ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rcx
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  sub    $0x8,%rsp
  lea    0x1c(%rsp),%rax
  push   %rax
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  movss  0x1c(%rsp),%xmm0
  movss  0x18(%rsp),%xmm2
  movss  0x24(%rsp),%xmm1
  subss  %xmm0,%xmm1
  movss  0x20(%rsp),%xmm3
  subss  %xmm2,%xmm3
  divss  %xmm3,%xmm1
  subss  0x14(%rsp),%xmm0
  subss  0x10(%rsp),%xmm2
  divss  %xmm2,%xmm0
  add    $0x10,%rsp
  ucomiss %xmm0,%xmm1
  jp     L20
  jne    L20
  lea    D13(%rip),%rdi
  call   <printf@plt>
L5:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L20:
  lea    D28(%rip),%rdi
  call   <printf@plt>
  jmp    L5
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D22
datatype: string
name: read_data
arguments: data_tag: D13
datatype: string
name: read_data
arguments: data_tag: D28
datatype: string

tool: D22: %f %f %f %f %f %f

tool: D13: All the points are on same line.

tool: D28: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%f %f %f %f %f %f", &x1, &y1, &x2, &y2, &x3, &y3);
  if ((y3 - y2) / (x3 - x2) == (y2 - y1) / (x2 - x1)) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  lea    0x8(%rsp),%rcx
  mov    %rsp,%rsi
  sub    $0x8,%rsp
  lea    0x1c(%rsp),%rax
  push   %rax
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  call   <scanf@plt>
  movss  0x1c(%rsp),%xmm0
  movss  0x18(%rsp),%xmm2
  movss  0x24(%rsp),%xmm1
  movss  0x20(%rsp),%xmm3
  subss  %xmm0,%xmm1
  subss  %xmm2,%xmm3
  subss  0x14(%rsp),%xmm0
  subss  0x10(%rsp),%xmm2
  pop    %rax
  pop    %rdx
  divss  %xmm3,%xmm1
  divss  %xmm2,%xmm0
  ucomiss %xmm0,%xmm1
  jp     L11
  jne    L11
  lea    D20(%rip),%rdi
  call   <printf@plt>
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L11:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  jmp    L14
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D2
type: string
name: retrieve_data
arguments: label: D20
type: string
name: retrieve_data
arguments: label: D11
type: string

tool: D2: %f %f %f %f %f %f

tool: D20: All the points are on same line.

tool: D11: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%f %f %f %f %f %f", &x1, &y1, &x2, &y2, &x3, &y3);
  if ((y3 - y2) / (x3 - x2) == (y2 - y1) / (x2 - x1)) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  lea    0x8(%rsp),%rcx
  mov    %rsp,%rsi
  sub    $0x8,%rsp
  lea    0x1c(%rsp),%rax
  push   %rax
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  call   <scanf@plt>
  movss  0x1c(%rsp),%xmm0
  movss  0x18(%rsp),%xmm2
  movss  0x24(%rsp),%xmm1
  movss  0x20(%rsp),%xmm3
  subss  %xmm0,%xmm1
  subss  %xmm2,%xmm3
  subss  0x14(%rsp),%xmm0
  subss  0x10(%rsp),%xmm2
  pop    %rax
  pop    %rdx
  divss  %xmm3,%xmm1
  divss  %xmm2,%xmm0
  ucomiss %xmm0,%xmm1
  jp     L0
  jne    L0
  lea    D24(%rip),%rdi
  call   <printf@plt>
L12:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L12
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
type_guessed: string
name: extract_symbolic_data
arguments: label: D24
type_guessed: string
name: extract_symbolic_data
arguments: label: D4
type_guessed: string

tool: D14: %f %f %f %f %f %f

tool: D24: All the points are on same line.

tool: D4: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%f %f %f %f %f %f", &x1, &y1, &x2, &y2, &x3, &y3);
  if ((y3 - y2) / (x3 - x2) == (y2 - y1) / (x2 - x1)) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<skip_space>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  jmp    L7
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L1
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  jne    L25
L1:
  addq   $0x1,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  jmp    L26
L25:
  nop
L26:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *skip_space(char *ptr) {
  for (; *ptr; ptr++)
    if (*ptr != ' ' && *ptr != '\t')
      break;
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<skip_space>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L3
  jmp    L26
L41:
  add    $0x1,%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L26
L3:
  cmp    $0x20,%dl
  je     L41
  cmp    $0x9,%dl
  je     L41
L26:
  ret
```

assistant: ```c
char *skip_space(char *ptr) {
  for (; *ptr; ptr++)
    if (*ptr != ' ' && *ptr != '\t')
      break;
  return ptr;
}
```
user: What is the c source code of the assembly code below?

```asm
<skip_space>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  jne    L9
  jmp    L7
L3:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L7
L9:
  cmp    $0x20,%dl
  je     L3
  cmp    $0x9,%dl
  je     L3
L7:
  ret
```

assistant: ```c
char *skip_space(char *ptr) {
  for (; *ptr; ptr++)
    if (*ptr != ' ' && *ptr != '\t')
      break;
  return ptr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<skip_space>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  jne    L37
  jmp    L41
L18:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L41
L37:
  cmp    $0x20,%dl
  je     L18
  cmp    $0x9,%dl
  je     L18
L41:
  ret
```

assistant: ```c
char *skip_space(char *ptr) {
  for (; *ptr; ptr++)
    if (*ptr != ' ' && *ptr != '\t')
      break;
  return ptr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xd0(%rbp),%rax
  mov    %rax,%rdi
  call   <hash@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
data_type: string

tool: D7: %ld


assistant: ```c
int main() {
  char word[200];
  printf("%ld\n", hash(word));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <hash@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0xd8,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type: string

tool: D8: %ld


assistant: ```c
int main() {
  char word[200];
  printf("%ld\n", hash(word));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <hash@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0xd8,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_guessed: string

tool: D7: %ld


assistant: ```c
int main() {
  char word[200];
  printf("%ld\n", hash(word));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <hash@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0xd8,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D3
data_type: string

tool: D3: %ld


assistant: ```c
int main() {
  char word[200];
  printf("%ld\n", hash(word));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<palindrome_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    $0x1a,%edx
  mov    $0x0,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <memset@plt>
  movq   $0x0,-0x10(%rbp)
  jmp    L36
L0:
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  sub    $0x61,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  lea    D36(%rip),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    $0x1,%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  lea    D36(%rip),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
L36:
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
  movl   $0x0,-0x14(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L35
L48:
  lea    D36(%rip),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x1,%al
  jne    L55
  addl   $0x1,-0x14(%rbp)
L55:
  addq   $0x1,-0x10(%rbp)
L35:
  cmpq   $0x19,-0x10(%rbp)
  ja     L52
  cmpl   $0x1,-0x14(%rbp)
  jle    L48
L52:
  cmpl   $0x1,-0x14(%rbp)
  setle  %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int palindrome_permutation(const char *str) {
  static char freq[26];
  memset(freq, 0, sizeof(freq));
  size_t i;
  for (i = 0; str[i] != '\0'; i++) {
    size_t idx = (unsigned char)str[i] - 'a';
    freq[idx] = (freq[idx] + 1) % 2;
  }
  int odd_occurrences = 0;
  for (i = 0; i < sizeof(freq) && odd_occurrences < 2; i++) {
    if (freq[i] == 1) {
      odd_occurrences++;
    }
  }
  return odd_occurrences < 2;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<palindrome_permutation>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x1a,%edx
  mov    $0x0,%esi
  lea    D29(%rip),%rdi
  call   <memset@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L30
  lea    0x1(%rbx),%rdi
  lea    D29(%rip),%rsi
L19:
  movzbl %al,%edx
  sub    $0x61,%edx
  movslq %edx,%rdx
  movsbl (%rsi,%rdx,1),%eax
  add    $0x1,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  mov    %al,(%rsi,%rdx,1)
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L19
L30:
  mov    $0x0,%ecx
  mov    $0x0,%edx
  lea    D29(%rip),%rsi
L35:
  cmpb   $0x1,(%rsi,%rdx,1)
  sete   %al
  movzbl %al,%eax
  add    %eax,%ecx
  add    $0x1,%rdx
  cmp    $0x1,%ecx
  setle  %al
  cmp    $0x19,%rdx
  ja     L26
  test   %al,%al
  jne    L35
L26:
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int palindrome_permutation(const char *str) {
  static char freq[26];
  memset(freq, 0, sizeof(freq));
  size_t i;
  for (i = 0; str[i] != '\0'; i++) {
    size_t idx = (unsigned char)str[i] - 'a';
    freq[idx] = (freq[idx] + 1) % 2;
  }
  int odd_occurrences = 0;
  for (i = 0; i < sizeof(freq) && odd_occurrences < 2; i++) {
    if (freq[i] == 1) {
      odd_occurrences++;
    }
  }
  return odd_occurrences < 2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<palindrome_permutation>:
  endbr64
  push   %rbp
  mov    $0x1a,%edx
  mov    %rdi,%rbp
  xor    %esi,%esi
  push   %rbx
  lea    D9(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x8,%rsp
  call   <memset@plt>
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L9
  lea    0x1(%rbp),%rdi
  nop
L27:
  movzbl %al,%edx
  add    $0x1,%rdi
  sub    $0x61,%edx
  movslq %edx,%rdx
  movsbl (%rbx,%rdx,1),%eax
  add    $0x1,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  mov    %al,(%rbx,%rdx,1)
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L27
L9:
  xor    %ecx,%ecx
  xor    %edx,%edx
  nop
L15:
  xor    %eax,%eax
  cmpb   $0x1,(%rbx,%rdx,1)
  sete   %al
  add    $0x1,%rdx
  add    %eax,%ecx
  cmp    $0x1,%ecx
  setle  %al
  cmp    $0x19,%rdx
  ja     L12
  test   %al,%al
  jne    L15
L12:
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int palindrome_permutation(const char *str) {
  static char freq[26];
  memset(freq, 0, sizeof(freq));
  size_t i;
  for (i = 0; str[i] != '\0'; i++) {
    size_t idx = (unsigned char)str[i] - 'a';
    freq[idx] = (freq[idx] + 1) % 2;
  }
  int odd_occurrences = 0;
  for (i = 0; i < sizeof(freq) && odd_occurrences < 2; i++) {
    if (freq[i] == 1) {
      odd_occurrences++;
    }
  }
  return odd_occurrences < 2;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<palindrome_permutation>:
  endbr64
  push   %rbp
  mov    $0x1a,%edx
  mov    %rdi,%rbp
  xor    %esi,%esi
  push   %rbx
  lea    D0(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x8,%rsp
  call   <memset@plt>
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L15
  lea    0x1(%rbp),%rdi
  nop
L31:
  movzbl %al,%edx
  add    $0x1,%rdi
  sub    $0x61,%edx
  movslq %edx,%rdx
  movsbl (%rbx,%rdx,1),%eax
  add    $0x1,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  mov    %al,(%rbx,%rdx,1)
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L31
L15:
  xor    %ecx,%ecx
  xor    %edx,%edx
  nop
L32:
  xor    %eax,%eax
  cmpb   $0x1,(%rbx,%rdx,1)
  sete   %al
  add    $0x1,%rdx
  add    %eax,%ecx
  cmp    $0x1,%ecx
  setle  %al
  cmp    $0x19,%rdx
  ja     L18
  test   %al,%al
  jne    L32
L18:
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int palindrome_permutation(const char *str) {
  static char freq[26];
  memset(freq, 0, sizeof(freq));
  size_t i;
  for (i = 0; str[i] != '\0'; i++) {
    size_t idx = (unsigned char)str[i] - 'a';
    freq[idx] = (freq[idx] + 1) % 2;
  }
  int odd_occurrences = 0;
  for (i = 0; i < sizeof(freq) && odd_occurrences < 2; i++) {
    if (freq[i] == 1) {
      odd_occurrences++;
    }
  }
  return odd_occurrences < 2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L4:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  mov    $0x13fa,%edi
  call   <kill@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  jmp    L4
```

assistant: ```c
int main() {
  while (1) {
    kill(5114, SIGINT);
    sleep(2);
  }
  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   %rbx
  mov    D0(%rip),%rbx
L0:
  mov    (%rbx),%esi
  mov    $0x13fa,%edi
  call   <kill@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  jmp    L0
```

assistant: ```c
int main() {
  while (1) {
    kill(5114, SIGINT);
    sleep(2);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D0(%rip),%rbx
  nopl   0x0(%rax)
L5:
  mov    (%rbx),%esi
  mov    $0x13fa,%edi
  call   <kill@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  jmp    L5
```

assistant: ```c
int main() {
  while (1) {
    kill(5114, SIGINT);
    sleep(2);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D1(%rip),%rbx
  nopl   0x0(%rax)
L5:
  mov    (%rbx),%esi
  mov    $0x13fa,%edi
  call   <kill@plt>
  mov    $0x2,%edi
  call   <sleep@plt>
  jmp    L5
```

assistant: ```c
int main() {
  while (1) {
    kill(5114, SIGINT);
    sleep(2);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<jpip_write_prxy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    $0x0,%eax
  call   <cio_tell@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x4,%edi
  call   <cio_skip@plt>
  mov    $0x4,%esi
  mov    $0x70727879,%edi
  call   <cio_write@plt>
  mov    -0x14(%rbp),%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <cio_write@plt>
  mov    -0x18(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x6a703263,%edi
  call   <cio_write@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <cio_write@plt>
  mov    -0x1c(%rbp),%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <cio_write@plt>
  mov    -0x20(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x63696478,%edi
  call   <cio_write@plt>
  mov    $0x0,%eax
  call   <cio_tell@plt>
  sub    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <cio_seek@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <cio_write@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edi
  call   <cio_seek@plt>
  nop
  leave
  ret
```

assistant: ```c
void jpip_write_prxy(int offset_jp2c, int length_jp2c, int offset_idx, int length_idx) {
  int len, lenp;
  lenp = cio_tell();
  cio_skip(4);
  cio_write(0x70727879, 4);
  cio_write(offset_jp2c, 8);
  cio_write(length_jp2c, 4);
  cio_write(0x6a703263, 4);
  cio_write(1, 1);
  cio_write(offset_idx, 8);
  cio_write(length_idx, 4);
  cio_write(0x63696478, 4);
  len = cio_tell() - lenp;
  cio_seek(lenp);
  cio_write(len, 4);
  cio_seek(lenp + len);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<jpip_write_prxy>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r14d
  mov    %esi,%r13d
  mov    %edx,%r12d
  mov    %ecx,%ebx
  mov    $0x0,%eax
  call   <cio_tell@plt>
  mov    %eax,%ebp
  mov    $0x4,%edi
  call   <cio_skip@plt>
  mov    $0x4,%esi
  mov    $0x70727879,%edi
  call   <cio_write@plt>
  mov    $0x8,%esi
  mov    %r14d,%edi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    %r13d,%edi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x6a703263,%edi
  call   <cio_write@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <cio_write@plt>
  mov    $0x8,%esi
  mov    %r12d,%edi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    %ebx,%edi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x63696478,%edi
  call   <cio_write@plt>
  mov    $0x0,%eax
  call   <cio_tell@plt>
  mov    %eax,%ebx
  mov    %eax,%r12d
  sub    %ebp,%r12d
  mov    %ebp,%edi
  call   <cio_seek@plt>
  mov    $0x4,%esi
  mov    %r12d,%edi
  call   <cio_write@plt>
  mov    %ebx,%edi
  call   <cio_seek@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void jpip_write_prxy(int offset_jp2c, int length_jp2c, int offset_idx, int length_idx) {
  int len, lenp;
  lenp = cio_tell();
  cio_skip(4);
  cio_write(0x70727879, 4);
  cio_write(offset_jp2c, 8);
  cio_write(length_jp2c, 4);
  cio_write(0x6a703263, 4);
  cio_write(1, 1);
  cio_write(offset_idx, 8);
  cio_write(length_idx, 4);
  cio_write(0x63696478, 4);
  len = cio_tell() - lenp;
  cio_seek(lenp);
  cio_write(len, 4);
  cio_seek(lenp + len);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<jpip_write_prxy>:
  endbr64
  push   %r15
  xor    %eax,%eax
  mov    %edi,%r15d
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %ecx,%ebp
  call   <cio_tell@plt>
  mov    $0x4,%edi
  mov    %eax,%r12d
  call   <cio_skip@plt>
  mov    $0x4,%esi
  mov    $0x70727879,%edi
  call   <cio_write@plt>
  mov    %r15d,%edi
  mov    $0x8,%esi
  call   <cio_write@plt>
  mov    %r14d,%edi
  mov    $0x4,%esi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x6a703263,%edi
  call   <cio_write@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <cio_write@plt>
  mov    %r13d,%edi
  mov    $0x8,%esi
  call   <cio_write@plt>
  mov    %ebp,%edi
  mov    $0x4,%esi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x63696478,%edi
  call   <cio_write@plt>
  xor    %eax,%eax
  call   <cio_tell@plt>
  mov    %r12d,%edi
  mov    %eax,%r13d
  mov    %eax,%ebp
  call   <cio_seek@plt>
  sub    %r12d,%r13d
  mov    $0x4,%esi
  mov    %r13d,%edi
  call   <cio_write@plt>
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10a0 <cio_seek@plt>
```

assistant: ```c
void jpip_write_prxy(int offset_jp2c, int length_jp2c, int offset_idx, int length_idx) {
  int len, lenp;
  lenp = cio_tell();
  cio_skip(4);
  cio_write(0x70727879, 4);
  cio_write(offset_jp2c, 8);
  cio_write(length_jp2c, 4);
  cio_write(0x6a703263, 4);
  cio_write(1, 1);
  cio_write(offset_idx, 8);
  cio_write(length_idx, 4);
  cio_write(0x63696478, 4);
  len = cio_tell() - lenp;
  cio_seek(lenp);
  cio_write(len, 4);
  cio_seek(lenp + len);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jpip_write_prxy>:
  endbr64
  push   %r15
  xor    %eax,%eax
  mov    %edi,%r15d
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %ecx,%ebp
  call   <cio_tell@plt>
  mov    $0x4,%edi
  mov    %eax,%r12d
  call   <cio_skip@plt>
  mov    $0x4,%esi
  mov    $0x70727879,%edi
  call   <cio_write@plt>
  mov    %r15d,%edi
  mov    $0x8,%esi
  call   <cio_write@plt>
  mov    %r14d,%edi
  mov    $0x4,%esi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x6a703263,%edi
  call   <cio_write@plt>
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <cio_write@plt>
  mov    %r13d,%edi
  mov    $0x8,%esi
  call   <cio_write@plt>
  mov    %ebp,%edi
  mov    $0x4,%esi
  call   <cio_write@plt>
  mov    $0x4,%esi
  mov    $0x63696478,%edi
  call   <cio_write@plt>
  xor    %eax,%eax
  call   <cio_tell@plt>
  mov    %r12d,%edi
  mov    %eax,%r13d
  mov    %eax,%ebp
  call   <cio_seek@plt>
  sub    %r12d,%r13d
  mov    $0x4,%esi
  mov    %r13d,%edi
  call   <cio_write@plt>
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10a0 <cio_seek@plt>
```

assistant: ```c
void jpip_write_prxy(int offset_jp2c, int length_jp2c, int offset_idx, int length_idx) {
  int len, lenp;
  lenp = cio_tell();
  cio_skip(4);
  cio_write(0x70727879, 4);
  cio_write(offset_jp2c, 8);
  cio_write(length_jp2c, 4);
  cio_write(0x6a703263, 4);
  cio_write(1, 1);
  cio_write(offset_idx, 8);
  cio_write(length_idx, 4);
  cio_write(0x63696478, 4);
  len = cio_tell() - lenp;
  cio_seek(lenp);
  cio_write(len, 4);
  cio_seek(lenp + len);
}
```
user: Translate the below assembly code into C programming language:

```asm
<UARTRxErrorClear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <_UARTBaseValid@plt>
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movq   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void UARTRxErrorClear(uint32_t ui32Base) {
  ASSERT(_UARTBaseValid(ui32Base));
  (*((volatile uint32_t *)(ui32Base + 0x00000004))) = 0;
}
```