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

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: parse_data
arguments: label: D2
type_infered: string

tool: D2: Usage: ./pblogd -f <filename> -i <interval>


assistant: ```c
void usage(void) {
  printf("Usage: ./pblogd -f <filename> -i <interval>\n");
  exit(1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x41,%ecx
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi_2@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
type: string

tool: D9: ==============


assistant: ```c
int main(int argc, const char *argv[]) {
  hanoi_2(3, 'B', 'C', 'A');
  printf("==============\n");
  hanoi(3, 'B', 'C');
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x41,%ecx
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi_2@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type_infered: string

tool: D1: ==============


assistant: ```c
int main(int argc, const char *argv[]) {
  hanoi_2(3, 'B', 'C', 'A');
  printf("==============\n");
  hanoi(3, 'B', 'C');
  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
  mov    $0x41,%ecx
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi_2@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
datatype: string

tool: D7: ==============


assistant: ```c
int main(int argc, const char *argv[]) {
  hanoi_2(3, 'B', 'C', 'A');
  printf("==============\n");
  hanoi(3, 'B', 'C');
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x41,%ecx
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi_2@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x43,%edx
  mov    $0x42,%esi
  mov    $0x3,%edi
  call   <hanoi@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: ==============


assistant: ```c
int main(int argc, const char *argv[]) {
  hanoi_2(3, 'B', 'C', 'A');
  printf("==============\n");
  hanoi(3, 'B', 'C');
  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
<ft_strsub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0xffffffffffffffff,-0x10(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L20
  mov    $0x0,%eax
  jmp    L21
L20:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strnew@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L40
  mov    $0x0,%eax
  jmp    L21
L4:
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
L40:
  addq   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L4
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L21:
  leave
  ret
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  size_t i;
  char *new;
  i = -1;
  if (s == NULL)
    return (NULL);
  if (!(new = ft_strnew(len)))
    return (NULL);
  while (++i < len)
    new[i] = s[start + i];
  new[i] = '\0';
  return (new);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ft_strsub>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L18
  mov    %esi,%ebp
  mov    %rdx,%rbx
  mov    %rdx,%rdi
  call   <ft_strnew@plt>
  test   %rax,%rax
  je     L25
  test   %rbx,%rbx
  je     L26
  mov    $0x0,%edx
  mov    %ebp,%esi
  add    %r12,%rsi
L32:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%rbx
  jne    L32
L26:
  movb   $0x0,(%rax,%rbx,1)
L25:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    %rdi,%rax
  jmp    L25
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  size_t i;
  char *new;
  i = -1;
  if (s == NULL)
    return (NULL);
  if (!(new = ft_strnew(len)))
    return (NULL);
  while (++i < len)
    new[i] = s[start + i];
  new[i] = '\0';
  return (new);
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_strsub>:
  endbr64
  test   %rdi,%rdi
  je     L41
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  call   <ft_strnew@plt>
  test   %rax,%rax
  je     L3
  test   %rbx,%rbx
  je     L11
  mov    %ebp,%esi
  xor    %ecx,%ecx
  add    %r12,%rsi
  xchg   %ax,%ax
L42:
  movzbl (%rsi,%rcx,1),%edx
  mov    %dl,(%rax,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rbx
  jne    L42
  add    %rax,%rbx
L19:
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %rax,%rbx
  jmp    L19
  nopl   (%rax)
L41:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  size_t i;
  char *new;
  i = -1;
  if (s == NULL)
    return (NULL);
  if (!(new = ft_strnew(len)))
    return (NULL);
  while (++i < len)
    new[i] = s[start + i];
  new[i] = '\0';
  return (new);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_strsub>:
  endbr64
  test   %rdi,%rdi
  je     L116
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %esi,%ebx
  call   <ft_strnew@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L151
  test   %rbp,%rbp
  je     L98
  mov    %ebx,%edx
  mov    %r8,%rcx
  lea    -0x1(%rbp),%rax
  lea    0x1(%r12,%rdx,1),%rsi
  sub    %rsi,%rcx
  cmp    $0xe,%rcx
  jbe    L158
  cmp    $0x6,%rax
  jbe    L158
  cmp    $0xe,%rax
  jbe    L117
  mov    %rbp,%rcx
  add    %r12,%rdx
  xor    %eax,%eax
  and    $0xfffffffffffffff0,%rcx
  nopl   0x0(%rax)
L42:
  movdqu (%rdx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L42
  mov    %rbp,%rcx
  and    $0xfffffffffffffff0,%rcx
  test   $0xf,%bpl
  je     L129
  mov    %rbp,%rax
  sub    %rcx,%rax
  lea    -0x1(%rax),%rsi
  cmp    $0x6,%rsi
  jbe    L153
L164:
  mov    (%rdx,%rcx,1),%rsi
  mov    %rsi,(%r8,%rcx,1)
  mov    %rax,%rsi
  and    $0xfffffffffffffff8,%rsi
  add    %rsi,%rcx
  cmp    %rsi,%rax
  je     L129
L153:
  movzbl (%rdx,%rcx,1),%eax
  mov    %al,(%r8,%rcx,1)
  lea    0x1(%rcx),%rax
  cmp    %rax,%rbp
  jbe    L129
  movzbl 0x1(%rdx,%rcx,1),%eax
  mov    %al,0x1(%r8,%rcx,1)
  lea    0x2(%rcx),%rax
  cmp    %rax,%rbp
  jbe    L129
  movzbl 0x2(%rdx,%rcx,1),%eax
  mov    %al,0x2(%r8,%rcx,1)
  lea    0x3(%rcx),%rax
  cmp    %rax,%rbp
  jbe    L129
  movzbl 0x3(%rdx,%rcx,1),%eax
  mov    %al,0x3(%r8,%rcx,1)
  lea    0x4(%rcx),%rax
  cmp    %rax,%rbp
  jbe    L129
  movzbl 0x4(%rdx,%rcx,1),%eax
  mov    %al,0x4(%r8,%rcx,1)
  lea    0x5(%rcx),%rax
  cmp    %rax,%rbp
  jbe    L129
  movzbl 0x5(%rdx,%rcx,1),%eax
  mov    %al,0x5(%r8,%rcx,1)
  lea    0x6(%rcx),%rax
  cmp    %rax,%rbp
  jbe    L129
  movzbl 0x6(%rdx,%rcx,1),%eax
  mov    %al,0x6(%r8,%rcx,1)
L129:
  lea    (%r8,%rbp,1),%rax
L98:
  movb   $0x0,(%rax)
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L117:
  mov    %rbp,%rax
  xor    %ecx,%ecx
  add    %r12,%rdx
  jmp    L164
L151:
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L158:
  xor    %ecx,%ecx
  add    %r12,%rdx
  nopl   (%rax)
L87:
  movzbl (%rdx,%rcx,1),%eax
  mov    %al,(%r8,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rbp
  jne    L87
  jmp    L129
  nopl   0x0(%rax,%rax,1)
L116:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  size_t i;
  char *new;
  i = -1;
  if (s == NULL)
    return (NULL);
  if (!(new = ft_strnew(len)))
    return (NULL);
  while (++i < len)
    new[i] = s[start + i];
  new[i] = '\0';
  return (new);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mb_memdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L7
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <mb_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *mb_memdup(char *val, unsigned size) {
  char *result = 0;
  if (val != 0) {
    result = (char *)mb_malloc(size);
    memcpy(result, val, size);
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mb_memdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L8
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %esi,%edi
  call   <mb_malloc@plt>
  mov    %rax,%rbp
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L8:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *mb_memdup(char *val, unsigned size) {
  char *result = 0;
  if (val != 0) {
    result = (char *)mb_malloc(size);
    memcpy(result, val, size);
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mb_memdup>:
  endbr64
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  test   %rdi,%rdi
  je     L6
  mov    %rdi,%rbp
  mov    %esi,%edi
  mov    %esi,%r13d
  call   <mb_malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
L6:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *mb_memdup(char *val, unsigned size) {
  char *result = 0;
  if (val != 0) {
    result = (char *)mb_malloc(size);
    memcpy(result, val, size);
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mb_memdup>:
  endbr64
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  test   %rdi,%rdi
  je     L8
  mov    %rdi,%rbp
  mov    %esi,%edi
  mov    %esi,%r13d
  call   <mb_malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
L8:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *mb_memdup(char *val, unsigned size) {
  char *result = 0;
  if (val != 0) {
    result = (char *)mb_malloc(size);
    memcpy(result, val, size);
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parseMetaDataDurationUs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L57
  mov    $0xffffffff,%eax
  jmp    L42
L57:
  mov    -0x18(%rbp),%rax
  sub    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jns    L6
  mov    $0xffffffff,%eax
  jmp    L42
L6:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <parseDouble@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L24
  mov    -0x24(%rbp),%eax
  jmp    L42
L24:
  movsd  -0x20(%rbp),%xmm1
  movsd  D3(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
L42:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
type_guessed: f64

tool: D3: 1000000

assistant: ```c
int64_t parseMetaDataDurationUs(const char *line) {
  const char *match = strstr(line, ":");
  if (match == NULL) {
    return -1;
  }
  ssize_t colonPos = match - line;
  if (colonPos < 0) {
    return -1;
  }
  double x;
  int err = parseDouble(line + colonPos + 1, &x);
  if (err != 0) {
    return err;
  }
  return ((int64_t)(x * 1E6));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parseMetaDataDurationUs>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x3a,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L40
  sub    %rbx,%rax
  js     L15
  mov    %rsp,%rsi
  lea    0x1(%rbx,%rax,1),%rdi
  call   <parseDouble@plt>
  test   %eax,%eax
  jne    L20
  movsd  D4(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  cvttsd2si %xmm0,%eax
L20:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L58
  add    $0x10,%rsp
  pop    %rbx
  ret
L40:
  mov    $0xffffffff,%eax
  jmp    L20
L15:
  mov    $0xffffffff,%eax
  jmp    L20
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type: f64

tool: D4: 1000000

assistant: ```c
int64_t parseMetaDataDurationUs(const char *line) {
  const char *match = strstr(line, ":");
  if (match == NULL) {
    return -1;
  }
  ssize_t colonPos = match - line;
  if (colonPos < 0) {
    return -1;
  }
  double x;
  int err = parseDouble(line + colonPos + 1, &x);
  if (err != 0) {
    return err;
  }
  return ((int64_t)(x * 1E6));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<parseMetaDataDurationUs>:
  endbr64
  push   %rbx
  mov    $0x3a,%esi
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strchr@plt>
  test   %rax,%rax
  je     L30
  sub    %rbx,%rax
  js     L30
  mov    %rsp,%rsi
  lea    0x1(%rbx,%rax,1),%rdi
  call   <parseDouble@plt>
  test   %eax,%eax
  jne    L18
  movsd  D5(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  cvttsd2si %xmm0,%eax
L18:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  mov    $0xffffffff,%eax
  jmp    L18
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
datatype: f64

tool: D5: 1000000

assistant: ```c
int64_t parseMetaDataDurationUs(const char *line) {
  const char *match = strstr(line, ":");
  if (match == NULL) {
    return -1;
  }
  ssize_t colonPos = match - line;
  if (colonPos < 0) {
    return -1;
  }
  double x;
  int err = parseDouble(line + colonPos + 1, &x);
  if (err != 0) {
    return err;
  }
  return ((int64_t)(x * 1E6));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<parseMetaDataDurationUs>:
  endbr64
  push   %rbx
  mov    $0x3a,%esi
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strchr@plt>
  test   %rax,%rax
  je     L33
  sub    %rbx,%rax
  js     L33
  mov    %rsp,%rsi
  lea    0x1(%rbx,%rax,1),%rdi
  call   <parseDouble@plt>
  test   %eax,%eax
  jne    L20
  movsd  D7(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  cvttsd2si %xmm0,%eax
L20:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L33:
  mov    $0xffffffff,%eax
  jmp    L20
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_guessed: f64

tool: D7: 1000000

assistant: ```c
int64_t parseMetaDataDurationUs(const char *line) {
  const char *match = strstr(line, ":");
  if (match == NULL) {
    return -1;
  }
  ssize_t colonPos = match - line;
  if (colonPos < 0) {
    return -1;
  }
  double x;
  int err = parseDouble(line + colonPos + 1, &x);
  if (err != 0) {
    return err;
  }
  return ((int64_t)(x * 1E6));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cmd_initscr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L10
  mov    $0x1,%edi
  call   <report_count@plt>
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    %eax,%edi
  call   <report_ptr@plt>
  jmp    L1
L10:
  nop
L1:
  leave
  ret
```

assistant: ```c
void cmd_initscr(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_ptr(initscr());
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cmd_initscr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L1
  mov    $0x1,%edi
  call   <report_count@plt>
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    %eax,%edi
  call   <report_ptr@plt>
L1:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cmd_initscr(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_ptr(initscr());
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<cmd_initscr>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L5
  mov    $0x1,%edi
  call   <report_count@plt>
  xor    %eax,%eax
  call   <initscr@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    10a0 <report_ptr@plt>
L5:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cmd_initscr(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_ptr(initscr());
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cmd_initscr>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L5
  mov    $0x1,%edi
  call   <report_count@plt>
  xor    %eax,%eax
  call   <initscr@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    10a0 <report_ptr@plt>
L5:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cmd_initscr(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_ptr(initscr());
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<poke>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x2a,(%rax)
  movq   $0x0,-0x8(%rbp)
  jmp    L17
L9:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L17:
  cmpq   $0x9,-0x8(%rbp)
  jbe    L9
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void poke(int *b, size_t index) {
  size_t i;
  volatile int sum = 0;
  b[index] = 42;
  for (i = 0; i < 10; i++)
    sum += b[i];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<poke>:
  endbr64
  movl   $0x0,-0x4(%rsp)
  movl   $0x2a,(%rdi,%rsi,4)
  mov    %rdi,%rax
  add    $0x28,%rdi
L4:
  mov    -0x4(%rsp),%edx
  add    (%rax),%edx
  mov    %edx,-0x4(%rsp)
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L4
  ret
```

assistant: ```c
void poke(int *b, size_t index) {
  size_t i;
  volatile int sum = 0;
  b[index] = 42;
  for (i = 0; i < 10; i++)
    sum += b[i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<poke>:
  endbr64
  movl   $0x0,-0x4(%rsp)
  lea    0x28(%rdi),%rdx
  movl   $0x2a,(%rdi,%rsi,4)
  nopw   0x0(%rax,%rax,1)
L3:
  mov    -0x4(%rsp),%eax
  add    $0x4,%rdi
  add    -0x4(%rdi),%eax
  mov    %eax,-0x4(%rsp)
  cmp    %rdx,%rdi
  jne    L3
  ret
```

assistant: ```c
void poke(int *b, size_t index) {
  size_t i;
  volatile int sum = 0;
  b[index] = 42;
  for (i = 0; i < 10; i++)
    sum += b[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<poke>:
  endbr64
  movl   $0x0,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  movl   $0x2a,(%rdi,%rsi,4)
  add    (%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x4(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x8(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0xc(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x10(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x14(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x18(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x1c(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x20(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  mov    -0x4(%rsp),%edx
  add    0x24(%rdi),%edx
  mov    %edx,-0x4(%rsp)
  ret
```

assistant: ```c
void poke(int *b, size_t index) {
  size_t i;
  volatile int sum = 0;
  b[index] = 42;
  for (i = 0; i < 10; i++)
    sum += b[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rand_zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L7
  mov    $0xffffffff,%eax
  jmp    L42
L7:
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L24
  mov    $0x0,%eax
  call   <rand@plt>
  jmp    L42
L24:
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  mov    -0x14(%rbp),%edx
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L6
L20:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,-0xc(%rbp)
L6:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jg     L20
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0xc(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
L42:
  leave
  ret
```

assistant: ```c
int rand_zero(int max) {
  if (max > RAND_MAX) {
    return -1;
  } else if (max == RAND_MAX) {
    return rand();
  }
  int excess = RAND_MAX % (max + 1) + 1;
  int lim = RAND_MAX - excess;
  int ret;
  for (ret = rand(); ret > lim; ret = rand())
    ;
  return ret % (max + 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rand_zero>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rax
  mov    (%rax),%ebx
  cmp    %edi,%ebx
  jl     L35
  je     L33
  lea    0x1(%rdi),%ebp
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  add    $0x1,%edx
  sub    %edx,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  cmp    %eax,%ebx
  jge    L42
L2:
  mov    $0x0,%eax
  call   <rand@plt>
  cmp    %eax,%ebx
  jl     L2
L42:
  cltd
  idiv   %ebp
L19:
  mov    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L33:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  jmp    L19
L35:
  mov    $0xffffffff,%edx
  jmp    L19
```

assistant: ```c
int rand_zero(int max) {
  if (max > RAND_MAX) {
    return -1;
  } else if (max == RAND_MAX) {
    return rand();
  }
  int excess = RAND_MAX % (max + 1) + 1;
  int lim = RAND_MAX - excess;
  int ret;
  for (ret = rand(); ret > lim; ret = rand())
    ;
  return ret % (max + 1);
}
```