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
<main>:
  endbr64
  push   %rbp
  lea    D33(%rip),%rdi
  lea    D27(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm1
  movss  (%rsp),%xmm0
  call   <sum@plt>
  lea    D24(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D33
type: string
name: disassemble_data
arguments: data_label: D27
type: string
name: disassemble_data
arguments: data_label: D4
type: string
name: disassemble_data
arguments: data_label: D24
type: string

tool: D33: Enter the value of x:

tool: D27: %f

tool: D4: Enter the value of y:

tool: D24: The sum of x and y is %f.


assistant: ```c
main() {
  float sum(float, float);
  float x, y, a;
  printf("Enter the value of x:\n");
  scanf("%f", &x);
  printf("Enter the value of y:\n");
  scanf("%f", &y);
  a = x + y;
  a = sum(x, y);
  printf("The sum of x and y is %f.\n", a);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<read_nbytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  jmp    L32
L60:
  nop
L16:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L0
  mov    D26(%rip),%rax
  mov    (%rax),%rdx
  mov    D4(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L16
L0:
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L10
  mov    $0xffffffff,%eax
  jmp    L42
L10:
  cmpl   $0x0,-0x4(%rbp)
  jne    L74
  mov    D26(%rip),%rax
  movq   $0x0,(%rax)
  mov    $0xffffffff,%eax
  jmp    L42
L74:
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  sub    %rax,-0x28(%rbp)
L32:
  cmpq   $0x0,-0x28(%rbp)
  jne    L60
  mov    $0x0,%eax
L42:
  leave
  ret
```

assistant: ```c
int read_nbytes(int fd, void *p, size_t sz) {
  while (sz != 0) {
    ssize_t r;
    while ((r = read(fd, p, sz)) == -1 && errno == EINTR)
      ;
    if (r == -1) {
      return -1;
    } else if (r == 0) {
      errno = 0;
      return -1;
    }
    p = (char *)p + r;
    sz -= r;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_nbytes>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  test   %rdx,%rdx
  jne    L3
  mov    $0x0,%eax
  jmp    L78
L18:
  test   %eax,%eax
  je     L24
  cltq
  add    %rax,%rbp
  sub    %rax,%rbx
  je     L33
L3:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  cmp    $0xffffffff,%eax
  jne    L18
  mov    D25(%rip),%rcx
  mov    D11(%rip),%rdx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L3
L78:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L24:
  mov    D25(%rip),%rax
  movq   $0x0,(%rax)
  mov    $0xffffffff,%eax
  jmp    L78
L33:
  mov    $0x0,%eax
  jmp    L78
```

assistant: ```c
int read_nbytes(int fd, void *p, size_t sz) {
  while (sz != 0) {
    ssize_t r;
    while ((r = read(fd, p, sz)) == -1 && errno == EINTR)
      ;
    if (r == -1) {
      return -1;
    } else if (r == 0) {
      errno = 0;
      return -1;
    }
    p = (char *)p + r;
    sz -= r;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<read_nbytes>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  test   %rdx,%rdx
  jne    L24
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L11:
  mov    D29(%rip),%rdx
  mov    D22(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L58
L24:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  cmp    $0xffffffff,%eax
  je     L11
  test   %eax,%eax
  je     L40
  cltq
  add    %rax,%rbp
  sub    %rax,%rbx
  jne    L24
L46:
  xor    %eax,%eax
L58:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L40:
  mov    D22(%rip),%rax
  movq   $0x0,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int read_nbytes(int fd, void *p, size_t sz) {
  while (sz != 0) {
    ssize_t r;
    while ((r = read(fd, p, sz)) == -1 && errno == EINTR)
      ;
    if (r == -1) {
      return -1;
    } else if (r == 0) {
      errno = 0;
      return -1;
    }
    p = (char *)p + r;
    sz -= r;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_nbytes>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  test   %rdx,%rdx
  jne    L41
  jmp    L43
  nopl   0x0(%rax,%rax,1)
L10:
  mov    D14(%rip),%rdx
  mov    D4(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L58
L41:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  cmp    $0xffffffff,%eax
  je     L10
  test   %eax,%eax
  je     L57
  cltq
  add    %rax,%rbp
  sub    %rax,%rbx
  jne    L41
L43:
  xor    %eax,%eax
L58:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L57:
  mov    D4(%rip),%rax
  movq   $0x0,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int read_nbytes(int fd, void *p, size_t sz) {
  while (sz != 0) {
    ssize_t r;
    while ((r = read(fd, p, sz)) == -1 && errno == EINTR)
      ;
    if (r == -1) {
      return -1;
    } else if (r == 0) {
      errno = 0;
      return -1;
    }
    p = (char *)p + r;
    sz -= r;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Gif_CopyString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L12
  mov    $0x0,%eax
  jmp    L22
L12:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L22
L3:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
L22:
  leave
  ret
```

assistant: ```c
char *Gif_CopyString(const char *s) {
  int l;
  char *copy;
  if (!s)
    return 0;
  l = strlen(s);
  copy = ((char *)malloc(sizeof(char) * (l + 1)));
  if (!copy)
    return 0;
  memcpy(copy, s, l + 1);
  return copy;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Gif_CopyString>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L25
  call   <strlen@plt>
  lea    0x1(%rax),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    $0x0,%r12d
  test   %rax,%rax
  je     L10
  mov    %rax,%r12
  mov    %ebp,%edx
  mov    %rbx,%rsi
  call   <memcpy@plt>
L10:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  mov    %rdi,%r12
  jmp    L10
```

assistant: ```c
char *Gif_CopyString(const char *s) {
  int l;
  char *copy;
  if (!s)
    return 0;
  l = strlen(s);
  copy = ((char *)malloc(sizeof(char) * (l + 1)));
  if (!copy)
    return 0;
  memcpy(copy, s, l + 1);
  return copy;
}
```
user: What is the c source code of the assembly code below?

```asm
<Gif_CopyString>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  test   %rdi,%rdi
  je     L10
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L10:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *Gif_CopyString(const char *s) {
  int l;
  char *copy;
  if (!s)
    return 0;
  l = strlen(s);
  copy = ((char *)malloc(sizeof(char) * (l + 1)));
  if (!copy)
    return 0;
  memcpy(copy, s, l + 1);
  return copy;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Gif_CopyString>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  test   %rdi,%rdi
  je     L11
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L11
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L11:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *Gif_CopyString(const char *s) {
  int l;
  char *copy;
  if (!s)
    return 0;
  l = strlen(s);
  copy = ((char *)malloc(sizeof(char) * (l + 1)));
  if (!copy)
    return 0;
  memcpy(copy, s, l + 1);
  return copy;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<val>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L15:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L8
  mov    -0x4(%rbp),%eax
  jmp    L28
L8:
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  mov    $0x0,%eax
L28:
  pop    %rbp
  ret
```

assistant: ```c
int val(char *sym, char c) {
  int i;
  for (i = 0; sym[i] != '\0'; i++)
    if (c == sym[i])
      return i;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<val>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L16
  mov    $0x0,%eax
L37:
  mov    %eax,%ecx
  cmp    %dl,%sil
  je     L23
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L37
  mov    $0x0,%ecx
L23:
  mov    %ecx,%eax
  ret
L16:
  mov    $0x0,%ecx
  jmp    L23
```

assistant: ```c
int val(char *sym, char c) {
  int i;
  for (i = 0; sym[i] != '\0'; i++)
    if (c == sym[i])
      return i;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<val>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L4
  xor    %eax,%eax
  jmp    L35
  nop
L6:
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L4
L35:
  mov    %eax,%r8d
  cmp    %dl,%sil
  jne    L6
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int val(char *sym, char c) {
  int i;
  for (i = 0; sym[i] != '\0'; i++)
    if (c == sym[i])
      return i;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<val>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L0
  xor    %eax,%eax
  jmp    L20
  nop
L14:
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L0
L20:
  mov    %eax,%r8d
  cmp    %dl,%sil
  jne    L14
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int val(char *sym, char c) {
  int i;
  for (i = 0; sym[i] != '\0'; i++)
    if (c == sym[i])
      return i;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strclone>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strclone(const char *p_old) {
  size_t length = strlen(p_old);
  char *p_str = (char *)malloc(length + 1);
  memcpy(p_str, p_old, length);
  p_str[length] = '\0';
  return p_str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strclone>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  mov    %rax,%rbp
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strclone(const char *p_old) {
  size_t length = strlen(p_old);
  char *p_str = (char *)malloc(length + 1);
  memcpy(p_str, p_old, length);
  p_str[length] = '\0';
  return p_str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strclone>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strclone(const char *p_old) {
  size_t length = strlen(p_old);
  char *p_str = (char *)malloc(length + 1);
  memcpy(p_str, p_old, length);
  p_str[length] = '\0';
  return p_str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strclone>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strclone(const char *p_old) {
  size_t length = strlen(p_old);
  char *p_str = (char *)malloc(length + 1);
  memcpy(p_str, p_old, length);
  p_str[length] = '\0';
  return p_str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tmult_ok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L21
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  cmp    %eax,-0x18(%rbp)
  jne    L26
L21:
  mov    $0x1,%eax
  jmp    L2
L26:
  mov    $0x0,%eax
L2:
  pop    %rbp
  ret
```

assistant: ```c
int tmult_ok(int x, int y) {
  int p = x * y;
  return !x || p / x == y;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tmult_ok>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int tmult_ok(int x, int y) {
  int p = x * y;
  return !x || p / x == y;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tmult_ok>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int tmult_ok(int x, int y) {
  int p = x * y;
  return !x || p / x == y;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tmult_ok>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int tmult_ok(int x, int y) {
  int p = x * y;
  return !x || p / x == y;
}
```
user: Translate the below assembly code into C programming language:

```asm
<buspirate_wait_for_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <buspirate_sendrecv@plt>
  mov    %eax,-0x8(%rbp)
  jmp    L22
L38:
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L36
  mov    $0x0,%eax
  jmp    L13
L36:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <buspirate_sendrecv@plt>
  mov    %eax,-0x8(%rbp)
L22:
  cmpl   $0x0,-0x8(%rbp)
  je     L38
  mov    -0x8(%rbp),%eax
L13:
  leave
  ret
```

assistant: ```c
int buspirate_wait_for_string(unsigned char *buf, const char *key) {
  unsigned int keylen = strlen(key);
  int ret;
  ret = buspirate_sendrecv(buf, 0, keylen);
  while (!ret) {
    if (!memcmp(buf, key, keylen))
      return 0;
    memmove(buf, buf + 1, keylen - 1);
    ret = buspirate_sendrecv(buf + keylen - 1, 0, 1);
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<buspirate_wait_for_string>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <buspirate_sendrecv@plt>
  test   %eax,%eax
  jne    L21
  lea    -0x1(%rbp),%r14d
  mov    %ebp,%eax
  lea    -0x1(%rbx,%rax,1),%r13
L4:
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <memcmp@plt>
  test   %eax,%eax
  je     L21
  lea    0x1(%rbx),%rsi
  mov    %r14d,%edx
  mov    %rbx,%rdi
  call   <memmove@plt>
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <buspirate_sendrecv@plt>
  test   %eax,%eax
  je     L4
L21:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int buspirate_wait_for_string(unsigned char *buf, const char *key) {
  unsigned int keylen = strlen(key);
  int ret;
  ret = buspirate_sendrecv(buf, 0, keylen);
  while (!ret) {
    if (!memcmp(buf, key, keylen))
      return 0;
    memmove(buf, buf + 1, keylen - 1);
    ret = buspirate_sendrecv(buf + keylen - 1, 0, 1);
  }
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<buspirate_wait_for_string>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %r14,%rdi
  mov    %eax,%edx
  mov    %eax,%ebx
  call   <buspirate_sendrecv@plt>
  test   %eax,%eax
  jne    L35
  mov    %ebx,%eax
  lea    -0x1(%rbx),%r13d
  lea    -0x1(%r14,%rax,1),%r12
  jmp    L16
  nopl   0x0(%rax)
L27:
  lea    0x1(%r14),%rsi
  mov    %r13d,%edx
  mov    %r14,%rdi
  call   <memmove@plt>
  xor    %esi,%esi
  mov    $0x1,%edx
  mov    %r12,%rdi
  call   <buspirate_sendrecv@plt>
  test   %eax,%eax
  jne    L35
L16:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L27
L35:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int buspirate_wait_for_string(unsigned char *buf, const char *key) {
  unsigned int keylen = strlen(key);
  int ret;
  ret = buspirate_sendrecv(buf, 0, keylen);
  while (!ret) {
    if (!memcmp(buf, key, keylen))
      return 0;
    memmove(buf, buf + 1, keylen - 1);
    ret = buspirate_sendrecv(buf + keylen - 1, 0, 1);
  }
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<buspirate_wait_for_string>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %r14,%rdi
  mov    %eax,%edx
  mov    %eax,%ebx
  call   <buspirate_sendrecv@plt>
  test   %eax,%eax
  jne    L31
  mov    %ebx,%eax
  lea    -0x1(%rbx),%r13d
  lea    -0x1(%r14,%rax,1),%r12
  jmp    L1
  nopl   0x0(%rax)
L8:
  lea    0x1(%r14),%rsi
  mov    %r13d,%edx
  mov    %r14,%rdi
  call   <memmove@plt>
  xor    %esi,%esi
  mov    $0x1,%edx
  mov    %r12,%rdi
  call   <buspirate_sendrecv@plt>
  test   %eax,%eax
  jne    L31
L1:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L8
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int buspirate_wait_for_string(unsigned char *buf, const char *key) {
  unsigned int keylen = strlen(key);
  int ret;
  ret = buspirate_sendrecv(buf, 0, keylen);
  while (!ret) {
    if (!memcmp(buf, key, keylen))
      return 0;
    memmove(buf, buf + 1, keylen - 1);
    ret = buspirate_sendrecv(buf + keylen - 1, 0, 1);
  }
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getFlips>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L25
L17:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  je     L12
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  addl   $0x1,-0xc(%rbp)
L12:
  addl   $0x1,-0x8(%rbp)
L25:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L7
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L17
L7:
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  cmp    %eax,-0x8(%rbp)
  setl   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int getFlips(int flips, int framebits, int *randomFlips) {
  int count = 0;
  int totalCount = 0;
  while (count < flips && totalCount < flips * 5) {
    int next = rand() % framebits;
    if (randomFlips[next] != 1) {
      randomFlips[next] = 1;
      count++;
    }
    totalCount++;
  }
  return totalCount < flips * 10;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getFlips>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  test   %edi,%edi
  jle    L54
  mov    %esi,%r12d
  mov    %rdx,%r14
  lea    (%rdi,%rdi,4),%r15d
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  jmp    L39
L57:
  add    $0x1,%ebx
  cmp    %r13d,%ebp
  jge    L69
  cmp    %r15d,%ebx
  je     L6
L39:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idiv   %r12d
  movslq %edx,%rdx
  lea    (%r14,%rdx,4),%rax
  cmpl   $0x1,(%rax)
  je     L57
  movl   $0x1,(%rax)
  add    $0x1,%ebp
  jmp    L57
L6:
  mov    %r15d,%ebx
L69:
  lea    0x0(%r13,%r13,4),%eax
  add    %eax,%eax
  cmp    %ebx,%eax
  setg   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  mov    $0x0,%ebx
  jmp    L69
```

assistant: ```c
int getFlips(int flips, int framebits, int *randomFlips) {
  int count = 0;
  int totalCount = 0;
  while (count < flips && totalCount < flips * 5) {
    int next = rand() % framebits;
    if (randomFlips[next] != 1) {
      randomFlips[next] = 1;
      count++;
    }
    totalCount++;
  }
  return totalCount < flips * 10;
}
```