ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D2(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x1,-0x24(%rbp)
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  jmp    L24
L38:
  movsd  -0x20(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <power@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
L24:
  movsd  -0x20(%rbp),%xmm0
  comisd D48(%rip),%xmm0
  ja     L38
  movsd  -0x20(%rbp),%xmm1
  movsd  D25(%rip),%xmm0
  comisd %xmm1,%xmm0
  ja     L38
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: f64
name: disassemble_data
arguments: tag: D46
type_guessed: string
name: disassemble_data
arguments: tag: D54
type_guessed: string
name: disassemble_data
arguments: tag: D48
type_guessed: f64[2]
name: disassemble_data
arguments: tag: D25
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D39
type_guessed: string

tool: D2: 1

tool: D46: Please input a value (Between (-1) and (+1) ): 

tool: D54: %lf

tool: {"D48": [1e-6,-1e-6]}

tool: {"D25": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0xbe]}

tool: D39: 
 The value of 1-(1/x) will be: %lf 


assistant: ```c
int main(void) {
  double num, sum = 1;
  int i = 1;
  double temp;
  printf("Please input a value (Between (-1) and (+1) ): ");
  scanf("%lf", &num);
  temp = num;
  while ((num > 0.000001) || (num < -0.000001)) {
    sum = sum + num;
    i++;
    num = power(temp, i);
  }
  printf("\n The value of 1-(1/x) will be: %lf \n", sum);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rbp
  mov    $0x1,%ebx
  movsd  D36(%rip),%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L24
L19:
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  add    $0x1,%ebx
  mov    %ebx,%edi
  movq   %rbp,%xmm0
  call   <power@plt>
  movsd  %xmm0,0x10(%rsp)
L24:
  movsd  0x10(%rsp),%xmm0
  comisd D27(%rip),%xmm0
  ja     L19
  movsd  D43(%rip),%xmm2
  comisd %xmm0,%xmm2
  ja     L19
  movsd  0x8(%rsp),%xmm0
  lea    D34(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type: string
name: extract_symbolic_data
arguments: tag: D31
type: string
name: extract_symbolic_data
arguments: tag: D36
type: f64
name: extract_symbolic_data
arguments: tag: D27
type: f64[2]
name: extract_symbolic_data
arguments: tag: D43
type: byte[8]
name: extract_symbolic_data
arguments: tag: D34
type: string

tool: D16: Please input a value (Between (-1) and (+1) ): 

tool: D31: %lf

tool: D36: 1

tool: {"D27": [1e-6,-1e-6]}

tool: {"D43": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0xbe]}

tool: D34: 
 The value of 1-(1/x) will be: %lf 


assistant: ```c
int main(void) {
  double num, sum = 1;
  int i = 1;
  double temp;
  printf("Please input a value (Between (-1) and (+1) ): ");
  scanf("%lf", &num);
  temp = num;
  while ((num > 0.000001) || (num < -0.000001)) {
    sum = sum + num;
    i++;
    num = power(temp, i);
  }
  printf("\n The value of 1-(1/x) will be: %lf \n", sum);
  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   %r14
  lea    D42(%rip),%rdi
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm3
  movsd  D44(%rip),%xmm1
  movq   %xmm3,%r14
  movapd %xmm3,%xmm0
  jmp    L25
  cs nopw 0x0(%rax,%rax,1)
L38:
  addsd  %xmm0,%xmm1
  add    $0x1,%ebx
  movq   %r14,%xmm0
  mov    %ebx,%edi
  movsd  %xmm1,0x8(%rsp)
  call   <power@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  %xmm0,0x10(%rsp)
L25:
  comisd D41(%rip),%xmm0
  ja     L38
  movsd  D45(%rip),%xmm2
  comisd %xmm0,%xmm2
  ja     L38
  movapd %xmm1,%xmm0
  lea    D2(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D42
type_infered: string
name: retrieve_data
arguments: label: D23
type_infered: string
name: retrieve_data
arguments: label: D44
type_infered: f64
name: retrieve_data
arguments: label: D41
type_infered: f64[2]
name: retrieve_data
arguments: label: D45
type_infered: byte[8]
name: retrieve_data
arguments: label: D2
type_infered: string

tool: D42: Please input a value (Between (-1) and (+1) ): 

tool: D23: %lf

tool: D44: 1

tool: {"D41": [1e-6,-1e-6]}

tool: {"D45": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0xbe]}

tool: D2: 
 The value of 1-(1/x) will be: %lf 


assistant: ```c
int main(void) {
  double num, sum = 1;
  int i = 1;
  double temp;
  printf("Please input a value (Between (-1) and (+1) ): ");
  scanf("%lf", &num);
  temp = num;
  while ((num > 0.000001) || (num < -0.000001)) {
    sum = sum + num;
    i++;
    num = power(temp, i);
  }
  printf("\n The value of 1-(1/x) will be: %lf \n", sum);
  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
  push   %r14
  lea    D27(%rip),%rdi
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm3
  movsd  D37(%rip),%xmm1
  movq   %xmm3,%r14
  movapd %xmm3,%xmm0
  jmp    L12
  cs nopw 0x0(%rax,%rax,1)
L8:
  addsd  %xmm0,%xmm1
  add    $0x1,%ebx
  movq   %r14,%xmm0
  mov    %ebx,%edi
  movsd  %xmm1,0x8(%rsp)
  call   <power@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  %xmm0,0x10(%rsp)
L12:
  comisd D35(%rip),%xmm0
  ja     L8
  movsd  D36(%rip),%xmm2
  comisd %xmm0,%xmm2
  ja     L8
  movapd %xmm1,%xmm0
  lea    D3(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r14
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D27
datatype: string
name: read_data
arguments: label: D24
datatype: string
name: read_data
arguments: label: D37
datatype: f64
name: read_data
arguments: label: D35
datatype: f64[2]
name: read_data
arguments: label: D36
datatype: byte[8]
name: read_data
arguments: label: D3
datatype: string

tool: D27: Please input a value (Between (-1) and (+1) ): 

tool: D24: %lf

tool: D37: 1

tool: {"D35": [1e-6,-1e-6]}

tool: {"D36": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0xbe]}

tool: D3: 
 The value of 1-(1/x) will be: %lf 


assistant: ```c
int main(void) {
  double num, sum = 1;
  int i = 1;
  double temp;
  printf("Please input a value (Between (-1) and (+1) ): ");
  scanf("%lf", &num);
  temp = num;
  while ((num > 0.000001) || (num < -0.000001)) {
    sum = sum + num;
    i++;
    num = power(temp, i);
  }
  printf("\n The value of 1-(1/x) will be: %lf \n", sum);
  return 0;
}
```
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
  lea    -0x9000(%rsp),%r11
L46:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L46
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x9c50(%rbp),%rax
  mov    $0x9c44,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x2,-0x9c50(%rbp)
  movl   $0x3,-0x9c5c(%rbp)
  movl   $0x1,-0x9c58(%rbp)
  jmp    L16
L1:
  movl   $0x0,-0x9c54(%rbp)
  jmp    L53
L76:
  mov    -0x9c54(%rbp),%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%ecx
  mov    -0x9c5c(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  je     L40
  addl   $0x1,-0x9c54(%rbp)
L53:
  mov    -0x9c54(%rbp),%eax
  cmp    -0x9c58(%rbp),%eax
  jl     L76
  jmp    L54
L40:
  nop
L54:
  mov    -0x9c54(%rbp),%eax
  cmp    -0x9c58(%rbp),%eax
  jne    L35
  mov    -0x9c58(%rbp),%eax
  cltq
  mov    -0x9c5c(%rbp),%edx
  mov    %edx,-0x9c50(%rbp,%rax,4)
  addl   $0x1,-0x9c58(%rbp)
L35:
  addl   $0x1,-0x9c5c(%rbp)
L16:
  cmpl   $0x2710,-0x9c58(%rbp)
  jle    L1
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    $0x2711,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
data_type: string

tool: D1: The prime number %d is %d



assistant: ```c
int main() {
  int list[10001] = {};
  int n, i, j;
  list[0] = 2;
  for (n = 3, i = 1; i < 10001; n++) {
    for (j = 0; j < i; j++) {
      if (!(n % list[j])) {
        break;
      }
    }
    if (j == i) {
      list[i] = n;
      i++;
    }
  }
  printf("The prime number %d is %d\n\n", 10001, list[10001 - 1]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  lea    -0x9000(%rsp),%r11
L52:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L52
  sub    $0xc58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x9c44,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  movl   $0x2,(%rsp)
  mov    $0x1,%r8d
  mov    $0x3,%edi
  jmp    L35
L43:
  cmp    %esi,%r8d
  je     L61
L10:
  add    $0x1,%edi
  cmp    $0x2710,%r8d
  jg     L38
L35:
  mov    %rsp,%rcx
  mov    $0x0,%esi
  test   %r8d,%r8d
  jle    L43
L59:
  mov    %edi,%eax
  cltd
  idivl  (%rcx)
  test   %edx,%edx
  je     L43
  add    $0x1,%esi
  add    $0x4,%rcx
  cmp    %r8d,%esi
  jne    L59
L61:
  movslq %r8d,%rax
  mov    %edi,(%rsp,%rax,4)
  add    $0x1,%r8d
  jmp    L10
L38:
  mov    0x9c40(%rsp),%edx
  mov    $0x2711,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  mov    $0x0,%eax
  add    $0x9c58,%rsp
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D4
type: string

tool: D4: The prime number %d is %d



assistant: ```c
int main() {
  int list[10001] = {};
  int n, i, j;
  list[0] = 2;
  for (n = 3, i = 1; i < 10001; n++) {
    for (j = 0; j < i; j++) {
      if (!(n % list[j])) {
        break;
      }
    }
    if (j == i) {
      list[i] = n;
      i++;
    }
  }
  printf("The prime number %d is %d\n\n", 10001, list[10001 - 1]);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  lea    -0x9000(%rsp),%r11
L56:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L56
  sub    $0xc58,%rsp
  mov    $0x9c44,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r9
  mov    %r9,%rdi
  call   <memset@plt>
  mov    $0x1,%r8d
  mov    $0x3,%edi
  movl   $0x2,(%rsp)
  mov    %rax,%r9
  xchg   %ax,%ax
L35:
  mov    %r9,%rcx
  xor    %esi,%esi
  test   %r8d,%r8d
  jg     L74
  jmp    L54
  nopl   0x0(%rax)
L10:
  add    $0x1,%esi
  add    $0x4,%rcx
  cmp    %r8d,%esi
  je     L70
L74:
  mov    %edi,%eax
  cltd
  idivl  (%rcx)
  test   %edx,%edx
  jne    L10
L54:
  cmp    %r8d,%esi
  je     L70
  add    $0x1,%edi
  jmp    L35
  nop
L70:
  movslq %r8d,%rax
  add    $0x1,%r8d
  mov    %edi,(%rsp,%rax,4)
  add    $0x1,%edi
  cmp    $0x2711,%r8d
  jne    L35
  mov    0x9c40(%rsp),%edx
  mov    $0x2711,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  xor    %eax,%eax
  add    $0x9c58,%rsp
  ret
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D0
type_guessed: string

tool: D0: The prime number %d is %d



assistant: ```c
int main() {
  int list[10001] = {};
  int n, i, j;
  list[0] = 2;
  for (n = 3, i = 1; i < 10001; n++) {
    for (j = 0; j < i; j++) {
      if (!(n % list[j])) {
        break;
      }
    }
    if (j == i) {
      list[i] = n;
      i++;
    }
  }
  printf("The prime number %d is %d\n\n", 10001, list[10001 - 1]);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  lea    -0x9000(%rsp),%r11
L50:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L50
  sub    $0xc58,%rsp
  mov    $0x9c44,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r9
  mov    %r9,%rdi
  call   <memset@plt>
  mov    $0x1,%r8d
  mov    $0x3,%edi
  movl   $0x2,(%rsp)
  mov    %rax,%r9
  xchg   %ax,%ax
L84:
  mov    %r9,%rcx
  xor    %esi,%esi
  test   %r8d,%r8d
  jg     L65
  jmp    L64
  nopl   0x0(%rax)
L48:
  add    $0x1,%esi
  add    $0x4,%rcx
  cmp    %r8d,%esi
  je     L20
L65:
  mov    %edi,%eax
  cltd
  idivl  (%rcx)
  test   %edx,%edx
  jne    L48
L64:
  cmp    %r8d,%esi
  je     L20
  add    $0x1,%edi
  cmp    $0x2711,%r8d
  jne    L84
L31:
  mov    0x9c40(%rsp),%edx
  mov    $0x2711,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  xor    %eax,%eax
  add    $0x9c58,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  movslq %r8d,%rax
  add    $0x1,%r8d
  mov    %edi,(%rsp,%rax,4)
  add    $0x1,%edi
  cmp    $0x2711,%r8d
  jne    L84
  jmp    L31
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type: string

tool: D5: The prime number %d is %d



assistant: ```c
int main() {
  int list[10001] = {};
  int n, i, j;
  list[0] = 2;
  for (n = 3, i = 1; i < 10001; n++) {
    for (j = 0; j < i; j++) {
      if (!(n % list[j])) {
        break;
      }
    }
    if (j == i) {
      list[i] = n;
      i++;
    }
  }
  printf("The prime number %d is %d\n\n", 10001, list[10001 - 1]);
}
```
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
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  movl   $0x14,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %d


assistant: ```c
int main() {
  int x = 0;
  ({ x; }) = 20;
  printf("%d\n", x);
  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    $0x14,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D1: %d


assistant: ```c
int main() {
  int x = 0;
  ({ x; }) = 20;
  printf("%d\n", x);
  return 0;
}
```
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    $0x14,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type: string

tool: D8: %d


assistant: ```c
int main() {
  int x = 0;
  ({ x; }) = 20;
  printf("%d\n", x);
  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
  mov    $0x14,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D8: %d


assistant: ```c
int main() {
  int x = 0;
  ({ x; }) = 20;
  printf("%d\n", x);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movabs $0x2073692073696854,%rax
  movabs $0x2e747365742061,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x1,-0x34(%rbp)
  jmp    L29
L11:
  mov    -0x34(%rbp),%edx
  mov    -0x34(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_malloc@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %eax,%edx
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  and    %edx,%eax
  test   %eax,%eax
  je     L67
  mov    $0x0,%eax
  call   <abort@plt>
L67:
  cmpq   $0x0,-0x28(%rbp)
  je     L58
  mov    -0x34(%rbp),%eax
  mov    $0x10,%edx
  cmp    %edx,%eax
  cmova  %edx,%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x2c(%rbp),%edx
  lea    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x2c(%rbp),%edx
  lea    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L7
  addl   $0x1,-0x30(%rbp)
L7:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <_mm_free@plt>
  jmp    L31
L58:
  addl   $0x1,-0x30(%rbp)
L31:
  shll   -0x34(%rbp)
L29:
  cmpl   $0x100000,-0x34(%rbp)
  jle    L11
  cmpl   $0x0,-0x30(%rbp)
  je     L54
  mov    $0x0,%eax
  call   <abort@plt>
L54:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L76
  call   <__stack_chk_fail@plt>
L76:
  leave
  ret
```

assistant: ```c
int main() {
  int alignment, n;
  void *ptr;
  int errors = 0;
  const char test[] = "This is a test.";
  for (alignment = 1; alignment <= (1 << 20); alignment += alignment) {
    ptr = _mm_malloc(alignment, alignment);
    if (((ptrdiff_t)ptr) & (alignment - 1))
      abort();
    if (ptr) {
      n = alignment > sizeof test ? sizeof test : alignment;
      memcpy(ptr, test, n);
      if (memcmp(ptr, test, n) != 0)
        errors++;
      _mm_free(ptr);
    } else
      errors++;
  }
  if (errors != 0)
    abort();
  exit(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
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x2073692073696854,%rax
  movabs $0x2e747365742061,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <_mm_malloc@plt>
  mov    %rax,%rbx
  mov    $0x15,%r13d
  mov    $0x0,%r14d
  mov    $0x1,%ebp
  mov    %rsp,%r15
  jmp    L73
L66:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L73
L7:
  add    $0x1,%r14d
L74:
  add    %ebp,%ebp
  sub    $0x1,%r13d
  je     L57
  mov    %ebp,%esi
  mov    %ebp,%edi
  call   <_mm_malloc@plt>
  mov    %rax,%rbx
  lea    -0x1(%rbp),%eax
  test   %ebx,%eax
  jne    L66
L73:
  test   %rbx,%rbx
  je     L7
  mov    $0x10,%r12d
  cmp    %r12d,%ebp
  cmovbe %ebp,%r12d
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <memcmp@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%r14d
  mov    %rbx,%rdi
  call   <_mm_free@plt>
  jmp    L74
L57:
  test   %r14d,%r14d
  jne    L65
L39:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L65:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L39
L60:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int alignment, n;
  void *ptr;
  int errors = 0;
  const char test[] = "This is a test.";
  for (alignment = 1; alignment <= (1 << 20); alignment += alignment) {
    ptr = _mm_malloc(alignment, alignment);
    if (((ptrdiff_t)ptr) & (alignment - 1))
      abort();
    if (ptr) {
      n = alignment > sizeof test ? sizeof test : alignment;
      memcpy(ptr, test, n);
      if (memcmp(ptr, test, n) != 0)
        errors++;
      _mm_free(ptr);
    } else
      errors++;
  }
  if (errors != 0)
    abort();
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x1,%esi
  mov    $0x1,%edi
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x15,%r12d
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  movdqa D9(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  movaps %xmm0,(%rsp)
  call   <_mm_malloc@plt>
  mov    %rax,%rbp
  nopl   0x0(%rax)
L41:
  test   %rbp,%rbp
  je     L65
L64:
  mov    $0x10,%r15d
  mov    %r14,%rsi
  mov    %rbp,%rdi
  cmp    %r15d,%ebx
  cmovbe %ebx,%r15d
  mov    %r15d,%edx
  call   <memcpy@plt>
  mov    %r15d,%edx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <memcmp@plt>
  mov    %rbp,%rdi
  cmp    $0x1,%rax
  sbb    $0xffffffff,%r13d
  add    %ebx,%ebx
  call   <_mm_free@plt>
  sub    $0x1,%r12d
  je     L51
L5:
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <_mm_malloc@plt>
  mov    %rax,%rbp
  lea    -0x1(%rbx),%eax
  test   %ebp,%eax
  je     L41
  xor    %eax,%eax
  call   <abort@plt>
  test   %rbp,%rbp
  jne    L64
L65:
  add    $0x1,%r13d
  add    %ebx,%ebx
  sub    $0x1,%r12d
  jne    L5
L51:
  test   %r13d,%r13d
  jne    L74
L7:
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L63
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L74:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L7
L63:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type: byte[16]

tool: {"D9": [0x54,0x68,0x69,0x73,0x20,0x69,0x73,0x20,0x61,0x20,0x74,0x65,0x73,0x74,0x2e,0x00]}

assistant: ```c
int main() {
  int alignment, n;
  void *ptr;
  int errors = 0;
  const char test[] = "This is a test.";
  for (alignment = 1; alignment <= (1 << 20); alignment += alignment) {
    ptr = _mm_malloc(alignment, alignment);
    if (((ptrdiff_t)ptr) & (alignment - 1))
      abort();
    if (ptr) {
      n = alignment > sizeof test ? sizeof test : alignment;
      memcpy(ptr, test, n);
      if (memcmp(ptr, test, n) != 0)
        errors++;
      _mm_free(ptr);
    } else
      errors++;
  }
  if (errors != 0)
    abort();
  exit(0);
}
```