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

```asm
<ModulusPow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L21
L22:
  mov    -0x18(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L5
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,%ecx
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,%eax
  imul   %ecx,%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x4(%rbp)
L5:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,%ecx
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,%eax
  imul   %ecx,%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x14(%rbp)
  sarl   -0x18(%rbp)
L21:
  cmpl   $0x0,-0x18(%rbp)
  jg     L22
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t ModulusPow(uint64_t base, uint64_t exp, uint64_t mod) {
  base = base % mod;
  uint64_t result = 1;
  while (exp > 0) {
    if (exp & 1)
      result = ((result % mod) * (base % mod)) % mod;
    base = ((base % mod) * (base % mod)) % mod;
    exp >>= 1;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ModulusPow>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%r8d
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%esi
  test   %r8d,%r8d
  jle    L4
  mov    $0x1,%edi
  jmp    L29
L32:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  imul   %edx,%edx
  mov    %edx,%eax
  cltd
  idiv   %ecx
  mov    %edx,%esi
  sar    %r8d
  test   %r8d,%r8d
  jle    L25
L29:
  test   $0x1,%r8b
  je     L32
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%edi
  mov    %esi,%eax
  cltd
  idiv   %ecx
  mov    %edi,%eax
  imul   %edx,%eax
  cltd
  idiv   %ecx
  mov    %edx,%edi
  jmp    L32
L4:
  mov    $0x1,%edi
L25:
  mov    %edi,%eax
  ret
```

assistant: ```c
uint64_t ModulusPow(uint64_t base, uint64_t exp, uint64_t mod) {
  base = base % mod;
  uint64_t result = 1;
  while (exp > 0) {
    if (exp & 1)
      result = ((result % mod) * (base % mod)) % mod;
    base = ((base % mod) * (base % mod)) % mod;
    exp >>= 1;
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ModulusPow>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%r8d
  mov    %edx,%esi
  mov    $0x1,%r9d
  cltd
  idiv   %esi
  test   %r8d,%r8d
  jle    L29
  nopl   0x0(%rax)
L0:
  mov    %edx,%eax
  cltd
  idiv   %esi
  mov    %edx,%ecx
  test   $0x1,%r8b
  je     L25
  mov    %r9d,%eax
  cltd
  idiv   %esi
  mov    %edx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%r9d
L25:
  mov    %ecx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %esi
  sar    %r8d
  jne    L0
L29:
  mov    %r9d,%eax
  ret
```

assistant: ```c
uint64_t ModulusPow(uint64_t base, uint64_t exp, uint64_t mod) {
  base = base % mod;
  uint64_t result = 1;
  while (exp > 0) {
    if (exp & 1)
      result = ((result % mod) * (base % mod)) % mod;
    base = ((base % mod) * (base % mod)) % mod;
    exp >>= 1;
  }
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ModulusPow>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%r8d
  mov    %edx,%esi
  mov    $0x1,%r9d
  cltd
  idiv   %esi
  test   %r8d,%r8d
  jle    L9
  nopl   0x0(%rax)
L6:
  mov    %edx,%eax
  cltd
  idiv   %esi
  mov    %edx,%ecx
  test   $0x1,%r8b
  je     L18
  mov    %r9d,%eax
  cltd
  idiv   %esi
  mov    %edx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%r9d
L18:
  mov    %ecx,%eax
  imul   %ecx,%eax
  cltd
  idiv   %esi
  sar    %r8d
  jne    L6
L9:
  mov    %r9d,%eax
  ret
```

assistant: ```c
uint64_t ModulusPow(uint64_t base, uint64_t exp, uint64_t mod) {
  base = base % mod;
  uint64_t result = 1;
  while (exp > 0) {
    if (exp & 1)
      result = ((result % mod) * (base % mod)) % mod;
    base = ((base % mod) * (base % mod)) % mod;
    exp >>= 1;
  }
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%edi
  call   <rdpmc@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
type_guessed: string

tool: D5: %lld


assistant: ```c
int main(int argc, char **argv) {
  printf("%lld\n", rdpmc(0));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <rdpmc@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
datatype: string

tool: D8: %lld


assistant: ```c
int main(int argc, char **argv) {
  printf("%lld\n", rdpmc(0));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <rdpmc@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
datatype: string

tool: D2: %lld


assistant: ```c
int main(int argc, char **argv) {
  printf("%lld\n", rdpmc(0));
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <rdpmc@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D3: %lld


assistant: ```c
int main(int argc, char **argv) {
  printf("%lld\n", rdpmc(0));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<__simple_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x1,-0x28(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L23
  addq   $0x1,-0x38(%rbp)
  jmp    L23
L88:
  shlq   -0x28(%rbp)
L23:
  mov    -0x28(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jae    L90
  cmpq   $0xf,-0x28(%rbp)
  jbe    L88
L90:
  lea    D83(%rip),%rax
  mov    %rax,%rdi
  call   <__lock@plt>
  mov    D49(%rip),%rax
  neg    %rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  sub    $0x1,%rax
  and    %rdx,%rax
  mov    %rax,-0x20(%rbp)
  mov    D124(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  add    $0x10,%eax
  cltq
  cmp    %rax,-0x38(%rbp)
  ja     L27
  mov    -0x20(%rbp),%rax
  add    %rax,-0x38(%rbp)
L27:
  mov    D82(%rip),%rax
  mov    D49(%rip),%rdx
  sub    %rdx,%rax
  cmp    %rax,-0x38(%rbp)
  jbe    L8
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <__expand_heap@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L83
  lea    D83(%rip),%rax
  mov    %rax,%rdi
  call   <__unlock@plt>
  mov    $0x0,%eax
  jmp    L42
L83:
  mov    D82(%rip),%rax
  cmp    %rax,-0x18(%rbp)
  je     L58
  mov    -0x18(%rbp),%rax
  mov    %rax,D74(%rip)
  mov    -0x20(%rbp),%rax
  sub    %rax,-0x38(%rbp)
  movq   $0x0,-0x20(%rbp)
L58:
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,D74(%rip)
L8:
  mov    D74(%rip),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    D74(%rip),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,D74(%rip)
  lea    D83(%rip),%rax
  mov    %rax,%rdi
  call   <__unlock@plt>
  mov    -0x10(%rbp),%rax
L42:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void *__simple_malloc(size_t n) {
  static char *cur, *end;
  static volatile int lock[2];
  size_t align = 1, pad;
  void *p;
  if (!n)
    n++;
  while (align < n && align < 16)
    align += align;
  __lock(lock);
  pad = -(uintptr_t)cur & align - 1;
  if (n <= SIZE_MAX / 2 + 16)
    n += pad;
  if (n > end - cur) {
    size_t m = n;
    char *new = __expand_heap(&m);
    if (!new) {
      __unlock(lock);
      return 0;
    }
    if (new != end) {
      cur = new;
      n -= pad;
      pad = 0;
    }
    end = new + m;
  }
  p = cur + pad;
  cur += n;
  __unlock(lock);
  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
<__simple_malloc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L46
  mov    %rdi,%rbp
  cmp    $0x1,%rdi
  jbe    L37
  mov    $0x1,%ebx
L109:
  add    %rbx,%rbx
  cmp    %rbx,%rbp
  jbe    L42
  cmp    $0xf,%rbx
  jbe    L109
  jmp    L42
L37:
  mov    $0x1,%ebx
  jmp    L42
L46:
  mov    $0x1,%ebp
  mov    $0x1,%ebx
L42:
  lea    D115(%rip),%rdi
  call   <__lock@plt>
  mov    D102(%rip),%rdx
  mov    %rdx,%rax
  neg    %rax
  sub    $0x1,%rbx
  and    %rax,%rbx
  mov    D113(%rip),%rax
  mov    (%rax),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  add    $0x10,%eax
  cltq
  lea    0x0(%rbp,%rbx,1),%rcx
  cmp    %rbp,%rax
  cmovae %rcx,%rbp
  mov    D79(%rip),%rax
  sub    %rdx,%rax
  cmp    %rbp,%rax
  jae    L55
  mov    %rbp,(%rsp)
  mov    %rsp,%rdi
  call   <__expand_heap@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L108
  cmp    %rax,D1(%rip)
  je     L16
  mov    %rax,D1(%rip)
  sub    %rbx,%rbp
  mov    $0x0,%ebx
L16:
  add    (%rsp),%r12
  mov    %r12,D1(%rip)
L55:
  add    D1(%rip),%rbx
  mov    %rbx,%r12
  mov    %rbp,%rax
  add    D1(%rip),%rax
  mov    %rax,D1(%rip)
  lea    D115(%rip),%rdi
  call   <__unlock@plt>
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L71
  mov    %r12,%rax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L108:
  lea    D115(%rip),%rdi
  call   <__unlock@plt>
  jmp    L23
L71:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *__simple_malloc(size_t n) {
  static char *cur, *end;
  static volatile int lock[2];
  size_t align = 1, pad;
  void *p;
  if (!n)
    n++;
  while (align < n && align < 16)
    align += align;
  __lock(lock);
  pad = -(uintptr_t)cur & align - 1;
  if (n <= SIZE_MAX / 2 + 16)
    n += pad;
  if (n > end - cur) {
    size_t m = n;
    char *new = __expand_heap(&m);
    if (!new) {
      __unlock(lock);
      return 0;
    }
    if (new != end) {
      cur = new;
      n -= pad;
      pad = 0;
    }
    end = new + m;
  }
  p = cur + pad;
  cur += n;
  __unlock(lock);
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__simple_malloc>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  cmovne %rdi,%rbp
  cmp    $0x1,%rdi
  jbe    L40
  mov    $0x1,%eax
  nopl   0x0(%rax)
L39:
  add    %rax,%rax
  cmp    %rax,%rbp
  jbe    L67
  cmp    $0xf,%rax
  jbe    L39
L67:
  lea    -0x1(%rax),%rbx
L41:
  lea    D47(%rip),%r13
  mov    %r13,%rdi
  call   <__lock@plt>
  mov    D62(%rip),%rdx
  mov    %rdx,%rax
  neg    %rax
  and    %rax,%rbx
  mov    D43(%rip),%rax
  lea    (%rdx,%rbx,1),%r12
  mov    (%rax),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  lea    0x0(%rbp,%rbx,1),%rcx
  sar    %eax
  add    $0x10,%eax
  cltq
  cmp    %rbp,%rax
  mov    D59(%rip),%rax
  cmovae %rcx,%rbp
  sub    %rdx,%rax
  cmp    %rbp,%rax
  jae    L51
  mov    %rsp,%rdi
  mov    %rbp,(%rsp)
  call   <__expand_heap@plt>
  test   %rax,%rax
  je     L16
  mov    D62(%rip),%rdx
  cmp    %rax,D46(%rip)
  lea    (%rdx,%rbx,1),%r12
  je     L83
  sub    %rbx,%rbp
  mov    %rax,%r12
  mov    %rax,%rdx
L83:
  add    (%rsp),%rax
  mov    %rax,D46(%rip)
L51:
  add    %rbp,%rdx
  mov    %r13,%rdi
  mov    %rdx,D46(%rip)
  call   <__unlock@plt>
L65:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L40:
  xor    %ebx,%ebx
  jmp    L41
L16:
  mov    %r13,%rdi
  xor    %r12d,%r12d
  call   <__unlock@plt>
  jmp    L65
L72:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *__simple_malloc(size_t n) {
  static char *cur, *end;
  static volatile int lock[2];
  size_t align = 1, pad;
  void *p;
  if (!n)
    n++;
  while (align < n && align < 16)
    align += align;
  __lock(lock);
  pad = -(uintptr_t)cur & align - 1;
  if (n <= SIZE_MAX / 2 + 16)
    n += pad;
  if (n > end - cur) {
    size_t m = n;
    char *new = __expand_heap(&m);
    if (!new) {
      __unlock(lock);
      return 0;
    }
    if (new != end) {
      cur = new;
      n -= pad;
      pad = 0;
    }
    end = new + m;
  }
  p = cur + pad;
  cur += n;
  __unlock(lock);
  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
<__simple_malloc>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  cmovne %rdi,%rbp
  cmp    $0x1,%rdi
  jbe    L1
  mov    $0x1,%eax
  nopl   0x0(%rax)
L83:
  add    %rax,%rax
  cmp    %rax,%rbp
  jbe    L54
  cmp    $0xf,%rax
  jbe    L83
L54:
  lea    -0x1(%rax),%rbx
L42:
  lea    D56(%rip),%r13
  mov    %r13,%rdi
  call   <__lock@plt>
  mov    D38(%rip),%rdx
  mov    %rdx,%rax
  neg    %rax
  and    %rax,%rbx
  mov    D55(%rip),%rax
  lea    (%rdx,%rbx,1),%r12
  mov    (%rax),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  lea    0x0(%rbp,%rbx,1),%rcx
  sar    %eax
  add    $0x10,%eax
  cltq
  cmp    %rbp,%rax
  mov    D69(%rip),%rax
  cmovae %rcx,%rbp
  sub    %rdx,%rax
  cmp    %rbp,%rax
  jae    L15
  mov    %rsp,%rdi
  mov    %rbp,(%rsp)
  call   <__expand_heap@plt>
  test   %rax,%rax
  je     L81
  mov    D38(%rip),%rdx
  cmp    %rax,D15(%rip)
  lea    (%rdx,%rbx,1),%r12
  je     L84
  sub    %rbx,%rbp
  mov    %rax,%r12
  mov    %rax,%rdx
L84:
  add    (%rsp),%rax
  mov    %rax,D15(%rip)
L15:
  add    %rbp,%rdx
  mov    %r13,%rdi
  mov    %rdx,D15(%rip)
  call   <__unlock@plt>
L38:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  xor    %ebx,%ebx
  jmp    L42
L81:
  mov    %r13,%rdi
  xor    %r12d,%r12d
  call   <__unlock@plt>
  jmp    L38
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *__simple_malloc(size_t n) {
  static char *cur, *end;
  static volatile int lock[2];
  size_t align = 1, pad;
  void *p;
  if (!n)
    n++;
  while (align < n && align < 16)
    align += align;
  __lock(lock);
  pad = -(uintptr_t)cur & align - 1;
  if (n <= SIZE_MAX / 2 + 16)
    n += pad;
  if (n > end - cur) {
    size_t m = n;
    char *new = __expand_heap(&m);
    if (!new) {
      __unlock(lock);
      return 0;
    }
    if (new != end) {
      cur = new;
      n -= pad;
      pad = 0;
    }
    end = new + m;
  }
  p = cur + pad;
  cur += n;
  __unlock(lock);
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<zwavecmd_actuator_set>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %edi,-0x114(%rbp)
  mov    %esi,-0x118(%rbp)
  mov    %edx,-0x11c(%rbp)
  mov    %ecx,-0x120(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x118(%rbp),%eax
  mov    %al,-0x110(%rbp)
  mov    -0x120(%rbp),%eax
  mov    %al,-0x10f(%rbp)
  cmpl   $0xffffffff,-0x11c(%rbp)
  jne    L0
  movb   $0x0,-0x10e(%rbp)
  movb   $0x0,-0x10d(%rbp)
  jmp    L16
L0:
  movb   $0x80,-0x10e(%rbp)
  mov    -0x11c(%rbp),%eax
  mov    %al,-0x10d(%rbp)
L16:
  lea    -0x110(%rbp),%rdx
  mov    -0x114(%rbp),%eax
  mov    $0x4,%r8d
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    $0x2c,%esi
  mov    %eax,%edi
  call   <zwave_sendClassCommand@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
void zwavecmd_actuator_set(unsigned int id, unsigned int scene, int level, int duration) {
  unsigned char buf[255];
  buf[0] = scene;
  buf[1] = duration;
  if (level == -1) {
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[2] = 0x80;
    buf[3] = level;
  }
  zwave_sendClassCommand(id, 0x2C, 0x01, buf, 4);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<zwavecmd_actuator_set>:
  endbr64
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  cmp    $0xffffffff,%edx
  cmovne %edx,%eax
  mov    $0x0,%edx
  mov    $0xffffff80,%r8d
  cmovne %r8d,%edx
  mov    %sil,(%rsp)
  mov    %cl,0x1(%rsp)
  mov    %dl,0x2(%rsp)
  mov    %al,0x3(%rsp)
  mov    %rsp,%rcx
  mov    $0x4,%r8d
  mov    $0x1,%edx
  mov    $0x2c,%esi
  call   <zwave_sendClassCommand@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x118,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void zwavecmd_actuator_set(unsigned int id, unsigned int scene, int level, int duration) {
  unsigned char buf[255];
  buf[0] = scene;
  buf[1] = duration;
  if (level == -1) {
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[2] = 0x80;
    buf[3] = level;
  }
  zwave_sendClassCommand(id, 0x2C, 0x01, buf, 4);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<zwavecmd_actuator_set>:
  endbr64
  sub    $0x118,%rsp
  mov    $0xffffff80,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  cmp    $0xffffffff,%edx
  mov    %sil,(%rsp)
  mov    $0x2c,%esi
  mov    %cl,0x1(%rsp)
  mov    %rsp,%rcx
  cmove  %eax,%edx
  cmovne %r8d,%eax
  mov    $0x4,%r8d
  mov    %dl,0x3(%rsp)
  mov    $0x1,%edx
  mov    %al,0x2(%rsp)
  call   <zwave_sendClassCommand@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x118,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void zwavecmd_actuator_set(unsigned int id, unsigned int scene, int level, int duration) {
  unsigned char buf[255];
  buf[0] = scene;
  buf[1] = duration;
  if (level == -1) {
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[2] = 0x80;
    buf[3] = level;
  }
  zwave_sendClassCommand(id, 0x2C, 0x01, buf, 4);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<zwavecmd_actuator_set>:
  endbr64
  sub    $0x118,%rsp
  mov    $0x4,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %sil,%al
  mov    $0x2c,%esi
  mov    %cl,%ah
  mov    $0xffffff80,%ecx
  mov    %ax,(%rsp)
  xor    %eax,%eax
  cmp    $0xffffffff,%edx
  cmove  %eax,%edx
  cmovne %ecx,%eax
  mov    %rsp,%rcx
  mov    %dl,0x3(%rsp)
  mov    $0x1,%edx
  mov    %al,0x2(%rsp)
  call   <zwave_sendClassCommand@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x118,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void zwavecmd_actuator_set(unsigned int id, unsigned int scene, int level, int duration) {
  unsigned char buf[255];
  buf[0] = scene;
  buf[1] = duration;
  if (level == -1) {
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[2] = 0x80;
    buf[3] = level;
  }
  zwave_sendClassCommand(id, 0x2C, 0x01, buf, 4);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  D21(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  D10(%rip),%xmm0
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D16
type_guessed: string
name: retrieve_data
arguments: label: D21
type_guessed: f64
name: retrieve_data
arguments: label: D10
type_guessed: byte[8]

tool: D16: %d

tool: D21: 1.8

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0x40,0x40]}

assistant: ```c
void main() {
  int cel;
  int far;
  scanf("%d", &cel);
  far = cel * 1.8 + 32;
  printf("%d", far);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D15(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdl 0x4(%rsp),%xmm0
  mulsd  D18(%rip),%xmm0
  addsd  D6(%rip),%xmm0
  cvttsd2si %xmm0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
type: string
name: retrieve_data
arguments: tag: D18
type: f64
name: retrieve_data
arguments: tag: D6
type: byte[8]

tool: D15: %d

tool: D18: 1.8

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x40,0x40]}

assistant: ```c
void main() {
  int cel;
  int far;
  scanf("%d", &cel);
  far = cel * 1.8 + 32;
  printf("%d", far);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D12(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  cvtsi2sdl 0x4(%rsp),%xmm0
  mulsd  D0(%rip),%xmm0
  addsd  D25(%rip),%xmm0
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D12
data_type: string
name: retrieve_data
arguments: tag: D0
data_type: f64
name: retrieve_data
arguments: tag: D25
data_type: byte[8]

tool: D12: %d

tool: D0: 1.8

tool: {"D25": [0x00,0x00,0x00,0x00,0x00,0x00,0x40,0x40]}

assistant: ```c
void main() {
  int cel;
  int far;
  scanf("%d", &cel);
  far = cel * 1.8 + 32;
  printf("%d", far);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D24(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  cvtsi2sdl 0x4(%rsp),%xmm0
  mulsd  D20(%rip),%xmm0
  addsd  D15(%rip),%xmm0
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D24
type: string
name: parse_data
arguments: tag: D20
type: f64
name: parse_data
arguments: tag: D15
type: byte[8]

tool: D24: %d

tool: D20: 1.8

tool: {"D15": [0x00,0x00,0x00,0x00,0x00,0x00,0x40,0x40]}

assistant: ```c
void main() {
  int cel;
  int far;
  scanf("%d", &cel);
  far = cel * 1.8 + 32;
  printf("%d", far);
}
```