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
<bpf_map_update_elem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rcx
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sys_bpf@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int bpf_map_update_elem(int fd, const void *key, const void *value, __u64 flags) {
  union bpf_attr attr;
  bzero(&attr, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return sys_bpf(BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bpf_map_update_elem>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %ecx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    $0x8,%esi
  mov    %r14,%rdi
  call   <bzero@plt>
  mov    %r13d,(%rsp)
  mov    %r12,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,(%rsp)
  mov    %rbp,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,(%rsp)
  mov    %ebx,(%rsp)
  mov    $0x8,%edx
  mov    %r14,%rsi
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <sys_bpf@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_map_update_elem(int fd, const void *key, const void *value, __u64 flags) {
  union bpf_attr attr;
  bzero(&attr, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return sys_bpf(BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bpf_map_update_elem>:
  endbr64
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  mov    $0x8,%esi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <bzero@plt>
  mov    %r12,%rdi
  mov    %r13d,(%rsp)
  call   <ptr_to_u64@plt>
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  call   <ptr_to_u64@plt>
  mov    $0x8,%edx
  mov    %r14,%rsi
  mov    %rax,(%rsp)
  mov    D0(%rip),%rax
  mov    %ebx,(%rsp)
  mov    (%rax),%edi
  call   <sys_bpf@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_map_update_elem(int fd, const void *key, const void *value, __u64 flags) {
  union bpf_attr attr;
  bzero(&attr, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return sys_bpf(BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bpf_map_update_elem>:
  endbr64
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  mov    $0x8,%esi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <bzero@plt>
  mov    %r12,%rdi
  mov    %r13d,(%rsp)
  call   <ptr_to_u64@plt>
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  call   <ptr_to_u64@plt>
  mov    $0x8,%edx
  mov    %r14,%rsi
  mov    %rax,(%rsp)
  mov    D9(%rip),%rax
  mov    %ebx,(%rsp)
  mov    (%rax),%edi
  call   <sys_bpf@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_map_update_elem(int fd, const void *key, const void *value, __u64 flags) {
  union bpf_attr attr;
  bzero(&attr, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return sys_bpf(BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  call   <fork@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L15
  mov    $0x0,%edi
  call   <wait@plt>
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L29
L15:
  cmpl   $0x0,-0x4(%rbp)
  jne    L26
  mov    $0x2,%edi
  call   <sleep@plt>
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L29
L26:
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L29:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D28
type_guessed: string
name: disassemble_data
arguments: tag: D13
type_guessed: string
name: disassemble_data
arguments: tag: D3
type_guessed: string
name: disassemble_data
arguments: tag: D35
type_guessed: string

tool: D28: [%d] hello world


tool: D13: [%d] *parent*


tool: D3: [%d] *child*


tool: D35: [%d] *????*


assistant: ```c
int main() {
  printf("[%d] hello world\n", getpid());
  int rv = fork();
  if (rv > 0) {
    wait(NULL);
    printf("[%d] *parent*\n", getpid());
  } else if (rv == 0) {
    sleep(2);
    printf("[%d] *child*\n", getpid());
  } else {
    printf("[%d] *????*\n", getpid());
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  call   <fork@plt>
  test   %eax,%eax
  jg     L2
  jne    L35
  mov    $0x2,%edi
  call   <sleep@plt>
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  jmp    L7
L2:
  mov    $0x0,%edi
  call   <wait@plt>
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D27(%rip),%rdi
  call   <printf@plt>
L7:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L35:
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L7
```

tool_calls: name: read_data
arguments: data_label: D16
data_type: string
name: read_data
arguments: data_label: D2
data_type: string
name: read_data
arguments: data_label: D27
data_type: string
name: read_data
arguments: data_label: D6
data_type: string

tool: D16: [%d] hello world


tool: D2: [%d] *child*


tool: D27: [%d] *parent*


tool: D6: [%d] *????*


assistant: ```c
int main() {
  printf("[%d] hello world\n", getpid());
  int rv = fork();
  if (rv > 0) {
    wait(NULL);
    printf("[%d] *parent*\n", getpid());
  } else if (rv == 0) {
    sleep(2);
    printf("[%d] *child*\n", getpid());
  } else {
    printf("[%d] *????*\n", getpid());
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D20(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  call   <fork@plt>
  test   %eax,%eax
  jg     L24
  je     L28
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
L18:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L28:
  mov    $0x2,%edi
  call   <sleep@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  jmp    L18
L24:
  xor    %edi,%edi
  call   <wait@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  jmp    L18
```

tool_calls: name: parse_data
arguments: tag: D20
type_guessed: string
name: parse_data
arguments: tag: D11
type_guessed: string
name: parse_data
arguments: tag: D12
type_guessed: string
name: parse_data
arguments: tag: D2
type_guessed: string

tool: D20: [%d] hello world


tool: D11: [%d] *????*


tool: D12: [%d] *child*


tool: D2: [%d] *parent*


assistant: ```c
int main() {
  printf("[%d] hello world\n", getpid());
  int rv = fork();
  if (rv > 0) {
    wait(NULL);
    printf("[%d] *parent*\n", getpid());
  } else if (rv == 0) {
    sleep(2);
    printf("[%d] *child*\n", getpid());
  } else {
    printf("[%d] *????*\n", getpid());
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D23(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  call   <fork@plt>
  test   %eax,%eax
  jg     L14
  je     L13
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D26(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
L5:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L13:
  mov    $0x2,%edi
  call   <sleep@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D32(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  jmp    L5
L14:
  xor    %edi,%edi
  call   <wait@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D27(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  jmp    L5
```

tool_calls: name: parse_data
arguments: data_label: D23
type_guessed: string
name: parse_data
arguments: data_label: D26
type_guessed: string
name: parse_data
arguments: data_label: D32
type_guessed: string
name: parse_data
arguments: data_label: D27
type_guessed: string

tool: D23: [%d] hello world


tool: D26: [%d] *????*


tool: D32: [%d] *child*


tool: D27: [%d] *parent*


assistant: ```c
int main() {
  printf("[%d] hello world\n", getpid());
  int rv = fork();
  if (rv > 0) {
    wait(NULL);
    printf("[%d] *parent*\n", getpid());
  } else if (rv == 0) {
    sleep(2);
    printf("[%d] *child*\n", getpid());
  } else {
    printf("[%d] *????*\n", getpid());
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4,-0x30(%rbp)
  movl   $0x2,-0x2c(%rbp)
  movl   $0x4,-0x28(%rbp)
  movl   $0x5,-0x24(%rbp)
  movl   $0x2,-0x20(%rbp)
  movl   $0x3,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x7,-0x60(%rbp)
  movl   $0x5,-0x5c(%rbp)
  movl   $0x2,-0x58(%rbp)
  movl   $0x4,-0x54(%rbp)
  lea    -0x6c(%rbp),%rdi
  lea    -0x70(%rbp),%rcx
  mov    -0x5c(%rbp),%edx
  mov    -0x60(%rbp),%esi
  lea    -0x30(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <find_two_repeating_numbers_in_array@plt>
  mov    -0x6c(%rbp),%edx
  mov    -0x70(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x70(%rbp),%eax
  cmp    %eax,-0x58(%rbp)
  jne    L0
  mov    -0x6c(%rbp),%eax
  cmp    %eax,-0x54(%rbp)
  jne    L0
  mov    $0x1,%eax
  jmp    L35
L0:
  mov    $0x0,%eax
L35:
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0x2,-0x40(%rbp)
  movl   $0x1,-0x3c(%rbp)
  movl   $0x2,-0x38(%rbp)
  movl   $0x1,-0x34(%rbp)
  movl   $0x4,-0x50(%rbp)
  movl   $0x2,-0x4c(%rbp)
  movl   $0x1,-0x48(%rbp)
  movl   $0x2,-0x44(%rbp)
  lea    -0x64(%rbp),%rdi
  lea    -0x68(%rbp),%rcx
  mov    -0x4c(%rbp),%edx
  mov    -0x50(%rbp),%esi
  lea    -0x40(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <find_two_repeating_numbers_in_array@plt>
  mov    -0x68(%rbp),%eax
  cmp    %eax,-0x48(%rbp)
  jne    L17
  mov    -0x64(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  jne    L17
  mov    $0x1,%eax
  jmp    L46
L17:
  mov    $0x0,%eax
L46:
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type_guessed: string

tool: D9: num1 = %d, num2 = %d


assistant: ```c
int main() {
  int array0[] = {4, 2, 4, 5, 2, 3, 1};
  int len0 = sizeof(array0) / sizeof(int);
  int num0 = 5;
  int exp_num01 = 2, exp_num02 = 4;
  int act_num01, act_num02;
  find_two_repeating_numbers_in_array(array0, len0, num0, &act_num01, &act_num02);
  printf("num1 = %d, num2 = %d\n", act_num01, act_num02);
  assert((exp_num01 == act_num01) && (exp_num02 == act_num02));
  int array1[] = {2, 1, 2, 1};
  int len1 = sizeof(array1) / sizeof(int);
  int num1 = 2;
  int exp_num11 = 1, exp_num12 = 2;
  int act_num11, act_num12;
  find_two_repeating_numbers_in_array(array1, len1, num1, &act_num11, &act_num12);
  assert((exp_num11 == act_num11) && (exp_num12 == act_num12));
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x4,0x20(%rsp)
  movl   $0x2,0x24(%rsp)
  movl   $0x4,0x28(%rsp)
  movl   $0x5,0x2c(%rsp)
  movl   $0x2,0x30(%rsp)
  movl   $0x3,0x34(%rsp)
  movl   $0x1,0x38(%rsp)
  mov    %rsp,%rcx
  lea    0x20(%rsp),%rdi
  lea    0x4(%rsp),%r8
  mov    $0x5,%edx
  mov    $0x7,%esi
  call   <find_two_repeating_numbers_in_array@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  cmpl   $0x2,(%rsp)
  je     L19
L40:
  call   <assert@plt>
  movl   $0x2,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  movl   $0x2,0x18(%rsp)
  movl   $0x1,0x1c(%rsp)
  lea    0x8(%rsp),%rcx
  lea    0x10(%rsp),%rdi
  lea    0xc(%rsp),%r8
  mov    $0x2,%edx
  mov    $0x4,%esi
  call   <find_two_repeating_numbers_in_array@plt>
  mov    $0x0,%edi
  cmpl   $0x1,0x8(%rsp)
  je     L16
L32:
  call   <assert@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L19:
  cmpl   $0x4,0x4(%rsp)
  sete   %dil
  movzbl %dil,%edi
  jmp    L40
L16:
  cmpl   $0x2,0xc(%rsp)
  sete   %dil
  movzbl %dil,%edi
  jmp    L32
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D8
datatype: string

tool: D8: num1 = %d, num2 = %d


assistant: ```c
int main() {
  int array0[] = {4, 2, 4, 5, 2, 3, 1};
  int len0 = sizeof(array0) / sizeof(int);
  int num0 = 5;
  int exp_num01 = 2, exp_num02 = 4;
  int act_num01, act_num02;
  find_two_repeating_numbers_in_array(array0, len0, num0, &act_num01, &act_num02);
  printf("num1 = %d, num2 = %d\n", act_num01, act_num02);
  assert((exp_num01 == act_num01) && (exp_num02 == act_num02));
  int array1[] = {2, 1, 2, 1};
  int len1 = sizeof(array1) / sizeof(int);
  int num1 = 2;
  int exp_num11 = 1, exp_num12 = 2;
  int act_num11, act_num12;
  find_two_repeating_numbers_in_array(array1, len1, num1, &act_num11, &act_num12);
  assert((exp_num11 == act_num11) && (exp_num12 == act_num12));
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    $0x5,%edx
  mov    $0x7,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x200000004,%rax
  mov    %rsp,%rcx
  lea    0x20(%rsp),%rdi
  lea    0x4(%rsp),%r8
  movl   $0x1,0x38(%rsp)
  mov    %rax,0x20(%rsp)
  movabs $0x500000004,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x300000002,%rax
  mov    %rax,0x30(%rsp)
  call   <find_two_repeating_numbers_in_array@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  cmpl   $0x2,(%rsp)
  jne    L3
  xor    %edi,%edi
  cmpl   $0x4,0x4(%rsp)
  sete   %dil
L3:
  call   <assert@plt>
  lea    0x10(%rsp),%rdi
  lea    0x8(%rsp),%rcx
  movabs $0x100000002,%rax
  lea    0xc(%rsp),%r8
  mov    $0x2,%edx
  mov    $0x4,%esi
  mov    %rax,0x10(%rsp)
  mov    %rax,0x18(%rsp)
  call   <find_two_repeating_numbers_in_array@plt>
  xor    %edi,%edi
  cmpl   $0x1,0x8(%rsp)
  jne    L11
  xor    %edi,%edi
  cmpl   $0x2,0xc(%rsp)
  sete   %dil
L11:
  call   <assert@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

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

tool: D4: num1 = %d, num2 = %d


assistant: ```c
int main() {
  int array0[] = {4, 2, 4, 5, 2, 3, 1};
  int len0 = sizeof(array0) / sizeof(int);
  int num0 = 5;
  int exp_num01 = 2, exp_num02 = 4;
  int act_num01, act_num02;
  find_two_repeating_numbers_in_array(array0, len0, num0, &act_num01, &act_num02);
  printf("num1 = %d, num2 = %d\n", act_num01, act_num02);
  assert((exp_num01 == act_num01) && (exp_num02 == act_num02));
  int array1[] = {2, 1, 2, 1};
  int len1 = sizeof(array1) / sizeof(int);
  int num1 = 2;
  int exp_num11 = 1, exp_num12 = 2;
  int act_num11, act_num12;
  find_two_repeating_numbers_in_array(array1, len1, num1, &act_num11, &act_num12);
  assert((exp_num11 == act_num11) && (exp_num12 == act_num12));
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    $0x5,%edx
  mov    $0x7,%esi
  movdqa D19(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rcx
  lea    0x20(%rsp),%rdi
  lea    0x4(%rsp),%r8
  mov    D27(%rip),%rax
  movaps %xmm0,0x20(%rsp)
  movl   $0x1,0x38(%rsp)
  mov    %rax,0x30(%rsp)
  call   <find_two_repeating_numbers_in_array@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  lea    D37(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  cmpl   $0x2,(%rsp)
  jne    L22
  xor    %edi,%edi
  cmpl   $0x4,0x4(%rsp)
  sete   %dil
L22:
  call   <assert@plt>
  lea    0x10(%rsp),%rdi
  lea    0x8(%rsp),%rcx
  movdqa D34(%rip),%xmm0
  lea    0xc(%rsp),%r8
  mov    $0x2,%edx
  mov    $0x4,%esi
  movaps %xmm0,0x10(%rsp)
  call   <find_two_repeating_numbers_in_array@plt>
  xor    %edi,%edi
  cmpl   $0x1,0x8(%rsp)
  jne    L21
  xor    %edi,%edi
  cmpl   $0x2,0xc(%rsp)
  sete   %dil
L21:
  call   <assert@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D19
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D27
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D37
type_guessed: string
name: disassemble_data
arguments: tag: D34
type_guessed: byte[16]

tool: {"D19": [0x04,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: {"D27": [0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: D37: num1 = %d, num2 = %d


tool: {"D34": [0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int array0[] = {4, 2, 4, 5, 2, 3, 1};
  int len0 = sizeof(array0) / sizeof(int);
  int num0 = 5;
  int exp_num01 = 2, exp_num02 = 4;
  int act_num01, act_num02;
  find_two_repeating_numbers_in_array(array0, len0, num0, &act_num01, &act_num02);
  printf("num1 = %d, num2 = %d\n", act_num01, act_num02);
  assert((exp_num01 == act_num01) && (exp_num02 == act_num02));
  int array1[] = {2, 1, 2, 1};
  int len1 = sizeof(array1) / sizeof(int);
  int num1 = 2;
  int exp_num11 = 1, exp_num12 = 2;
  int act_num11, act_num12;
  find_two_repeating_numbers_in_array(array1, len1, num1, &act_num11, &act_num12);
  assert((exp_num11 == act_num11) && (exp_num12 == act_num12));
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<crypto_aead_decrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbp
  mov    %rcx,%r13
  mov    %r8,%rbx
  mov    %r8,0x8(%rsp)
  mov    %r9,%r15
  mov    0x90(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  mov    0x98(%rsp),%rsi
  mov    0xa0(%rsp),%rdi
  call   <morus_initialization@plt>
  cmp    $0xf,%rbx
  jbe    L2
  cmp    $0xf,%r12
  jbe    L95
  mov    $0x10,%ebx
  lea    0x10(%rsp),%rax
  mov    %rax,(%rsp)
L30:
  lea    0x30(%rsp),%rsi
  lea    -0x10(%r15,%rbx,1),%rdi
  mov    (%rsp),%rdx
  call   <morus_enc_aut_step@plt>
  mov    %rbx,%rax
  add    $0x10,%rbx
  cmp    %rbx,%r12
  jae    L30
L112:
  mov    %r12,%rdx
  and    $0xf,%edx
  jne    L57
L22:
  mov    0x8(%rsp),%rdx
  sub    $0x10,%rdx
  mov    %rdx,0x0(%rbp)
  cmp    $0xf,%rdx
  jbe    L48
  mov    $0x10,%ebx
  lea    0x10(%rsp),%r15
L59:
  lea    -0x10(%r13,%rbx,1),%rsi
  lea    -0x10(%r14,%rbx,1),%rdi
  mov    %r15,%rdx
  call   <morus_dec_aut_step@plt>
  mov    0x0(%rbp),%rdx
  mov    %rbx,%rax
  add    $0x10,%rbx
  cmp    %rbx,%rdx
  jae    L59
L4:
  and    $0xf,%edx
  jne    L63
L26:
  lea    0x10(%rsp),%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    0x0(%rbp),%rdi
  call   <morus_tag_verification@plt>
L94:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L95:
  mov    $0x0,%eax
  jmp    L112
L57:
  lea    0x10(%rsp),%rcx
  lea    0x30(%rsp),%rsi
  lea    (%r15,%rax,1),%rdi
  call   <morus_enc_aut_partialblock@plt>
  jmp    L22
L48:
  mov    $0x0,%eax
  jmp    L4
L63:
  lea    0x10(%rsp),%rcx
  lea    0x0(%r13,%rax,1),%rsi
  lea    (%r14,%rax,1),%rdi
  call   <morus_dec_aut_partialblock@plt>
  jmp    L26
L2:
  mov    $0xffffffff,%eax
  jmp    L94
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crypto_aead_decrypt(unsigned char *m, unsigned long long *mlen, unsigned char *nsec, const unsigned char *c, unsigned long long clen, const unsigned char *ad, unsigned long long adlen, const unsigned char *npub, const unsigned char *k) {
  unsigned long i, j;
  unsigned char plaintextblock[16], ciphertextblock[16];
  unsigned char tag[16];
  __m128i morus_state[5];
  morus_initialization(k, npub, morus_state);
  if (clen < 16)
    return -1;
  for (i = 0; (i + 16) <= adlen; i += 16) {
    morus_enc_aut_step(ad + i, ciphertextblock, morus_state);
  }
  if ((adlen & 0xf) != 0) {
    morus_enc_aut_partialblock(ad + i, ciphertextblock, adlen & 0xf, morus_state);
  }
  *mlen = clen - 16;
  for (i = 0; (i + 16) <= *mlen; i += 16) {
    morus_dec_aut_step(m + i, c + i, morus_state);
  }
  if ((*mlen & 0xf) != 0) {
    morus_dec_aut_partialblock(m + i, c + i, *mlen & 0xf, morus_state);
  }
  return morus_tag_verification(*mlen, adlen, c, morus_state);
}
```
user: Translate the below assembly code into C programming language:

```asm
<xprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L14
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L14:
  mov    -0xd0(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xd0(%rbp),%rcx
  mov    -0xc8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vfprintf@plt>
  cmp    $0xffffffff,%eax
  jne    L10
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x4a,%edi
  call   <exit@plt>
L10:
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
datatype: string

tool: D3: fprintf

assistant: ```c
void xprintf(FILE *out, const char *format, ...) {
  va_list args;
  va_start(args, format);
  if (-1 == vfprintf(out, format, args)) {
    perror("fprintf");
    exit(74);
  }
  va_end(args);
}
```