ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jg     L57
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  jmp    L66
L57:
  mov    -0x18(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L63
L28:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x14(%rbp)
L63:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L28
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quickSort@plt>
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L55
L10:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L55:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L10
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
L66:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D54
data_type: string
name: read_data
arguments: data_label: D50
data_type: string
name: read_data
arguments: data_label: D65
data_type: string
name: read_data
arguments: data_label: D42
data_type: string
name: read_data
arguments: data_label: D31
data_type: string
name: read_data
arguments: data_label: D64
data_type: string
name: read_data
arguments: data_label: D24
data_type: string

tool: D54: Quick Sort 

tool: D50: Total Elements : 

tool: D65: %d

tool: D42: Invalid Input 

tool: D31: Unsorted Input 

tool: D64: Sorted Output 

tool: D24: %d 


assistant: ```c
int main() {
  printf("Quick Sort \n");
  printf("\n");
  int i, n;
  printf("Total Elements : ");
  scanf("%d", &n);
  printf("\n");
  if (n < 1) {
    printf("Invalid Input \n");
    printf("\n");
    return 1;
  }
  int *A = (int *)calloc(n, sizeof(int));
  printf("Unsorted Input \n");
  for (i = 0; i < n; i++) {
    scanf("%d", (A + i));
  }
  printf("\n");
  quickSort(A, 0, (n - 1));
  printf("Sorted Output \n");
  for (i = 0; i < n; i++) {
    printf("%d \n", *(A + i));
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```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    D1(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L25
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%rbx
  lea    D42(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L37
  mov    %rbx,%r12
  mov    $0x0,%ebp
  lea    D14(%rip),%r13
L0:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebp
  add    $0x4,%r12
  cmp    %ebp,0x4(%rsp)
  jg     L0
L37:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <quickSort@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L39
  mov    $0x0,%ebp
  lea    D6(%rip),%r12
L12:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  add    $0x4,%rbx
  cmp    %ebp,0x4(%rsp)
  jg     L12
L39:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
L44:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  lea    D78(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  jmp    L44
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D1
data_type: string
name: read_data
arguments: label: D12
data_type: string
name: read_data
arguments: label: D14
data_type: string
name: read_data
arguments: label: D42
data_type: string
name: read_data
arguments: label: D16
data_type: string
name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D78
data_type: string

tool: D1: Quick Sort 

tool: D12: Total Elements : 

tool: D14: %d

tool: D42: Unsorted Input 

tool: D16: Sorted Output 

tool: D6: %d 


tool: D78: Invalid Input 

assistant: ```c
int main() {
  printf("Quick Sort \n");
  printf("\n");
  int i, n;
  printf("Total Elements : ");
  scanf("%d", &n);
  printf("\n");
  if (n < 1) {
    printf("Invalid Input \n");
    printf("\n");
    return 1;
  }
  int *A = (int *)calloc(n, sizeof(int));
  printf("Unsorted Input \n");
  for (i = 0; i < n; i++) {
    scanf("%d", (A + i));
  }
  printf("\n");
  quickSort(A, 0, (n - 1));
  printf("Sorted Output \n");
  for (i = 0; i < n; i++) {
    printf("%d \n", *(A + i));
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D20(%rip),%rdi
  lea    D22(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L48
  mov    $0x4,%esi
  call   <calloc@plt>
  lea    D66(%rip),%rdi
  mov    %rax,%rbx
  call   <puts@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L10
  mov    %rbx,%r12
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebp
  add    $0x4,%r12
  call   <scanf@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L3
L10:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  xor    %esi,%esi
  mov    %rbx,%rdi
  lea    -0x1(%rax),%edx
  call   <quickSort@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L32
  xor    %ebp,%ebp
  lea    D12(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L19
L32:
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
L65:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  jmp    L65
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D20
type_guessed: string
name: unpack_data
arguments: data_tag: D22
type_guessed: string
name: unpack_data
arguments: data_tag: D32
type_guessed: string
name: unpack_data
arguments: data_tag: D66
type_guessed: string
name: unpack_data
arguments: data_tag: D38
type_guessed: string
name: unpack_data
arguments: data_tag: D12
type_guessed: string
name: unpack_data
arguments: data_tag: D26
type_guessed: string

tool: D20: Quick Sort 

tool: D22: %d

tool: D32: Total Elements : 

tool: D66: Unsorted Input 

tool: D38: Sorted Output 

tool: D12: %d 


tool: D26: Invalid Input 

assistant: ```c
int main() {
  printf("Quick Sort \n");
  printf("\n");
  int i, n;
  printf("Total Elements : ");
  scanf("%d", &n);
  printf("\n");
  if (n < 1) {
    printf("Invalid Input \n");
    printf("\n");
    return 1;
  }
  int *A = (int *)calloc(n, sizeof(int));
  printf("Unsorted Input \n");
  for (i = 0; i < n; i++) {
    scanf("%d", (A + i));
  }
  printf("\n");
  quickSort(A, 0, (n - 1));
  printf("Sorted Output \n");
  for (i = 0; i < n; i++) {
    printf("%d \n", *(A + i));
  }
  printf("\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D22(%rip),%rdi
  lea    D46(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L46
  mov    $0x4,%esi
  call   <calloc@plt>
  lea    D56(%rip),%rdi
  mov    %rax,%rbx
  call   <puts@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L1
  mov    %rbx,%r12
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebp
  add    $0x4,%r12
  call   <scanf@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L28
L1:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  xor    %esi,%esi
  mov    %rbx,%rdi
  lea    -0x1(%rax),%edx
  call   <quickSort@plt>
  lea    D62(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L33
  xor    %ebp,%ebp
  lea    D8(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L34:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L34
L33:
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
L68:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L41
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  jmp    L68
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D22
data_type: string
name: parse_data
arguments: data_label: D46
data_type: string
name: parse_data
arguments: data_label: D4
data_type: string
name: parse_data
arguments: data_label: D56
data_type: string
name: parse_data
arguments: data_label: D62
data_type: string
name: parse_data
arguments: data_label: D8
data_type: string
name: parse_data
arguments: data_label: D23
data_type: string

tool: D22: Quick Sort 

tool: D46: %d

tool: D4: Total Elements : 

tool: D56: Unsorted Input 

tool: D62: Sorted Output 

tool: D8: %d 


tool: D23: Invalid Input 

assistant: ```c
int main() {
  printf("Quick Sort \n");
  printf("\n");
  int i, n;
  printf("Total Elements : ");
  scanf("%d", &n);
  printf("\n");
  if (n < 1) {
    printf("Invalid Input \n");
    printf("\n");
    return 1;
  }
  int *A = (int *)calloc(n, sizeof(int));
  printf("Unsorted Input \n");
  for (i = 0; i < n; i++) {
    scanf("%d", (A + i));
  }
  printf("\n");
  quickSort(A, 0, (n - 1));
  printf("Sorted Output \n");
  for (i = 0; i < n; i++) {
    printf("%d \n", *(A + i));
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<selectOrder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x1,-0x1c(%rbp)
  jle    L51
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L4:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L43
L57:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L23
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L23:
  addl   $0x1,-0x4(%rbp)
L43:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L57
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  jmp    L68
L51:
  nop
L68:
  leave
  ret
```

assistant: ```c
void selectOrder(int array[], int n) {
  if (n <= 1) {
    return;
  }
  for (int i = 0; i < n; i++) {
    int selectIndex = i;
    for (int j = i + 1; j < n; j++) {
      if (array[j] > array[i]) {
        selectIndex = j;
      }
    }
    swap(array, i, selectIndex);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<selectOrder>:
  endbr64
  cmp    $0x1,%esi
  jle    L16
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%ebp
  lea    -0x1(%rsi),%r14d
  mov    $0x0,%r12d
L14:
  mov    %r12d,%esi
  lea    0x1(%r12),%r13d
  mov    %r12,%rax
  add    $0x1,%r12
  mov    (%rbx,%rax,4),%ecx
  mov    %r12,%rax
  mov    %esi,%edx
L6:
  cmp    %ecx,(%rbx,%rax,4)
  cmovg  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%ebp
  jg     L6
  mov    %rbx,%rdi
  call   <swap@plt>
  cmp    %r12,%r14
  jne    L14
  mov    %r13d,%edx
  mov    %r13d,%esi
  mov    %rbx,%rdi
  call   <swap@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  ret
```

assistant: ```c
void selectOrder(int array[], int n) {
  if (n <= 1) {
    return;
  }
  for (int i = 0; i < n; i++) {
    int selectIndex = i;
    for (int j = i + 1; j < n; j++) {
      if (array[j] > array[i]) {
        selectIndex = j;
      }
    }
    swap(array, i, selectIndex);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<selectOrder>:
  endbr64
  cmp    $0x1,%esi
  jle    L25
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    -0x1(%rsi),%ebx
  mov    %rbx,%r12
L10:
  mov    %r14d,%esi
  mov    0x0(%rbp,%r14,4),%ecx
  add    $0x1,%r14
  mov    %r14,%rax
  mov    %esi,%edx
  nopl   0x0(%rax)
L27:
  cmp    %ecx,0x0(%rbp,%rax,4)
  cmovg  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%r13d
  jg     L27
  mov    %rbp,%rdi
  call   <swap@plt>
  cmp    %r14,%rbx
  jne    L10
  pop    %rbx
  mov    %r12d,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <swap@plt>
L25:
  ret
```

assistant: ```c
void selectOrder(int array[], int n) {
  if (n <= 1) {
    return;
  }
  for (int i = 0; i < n; i++) {
    int selectIndex = i;
    for (int j = i + 1; j < n; j++) {
      if (array[j] > array[i]) {
        selectIndex = j;
      }
    }
    swap(array, i, selectIndex);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<turn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %esi,%eax
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %edi,%edx
  mov    %dx,-0x14(%rbp)
  mov    %ax,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  movswl -0x18(%rbp),%edx
  movswl -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  je     L12
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <waitAngle@plt>
  mov    %eax,-0x4(%rbp)
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L12
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  jne    L4
L12:
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  jmp    L10
L4:
  mov    -0x4(%rbp),%eax
L10:
  leave
  ret
```

assistant: ```c
int turn(short vel, short rad, int angle, int interrupt) {
  int ret = 0;
  if (drive(vel, rad) == -1 || (ret = waitAngle(angle, interrupt)) == INT_MIN || drive(0, 0) == -1)
    return INT_MIN;
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<turn>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%ebx
  mov    %ecx,%ebp
  movswl %si,%esi
  movswl %di,%edi
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  je     L2
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <waitAngle@plt>
  mov    %eax,%ebx
  mov    D7(%rip),%rax
  cmp    %ebx,(%rax)
  je     L2
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  jne    L12
L2:
  mov    D7(%rip),%rax
  mov    (%rax),%ebx
L12:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int turn(short vel, short rad, int angle, int interrupt) {
  int ret = 0;
  if (drive(vel, rad) == -1 || (ret = waitAngle(angle, interrupt)) == INT_MIN || drive(0, 0) == -1)
    return INT_MIN;
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<turn>:
  endbr64
  push   %r12
  movswl %si,%esi
  movswl %di,%edi
  mov    %edx,%r12d
  push   %rbp
  mov    %ecx,%ebp
  sub    $0x8,%rsp
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  je     L39
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <waitAngle@plt>
  mov    D15(%rip),%rbp
  mov    %eax,%r12d
  cmp    %eax,0x0(%rbp)
  je     L14
  xor    %esi,%esi
  xor    %edi,%edi
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  je     L25
L14:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  mov    D15(%rip),%rax
  mov    (%rax),%r12d
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   (%rax)
L25:
  mov    0x0(%rbp),%r12d
  jmp    L14
```

assistant: ```c
int turn(short vel, short rad, int angle, int interrupt) {
  int ret = 0;
  if (drive(vel, rad) == -1 || (ret = waitAngle(angle, interrupt)) == INT_MIN || drive(0, 0) == -1)
    return INT_MIN;
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<turn>:
  endbr64
  push   %r12
  movswl %si,%esi
  movswl %di,%edi
  mov    %edx,%r12d
  push   %rbp
  mov    %ecx,%ebp
  sub    $0x8,%rsp
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  je     L38
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <waitAngle@plt>
  mov    D6(%rip),%rbp
  mov    %eax,%r12d
  cmp    %eax,0x0(%rbp)
  je     L39
  xor    %esi,%esi
  xor    %edi,%edi
  call   <drive@plt>
  cmp    $0xffffffff,%eax
  je     L25
L39:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  mov    D6(%rip),%rax
  mov    (%rax),%r12d
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   (%rax)
L25:
  mov    0x0(%rbp),%r12d
  jmp    L39
```

assistant: ```c
int turn(short vel, short rad, int angle, int interrupt) {
  int ret = 0;
  if (drive(vel, rad) == -1 || (ret = waitAngle(angle, interrupt)) == INT_MIN || drive(0, 0) == -1)
    return INT_MIN;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<str_to_i>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x1,-0x18(%rbp)
  movq   $0x1,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x2b,%eax
  je     L32
  cmp    $0x2d,%eax
  jne    L24
  movq   $0xffffffffffffffff,-0x18(%rbp)
L32:
  addq   $0x1,-0x38(%rbp)
L24:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <str_length@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%eax
  mov    %rax,-0x8(%rbp)
  jmp    L21
L8:
  mov    -0x8(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  imul   -0x10(%rbp),%rax
  add    %rax,-0x20(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,-0x10(%rbp)
  subq   $0x1,-0x8(%rbp)
L21:
  cmpq   $0x0,-0x8(%rbp)
  jns    L8
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  leave
  ret
```

assistant: ```c
long str_to_i(const char *s) {
  long final = 0, sign = 1, mult = 1;
  switch (*s) {
  case '-':
    sign = -1;
  case '+':
    s++;
  }
  unsigned nlen = str_length(s);
  for (long ch = nlen - 1; ch >= 0; ch--) {
    final += (long)(s[ch] - '0') * (long)mult;
    mult *= 10;
  }
  return sign * final;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_to_i>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movzbl (%rdi),%eax
  cmp    $0x2b,%al
  je     L24
  mov    $0x1,%ebp
  cmp    $0x2d,%al
  jne    L22
  mov    $0xffffffffffffffff,%rbp
L2:
  add    $0x1,%rbx
L22:
  mov    %rbx,%rdi
  call   <str_length@plt>
  lea    -0x1(%rax),%ecx
  mov    $0x1,%edx
  mov    $0x0,%esi
L37:
  movsbl (%rbx,%rcx,1),%eax
  sub    $0x30,%eax
  cltq
  imul   %rdx,%rax
  add    %rax,%rsi
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  sub    $0x1,%rcx
  cmp    $0xffffffffffffffff,%rcx
  jne    L37
  mov    %rbp,%rax
  imul   %rsi,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    $0x1,%ebp
  jmp    L2
```

assistant: ```c
long str_to_i(const char *s) {
  long final = 0, sign = 1, mult = 1;
  switch (*s) {
  case '-':
    sign = -1;
  case '+':
    s++;
  }
  unsigned nlen = str_length(s);
  for (long ch = nlen - 1; ch >= 0; ch--) {
    final += (long)(s[ch] - '0') * (long)mult;
    mult *= 10;
  }
  return sign * final;
}
```
user: What is the c source code of the assembly code below:

```asm
<str_to_i>:
  endbr64
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  cmp    $0x2b,%al
  je     L16
  cmp    $0x2d,%al
  jne    L21
  mov    $0xffffffffffffffff,%rbp
L16:
  add    $0x1,%rbx
L21:
  mov    %rbx,%rdi
  call   <str_length@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  lea    -0x1(%rax),%ecx
  add    %rbx,%rcx
  nopl   (%rax)
L28:
  movsbl (%rcx),%eax
  mov    %rcx,%r8
  sub    $0x1,%rcx
  sub    $0x30,%eax
  cltq
  imul   %rdx,%rax
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  add    %rax,%rsi
  cmp    %r8,%rbx
  jne    L28
  mov    %rbp,%rax
  add    $0x8,%rsp
  imul   %rsi,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
long str_to_i(const char *s) {
  long final = 0, sign = 1, mult = 1;
  switch (*s) {
  case '-':
    sign = -1;
  case '+':
    s++;
  }
  unsigned nlen = str_length(s);
  for (long ch = nlen - 1; ch >= 0; ch--) {
    final += (long)(s[ch] - '0') * (long)mult;
    mult *= 10;
  }
  return sign * final;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<str_to_i>:
  endbr64
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  cmp    $0x2b,%al
  je     L17
  cmp    $0x2d,%al
  jne    L1
  mov    $0xffffffffffffffff,%rbp
L17:
  add    $0x1,%rbx
L1:
  mov    %rbx,%rdi
  call   <str_length@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  lea    -0x1(%rax),%ecx
  add    %rbx,%rcx
  nopl   (%rax)
L6:
  movsbl (%rcx),%eax
  mov    %rcx,%r8
  sub    $0x1,%rcx
  sub    $0x30,%eax
  cltq
  imul   %rdx,%rax
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  add    %rax,%rsi
  cmp    %r8,%rbx
  jne    L6
  mov    %rbp,%rax
  add    $0x8,%rsp
  imul   %rsi,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
long str_to_i(const char *s) {
  long final = 0, sign = 1, mult = 1;
  switch (*s) {
  case '-':
    sign = -1;
  case '+':
    s++;
  }
  unsigned nlen = str_length(s);
  for (long ch = nlen - 1; ch >= 0; ch--) {
    final += (long)(s[ch] - '0') * (long)mult;
    mult *= 10;
  }
  return sign * final;
}
```