ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mod_normal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    $0x0,%edx
  divl   -0x20(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L22
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
L22:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  shr    %eax
  cmp    %eax,%edx
  jbe    L2
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    -0x20(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
L2:
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jae    L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void mod_normal(int *poly, unsigned d, unsigned p) {
  int i;
  for (i = 0; i <= d; i++) {
    poly[i] %= p;
    if (poly[i] < 0)
      poly[i] += p;
    if (poly[i] > p / 2)
      poly[i] -= p;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<mod_normal>:
  endbr64
  mov    %edx,%r8d
  mov    %edx,%r9d
  shr    %r9d
  mov    %rdi,%rcx
  mov    %esi,%esi
  lea    0x4(%rdi,%rsi,4),%rsi
L6:
  mov    (%rcx),%eax
  mov    $0x0,%edx
  div    %r8d
  lea    (%rdx,%r8,1),%eax
  test   %edx,%edx
  cmovs  %eax,%edx
  mov    %edx,%eax
  sub    %r8d,%eax
  cmp    %r9d,%edx
  cmovbe %edx,%eax
  mov    %eax,(%rcx)
  add    $0x4,%rcx
  cmp    %rsi,%rcx
  jne    L6
  ret
```

assistant: ```c
void mod_normal(int *poly, unsigned d, unsigned p) {
  int i;
  for (i = 0; i <= d; i++) {
    poly[i] %= p;
    if (poly[i] < 0)
      poly[i] += p;
    if (poly[i] > p / 2)
      poly[i] -= p;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<mod_normal>:
  endbr64
  mov    %edx,%r8d
  mov    %esi,%esi
  mov    %edx,%ecx
  shr    %r8d
  lea    0x4(%rdi,%rsi,4),%rsi
  nopl   0x0(%rax,%rax,1)
L4:
  mov    (%rdi),%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  jns    L11
  add    %ecx,%edx
L11:
  mov    %edx,(%rdi)
  cmp    %edx,%r8d
  jae    L1
  sub    %ecx,%edx
  mov    %edx,(%rdi)
L1:
  add    $0x4,%rdi
  cmp    %rdi,%rsi
  jne    L4
  ret
```

assistant: ```c
void mod_normal(int *poly, unsigned d, unsigned p) {
  int i;
  for (i = 0; i <= d; i++) {
    poly[i] %= p;
    if (poly[i] < 0)
      poly[i] += p;
    if (poly[i] > p / 2)
      poly[i] -= p;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mod_normal>:
  endbr64
  mov    %edx,%r8d
  mov    %esi,%esi
  mov    %edx,%ecx
  shr    %r8d
  lea    0x4(%rdi,%rsi,4),%rsi
  jmp    L10
  nopl   (%rax)
L6:
  mov    %edx,(%rdi)
L17:
  cmp    %edx,%r8d
  jae    L5
  sub    %ecx,%edx
  mov    %edx,(%rdi)
L5:
  add    $0x4,%rdi
  cmp    %rdi,%rsi
  je     L3
L10:
  mov    (%rdi),%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  jns    L6
  add    %ecx,%edx
  mov    %edx,(%rdi)
  jmp    L17
  nopl   0x0(%rax)
L3:
  ret
```

assistant: ```c
void mod_normal(int *poly, unsigned d, unsigned p) {
  int i;
  for (i = 0; i <= d; i++) {
    poly[i] %= p;
    if (poly[i] < 0)
      poly[i] += p;
    if (poly[i] > p / 2)
      poly[i] -= p;
  }
}
```
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    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D46
type_infered: string
name: retrieve_data
arguments: data_tag: D10
type_infered: string
name: retrieve_data
arguments: data_tag: D5
type_infered: string
name: retrieve_data
arguments: data_tag: D24
type_infered: string

tool: D46: Enter x+y:

tool: D10: %d

tool: D5: Enter x-y:

tool: D24: The value of x and y are %d and %d:

assistant: ```c
int main() {
  int x_plus_y, x_minus_y, x, y;
  printf("Enter x+y:");
  scanf("%d", &x_plus_y);
  printf("Enter x-y:");
  scanf("%d", &x_minus_y);
  x = (x_plus_y + x_minus_y) / 2;
  y = (x_plus_y - x_minus_y) / 2;
  printf("The value of x and y are %d and %d:", x, y);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D35(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  mov    %eax,%ecx
  sub    0x4(%rsp),%ecx
  mov    %ecx,%edx
  shr    $0x1f,%edx
  add    %ecx,%edx
  sar    %edx
  add    0x4(%rsp),%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  sar    %esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D4
datatype: string
name: retrieve_data
arguments: data_label: D35
datatype: string
name: retrieve_data
arguments: data_label: D33
datatype: string
name: retrieve_data
arguments: data_label: D23
datatype: string

tool: D4: Enter x+y:

tool: D35: %d

tool: D33: Enter x-y:

tool: D23: The value of x and y are %d and %d:

assistant: ```c
int main() {
  int x_plus_y, x_minus_y, x, y;
  printf("Enter x+y:");
  scanf("%d", &x_plus_y);
  printf("Enter x-y:");
  scanf("%d", &x_minus_y);
  x = (x_plus_y + x_minus_y) / 2;
  y = (x_plus_y - x_minus_y) / 2;
  printf("The value of x and y are %d and %d:", x, y);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D32(%rip),%rdi
  lea    D33(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    D0(%rip),%rdi
  mov    %eax,%ecx
  add    0x4(%rsp),%eax
  sub    0x4(%rsp),%ecx
  mov    %eax,%esi
  mov    %ecx,%edx
  shr    $0x1f,%edx
  shr    $0x1f,%esi
  add    %eax,%esi
  add    %ecx,%edx
  xor    %eax,%eax
  sar    %edx
  sar    %esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D32
type_infered: string
name: retrieve_data
arguments: label: D33
type_infered: string
name: retrieve_data
arguments: label: D9
type_infered: string
name: retrieve_data
arguments: label: D0
type_infered: string

tool: D32: Enter x+y:

tool: D33: %d

tool: D9: Enter x-y:

tool: D0: The value of x and y are %d and %d:

assistant: ```c
int main() {
  int x_plus_y, x_minus_y, x, y;
  printf("Enter x+y:");
  scanf("%d", &x_plus_y);
  printf("Enter x-y:");
  scanf("%d", &x_minus_y);
  x = (x_plus_y + x_minus_y) / 2;
  y = (x_plus_y - x_minus_y) / 2;
  printf("The value of x and y are %d and %d:", x, y);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rdi
  lea    D4(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    D20(%rip),%rdi
  mov    %eax,%ecx
  add    0x4(%rsp),%eax
  sub    0x4(%rsp),%ecx
  mov    %eax,%esi
  mov    %ecx,%edx
  shr    $0x1f,%edx
  shr    $0x1f,%esi
  add    %eax,%esi
  add    %ecx,%edx
  xor    %eax,%eax
  sar    %edx
  sar    %esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D19
datatype: string
name: retrieve_data
arguments: label: D4
datatype: string
name: retrieve_data
arguments: label: D0
datatype: string
name: retrieve_data
arguments: label: D20
datatype: string

tool: D19: Enter x+y:

tool: D4: %d

tool: D0: Enter x-y:

tool: D20: The value of x and y are %d and %d:

assistant: ```c
int main() {
  int x_plus_y, x_minus_y, x, y;
  printf("Enter x+y:");
  scanf("%d", &x_plus_y);
  printf("Enter x-y:");
  scanf("%d", &x_minus_y);
  x = (x_plus_y + x_minus_y) / 2;
  y = (x_plus_y - x_minus_y) / 2;
  printf("The value of x and y are %d and %d:", x, y);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<selectPivotIndex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x3c(%rbp),%edx
  mov    -0x40(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jns    L84
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jg     L95
  mov    -0x24(%rbp),%eax
  jmp    L102
L95:
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jns    L104
  mov    -0x40(%rbp),%eax
  jmp    L102
L104:
  mov    -0x3c(%rbp),%eax
  jmp    L102
L84:
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jns    L83
  mov    -0x3c(%rbp),%eax
  jmp    L102
L83:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jns    L71
  mov    -0x40(%rbp),%eax
  jmp    L102
L71:
  mov    -0x24(%rbp),%eax
L102:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int selectPivotIndex(void **vals, int left, int right) {
  int mid;
  char *choices[3];
  choices[0] = vals[left];
  mid = (left + right + 1) / 2;
  choices[1] = vals[mid];
  choices[2] = vals[right];
  if (strcmp(choices[0], choices[1]) < 0) {
    if (strcmp(choices[1], choices[2]) <= 0) {
      return mid;
    } else if (strcmp(choices[0], choices[2]) < 0) {
      return right;
    } else {
      return left;
    }
  }
  if (strcmp(choices[0], choices[2]) < 0) {
    return left;
  } else if (strcmp(choices[1], choices[2]) < 0) {
    return right;
  } else {
    return mid;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<selectPivotIndex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  mov    %edx,%ebx
  movslq %esi,%rax
  mov    (%rdi,%rax,8),%r12
  lea    0x1(%rsi,%rdx,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%r13d
  cltq
  mov    (%rdi,%rax,8),%r14
  movslq %ebx,%rax
  mov    (%rdi,%rax,8),%r15
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L27
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L32
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  cmovs  %ebx,%r13d
L16:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L16
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  cmovs  %ebx,%ebp
  mov    %ebp,%r13d
  jmp    L16
L32:
  mov    %ebp,%r13d
  jmp    L16
```

assistant: ```c
int selectPivotIndex(void **vals, int left, int right) {
  int mid;
  char *choices[3];
  choices[0] = vals[left];
  mid = (left + right + 1) / 2;
  choices[1] = vals[mid];
  choices[2] = vals[right];
  if (strcmp(choices[0], choices[1]) < 0) {
    if (strcmp(choices[1], choices[2]) <= 0) {
      return mid;
    } else if (strcmp(choices[0], choices[2]) < 0) {
      return right;
    } else {
      return left;
    }
  }
  if (strcmp(choices[0], choices[2]) < 0) {
    return left;
  } else if (strcmp(choices[1], choices[2]) < 0) {
    return right;
  } else {
    return mid;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<selectPivotIndex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  movslq %esi,%rdx
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  mov    (%rdi,%rdx,8),%r13
  lea    0x1(%rsi,%rbp,1),%edx
  mov    %edx,%r12d
  shr    $0x1f,%r12d
  add    %edx,%r12d
  sar    %r12d
  movslq %r12d,%rdx
  mov    (%rdi,%rdx,8),%r14
  movslq %ebp,%rdx
  mov    (%rdi,%rdx,8),%r15
  mov    %r13,%rdi
  mov    %r14,%rsi
  call   <strcmp@plt>
  mov    %r15,%rsi
  test   %rax,%rax
  js     L59
  mov    %r13,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L1
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L2
L64:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L59:
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L64
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L2
L1:
  mov    %ebx,%r12d
  jmp    L64
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %ebp,%r12d
  jmp    L64
```

assistant: ```c
int selectPivotIndex(void **vals, int left, int right) {
  int mid;
  char *choices[3];
  choices[0] = vals[left];
  mid = (left + right + 1) / 2;
  choices[1] = vals[mid];
  choices[2] = vals[right];
  if (strcmp(choices[0], choices[1]) < 0) {
    if (strcmp(choices[1], choices[2]) <= 0) {
      return mid;
    } else if (strcmp(choices[0], choices[2]) < 0) {
      return right;
    } else {
      return left;
    }
  }
  if (strcmp(choices[0], choices[2]) < 0) {
    return left;
  } else if (strcmp(choices[1], choices[2]) < 0) {
    return right;
  } else {
    return mid;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<selectPivotIndex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  movslq %esi,%rdx
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  mov    (%rdi,%rdx,8),%r13
  lea    0x1(%rsi,%rbp,1),%edx
  mov    %edx,%r12d
  shr    $0x1f,%r12d
  add    %edx,%r12d
  sar    %r12d
  movslq %r12d,%rdx
  mov    (%rdi,%rdx,8),%r14
  movslq %ebp,%rdx
  mov    (%rdi,%rdx,8),%r15
  mov    %r13,%rdi
  mov    %r14,%rsi
  call   <strcmp@plt>
  mov    %r15,%rsi
  test   %rax,%rax
  js     L28
  mov    %r13,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L64
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L14
L56:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L28:
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L56
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  js     L14
L64:
  mov    %ebx,%r12d
  jmp    L56
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %ebp,%r12d
  jmp    L56
```

assistant: ```c
int selectPivotIndex(void **vals, int left, int right) {
  int mid;
  char *choices[3];
  choices[0] = vals[left];
  mid = (left + right + 1) / 2;
  choices[1] = vals[mid];
  choices[2] = vals[right];
  if (strcmp(choices[0], choices[1]) < 0) {
    if (strcmp(choices[1], choices[2]) <= 0) {
      return mid;
    } else if (strcmp(choices[0], choices[2]) < 0) {
      return right;
    } else {
      return left;
    }
  }
  if (strcmp(choices[0], choices[2]) < 0) {
    return left;
  } else if (strcmp(choices[1], choices[2]) < 0) {
    return right;
  } else {
    return mid;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<tips>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
datatype: string
name: disassemble_data
arguments: data_tag: D7
datatype: string
name: disassemble_data
arguments: data_tag: D79
datatype: string
name: disassemble_data
arguments: data_tag: D14
datatype: string
name: disassemble_data
arguments: data_tag: D54
datatype: string
name: disassemble_data
arguments: data_tag: D39
datatype: string
name: disassemble_data
arguments: data_tag: D52
datatype: string
name: disassemble_data
arguments: data_tag: D65
datatype: string

tool: D4: Notice that during encrypt the input file should be your plaintext, 

tool: D7: and the output file will be the correspondent ciphertext.


tool: D79: But, during decryption, the input file must be the ciphertext, 

tool: D14: and the output file will be the correspondent plaintext.


tool: D54: The program will also search for a file called 'keyfile' to use as key.

tool: D39: If the keyfile can't be found, the program will generate a key for you.

tool: D52: One must use the same keyfile to encrypt and decrypt to get the algorithm

tool: D65: to work properly.


assistant: ```c
void tips() {
  printf("Notice that during encrypt the input file should be your plaintext, ");
  printf("and the output file will be the correspondent ciphertext.\n");
  printf("But, during decryption, the input file must be the ciphertext, ");
  printf("and the output file will be the correspondent plaintext.\n");
  printf("The program will also search for a file called 'keyfile' to use as key.");
  printf("If the keyfile can't be found, the program will generate a key for you.");
  printf("One must use the same keyfile to encrypt and decrypt to get the algorithm");
  printf("to work properly.\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<tips>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    D45(%rip),%rdi
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
data_type: string
name: read_data
arguments: data_label: D19
data_type: string
name: read_data
arguments: data_label: D14
data_type: string
name: read_data
arguments: data_label: D4
data_type: string
name: read_data
arguments: data_label: D40
data_type: string
name: read_data
arguments: data_label: D45
data_type: string
name: read_data
arguments: data_label: D38
data_type: string
name: read_data
arguments: data_label: D27
data_type: string

tool: D0: Notice that during encrypt the input file should be your plaintext, 

tool: D19: and the output file will be the correspondent ciphertext.


tool: D14: But, during decryption, the input file must be the ciphertext, 

tool: D4: and the output file will be the correspondent plaintext.


tool: D40: The program will also search for a file called 'keyfile' to use as key.

tool: D45: If the keyfile can't be found, the program will generate a key for you.

tool: D38: One must use the same keyfile to encrypt and decrypt to get the algorithm

tool: D27: to work properly.


assistant: ```c
void tips() {
  printf("Notice that during encrypt the input file should be your plaintext, ");
  printf("and the output file will be the correspondent ciphertext.\n");
  printf("But, during decryption, the input file must be the ciphertext, ");
  printf("and the output file will be the correspondent plaintext.\n");
  printf("The program will also search for a file called 'keyfile' to use as key.");
  printf("If the keyfile can't be found, the program will generate a key for you.");
  printf("One must use the same keyfile to encrypt and decrypt to get the algorithm");
  printf("to work properly.\n");
}
```