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

```asm
<acquireSynchronization>:
  endbr64
  push   %r13
  push   %r12
  movsbl %sil,%r12d
  push   %rbp
  mov    %edi,%ebp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <readPacket@plt>
  test   %eax,%eax
  je     L37
  cmp    $0x1,%eax
  jne    L19
  cmpb   $0x1,0x7(%rsp)
  jne    L19
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%ecx
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <writePacket@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L37:
  xor    %ecx,%ecx
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <writePacket@plt>
  jmp    L19
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
datatype: string

tool: D2: Sending confirmation packet.


assistant: ```c
void acquireSynchronization(int fd, char control) {
  int sync = 0;
  char buf;
  int size;
  while (!sync) {
    size = readPacket(fd, control, &buf);
    if (size == 0) {
      writePacket(fd, control, &buf, size);
    } else if (size == 1 && buf == 1) {
      printf("Sending confirmation packet.\n");
      sync = 1;
      writePacket(fd, control, &buf, size);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<perf_config_bool>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <perf_config_bool_or_int@plt>
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int perf_config_bool(const char *name, const char *value) {
  int discard;
  return !!perf_config_bool_or_int(name, value, &discard);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<perf_config_bool>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  call   <perf_config_bool_or_int@plt>
  test   %eax,%eax
  setne  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int perf_config_bool(const char *name, const char *value) {
  int discard;
  return !!perf_config_bool_or_int(name, value, &discard);
}
```
user: What is the c source code of the assembly code below:

```asm
<perf_config_bool>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  call   <perf_config_bool_or_int@plt>
  test   %eax,%eax
  setne  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int perf_config_bool(const char *name, const char *value) {
  int discard;
  return !!perf_config_bool_or_int(name, value, &discard);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<perf_config_bool>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  call   <perf_config_bool_or_int@plt>
  test   %eax,%eax
  setne  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int perf_config_bool(const char *name, const char *value) {
  int discard;
  return !!perf_config_bool_or_int(name, value, &discard);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<encrypt9p>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xc0,%rsp
  mov    %rdi,-0xa8(%rbp)
  mov    %rsi,-0xb0(%rbp)
  mov    %edx,-0xb4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x7,-0xb4(%rbp)
  jg     L48
  mov    $0x0,%eax
  jmp    L32
L48:
  lea    -0x90(%rbp),%rdx
  mov    -0xa8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <key_setup@plt>
  mov    -0xb0(%rbp),%rax
  mov    %rax,-0x98(%rbp)
  subl   $0x1,-0xb4(%rbp)
  mov    -0xb4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x3,%edx
  sub    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x9c(%rbp)
  mov    -0xb4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xb4(%rbp)
  movl   $0x0,-0xa0(%rbp)
  jmp    L40
L53:
  mov    -0x98(%rbp),%rcx
  lea    -0x90(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <block_cipher@plt>
  addq   $0x7,-0x98(%rbp)
  addl   $0x1,-0xa0(%rbp)
L40:
  mov    -0xa0(%rbp),%eax
  cmp    -0xb4(%rbp),%eax
  jl     L53
  cmpl   $0x0,-0x9c(%rbp)
  je     L56
  mov    -0x9c(%rbp),%eax
  cltq
  lea    -0x7(%rax),%rdx
  mov    -0x98(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x90(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <block_cipher@plt>
L56:
  mov    $0x1,%eax
L32:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

assistant: ```c
int encrypt9p(void *key, void *vbuf, int n) {
  char ekey[128], *buf;
  int i, r;
  if (n < 8)
    return 0;
  key_setup(key, ekey);
  buf = vbuf;
  n--;
  r = n % 7;
  n /= 7;
  for (i = 0; i < n; i++) {
    block_cipher(ekey, buf, 0);
    buf += 7;
  }
  if (r)
    block_cipher(ekey, buf - 7 + r, 0);
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<encrypt9p>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  cmp    $0x7,%edx
  jg     L4
L31:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L47
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  mov    %rsi,%rbp
  mov    %edx,%ebx
  mov    %rsp,%rsi
  call   <key_setup@plt>
  sub    $0x1,%ebx
  movslq %ebx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %ebx,%eax
  sar    $0x2,%eax
  mov    %ebx,%ecx
  sar    $0x1f,%ecx
  mov    %eax,%r12d
  sub    %ecx,%r12d
  lea    0x0(,%r12,8),%edx
  sub    %r12d,%edx
  sub    %edx,%ebx
  mov    %ebx,%r12d
  sub    %ecx,%eax
  lea    -0x1(%rax),%r13d
  mov    %eax,%eax
  lea    0x0(%rbp,%rax,8),%r14
  sub    %rax,%r14
  mov    %rbp,%rbx
  mov    %rsp,%r15
L22:
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <block_cipher@plt>
  add    $0x7,%rbx
  cmp    %r14,%rbx
  jne    L22
  lea    0x0(,%r13,8),%rax
  sub    %r13,%rax
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %r12d,%r12d
  je     L31
  movslq %r12d,%r12
  add    %rdx,%r12
  lea    0x0(%rbp,%r12,1),%rsi
  mov    %rsp,%rdi
  mov    $0x0,%edx
  call   <block_cipher@plt>
  mov    $0x1,%eax
  jmp    L31
L47:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int encrypt9p(void *key, void *vbuf, int n) {
  char ekey[128], *buf;
  int i, r;
  if (n < 8)
    return 0;
  key_setup(key, ekey);
  buf = vbuf;
  n--;
  r = n % 7;
  n /= 7;
  for (i = 0; i < n; i++) {
    block_cipher(ekey, buf, 0);
    buf += 7;
  }
  if (r)
    block_cipher(ekey, buf - 7 + r, 0);
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<encrypt9p>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  cmp    $0x7,%edx
  jg     L35
L46:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x98,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L35:
  mov    %rsp,%r15
  mov    %rsi,%rbp
  mov    %edx,%ebx
  mov    %r15,%rsi
  call   <key_setup@plt>
  lea    -0x1(%rbx),%edx
  mov    %rdx,%rbx
  imul   $0x24924925,%rdx,%rdx
  mov    %ebx,%eax
  shr    $0x20,%rdx
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  mov    %eax,%r12d
  shr    $0x2,%eax
  shr    $0x2,%r12d
  lea    0x0(%rbp,%rax,8),%r14
  lea    -0x1(%rax),%r13d
  lea    0x0(,%r12,8),%edx
  sub    %rax,%r14
  sub    %r12d,%edx
  sub    %edx,%ebx
  movslq %ebx,%r12
  mov    %rbp,%rbx
  nopw   0x0(%rax,%rax,1)
L45:
  mov    %rbx,%rsi
  xor    %edx,%edx
  mov    %r15,%rdi
  add    $0x7,%rbx
  call   <block_cipher@plt>
  cmp    %r14,%rbx
  jne    L45
  lea    0x0(,%r13,8),%rax
  sub    %r13,%rax
  mov    $0x1,%r13d
  test   %r12d,%r12d
  je     L46
  add    %rax,%r12
  xor    %edx,%edx
  mov    %r15,%rdi
  lea    0x0(%rbp,%r12,1),%rsi
  call   <block_cipher@plt>
  jmp    L46
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int encrypt9p(void *key, void *vbuf, int n) {
  char ekey[128], *buf;
  int i, r;
  if (n < 8)
    return 0;
  key_setup(key, ekey);
  buf = vbuf;
  n--;
  r = n % 7;
  n /= 7;
  for (i = 0; i < n; i++) {
    block_cipher(ekey, buf, 0);
    buf += 7;
  }
  if (r)
    block_cipher(ekey, buf - 7 + r, 0);
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<encrypt9p>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  cmp    $0x7,%edx
  jg     L28
L29:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x98,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L28:
  mov    %rsp,%r15
  mov    %rsi,%rbp
  mov    %edx,%ebx
  mov    %r15,%rsi
  call   <key_setup@plt>
  lea    -0x1(%rbx),%edx
  mov    %rdx,%rbx
  imul   $0x24924925,%rdx,%rdx
  mov    %ebx,%eax
  shr    $0x20,%rdx
  sub    %edx,%eax
  shr    %eax
  add    %edx,%eax
  mov    %eax,%r12d
  shr    $0x2,%eax
  shr    $0x2,%r12d
  lea    0x0(%rbp,%rax,8),%r14
  lea    -0x1(%rax),%r13d
  lea    0x0(,%r12,8),%edx
  sub    %rax,%r14
  sub    %r12d,%edx
  sub    %edx,%ebx
  movslq %ebx,%r12
  mov    %rbp,%rbx
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %rbx,%rsi
  xor    %edx,%edx
  mov    %r15,%rdi
  add    $0x7,%rbx
  call   <block_cipher@plt>
  cmp    %r14,%rbx
  jne    L5
  lea    0x0(,%r13,8),%rax
  sub    %r13,%rax
  mov    $0x1,%r13d
  test   %r12d,%r12d
  je     L29
  add    %rax,%r12
  xor    %edx,%edx
  mov    %r15,%rdi
  lea    0x0(%rbp,%r12,1),%rsi
  call   <block_cipher@plt>
  jmp    L29
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int encrypt9p(void *key, void *vbuf, int n) {
  char ekey[128], *buf;
  int i, r;
  if (n < 8)
    return 0;
  key_setup(key, ekey);
  buf = vbuf;
  n--;
  r = n % 7;
  n /= 7;
  for (i = 0; i < n; i++) {
    block_cipher(ekey, buf, 0);
    buf += 7;
  }
  if (r)
    block_cipher(ekey, buf - 7 + r, 0);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<do_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L9
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void do_free(void *ptr) {
  if (ptr != NULL)
    free(ptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<do_free>:
  endbr64
  test   %rdi,%rdi
  je     L3
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L3:
  ret
```

assistant: ```c
void do_free(void *ptr) {
  if (ptr != NULL)
    free(ptr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<do_free>:
  endbr64
  test   %rdi,%rdi
  je     L4
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L4:
  ret
```

assistant: ```c
void do_free(void *ptr) {
  if (ptr != NULL)
    free(ptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<do_free>:
  endbr64
  test   %rdi,%rdi
  je     L3
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L3:
  ret
```

assistant: ```c
void do_free(void *ptr) {
  if (ptr != NULL)
    free(ptr);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x3d0900,%edi
  call   <sumFibUnevenUntil@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
type_guessed: string

tool: D4: %lu

assistant: ```c
int main() {
  printf("%lu", sumFibUnevenUntil(4000000));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3d0900,%edi
  call   <sumFibUnevenUntil@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
datatype: string

tool: D4: %lu

assistant: ```c
int main() {
  printf("%lu", sumFibUnevenUntil(4000000));
  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    $0x8,%rsp
  mov    $0x3d0900,%edi
  call   <sumFibUnevenUntil@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type: string

tool: D4: %lu

assistant: ```c
int main() {
  printf("%lu", sumFibUnevenUntil(4000000));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3d0900,%edi
  call   <sumFibUnevenUntil@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D6: %lu

assistant: ```c
int main() {
  printf("%lu", sumFibUnevenUntil(4000000));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<vkill_trailing>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L10:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  imul   -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movsbl -0x24(%rbp),%eax
  mov    -0x1c(%rbp),%ecx
  imul   -0x4(%rbp),%ecx
  movslq %ecx,%rsi
  mov    -0x18(%rbp),%rcx
  add    %rsi,%rcx
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <kill_trailingn@plt>
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  cmp    %eax,-0x4(%rbp)
  jl     L10
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *vkill_trailing(char *cstr, int elem_len, int sizeofcstr, char t) {
  int i;
  for (i = 0; i < sizeofcstr / elem_len; i++)
    kill_trailingn(cstr + elem_len * i, t, cstr + elem_len * (i + 1) - 1);
  return cstr;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vkill_trailing>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %edx,%eax
  cltd
  idiv   %esi
  test   %eax,%eax
  jle    L3
  mov    %ecx,%r14d
  mov    %eax,%r13d
  movslq %esi,%r12
  mov    %rdi,%rbx
  mov    $0x0,%ebp
  lea    -0x1(%r12),%r15
L16:
  add    $0x1,%ebp
  lea    (%r15,%rbx,1),%rdx
  movsbl %r14b,%esi
  mov    %rbx,%rdi
  call   <kill_trailingn@plt>
  add    %r12,%rbx
  cmp    %r13d,%ebp
  jne    L16
L3:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *vkill_trailing(char *cstr, int elem_len, int sizeofcstr, char t) {
  int i;
  for (i = 0; i < sizeofcstr / elem_len; i++)
    kill_trailingn(cstr + elem_len * i, t, cstr + elem_len * (i + 1) - 1);
  return cstr;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vkill_trailing>:
  endbr64
  mov    %edx,%eax
  push   %r15
  cltd
  push   %r14
  idiv   %esi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  test   %eax,%eax
  jle    L15
  movslq %esi,%rbp
  mov    %eax,%r12d
  movsbl %cl,%r13d
  mov    %rdi,%rbx
  xor    %r15d,%r15d
  lea    -0x1(%rbp),%r14
  nopl   0x0(%rax)
L7:
  lea    (%r14,%rbx,1),%rdx
  mov    %rbx,%rdi
  mov    %r13d,%esi
  add    $0x1,%r15d
  call   <kill_trailingn@plt>
  add    %rbp,%rbx
  cmp    %r12d,%r15d
  jne    L7
L15:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *vkill_trailing(char *cstr, int elem_len, int sizeofcstr, char t) {
  int i;
  for (i = 0; i < sizeofcstr / elem_len; i++)
    kill_trailingn(cstr + elem_len * i, t, cstr + elem_len * (i + 1) - 1);
  return cstr;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vkill_trailing>:
  endbr64
  mov    %edx,%eax
  push   %r15
  cltd
  push   %r14
  idiv   %esi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  test   %eax,%eax
  jle    L8
  movslq %esi,%rbp
  mov    %eax,%r12d
  movsbl %cl,%r13d
  mov    %rdi,%rbx
  xor    %r15d,%r15d
  lea    -0x1(%rbp),%r14
  nopl   0x0(%rax)
L7:
  lea    (%r14,%rbx,1),%rdx
  mov    %rbx,%rdi
  mov    %r13d,%esi
  add    $0x1,%r15d
  call   <kill_trailingn@plt>
  add    %rbp,%rbx
  cmp    %r12d,%r15d
  jne    L7
L8:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *vkill_trailing(char *cstr, int elem_len, int sizeofcstr, char t) {
  int i;
  for (i = 0; i < sizeofcstr / elem_len; i++)
    kill_trailingn(cstr + elem_len * i, t, cstr + elem_len * (i + 1) - 1);
  return cstr;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main(void) {
  gen_tabs();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int main(void) {
  gen_tabs();
  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
  xor    %eax,%eax
  call   <gen_tabs@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  gen_tabs();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

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

```asm
<yalex_strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L0
  cmpq   $0x0,-0x20(%rbp)
  jne    L37
L0:
  mov    $0xffffffff,%eax
  jmp    L155
L37:
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L124
L33:
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L32
  mov    -0xc(%rbp),%eax
  jmp    L155
L32:
  addl   $0x1,-0x8(%rbp)
L124:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L117
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L33
L117:
  movl   $0x0,-0x4(%rbp)
L150:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L158
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L158
  mov    $0x0,%eax
  jmp    L155
L158:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L77
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jge    L62
  mov    $0xffffffff,%eax
  jmp    L155
L62:
  mov    $0x1,%eax
  jmp    L155
L77:
  addl   $0x1,-0x4(%rbp)
  jmp    L150
L155:
  pop    %rbp
  ret
```

assistant: ```c
int yalex_strcmp(const char *a, const char *b) {
  if (a == 0 || b == 0)
    return -1;
  int offset = 0;
  for (int i = 0; a[i] && a[i] != b[0]; i++) {
    offset++;
    if (a[offset] == b[0]) {
      return offset;
    }
  }
  for (int i = 0;; i++) {
    if (a[i] == '\0' && b[i] == '\0') {
      return 0;
    }
    if (a[i] != b[i]) {
      return a[i] < b[i] ? -1 : 1;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<yalex_strcmp>:
  endbr64
  test   %rdi,%rdi
  je     L116
  test   %rsi,%rsi
  je     L116
  movzbl (%rdi),%edx
  mov    $0x0,%eax
  test   %dl,%dl
  je     L10
  movzbl (%rsi),%ecx
  mov    $0x1,%eax
L9:
  cmp    %dl,%cl
  je     L28
  mov    %eax,%edx
  cmp    %cl,(%rdi,%rax,1)
  je     L1
  add    $0x1,%rax
  movzbl -0x1(%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L9
  mov    $0x0,%eax
  jmp    L10
L28:
  mov    $0x0,%eax
  jmp    L10
L17:
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L20
L10:
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L17
  movzbl (%rsi,%rax,1),%ecx
  test   %cl,%cl
  je     L15
L20:
  cmp    %dl,%cl
  setle  %dl
  movzbl %dl,%edx
  lea    -0x1(%rdx,%rdx,1),%edx
L1:
  mov    %edx,%eax
  ret
L116:
  mov    $0xffffffff,%edx
  jmp    L1
L15:
  mov    $0x0,%edx
  jmp    L1
```

assistant: ```c
int yalex_strcmp(const char *a, const char *b) {
  if (a == 0 || b == 0)
    return -1;
  int offset = 0;
  for (int i = 0; a[i] && a[i] != b[0]; i++) {
    offset++;
    if (a[offset] == b[0]) {
      return offset;
    }
  }
  for (int i = 0;; i++) {
    if (a[i] == '\0' && b[i] == '\0') {
      return 0;
    }
    if (a[i] != b[i]) {
      return a[i] < b[i] ? -1 : 1;
    }
  }
}
```