ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D48(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x1c(%rsp),%rsi
  call   <scanf@plt>
  mov    0x1c(%rsp),%eax
  test   %eax,%eax
  jle    L44
  mov    $0x1,%r12d
  lea    0x20(%rsp),%r13
  lea    0x80(%rsp),%rbp
  lea    D3(%rip),%rbx
  lea    D38(%rip),%r14
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %r13,%r15
  pxor   %xmm0,%xmm0
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %r15,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <scanf@plt>
  add    $0x8,%r15
  movsd  0x8(%rsp),%xmm0
  addsd  -0x8(%r15),%xmm0
  cmp    %rbp,%r15
  jne    L11
  movsd  D26(%rip),%xmm2
  divsd  D33(%rip),%xmm0
  comisd %xmm0,%xmm2
  ja     L35
  movapd %xmm0,%xmm1
  divsd  D26(%rip),%xmm1
  cvttsd2si %xmm1,%edx
  mov    %r12d,%esi
  pxor   %xmm1,%xmm1
  mov    %r14,%rdi
  add    $0x1,%r12d
  imul   $0x3e8,%edx,%eax
  cvtsi2sd %eax,%xmm1
  mov    $0x1,%eax
  subsd  %xmm1,%xmm0
  call   <printf@plt>
  cmp    %r12d,0x1c(%rsp)
  jge    L5
L44:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0xa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L35:
  cvttsd2si %xmm0,%edx
  mov    %r12d,%esi
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  cmp    %r12d,0x1c(%rsp)
  jge    L5
  jmp    L44
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D48
type_guessed: string
name: unpack_data
arguments: data_tag: D3
type_guessed: string
name: unpack_data
arguments: data_tag: D38
type_guessed: string
name: unpack_data
arguments: data_tag: D26
type_guessed: f32[2]
name: unpack_data
arguments: data_tag: D33
type_guessed: f64
name: unpack_data
arguments: data_tag: D20
type_guessed: string

tool: D48: %d

tool: D3: %lf

tool: D38: %d $%d,%.2lf


tool: {"D26": [0.0,4.4765625]}

tool: D33: 12

tool: D20: %d $%.2lf


assistant: ```c
int main() {
  int t, i;
  double ara[15];
  scanf("%d", &t);
  for (i = 1; i <= t; i++) {
    int j, temp;
    double sum = 0;
    for (j = 0; j < 12; j++) {
      scanf("%lf", &ara[j]);
      sum += ara[j];
    }
    sum = sum / 12.0;
    if (sum < 1000) {
      printf("%d $%.2lf\n", i, sum);
      continue;
    }
    temp = sum / 1000;
    sum = sum - (temp * 1000);
    printf("%d $%d,%.2lf\n", i, temp, sum);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<outline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x138,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x124(%rbp)
  mov    -0x138(%rbp),%rbx
L19:
  movzbl (%rbx),%edx
  mov    -0x124(%rbp),%eax
  cltq
  mov    %dl,-0x120(%rbp,%rax,1)
  addl   $0x1,-0x124(%rbp)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L19
  mov    -0x124(%rbp),%eax
  cltq
  movb   $0x0,-0x120(%rbp,%rax,1)
  lea    -0x120(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

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

tool: D4: %s

assistant: ```c
void outline(s) char *s;
{
  char buf[256];
  int i = 0;
  register char *rs;
  rs = s;
  do {
    buf[i] = *rs;
    i++;
  } while (*rs++ != 10);
  buf[i] = '\0';
  printf("%s", buf);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<outline>:
  endbr64
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
L8:
  movzbl -0x1(%rdi,%rax,1),%edx
  mov    %dl,(%rsi,%rax,1)
  mov    %eax,%ecx
  add    $0x1,%rax
  cmp    $0xa,%dl
  jne    L8
  movslq %ecx,%rcx
  movb   $0x0,(%rsp,%rcx,1)
  mov    %rsp,%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x118,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D3
datatype: string

tool: D3: %s

assistant: ```c
void outline(s) char *s;
{
  char buf[256];
  int i = 0;
  register char *rs;
  rs = s;
  do {
    buf[i] = *rs;
    i++;
  } while (*rs++ != 10);
  buf[i] = '\0';
  printf("%s", buf);
}
```
user: Translate the below assembly code into C programming language:

```asm
<outline>:
  endbr64
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  cs nopw 0x0(%rax,%rax,1)
L13:
  movzbl -0x1(%rdi,%rax,1),%edx
  movslq %eax,%rcx
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    $0xa,%dl
  jne    L13
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D4(%rip),%rdi
  movb   $0x0,(%rsp,%rcx,1)
  call   <printf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x118,%rsp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D4
data_type: string

tool: D4: %s

assistant: ```c
void outline(s) char *s;
{
  char buf[256];
  int i = 0;
  register char *rs;
  rs = s;
  do {
    buf[i] = *rs;
    i++;
  } while (*rs++ != 10);
  buf[i] = '\0';
  printf("%s", buf);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<outline>:
  endbr64
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  cs nopw 0x0(%rax,%rax,1)
L18:
  movzbl -0x1(%rdi,%rax,1),%edx
  movslq %eax,%rcx
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    $0xa,%dl
  jne    L18
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D6(%rip),%rdi
  movb   $0x0,(%rsp,%rcx,1)
  call   <printf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x118,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D6
type: string

tool: D6: %s

assistant: ```c
void outline(s) char *s;
{
  char buf[256];
  int i = 0;
  register char *rs;
  rs = s;
  do {
    buf[i] = *rs;
    i++;
  } while (*rs++ != 10);
  buf[i] = '\0';
  printf("%s", buf);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_addr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
data_type: string

tool: D2: %p


assistant: ```c
int print_addr(int x) {
  printf("%p\n", &x);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<print_addr>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
datatype: string

tool: D5: %p


assistant: ```c
int print_addr(int x) {
  printf("%p\n", &x);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print_addr>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type_infered: string

tool: D5: %p


assistant: ```c
int print_addr(int x) {
  printf("%p\n", &x);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<print_addr>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

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

tool: D6: %p


assistant: ```c
int print_addr(int x) {
  printf("%p\n", &x);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x4(%rbp)
  jmp    L1
L4:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L4
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
data_type: string

tool: D5: Hello, world!


assistant: ```c
int main(int argc, char *argv[]) {
  printf("Hello, world!\n");
  for (int i = 1; i < argc; ++i) {
    puts(argv[i]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    D8(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1,%ebp
  jle    L11
  lea    0x8(%r12),%rbx
  lea    -0x2(%rbp),%eax
  lea    0x10(%r12,%rax,8),%rbp
L2:
  mov    (%rbx),%rdi
  call   <puts@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L2
L11:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
datatype: string

tool: D8: Hello, world!


assistant: ```c
int main(int argc, char *argv[]) {
  printf("Hello, world!\n");
  for (int i = 1; i < argc; ++i) {
    puts(argv[i]);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  lea    D1(%rip),%rdi
  push   %rbx
  call   <printf@plt>
  cmp    $0x1,%ebp
  jle    L6
  lea    -0x2(%rbp),%eax
  lea    0x8(%r12),%rbx
  lea    0x10(%r12,%rax,8),%rbp
  nopl   0x0(%rax,%rax,1)
L15:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <puts@plt>
  cmp    %rbp,%rbx
  jne    L15
L6:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D1
type: string

tool: D1: Hello, world!


assistant: ```c
int main(int argc, char *argv[]) {
  printf("Hello, world!\n");
  for (int i = 1; i < argc; ++i) {
    puts(argv[i]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  lea    D0(%rip),%rdi
  push   %rbx
  call   <printf@plt>
  cmp    $0x1,%ebp
  jle    L10
  lea    -0x2(%rbp),%eax
  lea    0x8(%r12),%rbx
  lea    0x10(%r12,%rax,8),%rbp
  nopl   0x0(%rax,%rax,1)
L17:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <puts@plt>
  cmp    %rbp,%rbx
  jne    L17
L10:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: Hello, world!


assistant: ```c
int main(int argc, char *argv[]) {
  printf("Hello, world!\n");
  for (int i = 1; i < argc; ++i) {
    puts(argv[i]);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<alloc_zero_tailing_info>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,-0x34(%rbp)
  mov    -0x30(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L2
L3:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L28
L22:
  mov    -0x34(%rbp),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x30(%rbp),%eax
  sub    -0x34(%rbp),%eax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x10(%rbp),%rdx
  add    %rdx,%rcx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    %rcx,%rdi
  call   <memset@plt>
  mov    -0x30(%rbp),%eax
  cltq
  add    %rax,-0x28(%rbp)
  mov    -0x30(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
  addl   $0x1,-0x18(%rbp)
L28:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L22
  mov    -0x8(%rbp),%rax
L2:
  leave
  ret
```

assistant: ```c
void *alloc_zero_tailing_info(const void *orecord, __u32 cnt, __u32 actual_rec_size, __u32 expected_rec_size) {
  __u64 info_len = actual_rec_size * cnt;
  void *info, *nrecord;
  int i;
  info = malloc(info_len);
  if (!info)
    return NULL;
  nrecord = info;
  for (i = 0; i < cnt; i++) {
    memcpy(nrecord, orecord, expected_rec_size);
    memset(nrecord + expected_rec_size, 0, actual_rec_size - expected_rec_size);
    orecord += actual_rec_size;
    nrecord += actual_rec_size;
  }
  return info;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<alloc_zero_tailing_info>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r14d
  mov    %edx,%r15d
  mov    %ecx,%r13d
  mov    %edx,%edi
  imul   %esi,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %rax,%rax
  je     L16
  test   %r14d,%r14d
  jle    L16
  mov    %rax,%rbx
  mov    $0x0,%r12d
  mov    %r15d,%eax
  sub    %r13d,%eax
  mov    %eax,0x4(%rsp)
L1:
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  movslq %r13d,%rdi
  add    %rbx,%rdi
  mov    0x4(%rsp),%edx
  mov    $0x0,%esi
  call   <memset@plt>
  movslq %r15d,%rax
  add    %rax,%rbp
  add    %rax,%rbx
  add    $0x1,%r12d
  cmp    %r12d,%r14d
  jne    L1
L16:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void *alloc_zero_tailing_info(const void *orecord, __u32 cnt, __u32 actual_rec_size, __u32 expected_rec_size) {
  __u64 info_len = actual_rec_size * cnt;
  void *info, *nrecord;
  int i;
  info = malloc(info_len);
  if (!info)
    return NULL;
  nrecord = info;
  for (i = 0; i < cnt; i++) {
    memcpy(nrecord, orecord, expected_rec_size);
    memset(nrecord + expected_rec_size, 0, actual_rec_size - expected_rec_size);
    orecord += actual_rec_size;
    nrecord += actual_rec_size;
  }
  return info;
}
```
user: Translate the below assembly code into C programming language:

```asm
<alloc_zero_tailing_info>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  push   %rbx
  imul   %r13d,%edi
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %rax,%rax
  je     L20
  test   %r14d,%r14d
  jle    L20
  mov    %rax,%rbx
  movslq %r12d,%rax
  mov    %r13d,%r14d
  xor    %r15d,%r15d
  mov    %rax,0x10(%rsp)
  sub    %r12d,%r14d
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r12d,%edx
  add    %r13,%rbp
  call   <memcpy@plt>
  mov    0x10(%rsp),%rax
  mov    %r14d,%edx
  xor    %esi,%esi
  add    $0x1,%r15d
  lea    (%rbx,%rax,1),%rdi
  add    %r13,%rbx
  call   <memset@plt>
  cmp    %r15d,0xc(%rsp)
  jne    L6
L20:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void *alloc_zero_tailing_info(const void *orecord, __u32 cnt, __u32 actual_rec_size, __u32 expected_rec_size) {
  __u64 info_len = actual_rec_size * cnt;
  void *info, *nrecord;
  int i;
  info = malloc(info_len);
  if (!info)
    return NULL;
  nrecord = info;
  for (i = 0; i < cnt; i++) {
    memcpy(nrecord, orecord, expected_rec_size);
    memset(nrecord + expected_rec_size, 0, actual_rec_size - expected_rec_size);
    orecord += actual_rec_size;
    nrecord += actual_rec_size;
  }
  return info;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<alloc_zero_tailing_info>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  push   %rbx
  imul   %r13d,%edi
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %rax,%rax
  je     L23
  test   %r14d,%r14d
  jle    L23
  mov    %rax,%rbx
  movslq %r12d,%rax
  mov    %r13d,%r14d
  xor    %r15d,%r15d
  mov    %rax,0x10(%rsp)
  sub    %r12d,%r14d
L19:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r12d,%edx
  add    %r13,%rbp
  call   <memcpy@plt>
  mov    0x10(%rsp),%rax
  mov    %r14d,%edx
  xor    %esi,%esi
  add    $0x1,%r15d
  lea    (%rbx,%rax,1),%rdi
  add    %r13,%rbx
  call   <memset@plt>
  cmp    %r15d,0xc(%rsp)
  jne    L19
L23:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void *alloc_zero_tailing_info(const void *orecord, __u32 cnt, __u32 actual_rec_size, __u32 expected_rec_size) {
  __u64 info_len = actual_rec_size * cnt;
  void *info, *nrecord;
  int i;
  info = malloc(info_len);
  if (!info)
    return NULL;
  nrecord = info;
  for (i = 0; i < cnt; i++) {
    memcpy(nrecord, orecord, expected_rec_size);
    memset(nrecord + expected_rec_size, 0, actual_rec_size - expected_rec_size);
    orecord += actual_rec_size;
    nrecord += actual_rec_size;
  }
  return info;
}
```
user: What is the c source code of the assembly code below:

```asm
<asn1_get_tag>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,-0x1c(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jg     L0
  mov    $0x14,%eax
  jmp    L39
L0:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  je     L10
  mov    $0x16,%eax
  jmp    L39
L10:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <asn1_get_len@plt>
L39:
  leave
  ret
```

assistant: ```c
int asn1_get_tag(unsigned char **p, const unsigned char *end, int *len, int tag) {
  if ((end - *p) < 1)
    return (0x0014);
  if (**p != tag)
    return (0x0016);
  (*p)++;
  return (asn1_get_len(p, end, len));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<asn1_get_tag>:
  endbr64
  mov    (%rdi),%r8
  mov    %rsi,%rax
  sub    %r8,%rax
  test   %rax,%rax
  jle    L7
  movzbl (%r8),%r9d
  mov    $0x16,%eax
  cmp    %ecx,%r9d
  je     L12
  ret
L12:
  sub    $0x8,%rsp
  add    $0x1,%r8
  mov    %r8,(%rdi)
  call   <asn1_get_len@plt>
  add    $0x8,%rsp
  ret
L7:
  mov    $0x14,%eax
  ret
```

assistant: ```c
int asn1_get_tag(unsigned char **p, const unsigned char *end, int *len, int tag) {
  if ((end - *p) < 1)
    return (0x0014);
  if (**p != tag)
    return (0x0016);
  (*p)++;
  return (asn1_get_len(p, end, len));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<asn1_get_tag>:
  endbr64
  mov    (%rdi),%rax
  mov    %rsi,%r8
  sub    %rax,%r8
  test   %r8,%r8
  jle    L1
  movzbl (%rax),%r8d
  mov    $0x16,%r9d
  cmp    %ecx,%r8d
  je     L10
  mov    %r9d,%eax
  ret
  nopl   (%rax)
L10:
  add    $0x1,%rax
  mov    %rax,(%rdi)
  jmp    1050 <asn1_get_len@plt>
  nopl   0x0(%rax)
L1:
  mov    $0x14,%r9d
  mov    %r9d,%eax
  ret
```

assistant: ```c
int asn1_get_tag(unsigned char **p, const unsigned char *end, int *len, int tag) {
  if ((end - *p) < 1)
    return (0x0014);
  if (**p != tag)
    return (0x0016);
  (*p)++;
  return (asn1_get_len(p, end, len));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<asn1_get_tag>:
  endbr64
  mov    (%rdi),%rax
  mov    %rsi,%r8
  sub    %rax,%r8
  test   %r8,%r8
  jle    L6
  movzbl (%rax),%r8d
  mov    $0x16,%r9d
  cmp    %ecx,%r8d
  je     L5
  mov    %r9d,%eax
  ret
  nopl   (%rax)
L5:
  add    $0x1,%rax
  mov    %rax,(%rdi)
  jmp    1050 <asn1_get_len@plt>
  nopl   0x0(%rax)
L6:
  mov    $0x14,%r9d
  mov    %r9d,%eax
  ret
```

assistant: ```c
int asn1_get_tag(unsigned char **p, const unsigned char *end, int *len, int tag) {
  if ((end - *p) < 1)
    return (0x0014);
  if (**p != tag)
    return (0x0016);
  (*p)++;
  return (asn1_get_len(p, end, len));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<authnone_validate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
bool_t authnone_validate(AUTH *x, struct opaque_auth *x1) {
  (void)x;
  (void)x1;
  return (TRUE);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<authnone_validate>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
bool_t authnone_validate(AUTH *x, struct opaque_auth *x1) {
  (void)x;
  (void)x1;
  return (TRUE);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<authnone_validate>:
  endbr64
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
bool_t authnone_validate(AUTH *x, struct opaque_auth *x1) {
  (void)x;
  (void)x1;
  return (TRUE);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<authnone_validate>:
  endbr64
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
bool_t authnone_validate(AUTH *x, struct opaque_auth *x1) {
  (void)x;
  (void)x1;
  return (TRUE);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<litCmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  sar    %eax
  cmp    %eax,%edx
  jge    L2
  mov    $0x1,%eax
  jmp    L4
L2:
  mov    $0x0,%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int litCmp(int *a, int *b) {
  if (((*a) >> 1) < ((*b) >> 1))
    return 1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<litCmp>:
  endbr64
  mov    (%rdi),%edx
  sar    %edx
  mov    (%rsi),%eax
  sar    %eax
  cmp    %eax,%edx
  setl   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int litCmp(int *a, int *b) {
  if (((*a) >> 1) < ((*b) >> 1))
    return 1;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<litCmp>:
  endbr64
  mov    (%rdi),%edx
  mov    (%rsi),%eax
  sar    %edx
  sar    %eax
  cmp    %eax,%edx
  setl   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int litCmp(int *a, int *b) {
  if (((*a) >> 1) < ((*b) >> 1))
    return 1;
  else
    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
<litCmp>:
  endbr64
  mov    (%rdi),%edx
  mov    (%rsi),%eax
  sar    %edx
  sar    %eax
  cmp    %eax,%edx
  setl   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int litCmp(int *a, int *b) {
  if (((*a) >> 1) < ((*b) >> 1))
    return 1;
  else
    return 0;
}
```