ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<draw_head>:
  endbr64
  push   %r12
  push   %rbp
  lea    0x1(%rsi),%ebp
  imul   %edi,%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L12
  xor    %ebx,%ebx
  lea    D13(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebp,%ebx
  jne    L13
L12:
  pop    %rbx
  lea    D18(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: label: D13
type: string
name: unpack_data
arguments: label: D18
type: string

tool: D13: *

tool: D18: *


assistant: ```c
void draw_head(int c, int s) {
  for (int i = 0; i < c * (s + 1); ++i) {
    printf("*");
  }
  printf("*\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<CHECK_SCORE>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x44(%rbp)
  mov    %ecx,-0x48(%rbp)
  mov    %r8,-0x50(%rbp)
  mov    %r9,-0x58(%rbp)
  mov    $0x0,%r12d
  mov    %r12d,%ebx
  mov    %ebx,%r14d
  mov    %r14d,%r13d
  mov    %r13d,-0x28(%rbp)
  jmp    L33
L43:
  mov    -0x50(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x50(%rbp)
  mov    (%rax),%ebx
  test   %ebx,%ebx
  jne    L34
  add    $0x1,%r13d
  movslq %r13d,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  add    $0x1,%r14d
  movslq %r14d,%rcx
  mov    -0x40(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %eax,-0x28(%rbp)
  add    $0x1,%r12d
  jmp    L33
L34:
  test   %ebx,%ebx
  jle    L69
  mov    %ebx,%edx
  imul   0x18(%rbp),%edx
  mov    0x10(%rbp),%eax
  add    %edx,%eax
  neg    %eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %eax,-0x28(%rbp)
  add    %ebx,%r14d
  add    %ebx,%r12d
  jmp    L33
L69:
  mov    %ebx,%eax
  imul   0x18(%rbp),%eax
  sub    0x10(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %eax,-0x28(%rbp)
  sub    %ebx,%r13d
  sub    %ebx,%r12d
L33:
  cmp    -0x44(%rbp),%r13d
  jl     L43
  cmp    -0x48(%rbp),%r14d
  jl     L43
  mov    0x20(%rbp),%rax
  mov    %r12d,(%rax)
  mov    -0x28(%rbp),%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

assistant: ```c
int CHECK_SCORE(const unsigned char *A, const unsigned char *B, int M, int N, int *S, int **w, int qq, int rr, int *NC) {
  register int i, j, op, nc;
  int itmp, score;
  score = i = j = op = nc = 0;
  while (i < M || j < N) {
    op = *S++;
    if (op == 0) {
      itmp = w[A[++i]][B[++j]];
      score += itmp;
      nc++;
    } else if (op > 0) {
      itmp = -(qq + op * rr);
      score += itmp;
      j += op;
      nc += op;
    } else {
      itmp = -(qq - op * rr);
      score += itmp;
      i -= op;
      nc -= op;
    }
  }
  *NC = nc;
  return score;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<CHECK_SCORE>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r10d
  mov    %ecx,%r12d
  mov    0x30(%rsp),%r13d
  test   %edx,%edx
  jg     L34
  test   %ecx,%ecx
  jle    L60
L34:
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%ecx
  mov    $0x0,%edx
L5:
  add    $0x4,%r8
  mov    -0x4(%r8),%eax
  test   %eax,%eax
  jne    L89
  add    $0x1,%edx
  add    $0x1,%ecx
  movslq %edx,%rax
  movzbl (%rbx,%rax,1),%eax
  movslq %ecx,%r11
  movzbl 0x0(%rbp,%r11,1),%r11d
  mov    (%r9,%rax,8),%rax
  add    (%rax,%r11,4),%esi
  add    $0x1,%edi
L28:
  cmp    %edx,%r10d
  jg     L5
  cmp    %r12d,%ecx
  jl     L5
L88:
  mov    0x38(%rsp),%rax
  mov    %edi,(%rax)
  mov    %esi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L89:
  jle    L35
  mov    %eax,%r11d
  imul   %r13d,%r11d
  add    0x28(%rsp),%r11d
  sub    %r11d,%esi
  add    %eax,%ecx
  add    %eax,%edi
  jmp    L28
L35:
  mov    %eax,%r11d
  imul   %r13d,%r11d
  sub    0x28(%rsp),%r11d
  add    %r11d,%esi
  sub    %eax,%edx
  sub    %eax,%edi
  jmp    L28
L60:
  mov    $0x0,%esi
  mov    $0x0,%edi
  jmp    L88
```

assistant: ```c
int CHECK_SCORE(const unsigned char *A, const unsigned char *B, int M, int N, int *S, int **w, int qq, int rr, int *NC) {
  register int i, j, op, nc;
  int itmp, score;
  score = i = j = op = nc = 0;
  while (i < M || j < N) {
    op = *S++;
    if (op == 0) {
      itmp = w[A[++i]][B[++j]];
      score += itmp;
      nc++;
    } else if (op > 0) {
      itmp = -(qq + op * rr);
      score += itmp;
      j += op;
      nc += op;
    } else {
      itmp = -(qq - op * rr);
      score += itmp;
      i -= op;
      nc -= op;
    }
  }
  *NC = nc;
  return score;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CHECK_SCORE>:
  endbr64
  push   %r13
  mov    %rdi,%r11
  xor    %r10d,%r10d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  mov    0x30(%rsp),%r13d
  xor    %esi,%esi
  test   %edx,%edx
  jg     L67
  test   %ecx,%ecx
  jle    L6
L67:
  xor    %ecx,%ecx
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L27:
  mov    (%r8),%eax
  add    $0x4,%r8
  test   %eax,%eax
  jne    L34
  add    $0x1,%edx
  add    $0x1,%ecx
  add    $0x1,%esi
  movslq %edx,%rax
  movslq %ecx,%rdi
  movzbl (%r11,%rax,1),%eax
  movzbl (%rbx,%rdi,1),%edi
  mov    (%r9,%rax,8),%rax
  add    (%rax,%rdi,4),%r10d
L13:
  cmp    %ebp,%edx
  jl     L27
  cmp    %ecx,%r12d
  jg     L27
L6:
  mov    0x38(%rsp),%rax
  mov    %esi,(%rax)
  mov    %r10d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L34:
  mov    %eax,%edi
  imul   %r13d,%edi
  test   %eax,%eax
  jle    L31
  add    0x28(%rsp),%edi
  add    %eax,%ecx
  add    %eax,%esi
  sub    %edi,%r10d
  jmp    L13
  nop
L31:
  sub    0x28(%rsp),%edi
  sub    %eax,%edx
  sub    %eax,%esi
  add    %edi,%r10d
  jmp    L13
```

assistant: ```c
int CHECK_SCORE(const unsigned char *A, const unsigned char *B, int M, int N, int *S, int **w, int qq, int rr, int *NC) {
  register int i, j, op, nc;
  int itmp, score;
  score = i = j = op = nc = 0;
  while (i < M || j < N) {
    op = *S++;
    if (op == 0) {
      itmp = w[A[++i]][B[++j]];
      score += itmp;
      nc++;
    } else if (op > 0) {
      itmp = -(qq + op * rr);
      score += itmp;
      j += op;
      nc += op;
    } else {
      itmp = -(qq - op * rr);
      score += itmp;
      i -= op;
      nc -= op;
    }
  }
  *NC = nc;
  return score;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CHECK_SCORE>:
  endbr64
  push   %r13
  mov    %rdi,%r11
  xor    %r10d,%r10d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  mov    0x30(%rsp),%r13d
  xor    %esi,%esi
  test   %edx,%edx
  jg     L32
  test   %ecx,%ecx
  jle    L17
L32:
  xor    %ecx,%ecx
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L49:
  mov    (%r8),%eax
  add    $0x4,%r8
  test   %eax,%eax
  jne    L18
  add    $0x1,%edx
  add    $0x1,%ecx
  add    $0x1,%esi
  movslq %edx,%rax
  movslq %ecx,%rdi
  movzbl (%r11,%rax,1),%eax
  movzbl (%rbx,%rdi,1),%edi
  mov    (%r9,%rax,8),%rax
  add    (%rax,%rdi,4),%r10d
L25:
  cmp    %ebp,%edx
  jl     L49
  cmp    %ecx,%r12d
  jg     L49
L17:
  mov    0x38(%rsp),%rax
  mov    %esi,(%rax)
  mov    %r10d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L18:
  mov    %eax,%edi
  imul   %r13d,%edi
  test   %eax,%eax
  jle    L0
  add    0x28(%rsp),%edi
  add    %eax,%ecx
  add    %eax,%esi
  sub    %edi,%r10d
  jmp    L25
  nop
L0:
  sub    0x28(%rsp),%edi
  sub    %eax,%edx
  sub    %eax,%esi
  add    %edi,%r10d
  jmp    L25
```

assistant: ```c
int CHECK_SCORE(const unsigned char *A, const unsigned char *B, int M, int N, int *S, int **w, int qq, int rr, int *NC) {
  register int i, j, op, nc;
  int itmp, score;
  score = i = j = op = nc = 0;
  while (i < M || j < N) {
    op = *S++;
    if (op == 0) {
      itmp = w[A[++i]][B[++j]];
      score += itmp;
      nc++;
    } else if (op > 0) {
      itmp = -(qq + op * rr);
      score += itmp;
      j += op;
      nc += op;
    } else {
      itmp = -(qq - op * rr);
      score += itmp;
      i -= op;
      nc -= op;
    }
  }
  *NC = nc;
  return score;
}
```
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
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <random_cast_double@plt>
  mov    $0x0,%eax
  call   <rand_int@plt>
  mov    $0x0,%eax
  call   <count_int@plt>
  mov    $0x0,%eax
  call   <rand_int_add@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  random_cast_double();
  rand_int();
  count_int();
  rand_int_add();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <random_cast_double@plt>
  mov    $0x0,%eax
  call   <rand_int@plt>
  mov    $0x0,%eax
  call   <count_int@plt>
  mov    $0x0,%eax
  call   <rand_int_add@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  random_cast_double();
  rand_int();
  count_int();
  rand_int_add();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <random_cast_double@plt>
  xor    %eax,%eax
  call   <rand_int@plt>
  xor    %eax,%eax
  call   <count_int@plt>
  xor    %eax,%eax
  call   <rand_int_add@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  random_cast_double();
  rand_int();
  count_int();
  rand_int_add();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <random_cast_double@plt>
  xor    %eax,%eax
  call   <rand_int@plt>
  xor    %eax,%eax
  call   <count_int@plt>
  xor    %eax,%eax
  call   <rand_int_add@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  random_cast_double();
  rand_int();
  count_int();
  rand_int_add();
  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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x8(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int main() {
  char a[10];
  a[10] = 0;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  char a[10];
  a[10] = 0;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  char a[10];
  a[10] = 0;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  char a[10];
  a[10] = 0;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D28(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D19(%rip)
  mov    D19(%rip),%eax
  pop    %rbp
  ret
```

assistant: ```c
int foo(void) {
  static int foo = 0;
  __asm__ __volatile__("" : : : "memory");
  foo++;
  return foo;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  mov    D18(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D5(%rip)
  ret
```

assistant: ```c
int foo(void) {
  static int foo = 0;
  __asm__ __volatile__("" : : : "memory");
  foo++;
  return foo;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo>:
  endbr64
  mov    D14(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D7(%rip)
  ret
```

assistant: ```c
int foo(void) {
  static int foo = 0;
  __asm__ __volatile__("" : : : "memory");
  foo++;
  return foo;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<foo>:
  endbr64
  mov    D1(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D18(%rip)
  ret
```

assistant: ```c
int foo(void) {
  static int foo = 0;
  __asm__ __volatile__("" : : : "memory");
  foo++;
  return foo;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<EMACTxEnable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0xc18,%eax
  cltq
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  add    $0xc18,%edx
  movslq %edx,%rdx
  or     $0x20,%ah
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  or     $0x8,%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void EMACTxEnable(uint32_t ui32Base) {
  (*((volatile uint32_t *)(ui32Base + 0x00000C18))) |= 0x00002000;
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000008;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<EMACTxEnable>:
  endbr64
  lea    0xc18(%rdi),%eax
  cltq
  mov    (%rax),%edx
  or     $0x20,%dh
  mov    %edx,(%rax)
  movslq %edi,%rdi
  mov    (%rdi),%eax
  or     $0x8,%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void EMACTxEnable(uint32_t ui32Base) {
  (*((volatile uint32_t *)(ui32Base + 0x00000C18))) |= 0x00002000;
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000008;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<EMACTxEnable>:
  endbr64
  lea    0xc18(%rdi),%eax
  movslq %edi,%rdi
  cltq
  mov    (%rax),%edx
  or     $0x20,%dh
  mov    %edx,(%rax)
  mov    (%rdi),%eax
  or     $0x8,%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void EMACTxEnable(uint32_t ui32Base) {
  (*((volatile uint32_t *)(ui32Base + 0x00000C18))) |= 0x00002000;
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000008;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<EMACTxEnable>:
  endbr64
  lea    0xc18(%rdi),%eax
  movslq %edi,%rdi
  cltq
  mov    (%rax),%edx
  or     $0x20,%dh
  mov    %edx,(%rax)
  mov    (%rdi),%eax
  or     $0x8,%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void EMACTxEnable(uint32_t ui32Base) {
  (*((volatile uint32_t *)(ui32Base + 0x00000C18))) |= 0x00002000;
  (*((volatile uint32_t *)(ui32Base + 0x00000000))) |= 0x00000008;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D16(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strleng@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D16
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D12
type_guessed: string

tool: D16: aa


tool: D12: %d

assistant: ```c
int main(void) {
  char *string = "aa\n";
  printf("%d", strleng(string));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rdi
  call   <strleng@plt>
  mov    %eax,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D14
data_type: string
name: disassemble_data
arguments: tag: D11
data_type: string

tool: D14: aa


tool: D11: %d

assistant: ```c
int main(void) {
  char *string = "aa\n";
  printf("%d", strleng(string));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <strleng@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type_infered: string
name: read_data
arguments: tag: D0
type_infered: string

tool: D3: aa


tool: D0: %d

assistant: ```c
int main(void) {
  char *string = "aa\n";
  printf("%d", strleng(string));
}
```
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
  lea    D18(%rip),%rdi
  call   <strleng@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D18
type_infered: string
name: retrieve_data
arguments: tag: D7
type_infered: string

tool: D18: aa


tool: D7: %d

assistant: ```c
int main(void) {
  char *string = "aa\n";
  printf("%d", strleng(string));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x78(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x74(%rbp)
  jmp    L8
L21:
  lea    -0x70(%rbp),%rdx
  mov    -0x74(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x74(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x74(%rbp)
L8:
  mov    -0x78(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x74(%rbp)
  jl     L21
  mov    -0x78(%rbp),%eax
  lea    -0x70(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <str_rev@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D29
type_guessed: string
name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D29: %d


tool: D4: %c

assistant: ```c
int main() {
  char str[100];
  int n, i;
  scanf("%d\n", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%c", &str[i]);
    printf("%c", str[i]);
  }
  str_rev(n, str);
  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   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  cmp    $0x1,%edi
  jle    L7
  lea    0x10(%rsp),%rbx
  mov    $0x0,%ebp
  lea    D6(%rip),%r12
L26:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  mov    0xc(%rsp),%edi
  add    $0x1,%rbx
  lea    -0x1(%rdi),%eax
  cmp    %ebp,%eax
  jg     L26
L7:
  lea    0x10(%rsp),%rsi
  call   <str_rev@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D18
datatype: string
name: parse_data
arguments: label: D6
datatype: string

tool: D18: %d


tool: D6: %c

assistant: ```c
int main() {
  char str[100];
  int n, i;
  scanf("%d\n", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%c", &str[i]);
    printf("%c", str[i]);
  }
  str_rev(n, str);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D15(%rip),%rdi
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r13
  mov    %r13,%rbx
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  cmp    $0x1,%edi
  jle    L24
  nopl   0x0(%rax)
L25:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <scanf@plt>
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0xc(%rsp),%edi
  lea    -0x1(%rdi),%eax
  cmp    %ebp,%eax
  jg     L25
L24:
  mov    %r13,%rsi
  call   <str_rev@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D15
type_guessed: string
name: retrieve_data
arguments: data_label: D0
type_guessed: string

tool: D15: %d


tool: D0: %c

assistant: ```c
int main() {
  char str[100];
  int n, i;
  scanf("%d\n", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%c", &str[i]);
    printf("%c", str[i]);
  }
  str_rev(n, str);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D5(%rip),%rdi
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r13
  mov    %r13,%rbx
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  cmp    $0x1,%edi
  jle    L21
  nopl   0x0(%rax)
L25:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <scanf@plt>
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0xc(%rsp),%edi
  lea    -0x1(%rdi),%eax
  cmp    %ebp,%eax
  jg     L25
L21:
  mov    %r13,%rsi
  call   <str_rev@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_guessed: string
name: unpack_data
arguments: data_label: D18
type_guessed: string

tool: D5: %d


tool: D18: %c

assistant: ```c
int main() {
  char str[100];
  int n, i;
  scanf("%d\n", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%c", &str[i]);
    printf("%c", str[i]);
  }
  str_rev(n, str);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<expand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x28(%rbp),%rax
  movzwl (%rax),%eax
  mov    %ax,-0x12(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movswq -0x12(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <zero@plt>
  jmp    L3
L12:
  mov    -0x10(%rbp),%rax
  movzwl (%rax),%eax
  movswq %ax,%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  movzwl 0x2(%rax),%eax
  mov    %ax,(%rdx)
  addq   $0x2,-0x10(%rbp)
L3:
  addq   $0x2,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L12
  nop
  nop
  leave
  ret
```

assistant: ```c
void expand(short *p1, short *p2, int len) {
  short l, *p, *q;
  l = *p1;
  p = p1;
  q = p + l;
  zero(p2, p2 + len);
  while (++p < q) {
    p2[*p] = *(p + 1);
    p++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<expand>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movswq (%rdi),%rax
  lea    (%rdi,%rax,2),%r12
  movslq %edx,%rdx
  lea    (%rsi,%rdx,2),%rsi
  mov    %rbp,%rdi
  call   <zero@plt>
  lea    0x2(%rbx),%rax
  cmp    %rax,%r12
  jbe    L9
L5:
  movzwl 0x2(%rax),%ecx
  movswq (%rax),%rdx
  mov    %cx,0x0(%rbp,%rdx,2)
  add    $0x4,%rax
  cmp    %rax,%r12
  ja     L5
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void expand(short *p1, short *p2, int len) {
  short l, *p, *q;
  l = *p1;
  p = p1;
  q = p + l;
  zero(p2, p2 + len);
  while (++p < q) {
    p2[*p] = *(p + 1);
    p++;
  }
}
```