ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<FLASH_defaultmode>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf86,%edx
  mov    $0x19,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x1001,%edx
  mov    $0xf,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0xf84209,%edx
  xor    %esi,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x809,%edx
  mov    $0x1,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x1400,%edx
  mov    $0x2,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0xfa031,%edx
  mov    $0x3,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x3e83e8,%edx
  mov    $0x4,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x7cf,%edx
  mov    $0x5,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x1f3f,%edx
  mov    $0x6,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x31,%edx
  mov    $0x18,%esi
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1050 <mbus_remote_register_write@plt>
```

assistant: ```c
void FLASH_defaultmode() {
  mbus_remote_register_write(0x5, 0x19, 0x1 << 11 | 0xf << 7 | 0x0 << 3 | 0x6);
  mbus_remote_register_write(0x5, 0x0F, 0x10 << 8 | 0x01);
  mbus_remote_register_write(0x5, 0x00, 0x1f << 19 | 0x2 << 13 | 0x4 << 7 | 0x09);
  mbus_remote_register_write(0x5, 0x01, 0x0008 << 8 | 0x09);
  mbus_remote_register_write(0x5, 0x02, 0x1400);
  mbus_remote_register_write(0x5, 0x03, 0x03e8 << 10 | 0x031);
  mbus_remote_register_write(0x5, 0x04, 0x3e8 << 12 | 0x3e8);
  mbus_remote_register_write(0x5, 0x05, 0x007cf);
  mbus_remote_register_write(0x5, 0x06, 0x01f3f);
  mbus_remote_register_write(0x5, 0x18, 0xC << 2 | 0x1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<FLASH_defaultmode>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf86,%edx
  mov    $0x19,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x1001,%edx
  mov    $0xf,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0xf84209,%edx
  xor    %esi,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x809,%edx
  mov    $0x1,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x1400,%edx
  mov    $0x2,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0xfa031,%edx
  mov    $0x3,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x3e83e8,%edx
  mov    $0x4,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x7cf,%edx
  mov    $0x5,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x1f3f,%edx
  mov    $0x6,%esi
  mov    $0x5,%edi
  call   <mbus_remote_register_write@plt>
  mov    $0x31,%edx
  mov    $0x18,%esi
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1050 <mbus_remote_register_write@plt>
```

assistant: ```c
void FLASH_defaultmode() {
  mbus_remote_register_write(0x5, 0x19, 0x1 << 11 | 0xf << 7 | 0x0 << 3 | 0x6);
  mbus_remote_register_write(0x5, 0x0F, 0x10 << 8 | 0x01);
  mbus_remote_register_write(0x5, 0x00, 0x1f << 19 | 0x2 << 13 | 0x4 << 7 | 0x09);
  mbus_remote_register_write(0x5, 0x01, 0x0008 << 8 | 0x09);
  mbus_remote_register_write(0x5, 0x02, 0x1400);
  mbus_remote_register_write(0x5, 0x03, 0x03e8 << 10 | 0x031);
  mbus_remote_register_write(0x5, 0x04, 0x3e8 << 12 | 0x3e8);
  mbus_remote_register_write(0x5, 0x05, 0x007cf);
  mbus_remote_register_write(0x5, 0x06, 0x01f3f);
  mbus_remote_register_write(0x5, 0x18, 0xC << 2 | 0x1);
}
```
user: What is the c source code of the assembly code below?

```asm
<clean_buff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  movq   $0x0,(%rax,%rdx,8)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int clean_buff(const int fd, char **buff_p, char *(*buff_sp)[8192]) {
  free(*buff_p);
  *buff_p = 0;
  (*buff_sp)[fd] = 0;
  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
<clean_buff>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    (%rsi),%rdi
  call   <free@plt>
  movq   $0x0,0x0(%rbp)
  movslq %ebx,%rbx
  movq   $0x0,(%r12,%rbx,8)
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int clean_buff(const int fd, char **buff_p, char *(*buff_sp)[8192]) {
  free(*buff_p);
  *buff_p = 0;
  (*buff_sp)[fd] = 0;
  return (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<clean_buff>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movslq %edi,%rbx
  mov    (%rsi),%rdi
  call   <free@plt>
  movq   $0x0,0x0(%rbp)
  xor    %eax,%eax
  movq   $0x0,(%r12,%rbx,8)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int clean_buff(const int fd, char **buff_p, char *(*buff_sp)[8192]) {
  free(*buff_p);
  *buff_p = 0;
  (*buff_sp)[fd] = 0;
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<clean_buff>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movslq %edi,%rbx
  mov    (%rsi),%rdi
  call   <free@plt>
  movq   $0x0,0x0(%rbp)
  xor    %eax,%eax
  movq   $0x0,(%r12,%rbx,8)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int clean_buff(const int fd, char **buff_p, char *(*buff_sp)[8192]) {
  free(*buff_p);
  *buff_p = 0;
  (*buff_sp)[fd] = 0;
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<target>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <organism@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L4
  mov    $0x0,%eax
  call   <item@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L13
L4:
  nop
L13:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int target(void) {
  int state;
  do {
    state = organism();
    if (!state)
      break;
    state = item();
  } while (0);
  return state;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<target>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <organism@plt>
  test   %eax,%eax
  jne    L13
L12:
  add    $0x8,%rsp
  ret
L13:
  mov    $0x0,%eax
  call   <item@plt>
  jmp    L12
```

assistant: ```c
int target(void) {
  int state;
  do {
    state = organism();
    if (!state)
      break;
    state = item();
  } while (0);
  return state;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<target>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <organism@plt>
  test   %eax,%eax
  jne    L2
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <item@plt>
```

assistant: ```c
int target(void) {
  int state;
  do {
    state = organism();
    if (!state)
      break;
    state = item();
  } while (0);
  return state;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<target>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <organism@plt>
  test   %eax,%eax
  jne    L3
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <item@plt>
```

assistant: ```c
int target(void) {
  int state;
  do {
    state = organism();
    if (!state)
      break;
    state = item();
  } while (0);
  return state;
}
```
user: What is the c source code of the assembly code below:

```asm
<ToUpper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L38
L34:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L20
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L20
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    -0x20(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L20:
  addl   $0x1,-0x4(%rbp)
L38:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L34
  nop
  nop
  leave
  ret
```

assistant: ```c
void ToUpper(char string[]) {
  int counter;
  for (counter = 0; counter < strlen(string); counter++) {
    if (string[counter] >= 'a' && string[counter] <= 'z') {
      string[counter] -= 32;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ToUpper>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
  jmp    L11
L18:
  add    $0x1,%rbx
L11:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L35
  movzbl 0x0(%rbp,%rbx,1),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L18
  sub    $0x20,%eax
  mov    %al,0x0(%rbp,%rbx,1)
  jmp    L18
L35:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ToUpper(char string[]) {
  int counter;
  for (counter = 0; counter < strlen(string); counter++) {
    if (string[counter] >= 'a' && string[counter] <= 'z') {
      string[counter] -= 32;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ToUpper>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L5
  nopl   0x0(%rax)
L17:
  movzbl 0x0(%rbp,%rbx,1),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L11
  sub    $0x20,%eax
  mov    %al,0x0(%rbp,%rbx,1)
L11:
  add    $0x1,%rbx
L5:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ToUpper(char string[]) {
  int counter;
  for (counter = 0; counter < strlen(string); counter++) {
    if (string[counter] >= 'a' && string[counter] <= 'z') {
      string[counter] -= 32;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ToUpper>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L1
  nopl   0x0(%rax)
L16:
  movzbl 0x0(%rbp,%rbx,1),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L2
  sub    $0x20,%eax
  mov    %al,0x0(%rbp,%rbx,1)
L2:
  add    $0x1,%rbx
L1:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ToUpper(char string[]) {
  int counter;
  for (counter = 0; counter < strlen(string); counter++) {
    if (string[counter] >= 'a' && string[counter] <= 'z') {
      string[counter] -= 32;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmdCopyLeft>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L63
L44:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    $0x0,%edi
  call   <cmdSet@plt>
  addl   $0x1,-0x4(%rbp)
L63:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L44
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L21:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L21
  mov    $0x0,%eax
  call   <cmdLoopBegin@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L51
L52:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  addl   $0x1,-0x4(%rbp)
L51:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L52
  movl   $0x0,-0x4(%rbp)
  jmp    L42
L15:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  addl   $0x1,-0x4(%rbp)
L42:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L15
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  mov    $0x0,%eax
  call   <cmdLoopEnd@plt>
  nop
  leave
  ret
```

assistant: ```c
void cmdCopyLeft(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cmdCopyLeft>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L26
  mov    %edi,%r12d
  mov    $0x0,%ebx
L15:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    $0x0,%edi
  call   <cmdSet@plt>
  mov    %ebx,%ebp
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L15
  mov    $0x0,%ebx
L6:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L6
  mov    $0x0,%eax
  call   <cmdLoopBegin@plt>
  mov    $0x0,%ebx
L34:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L34
  mov    $0x0,%ebx
L31:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L31
L27:
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  mov    $0x0,%eax
  call   <cmdLoopEnd@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  mov    $0x0,%eax
  call   <cmdLoopBegin@plt>
  jmp    L27
```

assistant: ```c
void cmdCopyLeft(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cmdCopyLeft>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L15
  mov    %edi,%r12d
  xor    %ebp,%ebp
  nopl   (%rax)
L23:
  mov    $0x1,%edi
  mov    %ebp,%ebx
  lea    0x1(%rbp),%ebp
  call   <cmdMoveLeft@plt>
  xor    %edi,%edi
  call   <cmdSet@plt>
  cmp    %ebp,%r12d
  jne    L23
  xor    %ebp,%ebp
  nopl   (%rax)
L8:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L8
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <cmdLoopBegin@plt>
  nopl   0x0(%rax)
L25:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L25
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L53:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L53
L41:
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10e0 <cmdLoopEnd@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  call   <cmdLoopBegin@plt>
  jmp    L41
```

assistant: ```c
void cmdCopyLeft(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cmdCopyLeft>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L8
  mov    %edi,%r12d
  xor    %ebp,%ebp
  nopl   (%rax)
L19:
  mov    $0x1,%edi
  mov    %ebp,%ebx
  lea    0x1(%rbp),%ebp
  call   <cmdMoveLeft@plt>
  xor    %edi,%edi
  call   <cmdSet@plt>
  cmp    %ebp,%r12d
  jne    L19
  xor    %ebp,%ebp
  nopl   (%rax)
L31:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L31
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <cmdLoopBegin@plt>
  nopl   0x0(%rax)
L9:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L9
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L6:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L6
L32:
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10e0 <cmdLoopEnd@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  call   <cmdLoopBegin@plt>
  jmp    L32
```

assistant: ```c
void cmdCopyLeft(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveLeft(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveRight(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<current_time_ns>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x18(%rbp),%rax
  imul   $0x3b9aca00,%rax,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<current_time_ns>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %rsp,%rsi
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<current_time_ns>:
  endbr64
  sub    $0x28,%rsp
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D0(%rip),%rax
  mov    %rsp,%rsi
  movaps %xmm0,(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<current_time_ns>:
  endbr64
  sub    $0x28,%rsp
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  mov    %rsp,%rsi
  movaps %xmm0,(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hashids_init2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <hashids_init3@plt>
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
datatype: string

tool: D7: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890

assistant: ```c
struct hashids_t *hashids_init2(const char *salt, unsigned int min_hash_length) {
  return hashids_init3(salt, min_hash_length,
                       "abcdefghijklmnopqrstuvwxyz"
                       "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                       "1234567890");
}
```
user: Translate the below assembly code into C programming language:

```asm
<hashids_init2>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdx
  call   <hashids_init3@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: string

tool: D5: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890

assistant: ```c
struct hashids_t *hashids_init2(const char *salt, unsigned int min_hash_length) {
  return hashids_init3(salt, min_hash_length,
                       "abcdefghijklmnopqrstuvwxyz"
                       "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                       "1234567890");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hashids_init2>:
  endbr64
  lea    D8(%rip),%rdx
  jmp    1050 <hashids_init3@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D8
data_type: string

tool: D8: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890

assistant: ```c
struct hashids_t *hashids_init2(const char *salt, unsigned int min_hash_length) {
  return hashids_init3(salt, min_hash_length,
                       "abcdefghijklmnopqrstuvwxyz"
                       "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                       "1234567890");
}
```
user: Translate the below assembly code into C programming language:

```asm
<hashids_init2>:
  endbr64
  lea    D2(%rip),%rdx
  jmp    1050 <hashids_init3@plt>
```

tool_calls: name: parse_data
arguments: label: D2
datatype: string

tool: D2: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890

assistant: ```c
struct hashids_t *hashids_init2(const char *salt, unsigned int min_hash_length) {
  return hashids_init3(salt, min_hash_length,
                       "abcdefghijklmnopqrstuvwxyz"
                       "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                       "1234567890");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<I2CFlagStatusClear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%rcx
  mov    -0x10(%rbp),%rax
  not    %rax
  mov    %rax,%rsi
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    %rax,%rdx
  mov    %rcx,%rax
  and    %rsi,%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void I2CFlagStatusClear(unsigned long ulBase, unsigned long ulFlag) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) &= ~(ulFlag);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<I2CFlagStatusClear>:
  endbr64
  mov    0x14(%rdi),%rax
  not    %rsi
  and    %rax,%rsi
  mov    %rsi,0x14(%rdi)
  ret
```

assistant: ```c
void I2CFlagStatusClear(unsigned long ulBase, unsigned long ulFlag) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) &= ~(ulFlag);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<I2CFlagStatusClear>:
  endbr64
  mov    0x14(%rdi),%rax
  not    %rsi
  and    %rax,%rsi
  mov    %rsi,0x14(%rdi)
  ret
```

assistant: ```c
void I2CFlagStatusClear(unsigned long ulBase, unsigned long ulFlag) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) &= ~(ulFlag);
}
```