ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x2,(%rsp)
  movl   $0x3,0x4(%rsp)
  movl   $0x5,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  movl   $0x7,0x10(%rsp)
  movl   $0x8,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  movl   $0x4,0x1c(%rsp)
  movl   $0x2,0x20(%rsp)
  movl   $0x3,0x24(%rsp)
  movl   $0x6,0x28(%rsp)
  movl   $0x9,0x2c(%rsp)
  movl   $0xa,0x30(%rsp)
  movl   $0x4,0x34(%rsp)
  mov    %rsp,%rbx
  mov    $0xd,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <partition@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0xd,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <tail_recursive_quicksort@plt>
  lea    0x38(%rsp),%r12
  lea    D11(%rip),%rbp
L13:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L13
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type: string
name: extract_symbolic_data
arguments: data_label: D11
type: string

tool: D6: q = %d


tool: D11: %d


assistant: ```c
int main(void) {
  int i;
  int num[] = {2, 3, 5, 1, 7, 8, 6, 4, 2, 3, 6, 9, 10, 4};
  printf("q = %d\n", partition(num, 0, 13));
  tail_recursive_quicksort(num, 0, 13);
  for (i = 0; i < 14; i++)
    printf("%d\n", num[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  movabs $0x100000005,%rcx
  push   %r12
  xor    %esi,%esi
  movabs $0x800000007,%rdx
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x300000002,%rax
  mov    %rsp,%rbx
  mov    %rcx,0x8(%rsp)
  lea    0x38(%rsp),%r12
  movabs $0x400000006,%rcx
  mov    %rdx,0x10(%rsp)
  mov    %rbx,%rdi
  mov    $0xd,%edx
  mov    %rcx,0x18(%rsp)
  mov    %rax,(%rsp)
  mov    %rax,0x20(%rsp)
  movabs $0x900000006,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x40000000a,%rax
  mov    %rax,0x30(%rsp)
  call   <partition@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xd,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <tail_recursive_quicksort@plt>
  nopl   0x0(%rax,%rax,1)
L6:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L6
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D10
data_type: string
name: parse_data
arguments: label: D4
data_type: string

tool: D10: %d


tool: D4: q = %d


assistant: ```c
int main(void) {
  int i;
  int num[] = {2, 3, 5, 1, 7, 8, 6, 4, 2, 3, 6, 9, 10, 4};
  printf("q = %d\n", partition(num, 0, 13));
  tail_recursive_quicksort(num, 0, 13);
  for (i = 0; i < 14; i++)
    printf("%d\n", num[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xd,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D43(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  movdqa D46(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    D25(%rip),%rax
  lea    0x38(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D53(%rip),%xmm0
  mov    %rbx,%rdi
  mov    %rax,0x30(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D20(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <partition@plt>
  lea    D23(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xd,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <tail_recursive_quicksort@plt>
  nopl   0x0(%rax)
L8:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L8
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D43
datatype: string
name: read_data
arguments: data_label: D46
datatype: byte[16]
name: read_data
arguments: data_label: D25
datatype: byte[8]
name: read_data
arguments: data_label: D53
datatype: byte[16]
name: read_data
arguments: data_label: D20
datatype: byte[16]
name: read_data
arguments: data_label: D23
datatype: string

tool: D43: %d


tool: {"D46": [0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D25": [0x0a,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D53": [0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D20": [0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x09,0x00,0x00,0x00]}

tool: D23: q = %d


assistant: ```c
int main(void) {
  int i;
  int num[] = {2, 3, 5, 1, 7, 8, 6, 4, 2, 3, 6, 9, 10, 4};
  printf("q = %d\n", partition(num, 0, 13));
  tail_recursive_quicksort(num, 0, 13);
  for (i = 0; i < 14; i++)
    printf("%d\n", num[i]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_jump_impossible@plt>
  test   %eax,%eax
  jne    L1
  addl   $0x1,-0x4(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L1:
  mov    $0x0,%eax
  call   <test_min_jumps_ones@plt>
  test   %eax,%eax
  jne    L20
  addl   $0x1,-0x4(%rbp)
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L20:
  mov    $0x0,%eax
  call   <test_min_jumps@plt>
  test   %eax,%eax
  jne    L19
  addl   $0x1,-0x4(%rbp)
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L19:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D2
datatype: string
name: disassemble_data
arguments: label: D39
datatype: string
name: disassemble_data
arguments: label: D38
datatype: string
name: disassemble_data
arguments: label: D36
datatype: string

tool: D2: Impossible jump test failed!

tool: D39: Min jumps ones test failed!

tool: D38: Min jumps test failed!

tool: D36: %d tests failed.


assistant: ```c
int main() {
  int counter = 0;
  if (!test_jump_impossible()) {
    counter++;
    printf("Impossible jump test failed!\n");
  }
  if (!test_min_jumps_ones()) {
    counter++;
    printf("Min jumps ones test failed!\n");
  }
  if (!test_min_jumps()) {
    counter++;
    printf("Min jumps test failed!\n");
  }
  printf("%d tests failed.\n", counter);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <test_jump_impossible@plt>
  mov    $0x0,%ebx
  test   %eax,%eax
  je     L1
L44:
  mov    $0x0,%eax
  call   <test_min_jumps_ones@plt>
  test   %eax,%eax
  je     L45
L53:
  mov    $0x0,%eax
  call   <test_min_jumps@plt>
  test   %eax,%eax
  je     L28
L14:
  mov    %ebx,%esi
  lea    D37(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L1:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%ebx
  jmp    L44
L45:
  add    $0x1,%ebx
  lea    D27(%rip),%rdi
  call   <puts@plt>
  jmp    L53
L28:
  add    $0x1,%ebx
  lea    D3(%rip),%rdi
  call   <puts@plt>
  jmp    L14
```

tool_calls: name: parse_data
arguments: label: D37
data_type: string
name: parse_data
arguments: label: D16
data_type: string
name: parse_data
arguments: label: D27
data_type: string
name: parse_data
arguments: label: D3
data_type: string

tool: D37: %d tests failed.


tool: D16: Impossible jump test failed!

tool: D27: Min jumps ones test failed!

tool: D3: Min jumps test failed!

assistant: ```c
int main() {
  int counter = 0;
  if (!test_jump_impossible()) {
    counter++;
    printf("Impossible jump test failed!\n");
  }
  if (!test_min_jumps_ones()) {
    counter++;
    printf("Min jumps ones test failed!\n");
  }
  if (!test_min_jumps()) {
    counter++;
    printf("Min jumps test failed!\n");
  }
  printf("%d tests failed.\n", counter);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <test_jump_impossible@plt>
  test   %eax,%eax
  je     L58
L31:
  xor    %eax,%eax
  call   <test_min_jumps_ones@plt>
  test   %eax,%eax
  je     L17
L20:
  xor    %eax,%eax
  call   <test_min_jumps@plt>
  test   %eax,%eax
  je     L23
L14:
  mov    %ebp,%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L58:
  lea    D8(%rip),%rdi
  mov    $0x1,%ebp
  call   <puts@plt>
  jmp    L31
L23:
  lea    D0(%rip),%rdi
  add    $0x1,%ebp
  call   <puts@plt>
  jmp    L14
L17:
  lea    D13(%rip),%rdi
  add    $0x1,%ebp
  call   <puts@plt>
  jmp    L20
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D19
datatype: string
name: extract_symbolic_data
arguments: data_label: D8
datatype: string
name: extract_symbolic_data
arguments: data_label: D0
datatype: string
name: extract_symbolic_data
arguments: data_label: D13
datatype: string

tool: D19: %d tests failed.


tool: D8: Impossible jump test failed!

tool: D0: Min jumps test failed!

tool: D13: Min jumps ones test failed!

assistant: ```c
int main() {
  int counter = 0;
  if (!test_jump_impossible()) {
    counter++;
    printf("Impossible jump test failed!\n");
  }
  if (!test_min_jumps_ones()) {
    counter++;
    printf("Min jumps ones test failed!\n");
  }
  if (!test_min_jumps()) {
    counter++;
    printf("Min jumps test failed!\n");
  }
  printf("%d tests failed.\n", counter);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <test_jump_impossible@plt>
  test   %eax,%eax
  je     L44
L20:
  xor    %eax,%eax
  call   <test_min_jumps_ones@plt>
  test   %eax,%eax
  je     L4
L9:
  xor    %eax,%eax
  call   <test_min_jumps@plt>
  test   %eax,%eax
  je     L27
L45:
  mov    %ebp,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L44:
  lea    D8(%rip),%rdi
  mov    $0x2,%ebx
  mov    $0x1,%ebp
  call   <puts@plt>
  jmp    L20
L27:
  lea    D10(%rip),%rdi
  add    $0x1,%ebp
  call   <puts@plt>
  jmp    L45
L4:
  lea    D32(%rip),%rdi
  mov    %ebx,%ebp
  call   <puts@plt>
  jmp    L9
```

tool_calls: name: disassemble_data
arguments: tag: D13
type_infered: string
name: disassemble_data
arguments: tag: D8
type_infered: string
name: disassemble_data
arguments: tag: D10
type_infered: string
name: disassemble_data
arguments: tag: D32
type_infered: string

tool: D13: %d tests failed.


tool: D8: Impossible jump test failed!

tool: D10: Min jumps test failed!

tool: D32: Min jumps ones test failed!

assistant: ```c
int main() {
  int counter = 0;
  if (!test_jump_impossible()) {
    counter++;
    printf("Impossible jump test failed!\n");
  }
  if (!test_min_jumps_ones()) {
    counter++;
    printf("Min jumps ones test failed!\n");
  }
  if (!test_min_jumps()) {
    counter++;
    printf("Min jumps test failed!\n");
  }
  printf("%d tests failed.\n", counter);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x2,%edi
  call   <f@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: string

tool: D1: %lf


assistant: ```c
int main() {
  printf("%lf\n", f(2));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <f@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type: string

tool: D1: %lf


assistant: ```c
int main() {
  printf("%lf\n", f(2));
  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    $0x2,%edi
  call   <f@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: string

tool: D6: %lf


assistant: ```c
int main() {
  printf("%lf\n", f(2));
  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
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <f@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_guessed: string

tool: D5: %lf


assistant: ```c
int main() {
  printf("%lf\n", f(2));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_key>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L0:
  mov    $0x0,%eax
  call   <scan_keyboard@plt>
  mov    %eax,-0x4(%rbp)
L17:
  cmpl   $0x0,-0x4(%rbp)
  je     L0
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int get_key(void) {
  int c = 0;
  while (c == 0) {
    c = scan_keyboard();
  }
  return (c);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_key>:
  endbr64
  sub    $0x8,%rsp
L9:
  mov    $0x0,%eax
  call   <scan_keyboard@plt>
  test   %eax,%eax
  je     L9
  add    $0x8,%rsp
  ret
```

assistant: ```c
int get_key(void) {
  int c = 0;
  while (c == 0) {
    c = scan_keyboard();
  }
  return (c);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<get_key>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <scan_keyboard@plt>
  test   %eax,%eax
  je     L3
  add    $0x8,%rsp
  ret
```

assistant: ```c
int get_key(void) {
  int c = 0;
  while (c == 0) {
    c = scan_keyboard();
  }
  return (c);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_key>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  call   <scan_keyboard@plt>
  test   %eax,%eax
  je     L0
  add    $0x8,%rsp
  ret
```

assistant: ```c
int get_key(void) {
  int c = 0;
  while (c == 0) {
    c = scan_keyboard();
  }
  return (c);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <setup@plt>
  mov    $0x0,%eax
  call   <loop@plt>
  nop
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  setup();
  do {
    loop();
    break;
  } while (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <setup@plt>
  mov    $0x0,%eax
  call   <loop@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  setup();
  do {
    loop();
    break;
  } while (1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <setup@plt>
  xor    %eax,%eax
  call   <loop@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  setup();
  do {
    loop();
    break;
  } while (1);
}
```
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   <setup@plt>
  xor    %eax,%eax
  call   <loop@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  setup();
  do {
    loop();
    break;
  } while (1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<insert_string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %edx,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movslq %edx,%rcx
  cmpb   $0x0,(%rdi,%rcx,1)
  je     L137
  movzbl (%rdi),%r9d
  lea    0x1(%rdi),%r10
  mov    $0x0,%edx
  mov    $0x0,%r12d
  mov    %rsp,%rbp
  test   %r9b,%r9b
  jne    L105
L117:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  movzbl (%rsp),%edx
  test   %dl,%dl
  je     L53
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
L95:
  mov    %dl,-0x1(%rdi,%rax,1)
  mov    %eax,%ecx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L95
L7:
  movslq %ecx,%rcx
  movb   $0x0,(%rdi,%rcx,1)
L27:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L137:
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L119
  lea    0x1(%r8),%eax
  cltq
  sub    %rcx,%rsi
L32:
  mov    %dl,-0x1(%rdi,%rax,1)
  mov    %eax,%r8d
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L32
L119:
  movslq %r8d,%r8
  movb   $0x0,(%rdi,%r8,1)
  jmp    L27
L68:
  mov    %r12d,%ebx
L97:
  add    %ebx,%edx
L14:
  movslq %edx,%rax
  mov    %r9b,(%rsp,%rax,1)
  add    $0x1,%edx
  add    $0x1,%r10
  movzbl -0x1(%r10),%r9d
  test   %r9b,%r9b
  je     L117
L105:
  cmp    %edx,%r8d
  jne    L14
  movzbl (%rsi),%ecx
  test   %cl,%cl
  je     L68
  mov    $0x1,%eax
  movslq %edx,%r11
  add    %rbp,%r11
L100:
  mov    %cl,-0x1(%r11,%rax,1)
  mov    %eax,%ebx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  test   %cl,%cl
  jne    L100
  jmp    L97
L53:
  mov    $0x0,%ecx
  jmp    L7
L61:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void insert_string(char source[], char inject[], int index) {
  int i, j, k;
  char temp_string[81];
  if (source[index] == '\0') {
    for (i = index, j = 0; inject[j] != '\0'; ++i, ++j) {
      source[i] = inject[j];
    }
    source[i] = '\0';
  } else {
    for (i = 0, j = 0; source[i] != '\0'; ++i, ++j) {
      if (j == index) {
        for (k = 0; inject[k] != '\0'; ++k) {
          temp_string[j + k] = inject[k];
        }
        j += k;
      }
      temp_string[j] = source[i];
    }
    temp_string[j] = '\0';
    for (i = 0; temp_string[i] != '\0'; ++i) {
      source[i] = temp_string[i];
    }
    source[i] = '\0';
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<insert_string>:
  endbr64
  push   %rbp
  mov    %rsi,%r8
  movslq %edx,%rsi
  push   %rbx
  lea    (%rdi,%rsi,1),%rcx
  mov    %rsi,%rax
  sub    $0x68,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,0x58(%rsp)
  xor    %edx,%edx
  cmpb   $0x0,(%rcx)
  je     L99
  movzbl (%rdi),%r9d
  lea    0x1(%rdi),%r10
  xor    %ecx,%ecx
  mov    %rsp,%rbp
  test   %r9b,%r9b
  jne    L69
  cs nopw 0x0(%rax,%rax,1)
L20:
  movslq %ecx,%rcx
  movb   $0x0,(%rsp,%rcx,1)
  movzbl (%rsp),%edx
  test   %dl,%dl
  je     L24
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  nopl   0x0(%rax)
L39:
  mov    %dl,-0x1(%rdi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L39
  add    %rcx,%rdi
L24:
  movb   $0x0,(%rdi)
L50:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L116:
  movzbl (%r8),%esi
  test   %sil,%sil
  je     L126
  movslq %eax,%r11
  mov    $0x1,%edx
  add    %rbp,%r11
  nopl   0x0(%rax)
L15:
  mov    %sil,-0x1(%r11,%rdx,1)
  mov    %edx,%ebx
  add    $0x1,%rdx
  movzbl -0x1(%r8,%rdx,1),%esi
  test   %sil,%sil
  jne    L15
  add    %ebx,%ecx
  nopl   0x0(%rax,%rax,1)
L126:
  movslq %ecx,%rdx
  add    $0x1,%r10
  add    $0x1,%ecx
  mov    %r9b,(%rsp,%rdx,1)
  movzbl -0x1(%r10),%r9d
  test   %r9b,%r9b
  je     L20
L69:
  cmp    %ecx,%eax
  jne    L126
  jmp    L116
  nopw   0x0(%rax,%rax,1)
L99:
  movzbl (%r8),%edx
  test   %dl,%dl
  je     L104
  add    $0x1,%eax
  sub    %rsi,%r8
  cltq
  nopl   0x0(%rax,%rax,1)
L53:
  mov    %dl,-0x1(%rdi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  movzbl -0x1(%r8,%rax,1),%edx
  test   %dl,%dl
  jne    L53
  add    %rdi,%rcx
L104:
  movb   $0x0,(%rcx)
  jmp    L50
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void insert_string(char source[], char inject[], int index) {
  int i, j, k;
  char temp_string[81];
  if (source[index] == '\0') {
    for (i = index, j = 0; inject[j] != '\0'; ++i, ++j) {
      source[i] = inject[j];
    }
    source[i] = '\0';
  } else {
    for (i = 0, j = 0; source[i] != '\0'; ++i, ++j) {
      if (j == index) {
        for (k = 0; inject[k] != '\0'; ++k) {
          temp_string[j + k] = inject[k];
        }
        j += k;
      }
      temp_string[j] = source[i];
    }
    temp_string[j] = '\0';
    for (i = 0; temp_string[i] != '\0'; ++i) {
      source[i] = temp_string[i];
    }
    source[i] = '\0';
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<insert_string>:
  endbr64
  push   %rbp
  mov    %rsi,%r8
  movslq %edx,%rsi
  push   %rbx
  lea    (%rdi,%rsi,1),%rcx
  mov    %rsi,%rax
  sub    $0x68,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,0x58(%rsp)
  xor    %edx,%edx
  cmpb   $0x0,(%rcx)
  je     L35
  movzbl (%rdi),%r9d
  lea    0x1(%rdi),%r10
  xor    %ecx,%ecx
  mov    %rsp,%rbp
  test   %r9b,%r9b
  jne    L50
  cs nopw 0x0(%rax,%rax,1)
L89:
  movslq %ecx,%rcx
  movb   $0x0,(%rsp,%rcx,1)
  movzbl (%rsp),%edx
  test   %dl,%dl
  je     L102
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  nopl   0x0(%rax)
L37:
  mov    %dl,-0x1(%rdi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L37
  add    %rcx,%rdi
L102:
  movb   $0x0,(%rdi)
L90:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L127
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L111:
  movzbl (%r8),%esi
  test   %sil,%sil
  je     L110
  movslq %eax,%r11
  mov    $0x1,%edx
  add    %rbp,%r11
  nopl   0x0(%rax)
L77:
  mov    %sil,-0x1(%r11,%rdx,1)
  mov    %edx,%ebx
  add    $0x1,%rdx
  movzbl -0x1(%r8,%rdx,1),%esi
  test   %sil,%sil
  jne    L77
  add    %ebx,%ecx
  nopl   0x0(%rax,%rax,1)
L110:
  movslq %ecx,%rdx
  add    $0x1,%r10
  add    $0x1,%ecx
  mov    %r9b,(%rsp,%rdx,1)
  movzbl -0x1(%r10),%r9d
  test   %r9b,%r9b
  je     L89
L50:
  cmp    %ecx,%eax
  jne    L110
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L35:
  movzbl (%r8),%edx
  test   %dl,%dl
  je     L26
  add    $0x1,%eax
  sub    %rsi,%r8
  cltq
  nopl   0x0(%rax,%rax,1)
L59:
  mov    %dl,-0x1(%rdi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  movzbl -0x1(%r8,%rax,1),%edx
  test   %dl,%dl
  jne    L59
  add    %rdi,%rcx
L26:
  movb   $0x0,(%rcx)
  jmp    L90
L127:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void insert_string(char source[], char inject[], int index) {
  int i, j, k;
  char temp_string[81];
  if (source[index] == '\0') {
    for (i = index, j = 0; inject[j] != '\0'; ++i, ++j) {
      source[i] = inject[j];
    }
    source[i] = '\0';
  } else {
    for (i = 0, j = 0; source[i] != '\0'; ++i, ++j) {
      if (j == index) {
        for (k = 0; inject[k] != '\0'; ++k) {
          temp_string[j + k] = inject[k];
        }
        j += k;
      }
      temp_string[j] = source[i];
    }
    temp_string[j] = '\0';
    for (i = 0; temp_string[i] != '\0'; ++i) {
      source[i] = temp_string[i];
    }
    source[i] = '\0';
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x60(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rcx
  lea    -0x50(%rbp),%rax
  mov    $0x3f,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  char buf[64];
  strncpy(buf, argv[1], 64 - 1);
}
```