbl %cl,%ebx
  movzwl %si,%esi
  mov    %ebp,%edi
  shl    $0x8,%r9d
  sub    %esi,%edi
  add    $0x1,%ecx
  add    $0x1,%r8d
  lea    (%rbx,%r9,1),%eax
  movzbl %cl,%ecx
  movzbl %r8b,%r8d
  movzwl %dx,%edx
  cltq
  shl    $0x8,%r8d
  sub    %edx,%ebp
  mov    (%r10,%rax,4),%r11d
  imul   %edi,%r11d
  mov    %r11d,%eax
  lea    (%r9,%rcx,1),%r11d
  add    %r8d,%ecx
  movslq %r11d,%r11
  movslq %ecx,%rcx
  mov    (%r10,%r11,4),%r9d
  imul   %esi,%r9d
  imul   (%r10,%rcx,4),%esi
  add    %r9d,%eax
  lea    (%rbx,%r8,1),%r9d
  pop    %rbx
  movslq %r9d,%r9
  sar    $0x8,%eax
  imul   (%r10,%r9,4),%edi
  imul   %ebp,%eax
  pop    %rbp
  add    %esi,%edi
  sar    $0x8,%edi
  imul   %edx,%edi
  add    %edi,%eax
  sar    $0x18,%eax
  ret
```

assistant: ```c
int ipol(uint8_t *src, int x, int y) {
  int int_x = x >> 16;
  int int_y = y >> 16;
  int frac_x = x & 0xFFFF;
  int frac_y = y & 0xFFFF;
  int s00 = src[(int_x & 255) + 256 * (int_y & 255)];
  int s01 = src[((int_x + 1) & 255) + 256 * (int_y & 255)];
  int s10 = src[(int_x & 255) + 256 * ((int_y + 1) & 255)];
  int s11 = src[((int_x + 1) & 255) + 256 * ((int_y + 1) & 255)];
  int s0 = (((1 << 16) - frac_x) * s00 + frac_x * s01) >> 8;
  int s1 = (((1 << 16) - frac_x) * s10 + frac_x * s11) >> 8;
  return (((1 << 16) - frac_y) * s0 + frac_y * s1) >> 24;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<modPow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x1,-0x8(%rbp)
  jmp    L17
L10:
  mov    -0x20(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L7
  mov    -0x8(%rbp),%rax
  imul   -0x18(%rbp),%rax
  cqto
  idivq  -0x28(%rbp)
  mov    %rdx,-0x8(%rbp)
L7:
  mov    -0x18(%rbp),%rax
  imul   %rax,%rax
  cqto
  idivq  -0x28(%rbp)
  mov    %rdx,-0x18(%rbp)
  sarq   -0x20(%rbp)
L17:
  cmpq   $0x0,-0x20(%rbp)
  jg     L10
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long modPow(long b, long e, long m) {
  long result = 1;
  while (e > 0) {
    if ((e & 1) == 1) {
      result = (result * b) % m;
    }
    b = (b * b) % m;
    e >>= 1;
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<modPow>:
  endbr64
  mov    %rdx,%rcx
  test   %rsi,%rsi
  jle    L9
  mov    $0x1,%r8d
  jmp    L46
L20:
  imul   %rdi,%rdi
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rdi
  sar    %rsi
  test   %rsi,%rsi
  jle    L10
L46:
  test   $0x1,%sil
  je     L20
  mov    %r8,%rax
  imul   %rdi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  jmp    L20
L9:
  mov    $0x1,%r8d
L10:
  mov    %r8,%rax
  ret
```

assistant: ```c
long modPow(long b, long e, long m) {
  long result = 1;
  while (e > 0) {
    if ((e & 1) == 1) {
      result = (result * b) % m;
    }
    b = (b * b) % m;
    e >>= 1;
  }
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<modPow>:
  endbr64
  mov    %rdx,%rcx
  mov    $0x1,%r8d
  test   %rsi,%rsi
  jle    L25
  nopw   0x0(%rax,%rax,1)
L10:
  test   $0x1,%sil
  je     L27
  mov    %r8,%rax
  imul   %rdi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
L27:
  imul   %rdi,%rdi
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  sar    %rsi
  mov    %rdx,%rdi
  jne    L10
L25:
  mov    %r8,%rax
  ret
```

assistant: ```c
long modPow(long b, long e, long m) {
  long result = 1;
  while (e > 0) {
    if ((e & 1) == 1) {
      result = (result * b) % m;
    }
    b = (b * b) % m;
    e >>= 1;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<modPow>:
  endbr64
  mov    %rdx,%rcx
  mov    $0x1,%r8d
  test   %rsi,%rsi
  jle    L0
  nopw   0x0(%rax,%rax,1)
L22:
  test   $0x1,%sil
  je     L4
  mov    %r8,%rax
  imul   %rdi,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
L4:
  imul   %rdi,%rdi
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  sar    %rsi
  mov    %rdx,%rdi
  jne    L22
L0:
  mov    %r8,%rax
  ret
```

assistant: ```c
long modPow(long b, long e, long m) {
  long result = 1;
  while (e > 0) {
    if ((e & 1) == 1) {
      result = (result * b) % m;
    }
    b = (b * b) % m;
    e >>= 1;
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__msgwrite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  leave
  ret
```

assistant: ```c
int __msgwrite(int sock, void *data, size_t cnt) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__msgwrite>:
  endbr64
  sub    $0x8,%rsp
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int __msgwrite(int sock, void *data, size_t cnt) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<__msgwrite>:
  endbr64
  sub    $0x8,%rsp
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int __msgwrite(int sock, void *data, size_t cnt) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__msgwrite>:
  endbr64
  sub    $0x8,%rsp
  mov    D0(%rip),%rax
  mov    (%rax),%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int __msgwrite(int sock, void *data, size_t cnt) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fuse_mnt_add_mount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <mtab_needs_update@plt>
  test   %eax,%eax
  jne    L1
  mov    $0x0,%eax
  jmp    L23
L1:
  mov    -0x38(%rbp),%rdi
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <add_mount@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L10
  mov    -0x38(%rbp),%rdi
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <add_mount_legacy@plt>
  mov    %eax,-0x4(%rbp)
L10:
  mov    -0x4(%rbp),%eax
L23:
  leave
  ret
```

assistant: ```c
int fuse_mnt_add_mount(const char *progname, const char *fsname, const char *mnt, const char *type, const char *opts) {
  int res;
  if (!mtab_needs_update(mnt))
    return 0;
  res = add_mount(progname, fsname, mnt, type, opts);
  if (res == -1)
    res = add_mount_legacy(progname, fsname, mnt, type, opts);
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<fuse_mnt_add_mount>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  mov    %rcx,%r13
  mov    %r8,%r14
  mov    %rdx,%rdi
  call   <mtab_needs_update@plt>
  test   %eax,%eax
  jne    L0
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <add_mount@plt>
  cmp    $0xffffffff,%eax
  jne    L6
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <add_mount_legacy@plt>
  jmp    L6
```

assistant: ```c
int fuse_mnt_add_mount(const char *progname, const char *fsname, const char *mnt, const char *type, const char *opts) {
  int res;
  if (!mtab_needs_update(mnt))
    return 0;
  res = add_mount(progname, fsname, mnt, type, opts);
  if (res == -1)
    res = add_mount_legacy(progname, fsname, mnt, type, opts);
  return res;
}
```