ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<find_trailing_zero_num>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L20
  cmpl   $0x1,-0x14(%rbp)
  jne    L28
L20:
  mov    $0x0,%eax
  jmp    L53
L28:
  movl   $0x2,-0x10(%rbp)
  jmp    L59
L45:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L7
L62:
  addl   $0x1,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
L7:
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L62
  jmp    L26
L87:
  addl   $0x1,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L87
  addl   $0x1,-0x10(%rbp)
L59:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L45
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L11
  mov    -0x4(%rbp),%eax
  jmp    L53
L11:
  mov    -0x8(%rbp),%eax
L53:
  pop    %rbp
  ret
```

assistant: ```c
uint16_t find_trailing_zero_num(uint16_t num) {
  uint16_t i, tmp;
  uint16_t mod2 = 0, mod5 = 0;
  if (num == 0 || num == 1) {
    return 0;
  }
  for (i = 2; i <= num; i += 1) {
    tmp = i;
    while (tmp % 5 == 0) {
      mod5 += 1;
      tmp /= 5;
    }
    while (tmp % 2 == 0) {
      mod2 += 1;
      tmp /= 2;
    }
  }
  if (mod2 > mod5) {
    return mod5;
  } else {
    return mod2;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<find_trailing_zero_num>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%edi
  jbe    L35
  jle    L0
  lea    0x1(%rdi),%r9d
  mov    $0x0,%esi
  mov    $0x2,%r8d
  mov    $0x0,%edi
  mov    $0x2,%ecx
  jmp    L21
L48:
  mov    %r8d,%ecx
L19:
  add    $0x1,%edi
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%edx
  lea    (%rdx,%rdx,4),%edx
  cmp    %edx,%eax
  je     L19
L21:
  test   $0x1,%cl
  jne    L50
L61:
  add    $0x1,%esi
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  mov    %eax,%ecx
  test   $0x1,%al
  je     L61
L50:
  add    $0x1,%r8d
  cmp    %r8d,%r9d
  je     L45
  movslq %r8d,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %r8d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  cmp    %eax,%r8d
  je     L48
  mov    %r8d,%ecx
  jmp    L21
L0:
  mov    $0x0,%edi
  mov    $0x0,%esi
L45:
  cmp    %edi,%esi
  mov    %edi,%eax
  cmovle %esi,%eax
L35:
  ret
```

assistant: ```c
uint16_t find_trailing_zero_num(uint16_t num) {
  uint16_t i, tmp;
  uint16_t mod2 = 0, mod5 = 0;
  if (num == 0 || num == 1) {
    return 0;
  }
  for (i = 2; i <= num; i += 1) {
    tmp = i;
    while (tmp % 5 == 0) {
      mod5 += 1;
      tmp /= 5;
    }
    while (tmp % 2 == 0) {
      mod2 += 1;
      tmp /= 2;
    }
  }
  if (mod2 > mod5) {
    return mod5;
  } else {
    return mod2;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<find_trailing_zero_num>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%edi
  jbe    L3
  jle    L3
  lea    0x1(%rdi),%r9d
  xor    %ecx,%ecx
  mov    $0x2,%r8d
  xor    %esi,%esi
  mov    $0x2,%eax
L74:
  test   $0x1,%al
  jne    L47
  nopl   0x0(%rax)
L56:
  mov    %eax,%edx
  add    $0x1,%ecx
  shr    $0x1f,%edx
  add    %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x2,%edx
  je     L56
L47:
  add    $0x1,%r8d
  cmp    %r9d,%r8d
  je     L53
  imul   $0xcccccccd,%r8d,%eax
  cmp    $0x33333333,%eax
  mov    %r8d,%eax
  ja     L74
  nopl   (%rax)
L25:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  add    $0x1,%esi
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%edi
  sar    $0x21,%rdx
  mov    %edx,%eax
  sub    %edi,%eax
  imul   $0xcccccccd,%eax,%edx
  add    $0x19999999,%edx
  cmp    $0x33333332,%edx
  jbe    L25
  jmp    L74
L3:
  ret
  nopl   0x0(%rax)
L53:
  cmp    %esi,%ecx
  mov    %esi,%eax
  cmovle %ecx,%eax
  ret
```

assistant: ```c
uint16_t find_trailing_zero_num(uint16_t num) {
  uint16_t i, tmp;
  uint16_t mod2 = 0, mod5 = 0;
  if (num == 0 || num == 1) {
    return 0;
  }
  for (i = 2; i <= num; i += 1) {
    tmp = i;
    while (tmp % 5 == 0) {
      mod5 += 1;
      tmp /= 5;
    }
    while (tmp % 2 == 0) {
      mod2 += 1;
      tmp /= 2;
    }
  }
  if (mod2 > mod5) {
    return mod5;
  } else {
    return mod2;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<find_trailing_zero_num>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%edi
  jbe    L8
  jle    L8
  lea    0x1(%rdi),%r9d
  xor    %ecx,%ecx
  mov    $0x2,%r8d
  xor    %esi,%esi
  mov    $0x2,%eax
L78:
  test   $0x1,%al
  jne    L76
  nopl   0x0(%rax)
L44:
  mov    %eax,%edx
  add    $0x1,%ecx
  shr    $0x1f,%edx
  add    %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x2,%edx
  je     L44
L76:
  add    $0x1,%r8d
  cmp    %r9d,%r8d
  je     L65
  imul   $0xcccccccd,%r8d,%eax
  cmp    $0x33333333,%eax
  mov    %r8d,%eax
  ja     L78
  nopl   (%rax)
L42:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  add    $0x1,%esi
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%edi
  sar    $0x21,%rdx
  mov    %edx,%eax
  sub    %edi,%eax
  imul   $0xcccccccd,%eax,%edx
  add    $0x19999999,%edx
  cmp    $0x33333332,%edx
  jbe    L42
  jmp    L78
L8:
  ret
  nopl   0x0(%rax)
L65:
  cmp    %esi,%ecx
  mov    %esi,%eax
  cmovle %ecx,%eax
  ret
```

assistant: ```c
uint16_t find_trailing_zero_num(uint16_t num) {
  uint16_t i, tmp;
  uint16_t mod2 = 0, mod5 = 0;
  if (num == 0 || num == 1) {
    return 0;
  }
  for (i = 2; i <= num; i += 1) {
    tmp = i;
    while (tmp % 5 == 0) {
      mod5 += 1;
      tmp /= 5;
    }
    while (tmp % 2 == 0) {
      mod2 += 1;
      tmp /= 2;
    }
  }
  if (mod2 > mod5) {
    return mod5;
  } else {
    return mod2;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<proc_cache_hash_pid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rcx
  movabs $0x824a4e60b3262bc5,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x8,%rax
  imul   $0x1f7,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long proc_cache_hash_pid(const pid_t pid) {
  unsigned long h = (unsigned long)pid;
  return h % (503);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<proc_cache_hash_pid>:
  endbr64
  movabs $0x824a4e60b3262bc5,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x8,%rax
  imul   $0x1f7,%rax,%rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  ret
```

assistant: ```c
unsigned long proc_cache_hash_pid(const pid_t pid) {
  unsigned long h = (unsigned long)pid;
  return h % (503);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<proc_cache_hash_pid>:
  endbr64
  movabs $0x824a4e60b3262bc5,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x8,%rax
  imul   $0x1f7,%rax,%rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  ret
```

assistant: ```c
unsigned long proc_cache_hash_pid(const pid_t pid) {
  unsigned long h = (unsigned long)pid;
  return h % (503);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<proc_cache_hash_pid>:
  endbr64
  movabs $0x824a4e60b3262bc5,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x8,%rax
  imul   $0x1f7,%rax,%rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  ret
```

assistant: ```c
unsigned long proc_cache_hash_pid(const pid_t pid) {
  unsigned long h = (unsigned long)pid;
  return h % (503);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ranf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L4
  mov    $0x0,%eax
  jmp    L2
L4:
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
L2:
  leave
  ret
```

assistant: ```c
int ranf(int max) {
  if (max <= 0)
    return (0);
  return (random() % max);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ranf>:
  endbr64
  mov    $0x0,%edx
  test   %edi,%edi
  jle    L5
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  idiv   %ebx
  mov    %edx,%eax
  pop    %rbx
  ret
L5:
  mov    %edx,%eax
  ret
```

assistant: ```c
int ranf(int max) {
  if (max <= 0)
    return (0);
  return (random() % max);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ranf>:
  endbr64
  test   %edi,%edi
  jle    L3
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <random@plt>
  cltd
  idiv   %ebx
  pop    %rbx
  mov    %edx,%eax
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ranf(int max) {
  if (max <= 0)
    return (0);
  return (random() % max);
}
```
user: What is the c source code of the assembly code below?

```asm
<ranf>:
  endbr64
  test   %edi,%edi
  jle    L3
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <random@plt>
  cltd
  idiv   %ebx
  pop    %rbx
  mov    %edx,%eax
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ranf(int max) {
  if (max <= 0)
    return (0);
  return (random() % max);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L24
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L9
L24:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <is_prime@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L9:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D14
data_type: string
name: extract_symbolic_data
arguments: data_label: D4
data_type: string
name: extract_symbolic_data
arguments: data_label: D10
data_type: string

tool: D14: %d

tool: D4: -1

tool: D10: %d


assistant: ```c
int main() {
  int n, c;
  if (scanf("%d", &n) == 0) {
    printf("-1\n");
    return -1;
  }
  c = is_prime(n);
  printf("%d\n", c);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L5
  mov    0x4(%rsp),%edi
  call   <is_prime@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L5:
  lea    D9(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L10
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D6
type_infered: string
name: read_data
arguments: data_label: D1
type_infered: string
name: read_data
arguments: data_label: D9
type_infered: string

tool: D6: %d

tool: D1: %d


tool: D9: -1

assistant: ```c
int main() {
  int n, c;
  if (scanf("%d", &n) == 0) {
    printf("-1\n");
    return -1;
  }
  c = is_prime(n);
  printf("%d\n", c);
  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
  sub    $0x18,%rsp
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  test   %rax,%rax
  je     L9
  mov    0x4(%rsp),%edi
  call   <is_prime@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
L12:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L9:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L12
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D13
type: string
name: unpack_data
arguments: label: D6
type: string
name: unpack_data
arguments: label: D17
type: string

tool: D13: %d

tool: D6: %d


tool: D17: -1

assistant: ```c
int main() {
  int n, c;
  if (scanf("%d", &n) == 0) {
    printf("-1\n");
    return -1;
  }
  c = is_prime(n);
  printf("%d\n", c);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  test   %rax,%rax
  je     L12
  mov    0x4(%rsp),%edi
  call   <is_prime@plt>
  lea    D29(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
L0:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x18,%rsp
  ret
L12:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L0
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D13
datatype: string
name: extract_symbolic_data
arguments: data_label: D29
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string

tool: D13: %d

tool: D29: %d


tool: D23: -1

assistant: ```c
int main() {
  int n, c;
  if (scanf("%d", &n) == 0) {
    printf("-1\n");
    return -1;
  }
  c = is_prime(n);
  printf("%d\n", c);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <unknown1@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L2
L10:
  addl   $0x1,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L2
  addl   $0x1,-0xc(%rbp)
L2:
  mov    $0x0,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L10
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  nop
  leave
  ret
```

assistant: ```c
void main() {
  int flag = unknown1();
  int x = 0;
  int y = 0;
  int j = 0;
  int i = 0;
  while (unknown1()) {
    x++;
    y++;
    i += x;
    j += y;
    if (flag)
      j += 1;
    j = j;
  }
  if (j <= i - 1) {
    goto ERROR;
  ERROR:;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <unknown1@plt>
L0:
  mov    $0x0,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L0
  add    $0x8,%rsp
  ret
```

assistant: ```c
void main() {
  int flag = unknown1();
  int x = 0;
  int y = 0;
  int j = 0;
  int i = 0;
  while (unknown1()) {
    x++;
    y++;
    i += x;
    j += y;
    if (flag)
      j += 1;
    j = j;
  }
  if (j <= i - 1) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <unknown1@plt>
  nop
L1:
  xor    %eax,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L1
  add    $0x8,%rsp
  ret
```

assistant: ```c
void main() {
  int flag = unknown1();
  int x = 0;
  int y = 0;
  int j = 0;
  int i = 0;
  while (unknown1()) {
    x++;
    y++;
    i += x;
    j += y;
    if (flag)
      j += 1;
    j = j;
  }
  if (j <= i - 1) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <unknown1@plt>
  nop
L2:
  xor    %eax,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L2
  add    $0x8,%rsp
  ret
```

assistant: ```c
void main() {
  int flag = unknown1();
  int x = 0;
  int y = 0;
  int j = 0;
  int i = 0;
  while (unknown1()) {
    x++;
    y++;
    i += x;
    j += y;
    if (flag)
      j += 1;
    j = j;
  }
  if (j <= i - 1) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_random_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L41
L20:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L21
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L21:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L35
  mov    -0x4(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L35:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L41
  mov    -0x4(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L41:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L20
  nop
  nop
  leave
  ret
```

assistant: ```c
void get_random_bytes(char *d, int dsize) {
  int i = 0, v;
  while (i < dsize) {
    v = rand();
    d[i++] = v & 0xff;
    if (i < dsize)
      d[i++] = (v >> 8) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 16) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 24) & 0xff;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_random_bytes>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x0,%ebx
  jmp    L9
L37:
  add    $0x4,%ebx
  movslq %edx,%rdx
  shr    $0x18,%eax
  mov    %al,(%r12,%rdx,1)
  jmp    L9
L41:
  mov    %edi,%edx
L65:
  mov    %edx,%ebx
L9:
  cmp    %ebp,%ebx
  jge    L25
  mov    $0x0,%eax
  call   <rand@plt>
  lea    0x1(%rbx),%edx
  movslq %ebx,%rcx
  mov    %al,(%r12,%rcx,1)
  cmp    %edx,%ebp
  jle    L65
  lea    0x2(%rbx),%edi
  movslq %edx,%rdx
  movzbl %ah,%ecx
  mov    %cl,(%r12,%rdx,1)
  cmp    %edi,%ebp
  jle    L41
  lea    0x3(%rbx),%edx
  movslq %edi,%rcx
  mov    %eax,%esi
  sar    $0x10,%esi
  mov    %sil,(%r12,%rcx,1)
  cmp    %edx,%ebp
  jg     L37
  mov    %edx,%ebx
  jmp    L9
L25:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void get_random_bytes(char *d, int dsize) {
  int i = 0, v;
  while (i < dsize) {
    v = rand();
    d[i++] = v & 0xff;
    if (i < dsize)
      d[i++] = (v >> 8) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 16) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 24) & 0xff;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_random_bytes>:
  endbr64
  push   %r12
  xor    %edx,%edx
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  nop
L55:
  mov    %edx,%ebx
  cmp    %ebp,%edx
  jge    L17
L47:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %ebx,%rcx
  lea    0x1(%rbx),%edx
  mov    %al,(%r12,%rcx,1)
  cmp    %edx,%ebp
  jle    L55
  movslq %edx,%rdx
  movzbl %ah,%ecx
  lea    0x2(%rbx),%edi
  mov    %cl,(%r12,%rdx,1)
  cmp    %edi,%ebp
  jle    L9
  mov    %eax,%esi
  movslq %edi,%rcx
  lea    0x3(%rbx),%edx
  sar    $0x10,%esi
  mov    %sil,(%r12,%rcx,1)
  cmp    %edx,%ebp
  jle    L55
  movslq %edx,%rdx
  shr    $0x18,%eax
  mov    %al,(%r12,%rdx,1)
  lea    0x4(%rbx),%edx
  mov    %edx,%ebx
  cmp    %ebp,%edx
  jl     L47
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  mov    %edi,%edx
  jmp    L55
```

assistant: ```c
void get_random_bytes(char *d, int dsize) {
  int i = 0, v;
  while (i < dsize) {
    v = rand();
    d[i++] = v & 0xff;
    if (i < dsize)
      d[i++] = (v >> 8) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 16) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 24) & 0xff;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_random_bytes>:
  endbr64
  push   %r12
  xor    %edx,%edx
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  nop
L6:
  mov    %edx,%ebx
  cmp    %ebp,%edx
  jge    L27
L13:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %ebx,%rcx
  lea    0x1(%rbx),%edx
  mov    %al,(%r12,%rcx,1)
  cmp    %edx,%ebp
  jle    L6
  movslq %edx,%rdx
  lea    0x2(%rbx),%edi
  movzbl %ah,%ecx
  mov    %cl,(%r12,%rdx,1)
  mov    %edi,%edx
  cmp    %edi,%ebp
  jle    L6
  mov    %eax,%esi
  movslq %edi,%rcx
  lea    0x3(%rbx),%edx
  sar    $0x10,%esi
  mov    %sil,(%r12,%rcx,1)
  cmp    %edx,%ebp
  jle    L6
  movslq %edx,%rdx
  shr    $0x18,%eax
  mov    %al,(%r12,%rdx,1)
  lea    0x4(%rbx),%edx
  mov    %edx,%ebx
  cmp    %ebp,%edx
  jl     L13
L27:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void get_random_bytes(char *d, int dsize) {
  int i = 0, v;
  while (i < dsize) {
    v = rand();
    d[i++] = v & 0xff;
    if (i < dsize)
      d[i++] = (v >> 8) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 16) & 0xff;
    if (i < dsize)
      d[i++] = (v >> 24) & 0xff;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sub_vec_i_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L28
L8:
  movl   $0x0,-0x4(%rbp)
  jmp    L20
L29:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%rsi
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rsi,%rax
  sub    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L20:
  mov    0x10(%rbp),%eax
  sub    -0x34(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jle    L29
  addl   $0x1,-0x8(%rbp)
L28:
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L8
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int **sub_vec_i_2(int **res, int **vec1, int **vec2, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2 + 1; j++)
      res[i][j] = vec1[i][j] - vec2[i][j];
  return res;
}
```