ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <pull@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 1; i < 10; i++)
    printf("%d\n", pull(i));
  return 0;
}
```
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
  lea    D4(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <pull@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
datatype: string

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 1; i < 10; i++)
    printf("%d\n", pull(i));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xa,%edi
  call   <pell@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D2
datatype: string

tool: D2: O numero de Pell para n = 10 eh: %d

assistant: ```c
int main() {
  printf("O numero de Pell para n = 10 eh: %d", pell(10));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <pell@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
type_infered: string

tool: D4: O numero de Pell para n = 10 eh: %d

assistant: ```c
int main() {
  printf("O numero de Pell para n = 10 eh: %d", pell(10));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <pell@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type: string

tool: D3: O numero de Pell para n = 10 eh: %d

assistant: ```c
int main() {
  printf("O numero de Pell para n = 10 eh: %d", pell(10));
  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
  mov    $0xa,%edi
  call   <pell@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type_guessed: string

tool: D6: O numero de Pell para n = 10 eh: %d

assistant: ```c
int main() {
  printf("O numero de Pell para n = 10 eh: %d", pell(10));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ftok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  jns    L4
  mov    $0xffffffff,%eax
  jmp    L11
L4:
  mov    -0x10(%rbp),%eax
  movzwl %ax,%edx
  mov    -0xc(%rbp),%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  shl    $0x18,%eax
  or     %edx,%eax
L11:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return -1;
  return ((st.st_ino & 0xffff) | ((st.st_dev & 0xff) << 16) | ((id & 0xffu) << 24));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ftok>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L17
  mov    0x4(%rsp),%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  movzwl (%rsp),%edx
  or     %edx,%eax
  shl    $0x18,%ebx
  or     %ebx,%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x10,%rsp
  pop    %rbx
  ret
L17:
  mov    $0xffffffff,%eax
  jmp    L9
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return -1;
  return ((st.st_ino & 0xffff) | ((st.st_dev & 0xff) << 16) | ((id & 0xffu) << 24));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ftok>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L3
  mov    0x4(%rsp),%eax
  movzwl (%rsp),%edx
  shl    $0x18,%ebx
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %edx,%eax
  or     %ebx,%eax
L21:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  mov    $0xffffffff,%eax
  jmp    L21
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return -1;
  return ((st.st_ino & 0xffff) | ((st.st_dev & 0xff) << 16) | ((id & 0xffu) << 24));
}
```
user: What is the c source code of the assembly code below?

```asm
<ftok>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L3
  mov    0x4(%rsp),%eax
  movzwl (%rsp),%edx
  shl    $0x18,%ebx
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %edx,%eax
  or     %ebx,%eax
L20:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  mov    $0xffffffff,%eax
  jmp    L20
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return -1;
  return ((st.st_ino & 0xffff) | ((st.st_dev & 0xff) << 16) | ((id & 0xffu) << 24));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<quick_sort_int_recursive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L51
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L90
L33:
  addl   $0x1,-0x10(%rbp)
L56:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  jle    L3
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L33
  jmp    L3
L53:
  subl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  jg     L28
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L53
L28:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap_int@plt>
L90:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L56
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jl     L41
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap_int@plt>
  jmp    L16
L41:
  addl   $0x1,-0x10(%rbp)
L16:
  cmpl   $0x0,-0x10(%rbp)
  je     L19
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort_int_recursive>
L19:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort_int_recursive>
  jmp    L70
L51:
  nop
L70:
  leave
  ret
```

assistant: ```c
void quick_sort_int_recursive(int64_t arr[], int start, int end) {
  if (start >= end)
    return;
  int64_t mid = arr[end];
  int left = start, right = end - 1;
  while (left < right) {
    while (arr[left] < mid && left < right)
      left++;
    while (arr[right] >= mid && left < right)
      right--;
    swap_int(&arr[left], &arr[right]);
  }
  if (arr[left] >= arr[end])
    swap_int(&arr[left], &arr[end]);
  else
    left++;
  if (left)
    quick_sort_int_recursive(arr, start, left - 1);
  quick_sort_int_recursive(arr, left + 1, end);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<quick_sort_int_recursive>:
  endbr64
  cmp    %edx,%esi
  jge    L151
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %esi,%r15d
  mov    %edx,%r14d
  movslq %edx,%rax
  lea    (%rdi,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  mov    (%rax),%r12
  lea    -0x1(%rdx),%ebp
  mov    %esi,%ebx
  cmp    %ebp,%esi
  jl     L123
L32:
  movslq %ebx,%rax
  lea    0x0(%r13,%rax,8),%rdi
  mov    0x8(%rsp),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdi)
  jge    L16
  add    $0x1,%ebx
L137:
  test   %ebx,%ebx
  jne    L52
L53:
  lea    0x1(%rbx),%esi
  mov    %r14d,%edx
  mov    %r13,%rdi
  call   <quick_sort_int_recursive>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L46:
  call   <swap_int@plt>
  cmp    %ebp,%ebx
  jge    L32
L123:
  movslq %ebx,%rax
  shl    $0x3,%rax
  lea    0x0(%r13,%rax,1),%rdi
  cmp    (%rdi),%r12
  jle    L56
  lea    0x8(%r13,%rax,1),%rax
  cmp    %ebp,%ebx
  jge    L56
L110:
  add    $0x1,%ebx
  mov    %rax,%rdi
  add    $0x8,%rax
  cmp    %r12,(%rdi)
  jge    L56
  cmp    %ebp,%ebx
  jl     L110
L56:
  movslq %ebp,%rax
  lea    0x0(%r13,%rax,8),%rsi
  cmp    (%rsi),%r12
  jg     L46
  cmp    %ebp,%ebx
  jge    L46
  lea    -0x8(%r13,%rax,8),%rax
L3:
  sub    $0x1,%ebp
  mov    %rax,%rsi
  sub    $0x8,%rax
  cmp    %r12,(%rsi)
  jl     L46
  cmp    %ebx,%ebp
  jg     L3
  jmp    L46
L16:
  mov    0x8(%rsp),%rsi
  call   <swap_int@plt>
  jmp    L137
L52:
  lea    -0x1(%rbx),%edx
  mov    %r15d,%esi
  mov    %r13,%rdi
  call   <quick_sort_int_recursive>
  jmp    L53
L151:
  ret
```

assistant: ```c
void quick_sort_int_recursive(int64_t arr[], int start, int end) {
  if (start >= end)
    return;
  int64_t mid = arr[end];
  int left = start, right = end - 1;
  while (left < right) {
    while (arr[left] < mid && left < right)
      left++;
    while (arr[right] >= mid && left < right)
      right--;
    swap_int(&arr[left], &arr[right]);
  }
  if (arr[left] >= arr[end])
    swap_int(&arr[left], &arr[end]);
  else
    left++;
  if (left)
    quick_sort_int_recursive(arr, start, left - 1);
  quick_sort_int_recursive(arr, left + 1, end);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<quick_sort_int_recursive>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0x8(%rsp)
  cmp    %edx,%esi
  jge    L173
  movslq %edx,%rax
  mov    %rdi,%r14
  mov    %esi,%r12d
  lea    (%rdi,%rax,8),%rax
  mov    %rax,(%rsp)
  lea    -0x1(%rdx),%eax
  mov    %eax,0xc(%rsp)
L63:
  mov    (%rsp),%rax
  mov    %r12d,%ebx
  mov    (%rax),%r13
  mov    0xc(%rsp),%eax
  mov    %eax,%ebp
  cmp    %eax,%r12d
  jge    L87
  nop
L50:
  movslq %ebx,%rax
  shl    $0x3,%rax
  cmp    %ebp,%ebx
  lea    (%r14,%rax,1),%r15
  setl   %r8b
  cmp    (%r15),%r13
  jle    L103
  lea    0x8(%r14,%rax,1),%rax
  jmp    L23
  nopl   (%rax)
L95:
  add    $0x1,%ebx
  mov    %rax,%r15
  cmp    %ebp,%ebx
  setl   %r8b
  add    $0x8,%rax
  cmp    %r13,(%r15)
  jge    L103
L23:
  test   %r8b,%r8b
  jne    L95
L103:
  movslq %ebp,%rax
  lea    (%r14,%rax,8),%rsi
  cmp    (%rsi),%r13
  jg     L40
  test   %r8b,%r8b
  je     L40
  lea    -0x8(%r14,%rax,8),%rax
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L157:
  cmp    %ebx,%ebp
  jle    L40
L28:
  mov    %rax,%rsi
  sub    $0x1,%ebp
  sub    $0x8,%rax
  cmp    %r13,(%rsi)
  jge    L157
L40:
  mov    %r15,%rdi
  call   <swap_int@plt>
  cmp    %ebp,%ebx
  jl     L50
  mov    (%rsp),%rax
  mov    (%rax),%r13
L25:
  cmp    %r13,(%r15)
  jge    L165
  add    $0x1,%ebx
  test   %ebx,%ebx
  jne    L4
L114:
  lea    0x1(%rbx),%r12d
  cmp    0x8(%rsp),%r12d
  jl     L63
L173:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L165:
  mov    (%rsp),%rsi
  mov    %r15,%rdi
  call   <swap_int@plt>
  test   %ebx,%ebx
  je     L114
L4:
  lea    -0x1(%rbx),%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  call   <quick_sort_int_recursive>
  jmp    L114
L87:
  movslq %r12d,%rax
  lea    (%r14,%rax,8),%r15
  jmp    L25
```

assistant: ```c
void quick_sort_int_recursive(int64_t arr[], int start, int end) {
  if (start >= end)
    return;
  int64_t mid = arr[end];
  int left = start, right = end - 1;
  while (left < right) {
    while (arr[left] < mid && left < right)
      left++;
    while (arr[right] >= mid && left < right)
      right--;
    swap_int(&arr[left], &arr[right]);
  }
  if (arr[left] >= arr[end])
    swap_int(&arr[left], &arr[end]);
  else
    left++;
  if (left)
    quick_sort_int_recursive(arr, start, left - 1);
  quick_sort_int_recursive(arr, left + 1, end);
}
```
user: Translate the below assembly code into C programming language:

```asm
<quick_sort_int_recursive>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  cmp    %edx,%esi
  jge    L99
  movslq %edx,%rax
  mov    %rdi,%r15
  mov    %esi,%r12d
  lea    (%rdi,%rax,8),%rax
  mov    %rax,(%rsp)
  lea    -0x1(%rdx),%eax
  mov    %eax,0x8(%rsp)
L131:
  mov    (%rsp),%rax
  mov    %r12d,%ebx
  mov    (%rax),%r13
  mov    0x8(%rsp),%eax
  mov    %eax,%ebp
  cmp    %eax,%r12d
  jge    L107
  nop
L174:
  movslq %ebx,%rax
  shl    $0x3,%rax
  cmp    %ebp,%ebx
  lea    (%r15,%rax,1),%r14
  setl   %r8b
  cmp    (%r14),%r13
  jle    L172
  lea    0x8(%r15,%rax,1),%rax
  jmp    L61
  nopl   (%rax)
L114:
  add    $0x1,%ebx
  mov    %rax,%r14
  cmp    %ebp,%ebx
  setl   %r8b
  add    $0x8,%rax
  cmp    %r13,(%r14)
  jge    L172
L61:
  test   %r8b,%r8b
  jne    L114
L172:
  movslq %ebp,%rax
  lea    (%r15,%rax,8),%rsi
  cmp    (%rsi),%r13
  jg     L136
  test   %r8b,%r8b
  je     L136
  lea    -0x8(%r15,%rax,8),%rax
  jmp    L105
  nopw   0x0(%rax,%rax,1)
L187:
  cmp    %ebx,%ebp
  jle    L136
L105:
  mov    %rax,%rsi
  sub    $0x1,%ebp
  sub    $0x8,%rax
  cmp    %r13,(%rsi)
  jge    L187
L136:
  mov    %r14,%rdi
  call   <swap_int@plt>
  cmp    %ebp,%ebx
  jl     L174
  mov    (%rsp),%rax
  mov    (%rax),%r13
L116:
  cmp    %r13,(%r14)
  jge    L47
  add    $0x1,%ebx
  test   %ebx,%ebx
  jne    L20
L166:
  cmpl   $0x1,0xc(%rsp)
  jle    L99
  mov    $0x1,%r12d
  jmp    L131
L47:
  mov    (%rsp),%rsi
  mov    %r14,%rdi
  call   <swap_int@plt>
  test   %ebx,%ebx
  je     L166
L20:
  mov    %r12d,%esi
  lea    -0x1(%rbx),%edx
  lea    0x1(%rbx),%r12d
  mov    %r15,%rdi
  call   <quick_sort_int_recursive>
  cmp    %r12d,0xc(%rsp)
  jg     L131
L99:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L107:
  movslq %r12d,%rax
  lea    (%r15,%rax,8),%r14
  jmp    L116
```

assistant: ```c
void quick_sort_int_recursive(int64_t arr[], int start, int end) {
  if (start >= end)
    return;
  int64_t mid = arr[end];
  int left = start, right = end - 1;
  while (left < right) {
    while (arr[left] < mid && left < right)
      left++;
    while (arr[right] >= mid && left < right)
      right--;
    swap_int(&arr[left], &arr[right]);
  }
  if (arr[left] >= arr[end])
    swap_int(&arr[left], &arr[end]);
  else
    left++;
  if (left)
    quick_sort_int_recursive(arr, start, left - 1);
  quick_sort_int_recursive(arr, left + 1, end);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<alloc_replace_initial_substrings>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x40(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <count_initial_substrings@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  sub    -0x28(%rbp),%rax
  imul   -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x48(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  mov    -0x38(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <replace_initial_substrings@plt>
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *alloc_replace_initial_substrings(const char *src_str, const char *old_sub_str, const char *new_sub_str) {
  char *dest_str;
  size_t old_sub_str_len = strlen(old_sub_str);
  size_t new_sub_str_len = strlen(new_sub_str);
  size_t sub_str_occs = count_initial_substrings(src_str, old_sub_str);
  size_t size_dest_str_len = strlen(src_str) - sub_str_occs * old_sub_str_len + sub_str_occs * new_sub_str_len;
  dest_str = (char *)malloc(size_dest_str_len + 1);
  if (dest_str != NULL)
    replace_initial_substrings(dest_str, src_str, old_sub_str, new_sub_str);
  return (dest_str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<alloc_replace_initial_substrings>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%r15
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <count_initial_substrings@plt>
  mov    %rax,%r14
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    %r15,%rbx
  imul   %r14,%rbx
  lea    0x1(%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L19
L4:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L19:
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <replace_initial_substrings@plt>
  jmp    L4
```

assistant: ```c
char *alloc_replace_initial_substrings(const char *src_str, const char *old_sub_str, const char *new_sub_str) {
  char *dest_str;
  size_t old_sub_str_len = strlen(old_sub_str);
  size_t new_sub_str_len = strlen(new_sub_str);
  size_t sub_str_occs = count_initial_substrings(src_str, old_sub_str);
  size_t size_dest_str_len = strlen(src_str) - sub_str_occs * old_sub_str_len + sub_str_occs * new_sub_str_len;
  dest_str = (char *)malloc(size_dest_str_len + 1);
  if (dest_str != NULL)
    replace_initial_substrings(dest_str, src_str, old_sub_str, new_sub_str);
  return (dest_str);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<alloc_replace_initial_substrings>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r15
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <count_initial_substrings@plt>
  sub    %r15,%rbx
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  imul   %r12,%rbx
  lea    0x1(%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L6
L15:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <replace_initial_substrings@plt>
  jmp    L15
```

assistant: ```c
char *alloc_replace_initial_substrings(const char *src_str, const char *old_sub_str, const char *new_sub_str) {
  char *dest_str;
  size_t old_sub_str_len = strlen(old_sub_str);
  size_t new_sub_str_len = strlen(new_sub_str);
  size_t sub_str_occs = count_initial_substrings(src_str, old_sub_str);
  size_t size_dest_str_len = strlen(src_str) - sub_str_occs * old_sub_str_len + sub_str_occs * new_sub_str_len;
  dest_str = (char *)malloc(size_dest_str_len + 1);
  if (dest_str != NULL)
    replace_initial_substrings(dest_str, src_str, old_sub_str, new_sub_str);
  return (dest_str);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<alloc_replace_initial_substrings>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r15
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <count_initial_substrings@plt>
  sub    %r15,%rbx
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  imul   %r12,%rbx
  lea    0x1(%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L10
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <replace_initial_substrings@plt>
  jmp    L8
```

assistant: ```c
char *alloc_replace_initial_substrings(const char *src_str, const char *old_sub_str, const char *new_sub_str) {
  char *dest_str;
  size_t old_sub_str_len = strlen(old_sub_str);
  size_t new_sub_str_len = strlen(new_sub_str);
  size_t sub_str_occs = count_initial_substrings(src_str, old_sub_str);
  size_t size_dest_str_len = strlen(src_str) - sub_str_occs * old_sub_str_len + sub_str_occs * new_sub_str_len;
  dest_str = (char *)malloc(size_dest_str_len + 1);
  if (dest_str != NULL)
    replace_initial_substrings(dest_str, src_str, old_sub_str, new_sub_str);
  return (dest_str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sock_close>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <shutdown@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int sock_close(int fd) {
  int err;
  err = shutdown(fd, 2);
  return err;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sock_close>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  call   <shutdown@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int sock_close(int fd) {
  int err;
  err = shutdown(fd, 2);
  return err;
}
```