ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<split_path>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %sil,0x6(%rsp)
  movb   $0x0,0x7(%rsp)
  call   <alloc_entry@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L19
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L59
  lea    0x6(%rsp),%rbx
  mov    %rax,%rdi
  mov    %rbx,%rsi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L77
  jmp    L12
L89:
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L12
L77:
  mov    %r12,%rdi
  call   <insert_entry@plt>
  test   %rax,%rax
  je     L89
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  xor    %r12d,%r12d
  call   <free_entry@plt>
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L12:
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L19
L35:
  call   <__stack_chk_fail@plt>
L59:
  lea    D6(%rip),%rdi
  call   <perror@plt>
  mov    %r12,%rdi
  xor    %r12d,%r12d
  call   <free_entry@plt>
  jmp    L19
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string

tool: D6: strdup

assistant: ```c
struct path_list *split_path(const char *list, char sep) {
  struct path_list *pl;
  char *p, *q;
  char ss[2];
  ss[0] = sep;
  ss[1] = '\0';
  pl = alloc_entry();
  if (!pl)
    return NULL;
  p = strdup(list);
  if (!p) {
    perror("strdup");
    free_entry(pl);
    return NULL;
  }
  q = strtok(p, ss);
  while (q) {
    if (insert_entry(pl, q)) {
      free(p);
      free_entry(pl);
      return NULL;
    }
    q = strtok(NULL, ss);
  }
  free(p);
  return pl;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ri_mem_alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *ri_mem_alloc(size_t byte) {
  void *p;
  assert(byte != 0);
  p = malloc(byte);
  assert(p != NULL);
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ri_mem_alloc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
void *ri_mem_alloc(size_t byte) {
  void *p;
  assert(byte != 0);
  p = malloc(byte);
  assert(p != NULL);
  return p;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ri_mem_alloc>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  test   %r12,%r12
  setne  %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <malloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *ri_mem_alloc(size_t byte) {
  void *p;
  assert(byte != 0);
  p = malloc(byte);
  assert(p != NULL);
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ri_mem_alloc>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  test   %r12,%r12
  setne  %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <malloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *ri_mem_alloc(size_t byte) {
  void *p;
  assert(byte != 0);
  p = malloc(byte);
  assert(p != NULL);
  return p;
}
```
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    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <get_memory_size@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_infered: string

tool: D8: Total physical memory: %lu


assistant: ```c
int main(int argc, char *argv[]) {
  uint64_t total_memory = get_memory_size();
  printf("Total physical memory: %lu\n", total_memory);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_memory_size@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type: string

tool: D7: Total physical memory: %lu


assistant: ```c
int main(int argc, char *argv[]) {
  uint64_t total_memory = get_memory_size();
  printf("Total physical memory: %lu\n", total_memory);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_memory_size@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string

tool: D6: Total physical memory: %lu


assistant: ```c
int main(int argc, char *argv[]) {
  uint64_t total_memory = get_memory_size();
  printf("Total physical memory: %lu\n", total_memory);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_memory_size@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D4
type_infered: string

tool: D4: Total physical memory: %lu


assistant: ```c
int main(int argc, char *argv[]) {
  uint64_t total_memory = get_memory_size();
  printf("Total physical memory: %lu\n", total_memory);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<writesn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <writes@plt>
  mov    $0x0,%eax
  call   <writenl@plt>
  nop
  leave
  ret
```

assistant: ```c
void writesn(const char *s) {
  writes(s);
  writenl();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<writesn>:
  endbr64
  sub    $0x8,%rsp
  call   <writes@plt>
  mov    $0x0,%eax
  call   <writenl@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void writesn(const char *s) {
  writes(s);
  writenl();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<writesn>:
  endbr64
  sub    $0x8,%rsp
  call   <writes@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <writenl@plt>
```

assistant: ```c
void writesn(const char *s) {
  writes(s);
  writenl();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<writesn>:
  endbr64
  sub    $0x8,%rsp
  call   <writes@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <writenl@plt>
```

assistant: ```c
void writesn(const char *s) {
  writes(s);
  writenl();
}
```
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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <GCD@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: string
name: unpack_data
arguments: data_tag: D3
type: string

tool: D8: %d %d

tool: D3: %d


assistant: ```c
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%d\n", GCD(a, b));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <GCD@plt>
  mov    %eax,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D18
type_guessed: string

tool: D0: %d %d

tool: D18: %d


assistant: ```c
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%d\n", GCD(a, b));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D11(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <GCD@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D11
type: string
name: unpack_data
arguments: data_label: D3
type: string

tool: D11: %d %d

tool: D3: %d


assistant: ```c
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%d\n", GCD(a, b));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D8(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <GCD@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D8
type: string
name: parse_data
arguments: tag: D13
type: string

tool: D8: %d %d

tool: D13: %d


assistant: ```c
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  printf("%d\n", GCD(a, b));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<addOK>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x1f,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  sar    $0x1f,%eax
  cmp    %eax,-0x4(%rbp)
  setne  %dl
  mov    -0x8(%rbp),%eax
  sar    $0x1f,%eax
  cmp    %eax,-0x4(%rbp)
  sete   %al
  or     %edx,%eax
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int addOK(int x, int y) {
  int sum = x + y;
  int a = x >> 31;
  return !!(a ^ y >> 31) | !(a ^ sum >> 31);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<addOK>:
  endbr64
  mov    %edi,%edx
  sar    $0x1f,%edx
  add    %esi,%edi
  sar    $0x1f,%edi
  cmp    %edx,%edi
  sete   %al
  sar    $0x1f,%esi
  cmp    %edx,%esi
  setne  %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int addOK(int x, int y) {
  int sum = x + y;
  int a = x >> 31;
  return !!(a ^ y >> 31) | !(a ^ sum >> 31);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<addOK>:
  endbr64
  mov    %edi,%edx
  add    %esi,%edi
  sar    $0x1f,%edx
  sar    $0x1f,%edi
  cmp    %edx,%edi
  sete   %al
  sar    $0x1f,%esi
  cmp    %edx,%esi
  setne  %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int addOK(int x, int y) {
  int sum = x + y;
  int a = x >> 31;
  return !!(a ^ y >> 31) | !(a ^ sum >> 31);
}
```
user: Translate the below assembly code into C programming language:

```asm
<addOK>:
  endbr64
  mov    %edi,%edx
  add    %esi,%edi
  sar    $0x1f,%edx
  sar    $0x1f,%edi
  cmp    %edx,%edi
  sete   %al
  sar    $0x1f,%esi
  cmp    %edx,%esi
  setne  %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int addOK(int x, int y) {
  int sum = x + y;
  int a = x >> 31;
  return !!(a ^ y >> 31) | !(a ^ sum >> 31);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_bits_of_int_v1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2a,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    $0x20,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
size_t get_bits_of_int_v1(void) {
  int zen_of_life[] = {42, 0};
  return ((void *)(zen_of_life + 1) - (void *)zen_of_life) * 8;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_bits_of_int_v1>:
  endbr64
  mov    $0x20,%eax
  ret
```

assistant: ```c
size_t get_bits_of_int_v1(void) {
  int zen_of_life[] = {42, 0};
  return ((void *)(zen_of_life + 1) - (void *)zen_of_life) * 8;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_bits_of_int_v1>:
  endbr64
  mov    $0x20,%eax
  ret
```

assistant: ```c
size_t get_bits_of_int_v1(void) {
  int zen_of_life[] = {42, 0};
  return ((void *)(zen_of_life + 1) - (void *)zen_of_life) * 8;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_bits_of_int_v1>:
  endbr64
  mov    $0x20,%eax
  ret
```

assistant: ```c
size_t get_bits_of_int_v1(void) {
  int zen_of_life[] = {42, 0};
  return ((void *)(zen_of_life + 1) - (void *)zen_of_life) * 8;
}
```
user: What is the c source code of the assembly code below?

```asm
<gets>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
L6:
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L29
  cmpl   $0xa,-0x4(%rbp)
  je     L29
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L6
L29:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  cmpl   $0x0,-0x8(%rbp)
  jg     L42
  cmpl   $0x0,-0x4(%rbp)
  js     L44
L42:
  mov    -0x18(%rbp),%rax
  jmp    L21
L44:
  mov    $0x0,%eax
L21:
  leave
  ret
```

assistant: ```c
char *gets(char *buf) {
  int c;
  int count = 0;
  for (;;) {
    c = fgetc(stdin);
    if (c < 0 || c == '\n')
      break;
    buf[count++] = c;
  }
  buf[count] = 0;
  return (count > 0 || c >= 0) ? buf : NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gets>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
  mov    D2(%rip),%r13
  jmp    L20
L10:
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
L20:
  mov    %ebx,%r12d
  mov    0x0(%r13),%edi
  call   <fgetc@plt>
  test   %eax,%eax
  js     L1
  cmp    $0xa,%eax
  jne    L10
L1:
  movslq %r12d,%rdx
  movb   $0x0,0x0(%rbp,%rdx,1)
  test   %r12d,%r12d
  jg     L27
  test   %eax,%eax
  mov    $0x0,%eax
  cmovs  %rax,%rbp
L27:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *gets(char *buf) {
  int c;
  int count = 0;
  for (;;) {
    c = fgetc(stdin);
    if (c < 0 || c == '\n')
      break;
    buf[count++] = c;
  }
  buf[count] = 0;
  return (count > 0 || c >= 0) ? buf : NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gets>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    D4(%rip),%r12
  jmp    L22
  nop
L21:
  mov    %al,(%rbx)
  add    $0x1,%ebp
  add    $0x1,%rbx
L22:
  mov    (%r12),%edi
  call   <fgetc@plt>
  test   %eax,%eax
  js     L19
  cmp    $0xa,%eax
  jne    L21
L19:
  movb   $0x0,(%rbx)
  test   %ebp,%ebp
  jg     L17
  test   %eax,%eax
  mov    $0x0,%eax
  cmovs  %rax,%r13
L17:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *gets(char *buf) {
  int c;
  int count = 0;
  for (;;) {
    c = fgetc(stdin);
    if (c < 0 || c == '\n')
      break;
    buf[count++] = c;
  }
  buf[count] = 0;
  return (count > 0 || c >= 0) ? buf : NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<gets>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    D4(%rip),%r12
  jmp    L22
  nop
L35:
  mov    %al,(%rbx)
  add    $0x1,%ebp
  add    $0x1,%rbx
L22:
  mov    (%r12),%edi
  call   <fgetc@plt>
  test   %eax,%eax
  js     L33
  cmp    $0xa,%eax
  jne    L35
L33:
  movb   $0x0,(%rbx)
  test   %ebp,%ebp
  jg     L17
  test   %eax,%eax
  mov    $0x0,%eax
  cmovs  %rax,%r13
L17:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *gets(char *buf) {
  int c;
  int count = 0;
  for (;;) {
    c = fgetc(stdin);
    if (c < 0 || c == '\n')
      break;
    buf[count++] = c;
  }
  buf[count] = 0;
  return (count > 0 || c >= 0) ? buf : NULL;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<calcFP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  D6(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x10(%rbp),%xmm0
  divsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
type_guessed: f64

tool: D6: 29.53

assistant: ```c
double calcFP(double Pressure) {
  double Pstd = 29.53;
  double FP = 0;
  FP = (Pressure - Pstd) / (Pstd);
  return FP;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calcFP>:
  endbr64
  movsd  D1(%rip),%xmm1
  subsd  %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D1
data_type: f64

tool: D1: 29.53

assistant: ```c
double calcFP(double Pressure) {
  double Pstd = 29.53;
  double FP = 0;
  FP = (Pressure - Pstd) / (Pstd);
  return FP;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<calcFP>:
  endbr64
  movsd  D6(%rip),%xmm1
  subsd  %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
datatype: f64

tool: D6: 29.53

assistant: ```c
double calcFP(double Pressure) {
  double Pstd = 29.53;
  double FP = 0;
  FP = (Pressure - Pstd) / (Pstd);
  return FP;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<calcFP>:
  endbr64
  movsd  D9(%rip),%xmm1
  subsd  %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
type: f64

tool: D9: 29.53

assistant: ```c
double calcFP(double Pressure) {
  double Pstd = 29.53;
  double FP = 0;
  FP = (Pressure - Pstd) / (Pstd);
  return FP;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  mov    %eax,0x4(%rsp)
  mov    0x4(%rdi),%edi
  mov    $0x0,%esi
  lea    0x4(%rsp),%r9
  test   %edi,%edi
  jne    L58
L4:
  mov    0x4(%rsp),%eax
  and    $0x369937ec,%eax
  mov    %eax,(%r10)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x18,%rsp
  ret
L23:
  movzbl (%rax),%ecx
  movzbl 0x1(%rax),%r8d
  mov    %r8b,(%rax)
  mov    %cl,0x1(%rax)
  movzbl 0x6(%rsp),%ecx
  movzbl 0x7(%rsp),%r8d
  mov    %r8b,0x6(%rsp)
  mov    %cl,0x7(%rsp)
L57:
  add    $0x2,%edx
  cmp    %edx,%edi
  jbe    L24
L3:
  mov    %edx,%ecx
  and    %esi,%ecx
  cmp    %ecx,0x4(%rsp)
  jbe    L23
  movzbl 0x1(%rax),%ecx
  movzbl 0x2(%rax),%r8d
  mov    %r8b,0x1(%rax)
  movzbl (%rax),%r8d
  mov    %r8b,0x2(%rax)
  mov    %cl,(%rax)
  jmp    L57
L24:
  add    $0x2,%esi
  cmp    %edi,%esi
  jae    L4
L58:
  mov    %esi,%eax
  lea    (%r9,%rax,4),%rax
  mov    $0x0,%edx
  jmp    L3
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  char copy11;
  char copy12;
  char copy13;
  char copy14;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        if (state[0UL] > (local2 & local1)) {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy11;
          copy11 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy12;
          copy12 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy14 = *((char *)(&state[0UL]) + 2);
          *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy14;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] & 916010988UL;
  }
}
```