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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_create_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_get_head_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_get_tail_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_get_element_at_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_is_equal_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_shallow_copy_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_prepend_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_append_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_head_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_middle_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_tail_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_none_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_from_empty_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_all_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(test_create_singly_linked_list() == 0);
  assert(test_get_head_singly_linked_list() == 0);
  assert(test_get_tail_singly_linked_list() == 0);
  assert(test_get_element_at_singly_linked_list() == 0);
  assert(test_is_equal_singly_linked_list() == 0);
  assert(test_shallow_copy_singly_linked_list() == 0);
  assert(test_prepend_singly_linked_list() == 0);
  assert(test_append_singly_linked_list() == 0);
  assert(test_remove_head_singly_linked_list() == 0);
  assert(test_remove_middle_singly_linked_list() == 0);
  assert(test_remove_tail_singly_linked_list() == 0);
  assert(test_remove_none_singly_linked_list() == 0);
  assert(test_remove_from_empty_singly_linked_list() == 0);
  assert(test_remove_all_singly_linked_list() == 0);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_create_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_get_head_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_get_tail_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_get_element_at_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_is_equal_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_shallow_copy_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_prepend_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_append_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_head_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_middle_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_tail_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_none_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_from_empty_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  call   <test_remove_all_singly_linked_list@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(test_create_singly_linked_list() == 0);
  assert(test_get_head_singly_linked_list() == 0);
  assert(test_get_tail_singly_linked_list() == 0);
  assert(test_get_element_at_singly_linked_list() == 0);
  assert(test_is_equal_singly_linked_list() == 0);
  assert(test_shallow_copy_singly_linked_list() == 0);
  assert(test_prepend_singly_linked_list() == 0);
  assert(test_append_singly_linked_list() == 0);
  assert(test_remove_head_singly_linked_list() == 0);
  assert(test_remove_middle_singly_linked_list() == 0);
  assert(test_remove_tail_singly_linked_list() == 0);
  assert(test_remove_none_singly_linked_list() == 0);
  assert(test_remove_from_empty_singly_linked_list() == 0);
  assert(test_remove_all_singly_linked_list() == 0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L27
L19:
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <euclidianMDC@plt>
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L27:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L19
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string
name: parse_data
arguments: data_label: D18
type_infered: string
name: parse_data
arguments: data_label: D13
type_infered: string

tool: D2: %d

tool: D18: %d %d

tool: D13: %d


assistant: ```c
int main(void) {
  int n, i;
  int f1, f2;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &f1, &f2);
    printf("%d\n", euclidianMDC(f1, f2));
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L1
  mov    $0x0,%ebx
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D0(%rip),%r12
  lea    D14(%rip),%rbp
L8:
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  call   <euclidianMDC@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jg     L8
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D20
type_guessed: string
name: extract_symbolic_data
arguments: label: D0
type_guessed: string
name: extract_symbolic_data
arguments: label: D14
type_guessed: string

tool: D20: %d

tool: D0: %d %d

tool: D14: %d


assistant: ```c
int main(void) {
  int n, i;
  int f1, f2;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &f1, &f2);
    printf("%d\n", euclidianMDC(f1, f2));
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  lea    D0(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L4
  xor    %r14d,%r14d
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D8(%rip),%rbp
  lea    D19(%rip),%rbx
  nopl   0x0(%rax)
L11:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  add    $0x1,%r14d
  call   <euclidianMDC@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    %r14d,0xc(%rsp)
  jg     L11
L4:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
datatype: string
name: retrieve_data
arguments: data_label: D8
datatype: string
name: retrieve_data
arguments: data_label: D19
datatype: string

tool: D0: %d

tool: D8: %d %d

tool: D19: %d


assistant: ```c
int main(void) {
  int n, i;
  int f1, f2;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &f1, &f2);
    printf("%d\n", euclidianMDC(f1, f2));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D17(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L24
  xor    %r14d,%r14d
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D13(%rip),%rbp
  lea    D24(%rip),%rbx
  nopl   0x0(%rax)
L7:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  add    $0x1,%r14d
  call   <euclidianMDC@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    %r14d,0xc(%rsp)
  jg     L7
L24:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D17
data_type: string
name: disassemble_data
arguments: data_tag: D13
data_type: string
name: disassemble_data
arguments: data_tag: D24
data_type: string

tool: D17: %d

tool: D13: %d %d

tool: D24: %d


assistant: ```c
int main(void) {
  int n, i;
  int f1, f2;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &f1, &f2);
    printf("%d\n", euclidianMDC(f1, f2));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bbz_int_log2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  jmp    L19
L5:
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L19:
  cmpl   $0x20,-0x8(%rbp)
  jg     L27
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L5
L27:
  cmpl   $0x20,-0x8(%rbp)
  setle  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int bbz_int_log2(int u) {
  int i = 0;
  int r = 1;
  assert(u > 0);
  while ((i <= 32) && (r < u)) {
    r = 2 * r + 1;
    i = i + 1;
  }
  assert(i <= 32);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bbz_int_log2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  test   %edi,%edi
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x1,%ebp
  jle    L23
  mov    $0x1,%eax
  mov    $0x0,%ebx
L9:
  lea    0x1(%rax,%rax,1),%eax
  add    $0x1,%ebx
  cmp    $0x20,%ebx
  setle  %dil
  cmp    %eax,%ebp
  jle    L3
  test   %dil,%dil
  jne    L9
L3:
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  mov    $0x0,%ebx
  mov    $0x1,%edi
  jmp    L3
```

assistant: ```c
int bbz_int_log2(int u) {
  int i = 0;
  int r = 1;
  assert(u > 0);
  while ((i <= 32) && (r < u)) {
    r = 2 * r + 1;
    i = i + 1;
  }
  assert(i <= 32);
  return i;
}
```
user: What is the c source code of the assembly code below:

```asm
<bbz_int_log2>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x1,%ebx
  jle    L5
  mov    $0x1,%eax
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L1:
  add    $0x1,%r12d
  lea    0x1(%rax,%rax,1),%eax
  cmp    $0x20,%r12d
  setle  %dil
  cmp    %eax,%ebx
  jle    L7
  test   %dil,%dil
  jne    L1
L7:
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L5:
  mov    $0x1,%edi
  xor    %r12d,%r12d
  call   <assert@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int bbz_int_log2(int u) {
  int i = 0;
  int r = 1;
  assert(u > 0);
  while ((i <= 32) && (r < u)) {
    r = 2 * r + 1;
    i = i + 1;
  }
  assert(i <= 32);
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bbz_int_log2>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x1,%ebx
  jle    L22
  mov    $0x1,%eax
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L19:
  add    $0x1,%r12d
  lea    0x1(%rax,%rax,1),%eax
  cmp    $0x20,%r12d
  setle  %dil
  cmp    %eax,%ebx
  jle    L18
  test   %dil,%dil
  jne    L19
L18:
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L22:
  mov    $0x1,%edi
  xor    %r12d,%r12d
  call   <assert@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int bbz_int_log2(int u) {
  int i = 0;
  int r = 1;
  assert(u > 0);
  while ((i <= 32) && (r < u)) {
    r = 2 * r + 1;
    i = i + 1;
  }
  assert(i <= 32);
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<del_set>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <idx@plt>
  mov    %eax,%ebx
  test   %ebx,%ebx
  je     L7
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <len@plt>
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <val@plt>
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  mov    %ebx,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <str@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <pop@plt>
L7:
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void del_set(a, v) {
  register int jj;
  jj = idx(a, v);
  if (!(jj == 0)) {
    str(a, jj, val(a, len(a)));
    pop(a);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<del_set>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <idx@plt>
  test   %eax,%eax
  jne    L2
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  mov    %eax,%ebp
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <len@plt>
  mov    %eax,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <val@plt>
  mov    %eax,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <str@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <pop@plt>
  jmp    L12
```

assistant: ```c
void del_set(a, v) {
  register int jj;
  jj = idx(a, v);
  if (!(jj == 0)) {
    str(a, jj, val(a, len(a)));
    pop(a);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<del_set>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <idx@plt>
  test   %eax,%eax
  jne    L6
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    %eax,%r12d
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <len@plt>
  mov    %ebp,%edi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <val@plt>
  mov    %ebp,%edi
  mov    %r12d,%esi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <str@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10b0 <pop@plt>
```

assistant: ```c
void del_set(a, v) {
  register int jj;
  jj = idx(a, v);
  if (!(jj == 0)) {
    str(a, jj, val(a, len(a)));
    pop(a);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<del_set>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <idx@plt>
  test   %eax,%eax
  jne    L8
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L8:
  mov    %eax,%r12d
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <len@plt>
  mov    %ebp,%edi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <val@plt>
  mov    %ebp,%edi
  mov    %r12d,%esi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <str@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10b0 <pop@plt>
```

assistant: ```c
void del_set(a, v) {
  register int jj;
  jj = idx(a, v);
  if (!(jj == 0)) {
    str(a, jj, val(a, len(a)));
    pop(a);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<anetRead>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L77
L25:
  mov    -0x28(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L80
  mov    -0x10(%rbp),%rax
  jmp    L85
L80:
  cmpq   $0xffffffffffffffff,-0x8(%rbp)
  jne    L76
  mov    D15(%rip),%rax
  mov    (%rax),%rdx
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L76
  jmp    L77
L76:
  cmpq   $0xffffffffffffffff,-0x8(%rbp)
  jne    L82
  mov    $0xffffffffffffffff,%rax
  jmp    L85
L82:
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    %rax,-0x20(%rbp)
L77:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jne    L25
  mov    -0x10(%rbp),%rax
L85:
  leave
  ret
```

assistant: ```c
size_t anetRead(int fd, char *buf, size_t count) {
  size_t nread;
  size_t totlen = 0;
  while (totlen != count) {
    nread = read(fd, buf, count - totlen);
    if (nread == 0)
      return totlen;
    if (nread == -1 && errno == EINTR) {
      continue;
    }
    if (nread == -1)
      return -1;
    totlen += nread;
    buf += nread;
  }
  return totlen;
}
```
user: Translate the below assembly code into C programming language:

```asm
<anetRead>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%r12
  test   %rdx,%rdx
  je     L77
  mov    %edi,%r13d
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  jmp    L64
L2:
  add    %rax,%rbx
  add    %rax,%rbp
L80:
  cmp    %rbx,%r12
  je     L37
L64:
  mov    %r12,%rdx
  sub    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  test   %rax,%rax
  je     L82
  cmp    $0xffffffffffffffff,%rax
  jne    L2
  mov    D8(%rip),%rcx
  mov    D17(%rip),%rdx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L80
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  mov    %rbx,%rax
  jmp    L19
L77:
  mov    %rdx,%rax
  jmp    L19
L82:
  mov    %rbx,%rax
  jmp    L19
```

assistant: ```c
size_t anetRead(int fd, char *buf, size_t count) {
  size_t nread;
  size_t totlen = 0;
  while (totlen != count) {
    nread = read(fd, buf, count - totlen);
    if (nread == 0)
      return totlen;
    if (nread == -1 && errno == EINTR) {
      continue;
    }
    if (nread == -1)
      return -1;
    totlen += nread;
    buf += nread;
  }
  return totlen;
}
```
user: Translate the below assembly code into C programming language:

```asm
<anetRead>:
  endbr64
  test   %rdx,%rdx
  je     L41
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L56
  nopl   0x0(%rax)
L24:
  add    %rax,%rbx
  add    %rax,%rbp
L20:
  cmp    %rbx,%r13
  je     L48
L56:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  sub    %rbx,%rdx
  call   <read@plt>
  test   %rax,%rax
  je     L29
  cmp    $0xffffffffffffffff,%rax
  jne    L24
  mov    D3(%rip),%rdx
  mov    D11(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L20
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L48:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L41:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t anetRead(int fd, char *buf, size_t count) {
  size_t nread;
  size_t totlen = 0;
  while (totlen != count) {
    nread = read(fd, buf, count - totlen);
    if (nread == 0)
      return totlen;
    if (nread == -1 && errno == EINTR) {
      continue;
    }
    if (nread == -1)
      return -1;
    totlen += nread;
    buf += nread;
  }
  return totlen;
}
```
user: What is the c source code of the assembly code below?

```asm
<anetRead>:
  endbr64
  test   %rdx,%rdx
  je     L46
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L55
  nopl   0x0(%rax)
L7:
  add    %rax,%rbx
  add    %rax,%rbp
L32:
  cmp    %rbx,%r13
  je     L39
L55:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  sub    %rbx,%rdx
  call   <read@plt>
  test   %rax,%rax
  je     L44
  cmp    $0xffffffffffffffff,%rax
  jne    L7
  mov    D3(%rip),%rdx
  mov    D8(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L32
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L44:
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L39:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t anetRead(int fd, char *buf, size_t count) {
  size_t nread;
  size_t totlen = 0;
  while (totlen != count) {
    nread = read(fd, buf, count - totlen);
    if (nread == 0)
      return totlen;
    if (nread == -1 && errno == EINTR) {
      continue;
    }
    if (nread == -1)
      return -1;
    totlen += nread;
    buf += nread;
  }
  return totlen;
}
```