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

```asm
<main>:
  endbr64
  push   %rbp
  lea    D29(%rip),%rdi
  lea    D30(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  mov    0x4(%rsp),%esi
  lea    D27(%rip),%rdi
  add    %eax,%esi
  sub    0x4(%rsp),%eax
  imul   %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D29
type: string
name: retrieve_data
arguments: label: D30
type: string
name: retrieve_data
arguments: label: D15
type: string
name: retrieve_data
arguments: label: D27
type: string

tool: D29: x = 

tool: D30: %d

tool: D15: y = 

tool: D27: (x+y)(x-y) = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int x, y, z;
  printf("x = ");
  scanf("%d", &x);
  printf("y = ");
  scanf("%d", &y);
  z = ((x + y) * (x - y));
  printf("(x+y)(x-y) = %d\n", z);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D26(%rip),%rdi
  lea    D33(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D36(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  mov    0x4(%rsp),%esi
  lea    D29(%rip),%rdi
  add    %eax,%esi
  sub    0x4(%rsp),%eax
  imul   %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D26
datatype: string
name: retrieve_data
arguments: data_label: D33
datatype: string
name: retrieve_data
arguments: data_label: D36
datatype: string
name: retrieve_data
arguments: data_label: D29
datatype: string

tool: D26: x = 

tool: D33: %d

tool: D36: y = 

tool: D29: (x+y)(x-y) = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int x, y, z;
  printf("x = ");
  scanf("%d", &x);
  printf("y = ");
  scanf("%d", &y);
  z = ((x + y) * (x - y));
  printf("(x+y)(x-y) = %d\n", z);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L5:
  cmpq   $0x0,-0x20(%rbp)
  je     L10
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
L10:
  movl   $0x0,-0x8(%rbp)
  jmp    L23
L49:
  mov    -0xc(%rbp),%eax
  add    %eax,-0x10(%rbp)
  addl   $0x1,-0x8(%rbp)
L23:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L49
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L5
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((no)) int foo(int x, int y, int *z) {
  int a, b, c, d;
  a = b = 0;
  for (d = 0; d < y; d++) {
    if (z)
      b = d * *z;
    for (c = 0; c < x; c++)
      a += b;
  }
  return a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  test   %esi,%esi
  jle    L61
  mov    $0x0,%r8d
  mov    $0x0,%r10d
  mov    $0x0,%r9d
  jmp    L34
L2:
  test   %edi,%edi
  jle    L63
  mov    $0x0,%eax
L13:
  mov    %eax,%ecx
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L13
  add    %r10d,%r9d
  imul   %r10d,%ecx
  add    %ecx,%r9d
L63:
  add    $0x1,%r8d
  cmp    %r8d,%esi
  je     L4
L34:
  test   %rdx,%rdx
  je     L2
  mov    %r8d,%r10d
  imul   (%rdx),%r10d
  jmp    L2
L61:
  mov    $0x0,%r9d
L4:
  mov    %r9d,%eax
  ret
```

assistant: ```c
__attribute__((no)) int foo(int x, int y, int *z) {
  int a, b, c, d;
  a = b = 0;
  for (d = 0; d < y; d++) {
    if (z)
      b = d * *z;
    for (c = 0; c < x; c++)
      a += b;
  }
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  test   %esi,%esi
  jle    L33
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  lea    -0x1(%rdi),%r10d
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L39:
  test   %rdx,%rdx
  je     L5
  mov    (%rdx),%r8d
  imul   %ecx,%r8d
L5:
  test   %edi,%edi
  jle    L6
  mov    %r10d,%r9d
  add    %r8d,%eax
  imul   %r8d,%r9d
  add    %r9d,%eax
L6:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  jne    L39
  ret
L33:
  xor    %eax,%eax
  ret
```

assistant: ```c
__attribute__((no)) int foo(int x, int y, int *z) {
  int a, b, c, d;
  a = b = 0;
  for (d = 0; d < y; d++) {
    if (z)
      b = d * *z;
    for (c = 0; c < x; c++)
      a += b;
  }
  return a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<foo>:
  endbr64
  test   %esi,%esi
  jle    L23
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  lea    -0x1(%rdi),%r10d
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L17:
  test   %rdx,%rdx
  je     L10
  mov    (%rdx),%r8d
  imul   %ecx,%r8d
L10:
  test   %edi,%edi
  jle    L35
  mov    %r10d,%r9d
  add    %r8d,%eax
  imul   %r8d,%r9d
  add    %r9d,%eax
L35:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  jne    L17
  ret
L23:
  xor    %eax,%eax
  ret
```

assistant: ```c
__attribute__((no)) int foo(int x, int y, int *z) {
  int a, b, c, d;
  a = b = 0;
  for (d = 0; d < y; d++) {
    if (z)
      b = d * *z;
    for (c = 0; c < x; c++)
      a += b;
  }
  return a;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<simple_assertion_failure>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x4,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <simple_check@plt>
  nop
  leave
  ret
```

assistant: ```c
void simple_assertion_failure() {
  int x = 4;
  simple_check(x);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<simple_assertion_failure>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <simple_check@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void simple_assertion_failure() {
  int x = 4;
  simple_check(x);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<simple_assertion_failure>:
  endbr64
  mov    $0x4,%edi
  jmp    1050 <simple_check@plt>
```

assistant: ```c
void simple_assertion_failure() {
  int x = 4;
  simple_check(x);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<simple_assertion_failure>:
  endbr64
  mov    $0x4,%edi
  jmp    1050 <simple_check@plt>
```

assistant: ```c
void simple_assertion_failure() {
  int x = 4;
  simple_check(x);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlenChar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L28:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xc0,%eax
  cmp    $0x80,%eax
  je     L23
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strlenChar(const char *z) {
  int n = 0;
  while (*z) {
    if ((0xc0 & *(z++)) != 0x80)
      n++;
  }
  return n;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strlenChar>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L11
  mov    $0x0,%edx
L8:
  add    $0x1,%rdi
  and    $0xc0,%eax
  cmp    $0x80,%eax
  setne  %al
  movzbl %al,%eax
  add    %eax,%edx
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L8
L14:
  mov    %edx,%eax
  ret
L11:
  mov    $0x0,%edx
  jmp    L14
```

assistant: ```c
int strlenChar(const char *z) {
  int n = 0;
  while (*z) {
    if ((0xc0 & *(z++)) != 0x80)
      n++;
  }
  return n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strlenChar>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L14
  xchg   %ax,%ax
L2:
  and    $0xc0,%eax
  add    $0x1,%rdi
  cmp    $0x80,%eax
  setne  %al
  movzbl %al,%eax
  add    %eax,%r8d
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L2
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strlenChar(const char *z) {
  int n = 0;
  while (*z) {
    if ((0xc0 & *(z++)) != 0x80)
      n++;
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strlenChar>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L8
  xchg   %ax,%ax
L3:
  and    $0xc0,%eax
  add    $0x1,%rdi
  cmp    $0x80,%eax
  setne  %al
  movzbl %al,%eax
  add    %eax,%r8d
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L3
L8:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strlenChar(const char *z) {
  int n = 0;
  while (*z) {
    if ((0xc0 & *(z++)) != 0x80)
      n++;
  }
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<__test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D0
type: string

tool: D0: abc

assistant: ```c
int __test() {
  char *str = "abc";
  return str[0];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__test>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
int __test() {
  char *str = "abc";
  return str[0];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<__test>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
int __test() {
  char *str = "abc";
  return str[0];
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<__test>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
int __test() {
  char *str = "abc";
  return str[0];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fatal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void fatal(const char *format, ...) {
  va_list args;
  va_start(args, format);
  vprintf(format, args);
  va_end(args);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fatal>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <vprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void fatal(const char *format, ...) {
  va_list args;
  va_start(args, format);
  vprintf(format, args);
  va_end(args);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fatal>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void fatal(const char *format, ...) {
  va_list args;
  va_start(args, format);
  vprintf(format, args);
  va_end(args);
  exit(1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fatal>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void fatal(const char *format, ...) {
  va_list args;
  va_start(args, format);
  vprintf(format, args);
  va_end(args);
  exit(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cleanup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <exposeEventThread_stop@plt>
  mov    $0x0,%eax
  call   <window_close@plt>
  mov    $0x0,%eax
  call   <fb_uninit@plt>
  mov    $0x0,%eax
  call   <ts_uninit@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cleanup() {
  exposeEventThread_stop();
  window_close();
  fb_uninit();
  ts_uninit();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cleanup>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <exposeEventThread_stop@plt>
  mov    $0x0,%eax
  call   <window_close@plt>
  mov    $0x0,%eax
  call   <fb_uninit@plt>
  mov    $0x0,%eax
  call   <ts_uninit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cleanup() {
  exposeEventThread_stop();
  window_close();
  fb_uninit();
  ts_uninit();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cleanup>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <exposeEventThread_stop@plt>
  xor    %eax,%eax
  call   <window_close@plt>
  xor    %eax,%eax
  call   <fb_uninit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1080 <ts_uninit@plt>
```

assistant: ```c
void cleanup() {
  exposeEventThread_stop();
  window_close();
  fb_uninit();
  ts_uninit();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cleanup>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <exposeEventThread_stop@plt>
  xor    %eax,%eax
  call   <window_close@plt>
  xor    %eax,%eax
  call   <fb_uninit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1080 <ts_uninit@plt>
```

assistant: ```c
void cleanup() {
  exposeEventThread_stop();
  window_close();
  fb_uninit();
  ts_uninit();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ya_byteswap_u32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x10,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  and    $0xff00ff,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x8,%eax
  and    $0xff00ff00,%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((always_, unused)) uint32_t ya_byteswap_u32(uint32_t v) {
  v = (v >> 16) | (v << 16);
  return ((v & 0xFF00FF00) >> 8) | ((v & 0x00FF00FF) << 8);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ya_byteswap_u32>:
  endbr64
  mov    %edi,%eax
  sar    $0x10,%edi
  shl    $0x10,%eax
  or     %eax,%edi
  mov    %edi,%eax
  shr    $0x8,%eax
  and    $0xff00ff,%eax
  shl    $0x8,%edi
  and    $0xff00ff00,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
__attribute__((always_, unused)) uint32_t ya_byteswap_u32(uint32_t v) {
  v = (v >> 16) | (v << 16);
  return ((v & 0xFF00FF00) >> 8) | ((v & 0x00FF00FF) << 8);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ya_byteswap_u32>:
  endbr64
  mov    %edi,%eax
  sar    $0x10,%edi
  shl    $0x10,%eax
  or     %eax,%edi
  mov    %edi,%eax
  shl    $0x8,%edi
  shr    $0x8,%eax
  and    $0xff00ff00,%edi
  and    $0xff00ff,%eax
  or     %edi,%eax
  ret
```

assistant: ```c
__attribute__((always_, unused)) uint32_t ya_byteswap_u32(uint32_t v) {
  v = (v >> 16) | (v << 16);
  return ((v & 0xFF00FF00) >> 8) | ((v & 0x00FF00FF) << 8);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ya_byteswap_u32>:
  endbr64
  mov    %edi,%eax
  sar    $0x10,%edi
  shl    $0x10,%eax
  or     %eax,%edi
  mov    %edi,%eax
  shl    $0x8,%edi
  shr    $0x8,%eax
  and    $0xff00ff00,%edi
  and    $0xff00ff,%eax
  or     %edi,%eax
  ret
```

assistant: ```c
__attribute__((always_, unused)) uint32_t ya_byteswap_u32(uint32_t v) {
  v = (v >> 16) | (v << 16);
  return ((v & 0xFF00FF00) >> 8) | ((v & 0x00FF00FF) << 8);
}
```
user: Translate the below assembly code into C programming language:

```asm
<CMod_SurfaceContents>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L95
  orl    $0x1,-0x4(%rbp)
L95:
  mov    -0x14(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L64
  orl    $0x1,-0x4(%rbp)
  orl    $0x20000000,-0x4(%rbp)
L64:
  mov    -0x14(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L20
  orl    $0x8,-0x4(%rbp)
L20:
  mov    -0x14(%rbp),%eax
  and    $0x10,%eax
  test   %eax,%eax
  je     L71
  orl    $0x10,-0x4(%rbp)
L71:
  mov    -0x14(%rbp),%eax
  and    $0x20,%eax
  test   %eax,%eax
  je     L55
  orl    $0x20,-0x4(%rbp)
L55:
  mov    -0x14(%rbp),%eax
  and    $0x8000,%eax
  test   %eax,%eax
  je     L0
  orl    $0x8000,-0x4(%rbp)
L0:
  mov    -0x14(%rbp),%eax
  and    $0x10000,%eax
  test   %eax,%eax
  je     L80
  orl    $0x10000,-0x4(%rbp)
L80:
  mov    -0x14(%rbp),%eax
  and    $0x20000,%eax
  test   %eax,%eax
  je     L92
  orl    $0x20000,-0x4(%rbp)
L92:
  mov    -0x14(%rbp),%eax
  and    $0x1000000,%eax
  test   %eax,%eax
  je     L83
  orl    $0x1000000,-0x4(%rbp)
L83:
  mov    -0x14(%rbp),%eax
  and    $0x10000000,%eax
  test   %eax,%eax
  je     L54
  orl    $0x20000000,-0x4(%rbp)
L54:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int CMod_SurfaceContents(int oldcontents) {
  int contents = 0;
  if (oldcontents & 1)
    contents |= 1;
  if (oldcontents & 2) {
    contents |= 1;
    contents |= 0x20000000;
  }
  if (oldcontents & 8)
    contents |= 8;
  if (oldcontents & 16)
    contents |= 16;
  if (oldcontents & 32)
    contents |= 32;
  if (oldcontents & 0x8000)
    contents |= 0x8000;
  if (oldcontents & 0x10000)
    contents |= 0x10000;
  if (oldcontents & 0x20000)
    contents |= 0x20000;
  if (oldcontents & 0x1000000)
    contents |= 0x1000000;
  if (oldcontents & 0x10000000)
    contents |= 0x20000000;
  return contents;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<CMod_SurfaceContents>:
  endbr64
  mov    %edi,%eax
  and    $0x1,%eax
  test   $0x2,%dil
  mov    $0x20000001,%edx
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x8,%edx
  test   $0x8,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x10,%edx
  test   $0x10,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20,%edx
  test   $0x20,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x80,%dh
  test   $0x8000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x10000,%edx
  test   $0x10000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20000,%edx
  test   $0x20000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x1000000,%edx
  test   $0x1000000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20000000,%edx
  test   $0x10000000,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int CMod_SurfaceContents(int oldcontents) {
  int contents = 0;
  if (oldcontents & 1)
    contents |= 1;
  if (oldcontents & 2) {
    contents |= 1;
    contents |= 0x20000000;
  }
  if (oldcontents & 8)
    contents |= 8;
  if (oldcontents & 16)
    contents |= 16;
  if (oldcontents & 32)
    contents |= 32;
  if (oldcontents & 0x8000)
    contents |= 0x8000;
  if (oldcontents & 0x10000)
    contents |= 0x10000;
  if (oldcontents & 0x20000)
    contents |= 0x20000;
  if (oldcontents & 0x1000000)
    contents |= 0x1000000;
  if (oldcontents & 0x10000000)
    contents |= 0x20000000;
  return contents;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CMod_SurfaceContents>:
  endbr64
  mov    %edi,%eax
  mov    $0x20000001,%edx
  and    $0x1,%eax
  test   $0x2,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x8,%edx
  test   $0x8,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x10,%edx
  test   $0x10,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20,%edx
  test   $0x20,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x80,%dh
  test   $0x8000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x10000,%edx
  test   $0x10000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20000,%edx
  test   $0x20000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x1000000,%edx
  test   $0x1000000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20000000,%edx
  and    $0x10000000,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int CMod_SurfaceContents(int oldcontents) {
  int contents = 0;
  if (oldcontents & 1)
    contents |= 1;
  if (oldcontents & 2) {
    contents |= 1;
    contents |= 0x20000000;
  }
  if (oldcontents & 8)
    contents |= 8;
  if (oldcontents & 16)
    contents |= 16;
  if (oldcontents & 32)
    contents |= 32;
  if (oldcontents & 0x8000)
    contents |= 0x8000;
  if (oldcontents & 0x10000)
    contents |= 0x10000;
  if (oldcontents & 0x20000)
    contents |= 0x20000;
  if (oldcontents & 0x1000000)
    contents |= 0x1000000;
  if (oldcontents & 0x10000000)
    contents |= 0x20000000;
  return contents;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CMod_SurfaceContents>:
  endbr64
  mov    %edi,%eax
  mov    $0x20000001,%edx
  and    $0x1,%eax
  test   $0x2,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x8,%edx
  test   $0x8,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x10,%edx
  test   $0x10,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20,%edx
  test   $0x20,%dil
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x80,%dh
  test   $0x8000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x10000,%edx
  test   $0x10000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20000,%edx
  test   $0x20000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x1000000,%edx
  test   $0x1000000,%edi
  cmovne %edx,%eax
  mov    %eax,%edx
  or     $0x20000000,%edx
  and    $0x10000000,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int CMod_SurfaceContents(int oldcontents) {
  int contents = 0;
  if (oldcontents & 1)
    contents |= 1;
  if (oldcontents & 2) {
    contents |= 1;
    contents |= 0x20000000;
  }
  if (oldcontents & 8)
    contents |= 8;
  if (oldcontents & 16)
    contents |= 16;
  if (oldcontents & 32)
    contents |= 32;
  if (oldcontents & 0x8000)
    contents |= 0x8000;
  if (oldcontents & 0x10000)
    contents |= 0x10000;
  if (oldcontents & 0x20000)
    contents |= 0x20000;
  if (oldcontents & 0x1000000)
    contents |= 0x1000000;
  if (oldcontents & 0x10000000)
    contents |= 0x20000000;
  return contents;
}
```