ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<Malloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L7
L13:
  mov    %rbx,%rax
  pop    %rbx
  ret
L7:
  lea    D4(%rip),%rdi
  call   <unix_error@plt>
  jmp    L13
```

tool_calls: name: unpack_data
arguments: data_tag: D4
datatype: string

tool: D4: malloc failed

assistant: ```c
void *Malloc(size_t size) {
  void *p;
  if ((p = malloc(size)) == NULL) {
    unix_error("malloc failed");
  }
  return (p);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Malloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L17
L14:
  mov    %r12,%rax
  pop    %r12
  ret
L17:
  lea    D2(%rip),%rdi
  call   <unix_error@plt>
  jmp    L14
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
data_type: string

tool: D2: malloc failed

assistant: ```c
void *Malloc(size_t size) {
  void *p;
  if ((p = malloc(size)) == NULL) {
    unix_error("malloc failed");
  }
  return (p);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Malloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L7
L15:
  mov    %r12,%rax
  pop    %r12
  ret
L7:
  lea    D1(%rip),%rdi
  call   <unix_error@plt>
  jmp    L15
```

tool_calls: name: retrieve_data
arguments: label: D1
type: string

tool: D1: malloc failed

assistant: ```c
void *Malloc(size_t size) {
  void *p;
  if ((p = malloc(size)) == NULL) {
    unix_error("malloc failed");
  }
  return (p);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<insertionSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L24
L33:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L49
L11:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x8(%rbp)
L49:
  cmpl   $0x0,-0x8(%rbp)
  js     L36
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L11
L36:
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L24:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L33
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void insertionSort(int arr[], int n) {
  int i, key, j;
  for (i = 1; i < n; i++) {
    key = arr[i];
    j = i - 1;
    while (j >= 0 && arr[j] > key) {
      arr[j + 1] = arr[j];
      j = j - 1;
    }
    arr[j + 1] = key;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<insertionSort>:
  endbr64
  cmp    $0x1,%esi
  jle    L1
  mov    %rdi,%r9
  lea    -0x1(%rsi),%r10d
  mov    $0x0,%r8d
  jmp    L29
L39:
  movslq %edx,%rdx
  mov    %esi,0x4(%rdi,%rdx,4)
  add    $0x1,%r8
  add    $0x4,%r9
  cmp    %r10,%r8
  je     L1
L29:
  mov    0x4(%r9),%esi
  mov    %r8d,%edx
  mov    %r9,%rax
  test   %r8d,%r8d
  js     L39
L64:
  mov    (%rax),%ecx
  cmp    %esi,%ecx
  jle    L39
  mov    %ecx,0x4(%rax)
  sub    $0x1,%edx
  sub    $0x4,%rax
  cmp    $0xffffffff,%edx
  jne    L64
  jmp    L39
L1:
  ret
```

assistant: ```c
void insertionSort(int arr[], int n) {
  int i, key, j;
  for (i = 1; i < n; i++) {
    key = arr[i];
    j = i - 1;
    while (j >= 0 && arr[j] > key) {
      arr[j + 1] = arr[j];
      j = j - 1;
    }
    arr[j + 1] = key;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<insertionSort>:
  endbr64
  lea    -0x1(%rsi),%r9d
  xor    %r8d,%r8d
  cmp    $0x1,%esi
  jle    L19
L17:
  mov    0x4(%rdi,%r8,4),%esi
  mov    %r8,%rax
  nopl   0x0(%rax,%rax,1)
L5:
  mov    (%rdi,%rax,4),%edx
  mov    %eax,%ecx
  cmp    %esi,%edx
  jle    L3
  mov    %edx,0x4(%rdi,%rax,4)
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L5
  mov    %rdi,%rax
L29:
  add    $0x1,%r8
  mov    %esi,(%rax)
  cmp    %r8,%r9
  jne    L17
L19:
  ret
  nopl   (%rax)
L3:
  add    $0x1,%ecx
  movslq %ecx,%rcx
  lea    (%rdi,%rcx,4),%rax
  jmp    L29
```

assistant: ```c
void insertionSort(int arr[], int n) {
  int i, key, j;
  for (i = 1; i < n; i++) {
    key = arr[i];
    j = i - 1;
    while (j >= 0 && arr[j] > key) {
      arr[j + 1] = arr[j];
      j = j - 1;
    }
    arr[j + 1] = key;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<insertionSort>:
  endbr64
  lea    -0x1(%rsi),%r9d
  xor    %r8d,%r8d
  cmp    $0x1,%esi
  jle    L44
L36:
  mov    0x4(%rdi,%r8,4),%esi
  mov    %r8,%rax
  nopl   0x0(%rax,%rax,1)
L32:
  mov    (%rdi,%rax,4),%edx
  mov    %eax,%ecx
  cmp    %esi,%edx
  jle    L14
  mov    %edx,0x4(%rdi,%rax,4)
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L32
  mov    %rdi,%rax
L35:
  add    $0x1,%r8
  mov    %esi,(%rax)
  cmp    %r8,%r9
  jne    L36
L44:
  ret
  nopl   (%rax)
L14:
  add    $0x1,%ecx
  movslq %ecx,%rcx
  lea    (%rdi,%rcx,4),%rax
  jmp    L35
```

assistant: ```c
void insertionSort(int arr[], int n) {
  int i, key, j;
  for (i = 1; i < n; i++) {
    key = arr[i];
    j = i - 1;
    while (j >= 0 && arr[j] > key) {
      arr[j + 1] = arr[j];
      j = j - 1;
    }
    arr[j + 1] = key;
  }
}
```
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    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D23(%rip),%rax
  mov    %rax,-0x68(%rbp)
  movabs $0x6c6c6120726f6620,%rax
  movabs $0x727020646f6f6720,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x72656d6d6172676f,%rax
  mov    %rax,-0x20(%rbp)
  movw   $0x73,-0x18(%rbp)
  mov    -0x68(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <copy1@plt>
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <copy2@plt>
  lea    -0x50(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D23
datatype: string
name: disassemble_data
arguments: data_label: D27
datatype: string
name: disassemble_data
arguments: data_label: D13
datatype: string

tool: D23: Now is the time

tool: D27: s1:  %s


tool: D13: s3:  %s


assistant: ```c
int main() {
  char s1[16];
  char *s2 = "Now is the time";
  char s3[25];
  char s4[] = " for all good programmers";
  copy1(s1, s2);
  printf("s1:  %s\n", s1);
  copy2(s3, s4);
  printf("s3:  %s\n", s3);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movabs $0x6c6c6120726f6620,%rax
  movabs $0x727020646f6f6720,%rdx
  mov    %rax,0x30(%rsp)
  mov    %rdx,0x38(%rsp)
  movabs $0x72656d6d6172676f,%rax
  mov    %rax,0x40(%rsp)
  movw   $0x73,0x48(%rsp)
  mov    %rsp,%rbx
  lea    D25(%rip),%rsi
  mov    %rbx,%rdi
  call   <copy1@plt>
  mov    %rbx,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  call   <copy2@plt>
  mov    %rbx,%rsi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x1,%eax
  add    $0x60,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D25
type_infered: string
name: parse_data
arguments: data_label: D1
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string

tool: D25: Now is the time

tool: D1: s1:  %s


tool: D10: s3:  %s


assistant: ```c
int main() {
  char s1[16];
  char *s2 = "Now is the time";
  char s3[25];
  char s4[] = " for all good programmers";
  copy1(s1, s2);
  printf("s1:  %s\n", s1);
  copy2(s3, s4);
  printf("s3:  %s\n", s3);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D12(%rip),%rsi
  sub    $0x60,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movabs $0x72656d6d6172676f,%rax
  mov    %rbp,%rdi
  mov    %rax,0x40(%rsp)
  mov    $0x73,%eax
  movaps %xmm0,0x30(%rsp)
  mov    %ax,0x48(%rsp)
  call   <copy1@plt>
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  lea    0x10(%rsp),%rbp
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  mov    %rbp,%rdi
  call   <copy2@plt>
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x60,%rsp
  mov    $0x1,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: byte[16]
name: extract_symbolic_data
arguments: data_tag: D20
data_type: string
name: extract_symbolic_data
arguments: data_tag: D3
data_type: string

tool: D12: Now is the time

tool: {"D1": [0x20,0x66,0x6f,0x72,0x20,0x61,0x6c,0x6c,0x20,0x67,0x6f,0x6f,0x64,0x20,0x70,0x72]}

tool: D20: s1:  %s


tool: D3: s3:  %s


assistant: ```c
int main() {
  char s1[16];
  char *s2 = "Now is the time";
  char s3[25];
  char s4[] = " for all good programmers";
  copy1(s1, s2);
  printf("s1:  %s\n", s1);
  copy2(s3, s4);
  printf("s3:  %s\n", s3);
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D35(%rip),%rsi
  sub    $0x60,%rsp
  movdqa D34(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movabs $0x72656d6d6172676f,%rax
  mov    %rbp,%rdi
  mov    %rax,0x40(%rsp)
  mov    $0x73,%eax
  movaps %xmm0,0x30(%rsp)
  mov    %ax,0x48(%rsp)
  call   <copy1@plt>
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  lea    0x10(%rsp),%rbp
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  mov    %rbp,%rdi
  call   <copy2@plt>
  mov    %rbp,%rsi
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x60,%rsp
  mov    $0x1,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D35
type: string
name: unpack_data
arguments: data_tag: D34
type: byte[16]
name: unpack_data
arguments: data_tag: D9
type: string
name: unpack_data
arguments: data_tag: D38
type: string

tool: D35: Now is the time

tool: {"D34": [0x20,0x66,0x6f,0x72,0x20,0x61,0x6c,0x6c,0x20,0x67,0x6f,0x6f,0x64,0x20,0x70,0x72]}

tool: D9: s1:  %s


tool: D38: s3:  %s


assistant: ```c
int main() {
  char s1[16];
  char *s2 = "Now is the time";
  char s3[25];
  char s4[] = " for all good programmers";
  copy1(s1, s2);
  printf("s1:  %s\n", s1);
  copy2(s3, s4);
  printf("s3:  %s\n", s3);
  return 1;
}
```
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   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <report_start@plt>
  mov    $0x0,%eax
  call   <base_test@plt>
  mov    $0x0,%eax
  call   <report_end@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
main() {
  report_start();
  base_test();
  report_end();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <report_start@plt>
  mov    $0x0,%eax
  call   <base_test@plt>
  mov    $0x0,%eax
  call   <report_end@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  report_start();
  base_test();
  report_end();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <report_start@plt>
  xor    %eax,%eax
  call   <base_test@plt>
  xor    %eax,%eax
  call   <report_end@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  report_start();
  base_test();
  report_end();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <report_start@plt>
  xor    %eax,%eax
  call   <base_test@plt>
  xor    %eax,%eax
  call   <report_end@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  report_start();
  base_test();
  report_end();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  and    $0x3,%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  int *ptr = (int *)Mem_Alloc(sizeof(int));
  assert(ptr != NULL);
  assert((int)ptr % 4 == 0);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x4,%edi
  call   <Mem_Alloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  test   $0x3,%bl
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  int *ptr = (int *)Mem_Alloc(sizeof(int));
  assert(ptr != NULL);
  assert((int)ptr % 4 == 0);
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0x4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbx
  setne  %dil
  call   <assert@plt>
  xor    %edi,%edi
  and    $0x3,%ebx
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  int *ptr = (int *)Mem_Alloc(sizeof(int));
  assert(ptr != NULL);
  assert((int)ptr % 4 == 0);
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1000,%edi
  call   <Mem_Init@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    $0x4,%edi
  call   <Mem_Alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbx
  setne  %dil
  call   <assert@plt>
  xor    %edi,%edi
  and    $0x3,%ebx
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  assert(Mem_Init(4096) == 0);
  int *ptr = (int *)Mem_Alloc(sizeof(int));
  assert(ptr != NULL);
  assert((int)ptr % 4 == 0);
  exit(0);
}
```
user: What is the c source code of the assembly code below:

```asm
<get_damage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  jmp    L48
L27:
  movslq %ebx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <get_number@plt>
  mov    %eax,%r14d
  nop
L28:
  mov    %ebx,%eax
  lea    0x1(%rax),%ebx
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x64,%al
  jne    L28
  movslq %ebx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <get_number@plt>
  mov    %eax,%r15d
  jmp    L26
L19:
  add    $0x1,%ebx
L26:
  movslq %ebx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L77
  movslq %ebx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
L77:
  mov    $0x0,%r13d
  jmp    L82
L20:
  cmpl   $0x0,-0x3c(%rbp)
  je     L43
  mov    %r15d,%esi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <get_rand@plt>
  add    %eax,%r12d
  jmp    L102
L43:
  add    %r15d,%r12d
L102:
  add    $0x1,%r13d
L82:
  cmp    %r14d,%r13d
  jl     L20
  movslq %ebx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L48
  add    $0x1,%ebx
L48:
  movslq %ebx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L27
  mov    %r12d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
int get_damage(ds, r)
char *ds;
int r;
{
  register int i = 0, j, n, d, total = 0;
  while (ds[i]) {
    n = get_number(ds + i);
    while (ds[i++] != 'd')
      ;
    d = get_number(ds + i);
    while ((ds[i] != '/') && ds[i])
      i++;
    for (j = 0; j < n; j++) {
      if (r) {
        total += get_rand(1, d);
      } else {
        total += d;
      }
    }
    if (ds[i] == '/') {
      i++;
    }
  }
  return (total);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_damage>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  cmpb   $0x0,(%rdi)
  je     L43
  mov    %rdi,%r13
  mov    %esi,%r15d
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  jmp    L98
L41:
  add    0x1c(%rsp),%r12d
L46:
  add    $0x1,%ebx
  cmp    %ebx,%r14d
  je     L81
L74:
  test   %r15d,%r15d
  je     L41
  mov    0x1c(%rsp),%esi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <get_rand@plt>
  add    %eax,%r12d
  jmp    L46
L81:
  mov    0x8(%rsp),%rax
  cmpb   $0x2f,(%rax)
  sete   %al
  movzbl %al,%eax
  add    %eax,%ebp
  movslq %ebp,%rdi
  add    %r13,%rdi
  cmpb   $0x0,(%rdi)
  je     L35
L98:
  mov    $0x0,%eax
  call   <get_number@plt>
  mov    %eax,%r14d
  lea    0x1(%rbp),%eax
  cltq
L62:
  mov    %eax,%ebp
  mov    %rax,0x10(%rsp)
  add    $0x1,%rax
  cmpb   $0x64,-0x2(%r13,%rax,1)
  jne    L62
  movslq %ebp,%rbx
  lea    0x0(%r13,%rbx,1),%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <get_number@plt>
  mov    %eax,0x1c(%rsp)
  mov    0x8(%rsp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L7
  test   %al,%al
  je     L7
  lea    0x1(%r13,%rbx,1),%rdx
  mov    %rdx,%rax
  mov    0x10(%rsp),%ecx
  add    $0x1,%ecx
  sub    %edx,%ecx
L27:
  lea    (%rcx,%rax,1),%ebp
  mov    %rax,0x8(%rsp)
  movzbl (%rax),%edx
  add    $0x1,%rax
  cmp    $0x2f,%dl
  je     L7
  test   %dl,%dl
  jne    L27
L7:
  test   %r14d,%r14d
  jle    L81
  mov    $0x0,%ebx
  jmp    L74
L43:
  mov    $0x0,%r12d
L35:
  mov    %r12d,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int get_damage(ds, r)
char *ds;
int r;
{
  register int i = 0, j, n, d, total = 0;
  while (ds[i]) {
    n = get_number(ds + i);
    while (ds[i++] != 'd')
      ;
    d = get_number(ds + i);
    while ((ds[i] != '/') && ds[i])
      i++;
    for (j = 0; j < n; j++) {
      if (r) {
        total += get_rand(1, d);
      } else {
        total += d;
      }
    }
    if (ds[i] == '/') {
      i++;
    }
  }
  return (total);
}
```
user: What is the c source code of the assembly code below?

```asm
<get_damage>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  cmpb   $0x0,(%rdi)
  je     L0
  mov    %rdi,%r14
  mov    %esi,%ebp
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L23:
  xor    %eax,%eax
  call   <get_number@plt>
  mov    %eax,%ecx
  lea    0x1(%r12),%eax
  cltq
L54:
  mov    %eax,%ebx
  mov    %eax,%r12d
  add    $0x1,%rax
  cmpb   $0x64,-0x2(%r14,%rax,1)
  jne    L54
  movslq %ebx,%rdx
  xor    %eax,%eax
  mov    %ecx,0x4(%rsp)
  lea    (%r14,%rdx,1),%r15
  mov    %rdx,0x8(%rsp)
  mov    %r15,%rdi
  call   <get_number@plt>
  mov    0x4(%rsp),%ecx
  mov    %eax,%esi
  movzbl (%r15),%eax
  cmp    $0x2f,%al
  je     L2
  test   %al,%al
  je     L2
  mov    0x8(%rsp),%rdx
  lea    0x1(%rbx),%edi
  lea    0x1(%r14,%rdx,1),%rax
  sub    %eax,%edi
  nopl   0x0(%rax)
L103:
  movzbl (%rax),%edx
  lea    (%rdi,%rax,1),%r12d
  mov    %rax,%r15
  add    $0x1,%rax
  cmp    $0x2f,%dl
  je     L2
  test   %dl,%dl
  jne    L103
L2:
  test   %ecx,%ecx
  jle    L120
  xor    %ebx,%ebx
  jmp    L53
  nop
L63:
  add    $0x1,%ebx
  add    %esi,%r13d
  cmp    %ebx,%ecx
  je     L120
L53:
  test   %ebp,%ebp
  je     L63
  mov    $0x1,%edi
  xor    %eax,%eax
  mov    %ecx,0x8(%rsp)
  add    $0x1,%ebx
  mov    %esi,0x4(%rsp)
  call   <get_rand@plt>
  mov    0x8(%rsp),%ecx
  mov    0x4(%rsp),%esi
  add    %eax,%r13d
  cmp    %ebx,%ecx
  jne    L53
L120:
  xor    %eax,%eax
  cmpb   $0x2f,(%r15)
  sete   %al
  add    %eax,%r12d
  movslq %r12d,%rdi
  add    %r14,%rdi
  cmpb   $0x0,(%rdi)
  jne    L23
L9:
  add    $0x18,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  xor    %r13d,%r13d
  jmp    L9
```

assistant: ```c
int get_damage(ds, r)
char *ds;
int r;
{
  register int i = 0, j, n, d, total = 0;
  while (ds[i]) {
    n = get_number(ds + i);
    while (ds[i++] != 'd')
      ;
    d = get_number(ds + i);
    while ((ds[i] != '/') && ds[i])
      i++;
    for (j = 0; j < n; j++) {
      if (r) {
        total += get_rand(1, d);
      } else {
        total += d;
      }
    }
    if (ds[i] == '/') {
      i++;
    }
  }
  return (total);
}
```