ylfeng/ReF-Decompile-dataset huggingface dataset:
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    D16(%rip),%rdi
  lea    D1(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %esi,%esi
  test   %edx,%edx
  jle    L26
  mov    $0xcccccccd,%r8d
  nopl   0x0(%rax)
L4:
  mov    %edx,%eax
  mov    %edx,%edi
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edi
  mov    %edx,%ecx
  mov    %eax,%edx
  add    %edi,%esi
  cmp    $0x9,%ecx
  jg     L4
  mov    %eax,0x4(%rsp)
L26:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D16
data_type: string
name: unpack_data
arguments: data_label: D1
data_type: string

tool: D16: Enter the intnumber

tool: D1: %d

assistant: ```c
main() {
  int a, sum = 0, r;
  printf("Enter the intnumber");
  scanf("%d", &a);
  for (a; a > 0;) {
    r = a % 10;
    sum = sum + r;
    a = a / 10;
  }
  printf("%d", sum);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rdi
  lea    D5(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %esi,%esi
  test   %edx,%edx
  jle    L3
  mov    $0xcccccccd,%r8d
  nopl   0x0(%rax)
L29:
  mov    %edx,%eax
  mov    %edx,%edi
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edi
  mov    %edx,%ecx
  mov    %eax,%edx
  add    %edi,%esi
  cmp    $0x9,%ecx
  jg     L29
  mov    %eax,0x4(%rsp)
L3:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D15
datatype: string
name: parse_data
arguments: tag: D5
datatype: string

tool: D15: Enter the intnumber

tool: D5: %d

assistant: ```c
main() {
  int a, sum = 0, r;
  printf("Enter the intnumber");
  scanf("%d", &a);
  for (a; a > 0;) {
    r = a % 10;
    sum = sum + r;
    a = a / 10;
  }
  printf("%d", sum);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D0(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D13(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x30(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x20(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <atan@plt>
  imul   $0xb4,%eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  D31(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  movq   %rdx,%xmm0
  lea    D18(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: f64
name: unpack_data
arguments: data_tag: D13
datatype: f64
name: unpack_data
arguments: data_tag: D31
datatype: f64
name: unpack_data
arguments: data_tag: D18
datatype: string
name: unpack_data
arguments: data_tag: D12
datatype: string

tool: D0: 23.9

tool: D13: 52.1

tool: D31: 3.1428

tool: D18: %lf.	

tool: D12: %s

assistant: ```c
int main() {
  int16_t inints[] = {23.9, 52.1};
  unsigned char *msg = malloc(10);
  double angle = atan((double)inints[1] / (double)inints[0]) * 180 / 3.1428;
  sprintf(msg, "%lf.\t", angle);
  printf("%s", msg);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  movsd  D34(%rip),%xmm0
  call   <atan@plt>
  imul   $0xb4,%eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  D18(%rip),%xmm0
  lea    D22(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rsi
  lea    D36(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_tag: D34
type_guessed: byte[8]
name: read_data
arguments: data_tag: D18
type_guessed: f64
name: read_data
arguments: data_tag: D22
type_guessed: string
name: read_data
arguments: data_tag: D36
type_guessed: string

tool: {"D34": [0x8d,0x53,0x6e,0xf7,0x77,0x70,0x01,0x40]}

tool: D18: 3.1428

tool: D22: %lf.	

tool: D36: %s

assistant: ```c
int main() {
  int16_t inints[] = {23.9, 52.1};
  unsigned char *msg = malloc(10);
  double angle = atan((double)inints[1] / (double)inints[0]) * 180 / 3.1428;
  sprintf(msg, "%lf.\t", angle);
  printf("%s", msg);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xa,%edi
  call   <malloc@plt>
  movsd  D35(%rip),%xmm0
  mov    %rax,%rbp
  call   <atan@plt>
  mov    %rbp,%rdi
  lea    D4(%rip),%rsi
  pxor   %xmm0,%xmm0
  imul   $0xb4,%eax,%eax
  cvtsi2sd %eax,%xmm0
  divsd  D31(%rip),%xmm0
  call   <sprintf@plt>
  mov    %rbp,%rsi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D35
data_type: byte[8]
name: parse_data
arguments: tag: D4
data_type: string
name: parse_data
arguments: tag: D31
data_type: f64
name: parse_data
arguments: tag: D10
data_type: string

tool: {"D35": [0x8d,0x53,0x6e,0xf7,0x77,0x70,0x01,0x40]}

tool: D4: %lf.	

tool: D31: 3.1428

tool: D10: %s

assistant: ```c
int main() {
  int16_t inints[] = {23.9, 52.1};
  unsigned char *msg = malloc(10);
  double angle = atan((double)inints[1] / (double)inints[0]) * 180 / 3.1428;
  sprintf(msg, "%lf.\t", angle);
  printf("%s", msg);
}
```
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    $0xa,%edi
  call   <malloc@plt>
  movsd  D11(%rip),%xmm0
  mov    %rax,%rbp
  call   <atan@plt>
  mov    %rbp,%rdi
  lea    D4(%rip),%rsi
  pxor   %xmm0,%xmm0
  imul   $0xb4,%eax,%eax
  cvtsi2sd %eax,%xmm0
  divsd  D16(%rip),%xmm0
  call   <sprintf@plt>
  mov    %rbp,%rsi
  lea    D26(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D11
data_type: byte[8]
name: extract_symbolic_data
arguments: data_label: D4
data_type: string
name: extract_symbolic_data
arguments: data_label: D16
data_type: f64
name: extract_symbolic_data
arguments: data_label: D26
data_type: string

tool: {"D11": [0x8d,0x53,0x6e,0xf7,0x77,0x70,0x01,0x40]}

tool: D4: %lf.	

tool: D16: 3.1428

tool: D26: %s

assistant: ```c
int main() {
  int16_t inints[] = {23.9, 52.1};
  unsigned char *msg = malloc(10);
  double angle = atan((double)inints[1] / (double)inints[0]) * 180 / 3.1428;
  sprintf(msg, "%lf.\t", angle);
  printf("%s", msg);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L7:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <test@plt>
  jmp    L7
```

assistant: ```c
int main() {
  int mykeys[8];
  while (1) {
    test(mykeys);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
L6:
  mov    %rbx,%rdi
  call   <test@plt>
  jmp    L6
```

assistant: ```c
int main() {
  int mykeys[8];
  while (1) {
    test(mykeys);
  }
  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   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L0:
  mov    %rbx,%rdi
  call   <test@plt>
  jmp    L0
```

assistant: ```c
int main() {
  int mykeys[8];
  while (1) {
    test(mykeys);
  }
  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   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L8:
  mov    %rbx,%rdi
  call   <test@plt>
  jmp    L8
```

assistant: ```c
int main() {
  int mykeys[8];
  while (1) {
    test(mykeys);
  }
  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
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D17
type_infered: string
name: parse_data
arguments: tag: D2
type_infered: string

tool: D17: Location: http://bgstation0.com/classic1/


tool: D2: 


assistant: ```c
int main(void) {
  printf("Location: http://bgstation0.com/classic1/\n");
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
type_guessed: string
name: parse_data
arguments: data_label: D14
type_guessed: string

tool: D19: Location: http://bgstation0.com/classic1/


tool: D14: 


assistant: ```c
int main(void) {
  printf("Location: http://bgstation0.com/classic1/\n");
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
datatype: string
name: retrieve_data
arguments: data_label: D4
datatype: string

tool: D8: Location: http://bgstation0.com/classic1/


tool: D4: 


assistant: ```c
int main(void) {
  printf("Location: http://bgstation0.com/classic1/\n");
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D16
type_guessed: string
name: parse_data
arguments: data_tag: D8
type_guessed: string

tool: D16: Location: http://bgstation0.com/classic1/


tool: D8: 


assistant: ```c
int main(void) {
  printf("Location: http://bgstation0.com/classic1/\n");
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<automaticArrayInit>:
  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    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L23
L5:
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0x14(%rbp,%rax,4),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L23:
  cmpl   $0x2,-0x18(%rbp)
  jle    L5
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L4
L0:
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0x14(%rbp,%rax,4),%eax
  lea    0x5(%rax),%edx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %edx,-0x14(%rbp,%rax,4)
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0x14(%rbp,%rax,4),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L4:
  cmpl   $0x2,-0x18(%rbp)
  jle    L0
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D36
type: string
name: read_data
arguments: tag: D16
type: string
name: read_data
arguments: tag: D6
type: string

tool: D36: 

Values on entering automaticArrayInit:

tool: D16: array2[ %d ] = %d  

tool: D6: 
Values on exiting automaticArrayInit:

assistant: ```c
void automaticArrayInit(void) {
  int array2[3] = {1, 2, 3};
  int i;
  printf("\n\nValues on entering automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i]);
  }
  printf("\nValues on exiting automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i] += 5);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<automaticArrayInit>:
  endbr64
  push   %r12
  push   %rbp
  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    D30(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebx
  lea    D29(%rip),%rbp
L11:
  mov    0xc(%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x3,%rbx
  jne    L11
  lea    D27(%rip),%rdi
  call   <puts@plt>
  lea    0xc(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D29(%rip),%r12
L7:
  mov    0x0(%rbp),%eax
  lea    0x5(%rax),%edx
  mov    %edx,0x0(%rbp)
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    $0x3,%ebx
  jne    L7
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D30
datatype: string
name: extract_symbolic_data
arguments: label: D29
datatype: string
name: extract_symbolic_data
arguments: label: D27
datatype: string

tool: D30: 

Values on entering automaticArrayInit:

tool: D29: array2[ %d ] = %d  

tool: D27: 
Values on exiting automaticArrayInit:

assistant: ```c
void automaticArrayInit(void) {
  int array2[3] = {1, 2, 3};
  int i;
  printf("\n\nValues on entering automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i]);
  }
  printf("\nValues on exiting automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i] += 5);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<automaticArrayInit>:
  endbr64
  push   %r12
  lea    D16(%rip),%rdi
  push   %rbp
  lea    D27(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x300000002,%rax
  lea    0xc(%rsp),%r12
  mov    %rax,0x10(%rsp)
  call   <puts@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x1,%esi
  call   <printf@plt>
  mov    $0x3,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x2,%esi
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edx
L17:
  add    $0x5,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %edx,(%r12,%rbx,4)
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x3,%rbx
  je     L8
  mov    (%r12,%rbx,4),%edx
  jmp    L17
L8:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D27
type_infered: string
name: unpack_data
arguments: data_label: D6
type_infered: string

tool: D16: 

Values on entering automaticArrayInit:

tool: D27: array2[ %d ] = %d  

tool: D6: 
Values on exiting automaticArrayInit:

assistant: ```c
void automaticArrayInit(void) {
  int array2[3] = {1, 2, 3};
  int i;
  printf("\n\nValues on entering automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i]);
  }
  printf("\nValues on exiting automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i] += 5);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<automaticArrayInit>:
  endbr64
  push   %r12
  lea    D33(%rip),%rdi
  lea    D23(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  lea    0x8(%rsp),%rbx
  movl   $0x3,0x10(%rsp)
  mov    %rax,0x8(%rsp)
  call   <puts@plt>
L23:
  mov    (%rbx,%rbp,4),%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x3,%rbp
  jne    L23
  lea    D18(%rip),%rdi
  xor    %ebp,%ebp
  call   <puts@plt>
L27:
  mov    (%rbx),%eax
  mov    %ebp,%esi
  mov    %r12,%rdi
  add    $0x1,%ebp
  add    $0x4,%rbx
  lea    0x5(%rax),%edx
  xor    %eax,%eax
  mov    %edx,-0x4(%rbx)
  call   <printf@plt>
  cmp    $0x3,%ebp
  jne    L27
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D33
type_guessed: string
name: read_data
arguments: label: D23
type_guessed: string
name: read_data
arguments: label: D2
type_guessed: byte[8]
name: read_data
arguments: label: D18
type_guessed: string

tool: D33: 

Values on entering automaticArrayInit:

tool: D23: array2[ %d ] = %d  

tool: {"D2": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: D18: 
Values on exiting automaticArrayInit:

assistant: ```c
void automaticArrayInit(void) {
  int array2[3] = {1, 2, 3};
  int i;
  printf("\n\nValues on entering automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i]);
  }
  printf("\nValues on exiting automaticArrayInit:\n");
  for (i = 0; i <= 2; i++) {
    printf("array2[ %d ] = %d  ", i, array2[i] += 5);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<parse_uint32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x2f,-0x9(%rbp)
  jle    L93
  cmpb   $0x39,-0x9(%rbp)
  jle    L104
L93:
  mov    D56(%rip),%rax
  mov    (%rax),%eax
  jmp    L136
L104:
  movsbl -0x9(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x8(%rbp)
  jg     L135
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x8(%rbp)
  jne    L159
  mov    D20(%rip),%rax
  mov    (%rax),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,-0x4(%rbp)
  jle    L159
L135:
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  jmp    L136
L159:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jle    L44
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  jmp    L136
L44:
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x2f,-0x9(%rbp)
  jle    L66
  cmpb   $0x39,-0x9(%rbp)
  jle    L104
L66:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jge    L154
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  jmp    L136
L154:
  cmpq   $0x0,-0x30(%rbp)
  je     L14
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L52
L14:
  cmpb   $0x0,-0x9(%rbp)
  je     L52
  mov    D56(%rip),%rax
  mov    (%rax),%eax
  jmp    L136
L52:
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  mov    $0x0,%eax
L136:
  pop    %rbp
  ret
```

assistant: ```c
int parse_uint32(const char *str, uint32_t *result, uint32_t min_value, uint32_t max_value, const char **end) {
  uint32_t value = 0;
  char c = *str;
  if (c < '0' || c > '9')
    return EINVAL;
  do {
    uint32_t d = c - '0';
    if (value > UINT32_MAX / 10 || (value == UINT32_MAX / 10 && d > UINT32_MAX % 10))
      return ERANGE;
    value = value * 10 + d;
    if (value > max_value)
      return ERANGE;
    ++str;
    c = *str;
  } while (c >= '0' && c <= '9');
  if (value < min_value)
    return ERANGE;
  if (end)
    *end = str;
  else if (c)
    return EINVAL;
  *result = value;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<parse_uint32>:
  endbr64
  mov    %edx,%r11d
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L62
  push   %rbx
  mov    %rsi,%r10
  mov    %ecx,%ebx
  mov    D14(%rip),%rdx
  mov    (%rdx),%r9d
  movslq %r9d,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %r9d,%edx
  sar    $0x1f,%edx
  sub    %edx,%ecx
  lea    (%rcx,%rcx,4),%edx
  add    %edx,%edx
  sub    %edx,%r9d
  mov    $0x0,%edx
  jmp    L79
L62:
  mov    D32(%rip),%rax
  mov    (%rax),%eax
  ret
L121:
  lea    (%rdx,%rdx,4),%edx
  lea    (%rax,%rdx,2),%edx
  cmp    %ebx,%edx
  jg     L126
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%esi
  cmp    $0x9,%sil
  ja     L20
L79:
  movsbl %al,%eax
  sub    $0x30,%eax
  cmp    %edx,%ecx
  jl     L16
  jne    L121
  cmp    %eax,%r9d
  jge    L121
L16:
  mov    D35(%rip),%rax
  mov    (%rax),%eax
L49:
  pop    %rbx
  ret
L126:
  mov    D35(%rip),%rax
  mov    (%rax),%eax
  jmp    L49
L20:
  cmp    %r11d,%edx
  jl     L101
  test   %r8,%r8
  je     L136
  mov    %rdi,(%r8)
L95:
  mov    %edx,(%r10)
  mov    $0x0,%eax
  jmp    L49
L101:
  mov    D35(%rip),%rax
  mov    (%rax),%eax
  jmp    L49
L136:
  test   %al,%al
  je     L95
  mov    D32(%rip),%rax
  mov    (%rax),%eax
  jmp    L49
```

assistant: ```c
int parse_uint32(const char *str, uint32_t *result, uint32_t min_value, uint32_t max_value, const char **end) {
  uint32_t value = 0;
  char c = *str;
  if (c < '0' || c > '9')
    return EINVAL;
  do {
    uint32_t d = c - '0';
    if (value > UINT32_MAX / 10 || (value == UINT32_MAX / 10 && d > UINT32_MAX % 10))
      return ERANGE;
    value = value * 10 + d;
    if (value > max_value)
      return ERANGE;
    ++str;
    c = *str;
  } while (c >= '0' && c <= '9');
  if (value < min_value)
    return ERANGE;
  if (end)
    *end = str;
  else if (c)
    return EINVAL;
  *result = value;
  return 0;
}
```