ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<comb_sort>:
  endbr64
  pxor   %xmm0,%xmm0
  movss  D4(%rip),%xmm1
  push   %r12
  mov    %edi,%r11d
  cvtsi2ss %edi,%xmm0
  push   %rbp
  lea    0x4(%rsi),%rbp
  push   %rbx
  mov    %rsi,%rbx
  divss  %xmm1,%xmm0
  cvttss2si %xmm0,%r10d
  nopl   0x0(%rax,%rax,1)
L78:
  test   %r10d,%r10d
  mov    $0x1,%r9d
  mov    %r11d,%edx
  cmovg  %r10d,%r9d
  sub    %r9d,%edx
  test   %edx,%edx
  jle    L73
  sub    $0x1,%edx
  mov    %rbx,%rax
  xor    %edi,%edi
  movslq %r9d,%rsi
  lea    0x0(%rbp,%rdx,4),%r8
  nopw   0x0(%rax,%rax,1)
L80:
  mov    (%rax),%edx
  mov    (%rax,%rsi,4),%ecx
  cmp    %ecx,%edx
  jle    L87
  mov    %ecx,(%rax)
  add    $0x1,%edi
  mov    %edx,(%rax,%rsi,4)
L87:
  add    $0x4,%rax
  cmp    %rax,%r8
  jne    L80
  test   %edi,%edi
  sete   %al
L2:
  cmp    $0x1,%r10d
  jg     L11
  test   %al,%al
  jne    L13
L11:
  pxor   %xmm0,%xmm0
  cvtsi2ss %r9d,%xmm0
  divss  %xmm1,%xmm0
  cvttss2si %xmm0,%r10d
  jmp    L78
  nopl   (%rax)
L73:
  mov    $0x1,%eax
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L13:
  test   %r11d,%r11d
  jle    L68
  lea    -0x1(%r11),%eax
  lea    D1(%rip),%rbp
  lea    0x4(%rbx,%rax,4),%r12
  nopl   (%rax)
L30:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L30
L68:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type_guessed: f32
name: unpack_data
arguments: data_tag: D1
type_guessed: string

tool: D4: 1.3

tool: D1: %d 

assistant: ```c
void comb_sort(int N, int arr[]) {
  float shrink = 1.3;
  int gap, t, c = 0;
  gap = N / shrink;
  while (1) {
    c = 0;
    if (gap < 1)
      gap = 1;
    for (int i = 0; i < N - gap; i++) {
      if (arr[i] > arr[i + gap]) {
        t = arr[i];
        arr[i] = arr[i + gap];
        arr[i + gap] = t;
        c++;
      }
    }
    if (c == 0 && gap == 1)
      break;
    gap = gap / shrink;
  }
  for (int i = 0; i < N; i++)
    printf("%d ", arr[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x2,%edi
  call   <square@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

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

tool: D7: %d


assistant: ```c
int main(void) {
  printf("%d\n", square(2));
  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
  mov    $0x2,%edi
  call   <square@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: %d


assistant: ```c
int main(void) {
  printf("%d\n", square(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
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <square@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D1
datatype: string

tool: D1: %d


assistant: ```c
int main(void) {
  printf("%d\n", square(2));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <square@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
datatype: string

tool: D9: %d


assistant: ```c
int main(void) {
  printf("%d\n", square(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
<Input>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L31
L26:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  sub    $0x30,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x8(%rbp)
L31:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jne    L26
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  movl   $0x0,-0x8(%rbp)
  jmp    L18
L15:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  sub    $0x30,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x8(%rbp)
L18:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jne    L15
  mov    -0x30(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void Input(int *m1, int *m2, int *n, int *m) {
  int i = 0;
  int c;
  while ((c = getchar()) != '\n') {
    m1[i] = c - '0';
    i++;
  }
  *n = i;
  i = 0;
  while ((c = getchar()) != '\n') {
    m2[i] = c - '0';
    i++;
  }
  *m = i;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Input>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%r15
  mov    %rcx,%r12
  mov    $0x0,%ebx
  jmp    L10
L2:
  sub    $0x30,%eax
  mov    %eax,0x0(%r13,%rbx,4)
  add    $0x1,%rbx
L10:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L2
  mov    %ebx,(%r15)
  mov    $0x0,%ebx
  jmp    L22
L27:
  sub    $0x30,%eax
  mov    %eax,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
L22:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L27
  mov    %ebx,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void Input(int *m1, int *m2, int *n, int *m) {
  int i = 0;
  int c;
  while ((c = getchar()) != '\n') {
    m1[i] = c - '0';
    i++;
  }
  *n = i;
  i = 0;
  while ((c = getchar()) != '\n') {
    m2[i] = c - '0';
    i++;
  }
  *m = i;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Input>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L32
  nopl   (%rax)
L1:
  sub    $0x30,%eax
  mov    %eax,(%r12,%r14,4)
  add    $0x1,%r14
L32:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L1
  mov    %r14d,0x0(%r13)
  xor    %r12d,%r12d
  jmp    L27
L37:
  sub    $0x30,%eax
  mov    %eax,(%rbx,%r12,4)
  add    $0x1,%r12
L27:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L37
  pop    %rbx
  mov    %r12d,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void Input(int *m1, int *m2, int *n, int *m) {
  int i = 0;
  int c;
  while ((c = getchar()) != '\n') {
    m1[i] = c - '0';
    i++;
  }
  *n = i;
  i = 0;
  while ((c = getchar()) != '\n') {
    m2[i] = c - '0';
    i++;
  }
  *m = i;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Input>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L18
  nopl   (%rax)
L1:
  sub    $0x30,%eax
  mov    %eax,(%r12,%r14,4)
  add    $0x1,%r14
L18:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L1
  mov    %r14d,0x0(%r13)
  xor    %r12d,%r12d
  jmp    L12
L7:
  sub    $0x30,%eax
  mov    %eax,(%rbx,%r12,4)
  add    $0x1,%r12
L12:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L7
  pop    %rbx
  mov    %r12d,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void Input(int *m1, int *m2, int *n, int *m) {
  int i = 0;
  int c;
  while ((c = getchar()) != '\n') {
    m1[i] = c - '0';
    i++;
  }
  *n = i;
  i = 0;
  while ((c = getchar()) != '\n') {
    m2[i] = c - '0';
    i++;
  }
  *m = i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  sub    $0xf0,%rsp
  mov    %edi,-0xf4(%rbp)
  mov    %rsi,-0xc8(%rbp)
  mov    %rdx,-0xc0(%rbp)
  mov    %rcx,-0xb8(%rbp)
  mov    %r8,-0xb0(%rbp)
  mov    %r9,-0xa8(%rbp)
  test   %al,%al
  je     L25
  movaps %xmm0,-0xa0(%rbp)
  movaps %xmm1,-0x90(%rbp)
  movaps %xmm2,-0x80(%rbp)
  movaps %xmm3,-0x70(%rbp)
  movaps %xmm4,-0x60(%rbp)
  movaps %xmm5,-0x50(%rbp)
  movaps %xmm6,-0x40(%rbp)
  movaps %xmm7,-0x30(%rbp)
L25:
  mov    %fs:0x28,%rax
  mov    %rax,-0xd8(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rsi
  mov    -0xf4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,-0x110(%rbp)
  movq   $0x0,-0x108(%rbp)
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0xf0(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  shl    $0x3,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  movslq %eax,%rdx
  shl    $0x3,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L23:
  cmp    %rdx,%rsp
  je     L31
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L23
L31:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L41
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L41:
  mov    %rsp,%rax
  add    $0x7,%rax
  shr    $0x3,%rax
  shl    $0x3,%rax
  mov    %rax,-0xe8(%rbp)
  mov    -0xe8(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0xe0(%rbp)
  mov    %rsi,%rsp
  nop
  mov    -0xd8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  lea    -0x20(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void test10(int n, ...) {
  struct S {
    double a[n];
  } s;
  double x = s.a[0];
}
```
user: What is the c source code of the assembly code below?

```asm
<test10>:
  endbr64
  ret
```

assistant: ```c
void test10(int n, ...) {
  struct S {
    double a[n];
  } s;
  double x = s.a[0];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test10>:
  endbr64
  ret
```

assistant: ```c
void test10(int n, ...) {
  struct S {
    double a[n];
  } s;
  double x = s.a[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
<test10>:
  endbr64
  ret
```

assistant: ```c
void test10(int n, ...) {
  struct S {
    double a[n];
  } s;
  double x = s.a[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
<board_aopen_i975xa_ydg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x1,%esi
  mov    $0x21,%edi
  call   <intel_ich_gpio_set@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int board_aopen_i975xa_ydg(void) {
  int ret;
  ret = intel_ich_gpio_set(33, 1);
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<board_aopen_i975xa_ydg>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%esi
  mov    $0x21,%edi
  call   <intel_ich_gpio_set@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int board_aopen_i975xa_ydg(void) {
  int ret;
  ret = intel_ich_gpio_set(33, 1);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<board_aopen_i975xa_ydg>:
  endbr64
  mov    $0x1,%esi
  mov    $0x21,%edi
  jmp    1050 <intel_ich_gpio_set@plt>
```

assistant: ```c
int board_aopen_i975xa_ydg(void) {
  int ret;
  ret = intel_ich_gpio_set(33, 1);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<board_aopen_i975xa_ydg>:
  endbr64
  mov    $0x1,%esi
  mov    $0x21,%edi
  jmp    1050 <intel_ich_gpio_set@plt>
```

assistant: ```c
int board_aopen_i975xa_ydg(void) {
  int ret;
  ret = intel_ich_gpio_set(33, 1);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x400,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3f4(%rbp)
  jmp    L17
L1:
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse@plt>
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L17:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getline@plt>
  mov    %eax,-0x3f4(%rbp)
  cmpl   $0x0,-0x3f4(%rbp)
  jg     L1
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
type_guessed: string

tool: D7: %s


assistant: ```c
int main(void) {
  int len = 0;
  char inputLine[1000];
  while ((len = getline(inputLine, 1000)) > 0) {
    reverse(inputLine);
    printf("%s\n", inputLine);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rbp
  jmp    L16
L1:
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L16:
  mov    %rsp,%rdi
  mov    $0x3e8,%esi
  call   <getline@plt>
  test   %eax,%eax
  jg     L1
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x3f8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
datatype: string

tool: D3: %s


assistant: ```c
int main(void) {
  int len = 0;
  char inputLine[1000];
  while ((len = getline(inputLine, 1000)) > 0) {
    reverse(inputLine);
    printf("%s\n", inputLine);
  }
  return 0;
}
```
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    D9(%rip),%rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L25
  nopl   0x0(%rax)
L12:
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L25:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getline@plt>
  test   %eax,%eax
  jg     L12
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D9
data_type: string

tool: D9: %s


assistant: ```c
int main(void) {
  int len = 0;
  char inputLine[1000];
  while ((len = getline(inputLine, 1000)) > 0) {
    reverse(inputLine);
    printf("%s\n", inputLine);
  }
  return 0;
}
```
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    D9(%rip),%rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L28
  nopl   0x0(%rax)
L2:
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L28:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getline@plt>
  test   %eax,%eax
  jg     L2
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
data_type: string

tool: D9: %s


assistant: ```c
int main(void) {
  int len = 0;
  char inputLine[1000];
  while ((len = getline(inputLine, 1000)) > 0) {
    reverse(inputLine);
    printf("%s\n", inputLine);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ddr_debug>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movzwl %ax,%eax
  mov    %eax,%edi
  call   <SET_DEBUG_LEVEL@plt>
  mov    $0x0,%eax
  call   <SET_DEBUG_VALID@plt>
  mov    $0x0,%eax
  call   <CLR_DEBUG_VALID@plt>
  mov    -0x4(%rbp),%eax
  shr    $0x10,%eax
  mov    %eax,%edi
  call   <SET_DEBUG_LEVEL@plt>
  mov    $0x0,%eax
  call   <SET_DEBUG_VALID@plt>
  mov    $0x0,%eax
  call   <CLR_DEBUG_VALID@plt>
  nop
  leave
  ret
```

assistant: ```c
void ddr_debug(int data) {
  SET_DEBUG_LEVEL(data & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
  SET_DEBUG_LEVEL((data >> 16) & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ddr_debug>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  movzwl %di,%edi
  call   <SET_DEBUG_LEVEL@plt>
  mov    $0x0,%eax
  call   <SET_DEBUG_VALID@plt>
  mov    $0x0,%eax
  call   <CLR_DEBUG_VALID@plt>
  shr    $0x10,%ebx
  mov    %ebx,%edi
  call   <SET_DEBUG_LEVEL@plt>
  mov    $0x0,%eax
  call   <SET_DEBUG_VALID@plt>
  mov    $0x0,%eax
  call   <CLR_DEBUG_VALID@plt>
  pop    %rbx
  ret
```

assistant: ```c
void ddr_debug(int data) {
  SET_DEBUG_LEVEL(data & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
  SET_DEBUG_LEVEL((data >> 16) & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
}
```
user: What is the c source code of the assembly code below?

```asm
<ddr_debug>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  movzwl %di,%edi
  shr    $0x10,%ebx
  call   <SET_DEBUG_LEVEL@plt>
  xor    %eax,%eax
  call   <SET_DEBUG_VALID@plt>
  xor    %eax,%eax
  call   <CLR_DEBUG_VALID@plt>
  mov    %ebx,%edi
  call   <SET_DEBUG_LEVEL@plt>
  xor    %eax,%eax
  call   <SET_DEBUG_VALID@plt>
  xor    %eax,%eax
  pop    %rbx
  jmp    1090 <CLR_DEBUG_VALID@plt>
```

assistant: ```c
void ddr_debug(int data) {
  SET_DEBUG_LEVEL(data & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
  SET_DEBUG_LEVEL((data >> 16) & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
}
```
user: What is the c source code of the assembly code below:

```asm
<ddr_debug>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  movzwl %di,%edi
  shr    $0x10,%ebx
  call   <SET_DEBUG_LEVEL@plt>
  xor    %eax,%eax
  call   <SET_DEBUG_VALID@plt>
  xor    %eax,%eax
  call   <CLR_DEBUG_VALID@plt>
  mov    %ebx,%edi
  call   <SET_DEBUG_LEVEL@plt>
  xor    %eax,%eax
  call   <SET_DEBUG_VALID@plt>
  xor    %eax,%eax
  pop    %rbx
  jmp    1090 <CLR_DEBUG_VALID@plt>
```

assistant: ```c
void ddr_debug(int data) {
  SET_DEBUG_LEVEL(data & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
  SET_DEBUG_LEVEL((data >> 16) & 0xffff);
  SET_DEBUG_VALID();
  CLR_DEBUG_VALID();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<SendCommand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x1,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <SendPacket@plt>
  test   %rax,%rax
  jns    L33
  mov    $0xffffffff,%eax
  jmp    L0
L33:
  movl   $0x23,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <SendPacket@plt>
  test   %rax,%rax
  jns    L60
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L0
L60:
  movl   $0x4,-0x1c(%rbp)
  lea    -0x1c(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <GetPacket@plt>
  test   %rax,%rax
  jns    L10
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L0
L10:
  mov    -0xc(%rbp),%eax
  cmp    $0x40,%eax
  je     L74
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L0
L74:
  mov    $0x0,%eax
L0:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
type: string
name: read_data
arguments: data_label: D8
type: string
name: read_data
arguments: data_label: D10
type: string

tool: D16: Failed to Get Status

tool: D8: Failed to Get Packet

tool: D10: Failed to get download command Return Code: %04x


assistant: ```c
int32_t SendCommand(uint8_t *pui8Command, uint8_t ui8Size) {
  uint8_t ui8Status;
  if (SendPacket(pui8Command, ui8Size, 1) < 0) {
    return (-1);
  }
  ui8Status = 0x23;
  if (SendPacket(&ui8Status, 1, 1) < 0) {
    printf("Failed to Get Status\n");
    return (-1);
  }
  ui8Size = sizeof(ui8Status);
  if (GetPacket(&ui8Status, &ui8Size) < 0) {
    printf("Failed to Get Packet\n");
    return (-1);
  }
  if (ui8Status != 0x40) {
    printf("Failed to get download command Return Code: %04x\n", ui8Status);
    return (-1);
  }
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<SendCommand>:
  endbr64
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  mov    %fs:0x28,%rdx
  mov    %rdx,0x18(%rsp)
  xor    %edx,%edx
  mov    $0x1,%edx
  call   <SendPacket@plt>
  test   %rax,%rax
  js     L42
  movl   $0x23,0x14(%rsp)
  lea    0x14(%rsp),%rdi
  mov    $0x1,%edx
  mov    $0x1,%esi
  call   <SendPacket@plt>
  test   %rax,%rax
  js     L53
  movl   $0x4,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rdi
  call   <GetPacket@plt>
  test   %rax,%rax
  js     L47
  mov    0x14(%rsp),%esi
  cmp    $0x40,%esi
  jne    L6
  mov    $0x0,%eax
L57:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x28,%rsp
  ret
L53:
  lea    D25(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L57
L47:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L57
L6:
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L57
L42:
  mov    $0xffffffff,%eax
  jmp    L57
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D25
type_infered: string
name: unpack_data
arguments: data_label: D26
type_infered: string
name: unpack_data
arguments: data_label: D12
type_infered: string

tool: D25: Failed to Get Status

tool: D26: Failed to Get Packet

tool: D12: Failed to get download command Return Code: %04x


assistant: ```c
int32_t SendCommand(uint8_t *pui8Command, uint8_t ui8Size) {
  uint8_t ui8Status;
  if (SendPacket(pui8Command, ui8Size, 1) < 0) {
    return (-1);
  }
  ui8Status = 0x23;
  if (SendPacket(&ui8Status, 1, 1) < 0) {
    printf("Failed to Get Status\n");
    return (-1);
  }
  ui8Size = sizeof(ui8Status);
  if (GetPacket(&ui8Status, &ui8Size) < 0) {
    printf("Failed to Get Packet\n");
    return (-1);
  }
  if (ui8Status != 0x40) {
    printf("Failed to get download command Return Code: %04x\n", ui8Status);
    return (-1);
  }
  return (0);
}
```