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

```asm
<is_equal>:
  endbr64
  test   %edx,%edx
  jle    L79
  lea    -0x1(%rcx),%r9d
  mov    %rsi,%r8
  xor    %r10d,%r10d
  xor    %r11d,%r11d
  add    $0x1,%r9
L80:
  test   %ecx,%ecx
  jle    L86
  push   %rbx
  xchg   %ax,%ax
L72:
  movslq %r10d,%rsi
  lea    0x0(,%rsi,4),%rax
  add    %r9,%rsi
  shl    $0x2,%rsi
  jmp    L27
  nopl   0x0(%rax)
L35:
  add    $0x4,%rax
  cmp    %rax,%rsi
  je     L64
L27:
  mov    (%r8,%rax,1),%ebx
  cmp    %ebx,(%rdi,%rax,1)
  je     L35
  xor    %eax,%eax
  pop    %rbx
  ret
  xchg   %ax,%ax
L64:
  add    $0x1,%r11d
  add    %ecx,%r10d
  cmp    %r11d,%edx
  je     L29
  test   %ecx,%ecx
  jg     L72
  jmp    L64
L86:
  add    $0x1,%r11d
  add    %ecx,%r10d
  cmp    %r11d,%edx
  jne    L80
L79:
  mov    $0x1,%eax
  ret
L29:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int is_equal(int *a, int *b, int h, int w) {
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      int a_val = *(a + i * w + j);
      int b_val = *(b + i * w + j);
      if (a_val != b_val)
        return 0;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_equal>:
  endbr64
  test   %edx,%edx
  jle    L34
  lea    -0x1(%rcx),%r9d
  mov    %rsi,%r8
  xor    %r10d,%r10d
  xor    %r11d,%r11d
  add    $0x1,%r9
L27:
  test   %ecx,%ecx
  jle    L9
  push   %rbx
  xchg   %ax,%ax
L46:
  movslq %r10d,%rsi
  lea    0x0(,%rsi,4),%rax
  add    %r9,%rsi
  shl    $0x2,%rsi
  jmp    L11
  nopl   0x0(%rax)
L2:
  add    $0x4,%rax
  cmp    %rax,%rsi
  je     L5
L11:
  mov    (%r8,%rax,1),%ebx
  cmp    %ebx,(%rdi,%rax,1)
  je     L2
  xor    %eax,%eax
  pop    %rbx
  ret
  xchg   %ax,%ax
L5:
  add    $0x1,%r11d
  add    %ecx,%r10d
  cmp    %r11d,%edx
  je     L48
  test   %ecx,%ecx
  jg     L46
  jmp    L5
L9:
  add    $0x1,%r11d
  add    %ecx,%r10d
  cmp    %r11d,%edx
  jne    L27
L34:
  mov    $0x1,%eax
  ret
L48:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int is_equal(int *a, int *b, int h, int w) {
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      int a_val = *(a + i * w + j);
      int b_val = *(b + i * w + j);
      if (a_val != b_val)
        return 0;
    }
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<atoi_push>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L35
  movl   $0xffffffff,-0xc(%rbp)
  addq   $0x1,-0x18(%rbp)
  jmp    L13
L35:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  jne    L13
  addq   $0x1,-0x18(%rbp)
  jmp    L13
L10:
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L13:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0xc(%rbp),%eax
  cltq
  imul   -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long int atoi_push(const char *str) {
  int s;
  long int r;
  r = 0;
  s = 1;
  if (*str == '-') {
    s = -1;
    str++;
  } else if (*str == '+')
    str++;
  while (*str) {
    r = r * 10 + (*str - '0');
    str++;
  }
  return ((r * s));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<atoi_push>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x2d,%dl
  je     L24
  mov    $0x1,%eax
  cmp    $0x2b,%dl
  je     L26
L19:
  movzbl (%rdi),%edx
  mov    $0x0,%ecx
  test   %dl,%dl
  je     L31
L16:
  lea    (%rcx,%rcx,4),%rcx
  movsbl %dl,%edx
  sub    $0x30,%edx
  movslq %edx,%rdx
  lea    (%rdx,%rcx,2),%rcx
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L16
L31:
  cltq
  imul   %rcx,%rax
  ret
L24:
  add    $0x1,%rdi
  mov    $0xffffffff,%eax
  jmp    L19
L26:
  add    $0x1,%rdi
  jmp    L19
```

assistant: ```c
long int atoi_push(const char *str) {
  int s;
  long int r;
  r = 0;
  s = 1;
  if (*str == '-') {
    s = -1;
    str++;
  } else if (*str == '+')
    str++;
  while (*str) {
    r = r * 10 + (*str - '0');
    str++;
  }
  return ((r * s));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<atoi_push>:
  endbr64
  movsbl (%rdi),%edx
  cmp    $0x2d,%dl
  je     L4
  mov    $0x1,%ecx
  cmp    $0x2b,%dl
  je     L48
L18:
  xor    %eax,%eax
  test   %dl,%dl
  je     L30
  nopl   0x0(%rax)
L17:
  sub    $0x30,%edx
  lea    (%rax,%rax,4),%rax
  add    $0x1,%rdi
  movslq %edx,%rdx
  lea    (%rdx,%rax,2),%rax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L17
  imul   %rcx,%rax
  ret
  xchg   %ax,%ax
L30:
  ret
  nopl   0x0(%rax)
L48:
  movsbl 0x1(%rdi),%edx
  add    $0x1,%rdi
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L4:
  movsbl 0x1(%rdi),%edx
  mov    $0xffffffffffffffff,%rcx
  add    $0x1,%rdi
  jmp    L18
```

assistant: ```c
long int atoi_push(const char *str) {
  int s;
  long int r;
  r = 0;
  s = 1;
  if (*str == '-') {
    s = -1;
    str++;
  } else if (*str == '+')
    str++;
  while (*str) {
    r = r * 10 + (*str - '0');
    str++;
  }
  return ((r * s));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<atoi_push>:
  endbr64
  movsbl (%rdi),%edx
  cmp    $0x2d,%dl
  je     L51
  mov    $0x1,%ecx
  cmp    $0x2b,%dl
  je     L26
L5:
  xor    %eax,%eax
  test   %dl,%dl
  je     L53
  nopl   0x0(%rax)
L28:
  sub    $0x30,%edx
  lea    (%rax,%rax,4),%rax
  add    $0x1,%rdi
  movslq %edx,%rdx
  lea    (%rdx,%rax,2),%rax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L28
  imul   %rcx,%rax
  ret
  xchg   %ax,%ax
L53:
  ret
  nopl   0x0(%rax)
L26:
  movsbl 0x1(%rdi),%edx
  add    $0x1,%rdi
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L51:
  movsbl 0x1(%rdi),%edx
  mov    $0xffffffffffffffff,%rcx
  add    $0x1,%rdi
  jmp    L5
```

assistant: ```c
long int atoi_push(const char *str) {
  int s;
  long int r;
  r = 0;
  s = 1;
  if (*str == '-') {
    s = -1;
    str++;
  } else if (*str == '+')
    str++;
  while (*str) {
    r = r * 10 + (*str - '0');
    str++;
  }
  return ((r * s));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hff1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L37
L35:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L37:
  cmpl   $0x2,-0xc(%rbp)
  jle    L35
  movsd  -0x8(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L23
  mov    $0x1,%eax
  jmp    L12
L23:
  mov    $0x0,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int hff1(const double *p, const double *q) {
  double tmp = 0;
#pragma omp simd reduction(+ : tmp)
  for (int i = 0; i < 3; i++)
    tmp += (p[i] * p[i] - p[i] * q[i]);
  if (tmp > 0)
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hff1>:
  endbr64
  mov    $0x0,%eax
  pxor   %xmm2,%xmm2
L3:
  movsd  (%rdi,%rax,1),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  mulsd  (%rsi,%rax,1),%xmm0
  subsd  %xmm0,%xmm1
  addsd  %xmm1,%xmm2
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L3
  comisd D2(%rip),%xmm2
  seta   %al
  movzbl %al,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_guessed: f64

tool: D2: 0

assistant: ```c
int hff1(const double *p, const double *q) {
  double tmp = 0;
#pragma omp simd reduction(+ : tmp)
  for (int i = 0; i < 3; i++)
    tmp += (p[i] * p[i] - p[i] * q[i]);
  if (tmp > 0)
    return 1;
  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
<hff1>:
  endbr64
  pxor   %xmm3,%xmm3
  xor    %eax,%eax
  movapd %xmm3,%xmm2
L8:
  movsd  (%rdi,%rax,1),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  mulsd  (%rsi,%rax,1),%xmm0
  add    $0x8,%rax
  subsd  %xmm0,%xmm1
  addsd  %xmm1,%xmm2
  cmp    $0x18,%rax
  jne    L8
  xor    %eax,%eax
  comisd %xmm3,%xmm2
  seta   %al
  ret
```

assistant: ```c
int hff1(const double *p, const double *q) {
  double tmp = 0;
#pragma omp simd reduction(+ : tmp)
  for (int i = 0; i < 3; i++)
    tmp += (p[i] * p[i] - p[i] * q[i]);
  if (tmp > 0)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hff1>:
  endbr64
  movupd (%rdi),%xmm1
  movupd (%rsi),%xmm4
  pxor   %xmm3,%xmm3
  xor    %eax,%eax
  movsd  0x10(%rdi),%xmm2
  movapd %xmm1,%xmm0
  mulpd  %xmm1,%xmm0
  mulpd  %xmm4,%xmm1
  subpd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  addsd  %xmm3,%xmm1
  addsd  %xmm0,%xmm1
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  mulsd  0x10(%rsi),%xmm2
  subsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  comisd %xmm3,%xmm0
  seta   %al
  ret
```

assistant: ```c
int hff1(const double *p, const double *q) {
  double tmp = 0;
#pragma omp simd reduction(+ : tmp)
  for (int i = 0; i < 3; i++)
    tmp += (p[i] * p[i] - p[i] * q[i]);
  if (tmp > 0)
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test_three_elements>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x14(%rbp)
  movl   $0x2,-0x10(%rbp)
  movl   $0x3,-0xc(%rbp)
  lea    -0x14(%rbp),%rax
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <search@plt>
  test   %eax,%eax
  jne    L3
  lea    -0x14(%rbp),%rax
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <search@plt>
  cmp    $0x1,%eax
  jne    L3
  lea    -0x14(%rbp),%rax
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <search@plt>
  cmp    $0x2,%eax
  jne    L3
  lea    -0x14(%rbp),%rax
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  jne    L3
  lea    -0x14(%rbp),%rax
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0xffffffff,%esi
  mov    %rax,%rdi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  jne    L3
  mov    $0x1,%eax
  jmp    L34
L3:
  mov    $0x0,%eax
L34:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

assistant: ```c
int test_three_elements() {
  int array[] = {1, 2, 3};
  return 0 == search(array, 1, 0, 2) && 1 == search(array, 2, 0, 2) && 2 == search(array, 3, 0, 2) && -1 == search(array, 4, 0, 2) && -1 == search(array, -1, 0, 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<test_three_elements>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,0xc(%rsp)
  movl   $0x2,0x10(%rsp)
  movl   $0x3,0x14(%rsp)
  lea    0xc(%rsp),%rdi
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x1,%esi
  call   <search@plt>
  mov    $0x0,%ebx
  test   %eax,%eax
  je     L36
L58:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    %ebx,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L36:
  mov    %eax,%ebx
  lea    0xc(%rsp),%rdi
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x2,%esi
  call   <search@plt>
  cmp    $0x1,%eax
  jne    L58
  lea    0xc(%rsp),%rdi
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x3,%esi
  call   <search@plt>
  cmp    $0x2,%eax
  jne    L58
  lea    0xc(%rsp),%rdi
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x4,%esi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  jne    L58
  lea    0xc(%rsp),%rdi
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0xffffffff,%esi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  sete   %bl
  movzbl %bl,%ebx
  jmp    L58
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_three_elements() {
  int array[] = {1, 2, 3};
  return 0 == search(array, 1, 0, 2) && 1 == search(array, 2, 0, 2) && 2 == search(array, 3, 0, 2) && -1 == search(array, 4, 0, 2) && -1 == search(array, -1, 0, 2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_three_elements>:
  endbr64
  push   %rbp
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x1,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x200000001,%rax
  lea    0xc(%rsp),%rbp
  movl   $0x3,0x14(%rsp)
  mov    %rbp,%rdi
  mov    %rax,0xc(%rsp)
  call   <search@plt>
  test   %eax,%eax
  je     L41
L13:
  xor    %eax,%eax
L44:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L34
  add    $0x20,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0x1,%eax
  jne    L13
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0x2,%eax
  jne    L13
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  jne    L13
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0xffffffff,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L44
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_three_elements() {
  int array[] = {1, 2, 3};
  return 0 == search(array, 1, 0, 2) && 1 == search(array, 2, 0, 2) && 2 == search(array, 3, 0, 2) && -1 == search(array, 4, 0, 2) && -1 == search(array, -1, 0, 2);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test_three_elements>:
  endbr64
  push   %rbp
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x1,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rbp
  mov    D5(%rip),%rax
  movl   $0x3,0x10(%rsp)
  mov    %rbp,%rdi
  mov    %rax,0x8(%rsp)
  call   <search@plt>
  test   %eax,%eax
  je     L53
L40:
  xor    %eax,%eax
L47:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L31
  add    $0x20,%rsp
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L53:
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0x1,%eax
  jne    L40
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0x2,%eax
  jne    L40
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  jne    L40
  xor    %edx,%edx
  mov    $0x2,%ecx
  mov    $0xffffffff,%esi
  mov    %rbp,%rdi
  call   <search@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L47
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type: byte[8]

tool: {"D5": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

assistant: ```c
int test_three_elements() {
  int array[] = {1, 2, 3};
  return 0 == search(array, 1, 0, 2) && 1 == search(array, 2, 0, 2) && 2 == search(array, 3, 0, 2) && -1 == search(array, 4, 0, 2) && -1 == search(array, -1, 0, 2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L40
L37:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x18(%rbp)
L40:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L37
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quick_sort@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L10
L17:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L10:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L17
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D16
type: string
name: retrieve_data
arguments: label: D41
type: string
name: retrieve_data
arguments: label: D25
type: string
name: retrieve_data
arguments: label: D48
type: string

tool: D16: How many numbers? 

tool: D41: %d

tool: D25: Enter %d numbers: 

tool: D48: %d	

assistant: ```c
int main() {
  int n;
  printf("How many numbers? ");
  scanf("%d", &n);
  int *input = (int *)malloc(n * sizeof(int));
  printf("Enter %d numbers: ", n);
  for (int i = 0; i < n; i++) {
    scanf("%d", input + i);
  }
  quick_sort(input, 0, n - 1);
  for (int i = 0; i < n; i++) {
    printf("%d\t", *(input + i));
  }
  free(input);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    0x4(%rsp),%esi
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L7
  mov    %r12,%rbp
  mov    $0x0,%ebx
  lea    D1(%rip),%r13
L46:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%edx
  add    $0x4,%rbp
  cmp    %ebx,%edx
  jg     L46
L7:
  sub    $0x1,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <quick_sort@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L20
  mov    %r12,%rbp
  mov    $0x0,%ebx
  lea    D22(%rip),%r13
L13:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jg     L13
L20:
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D26
datatype: string
name: extract_symbolic_data
arguments: tag: D1
datatype: string
name: extract_symbolic_data
arguments: tag: D39
datatype: string
name: extract_symbolic_data
arguments: tag: D22
datatype: string

tool: D26: How many numbers? 

tool: D1: %d

tool: D39: Enter %d numbers: 

tool: D22: %d	

assistant: ```c
int main() {
  int n;
  printf("How many numbers? ");
  scanf("%d", &n);
  int *input = (int *)malloc(n * sizeof(int));
  printf("Enter %d numbers: ", n);
  for (int i = 0; i < n; i++) {
    scanf("%d", input + i);
  }
  quick_sort(input, 0, n - 1);
  for (int i = 0; i < n; i++) {
    printf("%d\t", *(input + i));
  }
  free(input);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D29(%rip),%rdi
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%esi
  lea    D30(%rip),%rdi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L45
  mov    %r12,%rbp
  xor    %ebx,%ebx
L23:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    %ebx,%edx
  jg     L23
L45:
  sub    $0x1,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <quick_sort@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L7
  mov    %r12,%rbp
  xor    %ebx,%ebx
  lea    D2(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L4:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L4
L7:
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D29
type_guessed: string
name: unpack_data
arguments: label: D7
type_guessed: string
name: unpack_data
arguments: label: D30
type_guessed: string
name: unpack_data
arguments: label: D2
type_guessed: string

tool: D29: How many numbers? 

tool: D7: %d

tool: D30: Enter %d numbers: 

tool: D2: %d	

assistant: ```c
int main() {
  int n;
  printf("How many numbers? ");
  scanf("%d", &n);
  int *input = (int *)malloc(n * sizeof(int));
  printf("Enter %d numbers: ", n);
  for (int i = 0; i < n; i++) {
    scanf("%d", input + i);
  }
  quick_sort(input, 0, n - 1);
  for (int i = 0; i < n; i++) {
    printf("%d\t", *(input + i));
  }
  free(input);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D3(%rip),%rdi
  lea    D1(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%esi
  lea    D39(%rip),%rdi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L32
  mov    %r12,%rbp
  xor    %ebx,%ebx
L19:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    %ebx,%edx
  jg     L19
L32:
  sub    $0x1,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <quick_sort@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L12
  mov    %r12,%rbp
  xor    %ebx,%ebx
  lea    D31(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L40:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L40
L12:
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D3
datatype: string
name: parse_data
arguments: data_tag: D1
datatype: string
name: parse_data
arguments: data_tag: D39
datatype: string
name: parse_data
arguments: data_tag: D31
datatype: string

tool: D3: How many numbers? 

tool: D1: %d

tool: D39: Enter %d numbers: 

tool: D31: %d	

assistant: ```c
int main() {
  int n;
  printf("How many numbers? ");
  scanf("%d", &n);
  int *input = (int *)malloc(n * sizeof(int));
  printf("Enter %d numbers: ", n);
  for (int i = 0; i < n; i++) {
    scanf("%d", input + i);
  }
  quick_sort(input, 0, n - 1);
  for (int i = 0; i < n; i++) {
    printf("%d\t", *(input + i));
  }
  free(input);
}
```