ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<set_sock_opt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0x1c(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x10(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%esi
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    $0x4,%r8d
  mov    %eax,%edi
  call   <setsockopt@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
bool set_sock_opt(int socket, int option, bool enable) {
  int yes = (enable ? 1 : 0);
  int err = setsockopt(socket, SOL_SOCKET, option, &yes, sizeof(yes));
  return !err;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<set_sock_opt>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  setne  %al
  movzbl %al,%eax
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rcx
  mov    $0x4,%r8d
  mov    %esi,%edx
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  test   %eax,%eax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool set_sock_opt(int socket, int option, bool enable) {
  int yes = (enable ? 1 : 0);
  int err = setsockopt(socket, SOL_SOCKET, option, &yes, sizeof(yes));
  return !err;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<set_sock_opt>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  xor    %eax,%eax
  test   %edx,%edx
  mov    %esi,%edx
  setne  %al
  lea    0x4(%rsp),%rcx
  mov    %eax,0x4(%rsp)
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  test   %eax,%eax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool set_sock_opt(int socket, int option, bool enable) {
  int yes = (enable ? 1 : 0);
  int err = setsockopt(socket, SOL_SOCKET, option, &yes, sizeof(yes));
  return !err;
}
```
user: What is the c source code of the assembly code below?

```asm
<set_sock_opt>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  xor    %eax,%eax
  test   %edx,%edx
  mov    %esi,%edx
  setne  %al
  lea    0x4(%rsp),%rcx
  mov    %eax,0x4(%rsp)
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  test   %eax,%eax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool set_sock_opt(int socket, int option, bool enable) {
  int yes = (enable ? 1 : 0);
  int err = setsockopt(socket, SOL_SOCKET, option, &yes, sizeof(yes));
  return !err;
}
```
user: Translate the below assembly code into C programming language:

```asm
<binsearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L50
L60:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jle    L53
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L50
L53:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jge    L20
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L50
L20:
  mov    -0x4(%rbp),%eax
  jmp    L6
L50:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L60
  mov    -0xc(%rbp),%eax
  sub    -0x24(%rbp),%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int binsearch(int *ary, int n, int key, int before) {
  int first = 0;
  int last = n - 1;
  int mid;
  while (first <= last) {
    mid = (first + last) / 2;
    if (key > ary[mid]) {
      first = mid + 1;
    } else if (key < ary[mid]) {
      last = mid - 1;
    } else {
      return mid;
    }
  }
  return first - before;
}
```
user: What is the c source code of the assembly code below?

```asm
<binsearch>:
  endbr64
  mov    %rdi,%r9
  sub    $0x1,%esi
  js     L16
  mov    $0x0,%r8d
  jmp    L14
L22:
  jle    L18
  lea    -0x1(%rax),%esi
L0:
  cmp    %esi,%r8d
  jg     L43
L14:
  lea    (%r8,%rsi,1),%edi
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  mov    %eax,%r10d
  movslq %eax,%rdi
  mov    (%r9,%rdi,4),%edi
  cmp    %edx,%edi
  jge    L22
  lea    0x1(%rax),%r8d
  jmp    L0
L16:
  mov    $0x0,%r8d
L43:
  mov    %r8d,%r10d
  sub    %ecx,%r10d
L18:
  mov    %r10d,%eax
  ret
```

assistant: ```c
int binsearch(int *ary, int n, int key, int before) {
  int first = 0;
  int last = n - 1;
  int mid;
  while (first <= last) {
    mid = (first + last) / 2;
    if (key > ary[mid]) {
      first = mid + 1;
    } else if (key < ary[mid]) {
      last = mid - 1;
    } else {
      return mid;
    }
  }
  return first - before;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<binsearch>:
  endbr64
  xor    %r8d,%r8d
  sub    $0x1,%esi
  jns    L12
  jmp    L16
  xchg   %ax,%ax
L24:
  lea    0x1(%rax),%r8d
  cmp    %r8d,%esi
  jl     L16
L12:
  lea    (%rsi,%r8,1),%eax
  sar    %eax
  movslq %eax,%r9
  cmp    %edx,(%rdi,%r9,4)
  jl     L24
  jle    L5
  lea    -0x1(%rax),%esi
  cmp    %r8d,%esi
  jge    L12
L16:
  mov    %r8d,%eax
  sub    %ecx,%eax
L5:
  ret
```

assistant: ```c
int binsearch(int *ary, int n, int key, int before) {
  int first = 0;
  int last = n - 1;
  int mid;
  while (first <= last) {
    mid = (first + last) / 2;
    if (key > ary[mid]) {
      first = mid + 1;
    } else if (key < ary[mid]) {
      last = mid - 1;
    } else {
      return mid;
    }
  }
  return first - before;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<binsearch>:
  endbr64
  xor    %r8d,%r8d
  sub    $0x1,%esi
  jns    L29
  jmp    L46
  xchg   %ax,%ax
L14:
  lea    0x1(%rax),%r8d
  cmp    %r8d,%esi
  jl     L46
L29:
  lea    (%rsi,%r8,1),%eax
  sar    %eax
  movslq %eax,%r9
  cmp    %edx,(%rdi,%r9,4)
  jl     L14
  jle    L25
  lea    -0x1(%rax),%esi
  cmp    %r8d,%esi
  jge    L29
L46:
  mov    %r8d,%eax
  sub    %ecx,%eax
L25:
  ret
```

assistant: ```c
int binsearch(int *ary, int n, int key, int before) {
  int first = 0;
  int last = n - 1;
  int mid;
  while (first <= last) {
    mid = (first + last) / 2;
    if (key > ary[mid]) {
      first = mid + 1;
    } else if (key < ary[mid]) {
      last = mid - 1;
    } else {
      return mid;
    }
  }
  return first - before;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dvbcfg_parse_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L56
L53:
  addq   $0x1,-0x18(%rbp)
L56:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L2
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L53
L2:
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x40(%rbp),%rax
  ja     L48
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L44
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L66
L44:
  mov    -0x28(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
L66:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  jmp    L14
L48:
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  nop
L14:
  leave
  ret
```

assistant: ```c
void dvbcfg_parse_string(char **text, char *tokens, char *dest, unsigned long size) {
  char *start = *text;
  char *stop = *text;
  unsigned long length;
  while ((*stop != '\0') && (strchr(tokens, *stop) == NULL))
    stop++;
  length = (stop - start) + 1;
  if (length <= size) {
    if (strchr(tokens, *stop) != NULL) {
      *stop = '\0';
      *text = stop + 1;
    } else
      *text = stop;
    memcpy(dest, start, length);
    return;
  }
  *text = NULL;
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dvbcfg_parse_string>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,0x8(%rsp)
  mov    %rcx,%r15
  mov    (%rdi),%r13
  movzbl 0x0(%r13),%esi
  mov    %r13,%rbx
  test   %sil,%sil
  je     L43
L49:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L43
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L49
L43:
  mov    %rbx,%r14
  sub    %r13,%r14
  add    $0x1,%r14
  cmp    %r15,%r14
  jbe    L21
  movq   $0x0,(%r12)
L55:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L58
  movb   $0x0,(%rbx)
  add    $0x1,%rbx
L58:
  mov    %rbx,(%r12)
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    0x8(%rsp),%rdi
  call   <memcpy@plt>
  jmp    L55
```

assistant: ```c
void dvbcfg_parse_string(char **text, char *tokens, char *dest, unsigned long size) {
  char *start = *text;
  char *stop = *text;
  unsigned long length;
  while ((*stop != '\0') && (strchr(tokens, *stop) == NULL))
    stop++;
  length = (stop - start) + 1;
  if (length <= size) {
    if (strchr(tokens, *stop) != NULL) {
      *stop = '\0';
      *text = stop + 1;
    } else
      *text = stop;
    memcpy(dest, start, length);
    return;
  }
  *text = NULL;
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dvbcfg_parse_string>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x1,%edx
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    (%rdi),%rbp
  movsbl 0x0(%rbp),%esi
  mov    %rbp,%r14
  test   %sil,%sil
  jne    L18
  jmp    L27
  nopl   0x0(%rax)
L25:
  movsbl 0x1(%r14),%esi
  add    $0x1,%r14
  test   %sil,%sil
  je     L38
L18:
  mov    %r15,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L25
L38:
  mov    %r14,%rdx
  sub    %rbp,%rdx
  add    $0x1,%rdx
L27:
  cmp    %rdx,%r12
  jae    L48
  movq   $0x0,(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L48:
  movsbl (%r14),%esi
  mov    %r15,%rdi
  mov    %rdx,0x8(%rsp)
  call   <strchr@plt>
  mov    0x8(%rsp),%rdx
  test   %rax,%rax
  je     L8
  movb   $0x0,(%r14)
  add    $0x1,%r14
L8:
  mov    %r14,(%rbx)
  add    $0x18,%rsp
  mov    %rbp,%rsi
  mov    %r13,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <memcpy@plt>
```

assistant: ```c
void dvbcfg_parse_string(char **text, char *tokens, char *dest, unsigned long size) {
  char *start = *text;
  char *stop = *text;
  unsigned long length;
  while ((*stop != '\0') && (strchr(tokens, *stop) == NULL))
    stop++;
  length = (stop - start) + 1;
  if (length <= size) {
    if (strchr(tokens, *stop) != NULL) {
      *stop = '\0';
      *text = stop + 1;
    } else
      *text = stop;
    memcpy(dest, start, length);
    return;
  }
  *text = NULL;
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dvbcfg_parse_string>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x1,%edx
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    (%rdi),%rbp
  movsbl 0x0(%rbp),%esi
  mov    %rbp,%r14
  test   %sil,%sil
  jne    L54
  jmp    L7
  nopl   0x0(%rax)
L26:
  movsbl 0x1(%r14),%esi
  add    $0x1,%r14
  test   %sil,%sil
  je     L48
L54:
  mov    %r15,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L26
L48:
  mov    %r14,%rdx
  sub    %rbp,%rdx
  add    $0x1,%rdx
L7:
  cmp    %rdx,%r12
  jae    L25
  movq   $0x0,(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  movsbl (%r14),%esi
  mov    %r15,%rdi
  mov    %rdx,0x8(%rsp)
  call   <strchr@plt>
  mov    0x8(%rsp),%rdx
  test   %rax,%rax
  je     L58
  movb   $0x0,(%r14)
  add    $0x1,%r14
L58:
  mov    %r14,(%rbx)
  add    $0x18,%rsp
  mov    %rbp,%rsi
  mov    %r13,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <memcpy@plt>
```

assistant: ```c
void dvbcfg_parse_string(char **text, char *tokens, char *dest, unsigned long size) {
  char *start = *text;
  char *stop = *text;
  unsigned long length;
  while ((*stop != '\0') && (strchr(tokens, *stop) == NULL))
    stop++;
  length = (stop - start) + 1;
  if (length <= size) {
    if (strchr(tokens, *stop) != NULL) {
      *stop = '\0';
      *text = stop + 1;
    } else
      *text = stop;
    memcpy(dest, start, length);
    return;
  }
  *text = NULL;
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_dns_parse_uint16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <ntohs@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint16_t _dns_parse_uint16(char **p) {
  uint16_t *x, v;
  x = (uint16_t *)*p;
  v = ntohs(*x);
  *p += 2;
  return v;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_dns_parse_uint16>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rax
  mov    (%rax),%edi
  call   <ntohs@plt>
  addq   $0x2,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
uint16_t _dns_parse_uint16(char **p) {
  uint16_t *x, v;
  x = (uint16_t *)*p;
  v = ntohs(*x);
  *p += 2;
  return v;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_dns_parse_uint16>:
  endbr64
  push   %rbx
  mov    (%rdi),%rax
  mov    %rdi,%rbx
  mov    (%rax),%edi
  call   <ntohs@plt>
  addq   $0x2,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
uint16_t _dns_parse_uint16(char **p) {
  uint16_t *x, v;
  x = (uint16_t *)*p;
  v = ntohs(*x);
  *p += 2;
  return v;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_dns_parse_uint16>:
  endbr64
  push   %rbx
  mov    (%rdi),%rax
  mov    %rdi,%rbx
  mov    (%rax),%edi
  call   <ntohs@plt>
  addq   $0x2,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
uint16_t _dns_parse_uint16(char **p) {
  uint16_t *x, v;
  x = (uint16_t *)*p;
  v = ntohs(*x);
  *p += 2;
  return v;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int f() {
  const int *p;
  int v;
  v = *p;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int f() {
  const int *p;
  int v;
  v = *p;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<f>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f() {
  const int *p;
  int v;
  v = *p;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f() {
  const int *p;
  int v;
  v = *p;
  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
<events_timer_register_double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movsd  %xmm0,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  -0x38(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm2
  movsd  -0x38(%rbp),%xmm0
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  D1(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  lea    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <events_timer_register@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
data_type: f64

tool: D1: 1000000

assistant: ```c
void *events_timer_register_double(int (*func)(void *), void *cookie, double timeo) {
  struct timeval tv;
  tv.tv_sec = timeo;
  tv.tv_usec = (timeo - tv.tv_sec) * 1000000.0;
  return (events_timer_register(func, cookie, &tv));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<events_timer_register_double>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  subsd  %xmm0,%xmm0
  mulsd  D6(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  mov    %rsp,%rdx
  call   <events_timer_register@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D6
type_guessed: f64

tool: D6: 1000000

assistant: ```c
void *events_timer_register_double(int (*func)(void *), void *cookie, double timeo) {
  struct timeval tv;
  tv.tv_sec = timeo;
  tv.tv_usec = (timeo - tv.tv_sec) * 1000000.0;
  return (events_timer_register(func, cookie, &tv));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<events_timer_register_double>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  subsd  %xmm0,%xmm0
  mov    %rsp,%rdx
  mulsd  D5(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <events_timer_register@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type: f64

tool: D5: 1000000

assistant: ```c
void *events_timer_register_double(int (*func)(void *), void *cookie, double timeo) {
  struct timeval tv;
  tv.tv_sec = timeo;
  tv.tv_usec = (timeo - tv.tv_sec) * 1000000.0;
  return (events_timer_register(func, cookie, &tv));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<events_timer_register_double>:
  endbr64
  movapd %xmm0,%xmm1
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  subsd  %xmm0,%xmm1
  mulsd  D6(%rip),%xmm1
  mov    %rsp,%rdx
  unpcklpd %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  call   <events_timer_register@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type: f64

tool: D6: 1000000

assistant: ```c
void *events_timer_register_double(int (*func)(void *), void *cookie, double timeo) {
  struct timeval tv;
  tv.tv_sec = timeo;
  tv.tv_usec = (timeo - tv.tv_sec) * 1000000.0;
  return (events_timer_register(func, cookie, &tv));
}
```
user: What is the c source code of the assembly code below?

```asm
<memchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L26
L11:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L36
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  jmp    L22
L36:
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jb     L11
  mov    $0x0,%eax
L22:
  pop    %rbp
  ret
```

assistant: ```c
void *memchr(const void *_s, int c, unsigned int n) {
  const char *s = (const char *)_s;
  for (unsigned int i = 0; i < n; i++) {
    if (s[i] == c)
      return (void *)&s[i];
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memchr>:
  endbr64
  test   %edx,%edx
  je     L14
  mov    %rdi,%rax
  lea    -0x1(%rdx),%edx
  add    %rdx,%rdi
  jmp    L11
L19:
  mov    %rdx,%rax
L11:
  mov    %rax,%rcx
  movsbl (%rax),%edx
  cmp    %esi,%edx
  je     L39
  lea    0x1(%rax),%rdx
  cmp    %rdi,%rax
  jne    L19
  mov    $0x0,%ecx
L39:
  mov    %rcx,%rax
  ret
L14:
  mov    $0x0,%ecx
  jmp    L39
```

assistant: ```c
void *memchr(const void *_s, int c, unsigned int n) {
  const char *s = (const char *)_s;
  for (unsigned int i = 0; i < n; i++) {
    if (s[i] == c)
      return (void *)&s[i];
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L10
  lea    -0x1(%rdx),%ecx
  add    %rdi,%rcx
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L0:
  lea    0x1(%rax),%rdx
  cmp    %rcx,%rax
  je     L10
  mov    %rdx,%rax
L32:
  movsbl (%rax),%edx
  cmp    %esi,%edx
  jne    L0
  ret
  nopl   0x0(%rax)
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *_s, int c, unsigned int n) {
  const char *s = (const char *)_s;
  for (unsigned int i = 0; i < n; i++) {
    if (s[i] == c)
      return (void *)&s[i];
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L29
  lea    -0x1(%rdx),%ecx
  add    %rdi,%rcx
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L24:
  lea    0x1(%rax),%rdx
  cmp    %rcx,%rax
  je     L29
  mov    %rdx,%rax
L39:
  movsbl (%rax),%edx
  cmp    %esi,%edx
  jne    L24
  ret
  nopl   0x0(%rax)
L29:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *_s, int c, unsigned int n) {
  const char *s = (const char *)_s;
  for (unsigned int i = 0; i < n; i++) {
    if (s[i] == c)
      return (void *)&s[i];
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x5,%edi
  call   <t@plt>
  mov    $0x14,%edi
  call   <t@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  t(5);
  t(20);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <t@plt>
  mov    $0x14,%edi
  call   <t@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  t(5);
  t(20);
}
```
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    $0x5,%edi
  call   <t@plt>
  mov    $0x14,%edi
  call   <t@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  t(5);
  t(20);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <t@plt>
  mov    $0x14,%edi
  call   <t@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  t(5);
  t(20);
}
```