ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<partition>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r14
  mov    %rdi,0x20(%rsp)
  mov    %esi,%r15d
  mov    %esi,0x28(%rsp)
  mov    %rdx,%r13
  mov    %ecx,%ebx
  mov    %r8d,%ebp
  mov    %esi,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r15d,%r15d
  jle    L61
  mov    %r15d,%ecx
  imul   %ebx,%ecx
  movslq %ecx,%rcx
  mov    %r15d,%esi
  mov    $0x0,%eax
  add    %r14,%rcx
L58:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L58
L61:
  lea    -0x1(%rbx),%r15d
  add    $0x1,%ebp
  mov    0x28(%rsp),%eax
  movslq %eax,%rcx
  mov    %rcx,0x18(%rsp)
  imul   %eax,%ebx
  movslq %ebx,%r14
  add    0x20(%rsp),%r14
  neg    %eax
  mov    %eax,0x2c(%rsp)
  jmp    L30
L26:
  add    0x18(%rsp),%r14
L30:
  add    $0x1,%r15d
  mov    %r14,0x10(%rsp)
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   *%r13
  test   %eax,%eax
  js     L26
  movslq 0x2c(%rsp),%rax
  mov    %rax,(%rsp)
  lea    -0x1(%rbp),%ebx
  imul   0x28(%rsp),%ebx
  movslq %ebx,%rbx
  add    0x20(%rsp),%rbx
L23:
  sub    $0x1,%ebp
  mov    %rbx,0x8(%rsp)
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   *%r13
  add    (%rsp),%rbx
  test   %eax,%eax
  jg     L23
  cmp    %ebp,%r15d
  jge    L28
  mov    0x28(%rsp),%edx
  mov    0x8(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  call   <swap@plt>
  jmp    L26
L28:
  mov    %ebp,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int partition(void *base, int size, int (*compare)(const void *, const void *), int low, int high) {
  void *pivot = malloc(size);
  int i;
  for (i = 0; i < size; i++)
    *((char *)pivot + i) = *((char *)base + low * size + i);
  low--;
  high++;
  while (1) {
    do {
      low++;
    } while (compare(((char *)base) + low * size, pivot) < 0);
    do {
      high--;
    } while (compare(((char *)base) + high * size, pivot) > 0);
    if (low < high) {
      swap(((char *)base) + low * size, ((char *)base) + high * size, size);
    } else
      return high;
  }
  free(pivot);
}
```
user: What is the c source code of the assembly code below:

```asm
<partition>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %r8d,%ebp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,0x20(%rsp)
  mov    %esi,%edi
  mov    %esi,0x28(%rsp)
  call   <malloc@plt>
  mov    %r14d,%edx
  imul   %r13d,%edx
  mov    %rax,%rbx
  movslq %r14d,%rax
  mov    %rax,%rcx
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %edx,%rdx
  add    %rdx,%r15
  test   %ecx,%ecx
  jle    L58
  nopw   0x0(%rax,%rax,1)
L40:
  movzbl (%r15,%rax,1),%edx
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    0x18(%rsp),%rax
  jne    L40
L58:
  lea    -0x1(%r13),%eax
  mov    %r15,0x8(%rsp)
  add    $0x1,%ebp
  mov    %eax,0x14(%rsp)
  mov    0x28(%rsp),%eax
  neg    %eax
  mov    %eax,0x2c(%rsp)
  nopl   (%rax)
L49:
  addl   $0x1,0x14(%rsp)
  mov    0x8(%rsp),%rdi
  mov    %rbx,%rsi
  call   *%r12
  test   %eax,%eax
  js     L19
  mov    0x28(%rsp),%edx
  lea    -0x1(%rbp),%r15d
  movslq 0x2c(%rsp),%rax
  imul   %r15d,%edx
  mov    %rax,(%rsp)
  mov    0x20(%rsp),%rax
  movslq %edx,%rdx
  lea    (%rax,%rdx,1),%r13
  jmp    L67
  nop
L61:
  sub    $0x1,%r15d
L67:
  mov    %r13,%rdi
  mov    %rbx,%rsi
  mov    %r13,%r14
  mov    %r15d,%ebp
  call   *%r12
  add    (%rsp),%r13
  test   %eax,%eax
  jg     L61
  cmp    %r15d,0x14(%rsp)
  jge    L41
  mov    0x28(%rsp),%edx
  mov    0x8(%rsp),%rdi
  mov    %r14,%rsi
  call   <swap@plt>
L19:
  mov    0x18(%rsp),%rcx
  add    %rcx,0x8(%rsp)
  jmp    L49
  nop
L41:
  add    $0x38,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int partition(void *base, int size, int (*compare)(const void *, const void *), int low, int high) {
  void *pivot = malloc(size);
  int i;
  for (i = 0; i < size; i++)
    *((char *)pivot + i) = *((char *)base + low * size + i);
  low--;
  high++;
  while (1) {
    do {
      low++;
    } while (compare(((char *)base) + low * size, pivot) < 0);
    do {
      high--;
    } while (compare(((char *)base) + high * size, pivot) > 0);
    if (low < high) {
      swap(((char *)base) + low * size, ((char *)base) + high * size, size);
    } else
      return high;
  }
  free(pivot);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<diff_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   $0xf4240,%eax,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  add    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
long long diff_time(struct timespec *start) {
  long long diff;
  struct timespec curr_time;
  clock_gettime(CLOCK_MONOTONIC, &curr_time);
  diff = (curr_time.tv_sec - start->tv_sec) * 1000 * 1000;
  diff += (curr_time.tv_nsec - start->tv_nsec) / 1000;
  return diff;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<diff_time>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D1(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    (%rsp),%edx
  sub    (%rbx),%edx
  imul   $0xf4240,%edx,%edx
  movslq %edx,%rdx
  mov    0x4(%rsp),%ecx
  sub    0x4(%rbx),%ecx
  movslq %ecx,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  cltq
  add    %rdx,%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long diff_time(struct timespec *start) {
  long long diff;
  struct timespec curr_time;
  clock_gettime(CLOCK_MONOTONIC, &curr_time);
  diff = (curr_time.tv_sec - start->tv_sec) * 1000 * 1000;
  diff += (curr_time.tv_nsec - start->tv_nsec) / 1000;
  return diff;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<diff_time>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    0x4(%rsp),%ecx
  sub    0x4(%rbx),%ecx
  movslq %ecx,%rax
  mov    (%rsp),%edx
  sub    (%rbx),%edx
  sar    $0x1f,%ecx
  imul   $0x10624dd3,%rax,%rax
  imul   $0xf4240,%edx,%edx
  sar    $0x26,%rax
  sub    %ecx,%eax
  movslq %edx,%rdx
  cltq
  add    %rdx,%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long diff_time(struct timespec *start) {
  long long diff;
  struct timespec curr_time;
  clock_gettime(CLOCK_MONOTONIC, &curr_time);
  diff = (curr_time.tv_sec - start->tv_sec) * 1000 * 1000;
  diff += (curr_time.tv_nsec - start->tv_nsec) / 1000;
  return diff;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<diff_time>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    0x4(%rsp),%ecx
  sub    0x4(%rbx),%ecx
  movslq %ecx,%rax
  mov    (%rsp),%edx
  sub    (%rbx),%edx
  sar    $0x1f,%ecx
  imul   $0x10624dd3,%rax,%rax
  imul   $0xf4240,%edx,%edx
  sar    $0x26,%rax
  sub    %ecx,%eax
  movslq %edx,%rdx
  cltq
  add    %rdx,%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long diff_time(struct timespec *start) {
  long long diff;
  struct timespec curr_time;
  clock_gettime(CLOCK_MONOTONIC, &curr_time);
  diff = (curr_time.tv_sec - start->tv_sec) * 1000 * 1000;
  diff += (curr_time.tv_nsec - start->tv_nsec) / 1000;
  return diff;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_next>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  jmp    L29
L19:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
L29:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L19
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L19
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_next(const char **pp) {
  while (**pp == ' ' || **pp == '\t')
    (*pp)++;
  return **pp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_next>:
  endbr64
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  cmp    $0x20,%al
  je     L26
  cmp    $0x9,%al
  jne    L16
L26:
  add    $0x1,%rdx
L7:
  mov    %rdx,(%rdi)
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  cmp    $0x20,%al
  je     L7
  cmp    $0x9,%al
  je     L7
L16:
  movsbl %al,%eax
  ret
```

assistant: ```c
int get_next(const char **pp) {
  while (**pp == ' ' || **pp == '\t')
    (*pp)++;
  return **pp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_next>:
  endbr64
  mov    (%rdi),%rdx
  movsbl (%rdx),%eax
  cmp    $0x20,%al
  je     L6
  cmp    $0x9,%al
  jne    L9
L6:
  add    $0x1,%rdx
  cs nopw 0x0(%rax,%rax,1)
L20:
  mov    %rdx,(%rdi)
  movsbl (%rdx),%eax
  add    $0x1,%rdx
  cmp    $0x20,%al
  je     L20
  cmp    $0x9,%al
  je     L20
L9:
  ret
```

assistant: ```c
int get_next(const char **pp) {
  while (**pp == ' ' || **pp == '\t')
    (*pp)++;
  return **pp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_next>:
  endbr64
  mov    (%rdi),%rdx
  movsbl (%rdx),%eax
  cmp    $0x20,%al
  je     L14
  cmp    $0x9,%al
  jne    L32
L14:
  add    $0x1,%rdx
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    %rdx,(%rdi)
  movsbl (%rdx),%eax
  add    $0x1,%rdx
  cmp    $0x20,%al
  je     L22
  cmp    $0x9,%al
  je     L22
L32:
  ret
```

assistant: ```c
int get_next(const char **pp) {
  while (**pp == ' ' || **pp == '\t')
    (*pp)++;
  return **pp;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fib>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L31
  mov    $0x0,%eax
  jmp    L11
L31:
  cmpl   $0x1,-0x14(%rbp)
  jne    L8
  mov    $0x1,%eax
  jmp    L11
L8:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fib>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fib>
  add    %ebx,%eax
L11:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fib(int n) {
  if (n == 0)
    return 0;
  else if (n == 1)
    return 1;
  else
    return fib(n - 1) + fib(n - 2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fib>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  test   %edi,%edi
  je     L22
  cmp    $0x1,%edi
  jne    L20
L22:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  lea    -0x1(%rdi),%edi
  call   <fib>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fib>
  lea    0x0(%rbp,%rax,1),%ebx
  jmp    L22
```

assistant: ```c
int fib(int n) {
  if (n == 0)
    return 0;
  else if (n == 1)
    return 1;
  else
    return fib(n - 1) + fib(n - 2);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<write_all>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sf_write@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L17
  mov    -0x18(%rbp),%eax
  jmp    L0
L17:
  cmpl   $0x0,-0x4(%rbp)
  jns    L34
  mov    $0xffffffff,%eax
  jmp    L0
L34:
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
  jmp    L14
L0:
  leave
  ret
```

assistant: ```c
int write_all(int fd, const char *buf, int len) {
  int left = len;
  int written;
  for (;;) {
    written = sf_write(fd, buf, left);
    if (written == left) {
      return len;
    }
    if (written < 0) {
      return -1;
    }
    left -= written;
    buf += written;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<write_all>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %edx,%r13d
  mov    %edx,%ebx
  jmp    L20
L3:
  sub    %eax,%ebx
  cltq
  add    %rax,%rbp
L20:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <sf_write@plt>
  cmp    %eax,%ebx
  je     L13
  test   %eax,%eax
  jns    L3
  mov    $0xffffffff,%r13d
L13:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int write_all(int fd, const char *buf, int len) {
  int left = len;
  int written;
  for (;;) {
    written = sf_write(fd, buf, left);
    if (written == left) {
      return len;
    }
    if (written < 0) {
      return -1;
    }
    left -= written;
    buf += written;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<write_all>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L10:
  test   %eax,%eax
  js     L17
  sub    %eax,%ebx
  cltq
  add    %rax,%r13
L12:
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <sf_write@plt>
  cmp    %eax,%ebx
  jne    L10
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L17:
  add    $0x8,%rsp
  mov    $0xffffffff,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int write_all(int fd, const char *buf, int len) {
  int left = len;
  int written;
  for (;;) {
    written = sf_write(fd, buf, left);
    if (written == left) {
      return len;
    }
    if (written < 0) {
      return -1;
    }
    left -= written;
    buf += written;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<write_all>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L5:
  test   %eax,%eax
  js     L19
  sub    %eax,%ebx
  cltq
  add    %rax,%r13
L4:
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <sf_write@plt>
  cmp    %eax,%ebx
  jne    L5
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L19:
  add    $0x8,%rsp
  mov    $0xffffffff,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int write_all(int fd, const char *buf, int len) {
  int left = len;
  int written;
  for (;;) {
    written = sf_write(fd, buf, left);
    if (written == left) {
      return len;
    }
    if (written < 0) {
      return -1;
    }
    left -= written;
    buf += written;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x400,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L4
L10:
  mov    -0x3f4(%rbp),%edx
  lea    -0x3f0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L4:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getnextline@plt>
  mov    %eax,-0x3f4(%rbp)
  cmpl   $0x0,-0x3f4(%rbp)
  jg     L10
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_guessed: string

tool: D8: %s

assistant: ```c
int main(void) {
  char line[1000];
  int len;
  while ((len = getnextline(line, 1000)) > 0) {
    reverse(line, len);
    printf("%s", line);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rbp
  jmp    L4
L19:
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L4:
  mov    %rsp,%rdi
  mov    $0x3e8,%esi
  call   <getnextline@plt>
  mov    %eax,%esi
  test   %eax,%eax
  jg     L19
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x3f8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
data_type: string

tool: D1: %s

assistant: ```c
int main(void) {
  char line[1000];
  int len;
  while ((len = getnextline(line, 1000)) > 0) {
    reverse(line, len);
    printf("%s", line);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L10
  nopl   0x0(%rax)
L24:
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L10:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getnextline@plt>
  mov    %eax,%esi
  test   %eax,%eax
  jg     L24
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
datatype: string

tool: D0: %s

assistant: ```c
int main(void) {
  char line[1000];
  int len;
  while ((len = getnextline(line, 1000)) > 0) {
    reverse(line, len);
    printf("%s", line);
  }
}
```
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    D8(%rip),%rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  jmp    L6
  nopl   0x0(%rax)
L4:
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L6:
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <getnextline@plt>
  mov    %eax,%esi
  test   %eax,%eax
  jg     L4
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string

tool: D8: %s

assistant: ```c
int main(void) {
  char line[1000];
  int len;
  while ((len = getnextline(line, 1000)) > 0) {
    reverse(line, len);
    printf("%s", line);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<AMCRealRandomInt32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  mov    -0x48(%rbp),%rax
  cmp    -0x50(%rbp),%rax
  jne    L18
  mov    -0x48(%rbp),%rax
  jmp    L34
L18:
  mov    -0x48(%rbp),%rax
  cmp    -0x50(%rbp),%rax
  jle    L11
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x50(%rbp)
L11:
  movq   $0x0,-0x38(%rbp)
  lea    -0x38(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <AMCReadRandomByteArray@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdq -0x48(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdq -0x50(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    $0x80000000,%edx
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movapd %xmm0,%xmm1
  subsd  -0x28(%rbp),%xmm1
  movsd  D13(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  movsd  D6(%rip),%xmm2
  divsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x28(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
L34:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D13
data_type: f64
name: unpack_data
arguments: data_label: D6
data_type: byte[8]

tool: D13: 1

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x41]}

assistant: ```c
int32_t AMCRealRandomInt32(int32_t min, int32_t max) {
  int32_t ret = 0;
  if (min == max) {
    return min;
  } else {
    int32_t tmp;
    int64_t result;
    double resultFloat, minFloat, maxFloat, tmpFloat;
    if (min > max) {
      tmp = min;
      min = max;
      max = tmp;
    }
    tmp = 0;
    AMCReadRandomByteArray(&tmp, sizeof(tmp));
    minFloat = min;
    maxFloat = max;
    tmpFloat = (double)((int64_t)tmp + 0x80000000);
    resultFloat = minFloat + (maxFloat - minFloat + 1.0) * (tmpFloat / ((double)0x100000000));
    ret = (int32_t)resultFloat;
    ;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<AMCRealRandomInt32>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsi,%rax
  cmp    %rsi,%rdi
  je     L22
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  jle    L26
  mov    %rsi,%rbp
  mov    %rdi,%rbx
L26:
  movq   $0x0,(%rsp)
  mov    %rsp,%rdi
  mov    $0x8,%esi
  call   <AMCReadRandomByteArray@plt>
  pxor   %xmm2,%xmm2
  cvtsi2sd %rbp,%xmm2
  mov    $0x80000000,%eax
  add    (%rsp),%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  mulsd  D6(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %rbx,%xmm1
  subsd  %xmm2,%xmm1
  addsd  D0(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  addsd  %xmm2,%xmm0
  cvttsd2si %xmm0,%rax
L22:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
datatype: byte[8]
name: retrieve_data
arguments: data_label: D0
datatype: f64

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3d]}

tool: D0: 1

assistant: ```c
int32_t AMCRealRandomInt32(int32_t min, int32_t max) {
  int32_t ret = 0;
  if (min == max) {
    return min;
  } else {
    int32_t tmp;
    int64_t result;
    double resultFloat, minFloat, maxFloat, tmpFloat;
    if (min > max) {
      tmp = min;
      min = max;
      max = tmp;
    }
    tmp = 0;
    AMCReadRandomByteArray(&tmp, sizeof(tmp));
    minFloat = min;
    maxFloat = max;
    tmpFloat = (double)((int64_t)tmp + 0x80000000);
    resultFloat = minFloat + (maxFloat - minFloat + 1.0) * (tmpFloat / ((double)0x100000000));
    ret = (int32_t)resultFloat;
    ;
  }
  return ret;
}
```