ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<strncasehash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L17
L6:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x15(%rbp)
  mov    -0x8(%rbp),%rax
  imul   $0xf4243,%rax,%rdx
  movzbl -0x15(%rbp),%eax
  cmpb   $0x40,-0x15(%rbp)
  jbe    L26
  cmpb   $0x5a,-0x15(%rbp)
  ja     L26
  mov    $0x1,%ecx
  jmp    L39
L26:
  mov    $0x0,%ecx
L39:
  shl    $0x5,%ecx
  add    %ecx,%eax
  cltq
  xor    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0x14(%rbp)
L17:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L9
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
L9:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long strncasehash(const char *str, int length) {
  unsigned char *c = (unsigned char *)str;
  unsigned long hash = 0;
  int n = 0;
  while (n < length && *c != '\0') {
    unsigned char lc = *c++;
    hash = (hash * 0xF4243) ^ (lc + 32 * (lc >= 'A' && lc <= 'Z'));
    n++;
  }
  return hash;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strncasehash>:
  endbr64
  test   %esi,%esi
  jle    L26
  mov    %esi,%esi
  add    %rdi,%rsi
  mov    $0x0,%eax
L16:
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L25
  add    $0x1,%rdi
  imul   $0xf4243,%rax,%rax
  lea    -0x41(%rcx),%edx
  cmp    $0x19,%dl
  setbe  %dl
  movzbl %dl,%edx
  shl    $0x5,%edx
  movzbl %cl,%ecx
  add    %ecx,%edx
  movslq %edx,%rdx
  xor    %rdx,%rax
  cmp    %rsi,%rdi
  jne    L16
L27:
  ret
L25:
  ret
L26:
  mov    $0x0,%eax
  jmp    L27
```

assistant: ```c
unsigned long strncasehash(const char *str, int length) {
  unsigned char *c = (unsigned char *)str;
  unsigned long hash = 0;
  int n = 0;
  while (n < length && *c != '\0') {
    unsigned char lc = *c++;
    hash = (hash * 0xF4243) ^ (lc + 32 * (lc >= 'A' && lc <= 'Z'));
    n++;
  }
  return hash;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncasehash>:
  endbr64
  test   %esi,%esi
  jle    L5
  movslq %esi,%rsi
  xor    %eax,%eax
  add    %rdi,%rsi
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L16:
  imul   $0xf4243,%rax,%rax
  lea    -0x41(%rcx),%edx
  add    $0x1,%rdi
  cmp    $0x19,%dl
  setbe  %dl
  movzbl %dl,%edx
  shl    $0x5,%edx
  add    %ecx,%edx
  movslq %edx,%rdx
  xor    %rdx,%rax
  cmp    %rsi,%rdi
  je     L9
L7:
  movzbl (%rdi),%ecx
  test   %cl,%cl
  jne    L16
  ret
L5:
  xor    %eax,%eax
L9:
  ret
```

assistant: ```c
unsigned long strncasehash(const char *str, int length) {
  unsigned char *c = (unsigned char *)str;
  unsigned long hash = 0;
  int n = 0;
  while (n < length && *c != '\0') {
    unsigned char lc = *c++;
    hash = (hash * 0xF4243) ^ (lc + 32 * (lc >= 'A' && lc <= 'Z'));
    n++;
  }
  return hash;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strncasehash>:
  endbr64
  test   %esi,%esi
  jle    L29
  movslq %esi,%rsi
  xor    %eax,%eax
  add    %rdi,%rsi
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L0:
  imul   $0xf4243,%rax,%rax
  lea    -0x41(%rcx),%edx
  add    $0x1,%rdi
  cmp    $0x19,%dl
  setbe  %dl
  movzbl %dl,%edx
  shl    $0x5,%edx
  add    %ecx,%edx
  movslq %edx,%rdx
  xor    %rdx,%rax
  cmp    %rdi,%rsi
  je     L30
L17:
  movzbl (%rdi),%ecx
  test   %cl,%cl
  jne    L0
  ret
L29:
  xor    %eax,%eax
L30:
  ret
```

assistant: ```c
unsigned long strncasehash(const char *str, int length) {
  unsigned char *c = (unsigned char *)str;
  unsigned long hash = 0;
  int n = 0;
  while (n < length && *c != '\0') {
    unsigned char lc = *c++;
    hash = (hash * 0xF4243) ^ (lc + 32 * (lc >= 'A' && lc <= 'Z'));
    n++;
  }
  return hash;
}
```
user: Translate the below assembly code into C programming language:

```asm
<match_tz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0xa,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strtoul@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  add    $0x1,%rdx
  sub    %rdx,%rax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x1c(%rbp)
  cmpl   $0x4,-0x14(%rbp)
  jne    L87
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L7
L87:
  cmpl   $0x2,-0x14(%rbp)
  je     L73
  movl   $0x63,-0x1c(%rbp)
  jmp    L7
L73:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3a,%al
  jne    L7
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0xa,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strtoul@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  add    $0x1,%rdx
  sub    %rdx,%rax
  cmp    $0x5,%rax
  je     L7
  movl   $0x63,-0x1c(%rbp)
L7:
  cmpl   $0x3b,-0x1c(%rbp)
  jg     L21
  cmpl   $0x17,-0x20(%rbp)
  jg     L21
  mov    -0x20(%rbp),%eax
  imul   $0x3c,%eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L4
  negl   -0x18(%rbp)
L4:
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,(%rax)
L21:
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
int match_tz(const char *date, int *offp) {
  char *end;
  int hour = strtoul(date + 1, &end, 10);
  int n = end - (date + 1);
  int min = 0;
  if (n == 4) {
    min = hour % 100;
    hour = hour / 100;
  } else if (n != 2) {
    min = 99;
  } else if (*end == ':') {
    min = strtoul(end + 1, &end, 10);
    if (end - (date + 1) != 5)
      min = 99;
  }
  if (min < 60 && hour < 24) {
    int offset = hour * 60 + min;
    if (*date == '-')
      offset = -offset;
    *offp = offset;
  }
  return end - date;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<match_tz>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x1(%rdi),%r13
  mov    %rsp,%rsi
  mov    $0xa,%edx
  mov    %r13,%rdi
  call   <strtoul@plt>
  mov    %eax,%ebx
  mov    (%rsp),%rdi
  mov    %rdi,%rax
  sub    %r13,%rax
  cmp    $0x4,%eax
  je     L47
  cmp    $0x2,%eax
  jne    L5
  mov    $0x0,%eax
  cmpb   $0x3a,(%rdi)
  jne    L36
  mov    %rsp,%rsi
  add    $0x1,%rdi
  mov    $0xa,%edx
  call   <strtoul@plt>
  mov    (%rsp),%rdx
  sub    %r13,%rdx
  cmp    $0x5,%rdx
  jne    L5
  jmp    L36
L47:
  movslq %ebx,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  mov    %ebx,%ecx
  sar    $0x1f,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   $0x64,%eax,%esi
  mov    %ebx,%eax
  sub    %esi,%eax
  sub    %ecx,%edx
  mov    %edx,%ebx
L36:
  cmp    $0x3b,%eax
  jg     L5
  cmp    $0x17,%ebx
  jg     L5
  imul   $0x3c,%ebx,%ebx
  add    %eax,%ebx
  mov    %ebx,%eax
  neg    %eax
  cmpb   $0x2d,0x0(%rbp)
  cmove  %eax,%ebx
  mov    %ebx,(%r12)
L5:
  mov    (%rsp),%eax
  sub    %ebp,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_tz(const char *date, int *offp) {
  char *end;
  int hour = strtoul(date + 1, &end, 10);
  int n = end - (date + 1);
  int min = 0;
  if (n == 4) {
    min = hour % 100;
    hour = hour / 100;
  } else if (n != 2) {
    min = 99;
  } else if (*end == ':') {
    min = strtoul(end + 1, &end, 10);
    if (end - (date + 1) != 5)
      min = 99;
  }
  if (min < 60 && hour < 24) {
    int offset = hour * 60 + min;
    if (*date == '-')
      offset = -offset;
    *offp = offset;
  }
  return end - date;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<match_tz>:
  endbr64
  push   %r14
  mov    $0xa,%edx
  push   %r13
  lea    0x1(%rdi),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13,%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rsi
  call   <strtoul@plt>
  mov    %eax,%ebx
  mov    (%rsp),%rax
  mov    %rax,%rdx
  sub    %r13,%rdx
  cmp    $0x4,%edx
  je     L25
  cmp    $0x2,%edx
  je     L32
L1:
  sub    %ebp,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L93
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L32:
  xor    %ecx,%ecx
  cmpb   $0x3a,(%rax)
  mov    $0x1,%edx
  je     L24
L28:
  cmp    $0x17,%ebx
  jg     L1
  test   %dl,%dl
  je     L1
  imul   $0x3c,%ebx,%ebx
  add    %ecx,%ebx
  mov    %ebx,%edx
  neg    %edx
  cmpb   $0x2d,0x0(%rbp)
  cmove  %edx,%ebx
  mov    %ebx,(%r12)
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L25:
  movslq %ebx,%rdx
  mov    %ebx,%esi
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%esi
  sar    $0x25,%rdx
  mov    %edx,%ecx
  sub    %esi,%ecx
  imul   $0x64,%ecx,%edi
  mov    %ebx,%ecx
  mov    %edx,%ebx
  sub    %esi,%ebx
  sub    %edi,%ecx
  cmp    $0x3b,%ecx
  setle  %dl
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L24:
  mov    $0xa,%edx
  lea    0x1(%rax),%rdi
  mov    %r14,%rsi
  call   <strtoul@plt>
  mov    %eax,%ecx
  mov    (%rsp),%rax
  mov    %rax,%rdx
  sub    %r13,%rdx
  cmp    $0x5,%rdx
  jne    L1
  cmp    $0x3b,%ecx
  setle  %dl
  jmp    L28
L93:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_tz(const char *date, int *offp) {
  char *end;
  int hour = strtoul(date + 1, &end, 10);
  int n = end - (date + 1);
  int min = 0;
  if (n == 4) {
    min = hour % 100;
    hour = hour / 100;
  } else if (n != 2) {
    min = 99;
  } else if (*end == ':') {
    min = strtoul(end + 1, &end, 10);
    if (end - (date + 1) != 5)
      min = 99;
  }
  if (min < 60 && hour < 24) {
    int offset = hour * 60 + min;
    if (*date == '-')
      offset = -offset;
    *offp = offset;
  }
  return end - date;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<match_tz>:
  endbr64
  push   %r14
  mov    $0xa,%edx
  push   %r13
  lea    0x1(%rdi),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13,%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rsi
  call   <strtoul@plt>
  mov    %eax,%ebx
  mov    (%rsp),%rax
  mov    %rax,%rdx
  sub    %r13,%rdx
  cmp    $0x4,%edx
  je     L3
  cmp    $0x2,%edx
  je     L87
L45:
  sub    %ebp,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L87:
  xor    %ecx,%ecx
  cmpb   $0x3a,(%rax)
  mov    $0x1,%edx
  je     L83
L25:
  cmp    $0x17,%ebx
  jg     L45
  test   %dl,%dl
  je     L45
  imul   $0x3c,%ebx,%ebx
  add    %ecx,%ebx
  mov    %ebx,%edx
  neg    %edx
  cmpb   $0x2d,0x0(%rbp)
  cmove  %edx,%ebx
  mov    %ebx,(%r12)
  jmp    L45
  nopl   0x0(%rax,%rax,1)
L3:
  movslq %ebx,%rdx
  mov    %ebx,%esi
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%esi
  sar    $0x25,%rdx
  mov    %edx,%ecx
  sub    %esi,%ecx
  imul   $0x64,%ecx,%edi
  mov    %ebx,%ecx
  mov    %edx,%ebx
  sub    %esi,%ebx
  sub    %edi,%ecx
  cmp    $0x3b,%ecx
  setle  %dl
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L83:
  mov    $0xa,%edx
  lea    0x1(%rax),%rdi
  mov    %r14,%rsi
  call   <strtoul@plt>
  mov    %eax,%ecx
  mov    (%rsp),%rax
  mov    %rax,%rdx
  sub    %r13,%rdx
  cmp    $0x5,%rdx
  jne    L45
  cmp    $0x3b,%ecx
  setle  %dl
  jmp    L25
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_tz(const char *date, int *offp) {
  char *end;
  int hour = strtoul(date + 1, &end, 10);
  int n = end - (date + 1);
  int min = 0;
  if (n == 4) {
    min = hour % 100;
    hour = hour / 100;
  } else if (n != 2) {
    min = 99;
  } else if (*end == ':') {
    min = strtoul(end + 1, &end, 10);
    if (end - (date + 1) != 5)
      min = 99;
  }
  if (min < 60 && hour < 24) {
    int offset = hour * 60 + min;
    if (*date == '-')
      offset = -offset;
    *offp = offset;
  }
  return end - date;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<add_num>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L19
L7:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x4(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  jne    L7
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  and    $0x1,%eax
  lea    0x1(%rax),%ecx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  add    %ecx,%ecx
  mov    -0x4(%rbp),%eax
  cltq
  mov    %ecx,(%rdx,%rax,4)
  nop
  leave
  ret
```

assistant: ```c
void add_num(int num[4][4]) {
  int i = abs(rand() % 4);
  int j = abs(rand() % 4);
  while (num[i][j] != 0) {
    i = abs(rand() % 4);
    j = abs(rand() % 4);
  }
  num[i][j] = (abs(rand()) % 2 + 1) * 2;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<add_num>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%r12d
  neg    %r12d
  cmovs  %eax,%r12d
  mov    %ebx,%eax
  sar    $0x1f,%eax
  shr    $0x1e,%eax
  add    %eax,%ebx
  and    $0x3,%ebx
  sub    %eax,%ebx
  mov    %ebx,%eax
  neg    %eax
  cmovns %eax,%ebx
  movslq %ebx,%rbx
  shl    $0x4,%rbx
  add    %rbp,%rbx
  movslq %r12d,%rax
  cmpl   $0x0,(%rbx,%rax,4)
  je     L1
L15:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%r12d
  neg    %r12d
  cmovs  %eax,%r12d
  mov    %ebx,%eax
  sar    $0x1f,%eax
  shr    $0x1e,%eax
  add    %eax,%ebx
  and    $0x3,%ebx
  sub    %eax,%ebx
  mov    %ebx,%eax
  neg    %eax
  cmovns %eax,%ebx
  movslq %ebx,%rbx
  shl    $0x4,%rbx
  add    %rbp,%rbx
  movslq %r12d,%rax
  cmpl   $0x0,(%rbx,%rax,4)
  jne    L15
L1:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %r12d,%r12
  neg    %eax
  cmovs  %edx,%eax
  and    $0x1,%eax
  lea    0x2(%rax,%rax,1),%eax
  mov    %eax,(%rbx,%r12,4)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void add_num(int num[4][4]) {
  int i = abs(rand() % 4);
  int j = abs(rand() % 4);
  while (num[i][j] != 0) {
    i = abs(rand() % 4);
    j = abs(rand() % 4);
  }
  num[i][j] = (abs(rand()) % 2 + 1) * 2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<add_num>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebx,%ecx
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  sar    $0x1f,%ecx
  shr    $0x1e,%ecx
  lea    (%rbx,%rcx,1),%eax
  movslq %edx,%r12
  and    $0x3,%eax
  sub    %ecx,%eax
  mov    %eax,%ebx
  neg    %ebx
  cmovs  %eax,%ebx
  movslq %ebx,%rbx
  shl    $0x4,%rbx
  add    %rbp,%rbx
  mov    (%rbx,%r12,4),%eax
  test   %eax,%eax
  jne    L5
  xor    %eax,%eax
  call   <rand@plt>
  mov    %eax,%r8d
  neg    %eax
  cmovs  %r8d,%eax
  and    $0x1,%eax
  lea    0x2(%rax,%rax,1),%eax
  mov    %eax,(%rbx,%r12,4)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void add_num(int num[4][4]) {
  int i = abs(rand() % 4);
  int j = abs(rand() % 4);
  while (num[i][j] != 0) {
    i = abs(rand() % 4);
    j = abs(rand() % 4);
  }
  num[i][j] = (abs(rand()) % 2 + 1) * 2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<add_num>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  call   <rand@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebx,%ecx
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  sar    $0x1f,%ecx
  shr    $0x1e,%ecx
  lea    (%rbx,%rcx,1),%eax
  movslq %edx,%r12
  and    $0x3,%eax
  sub    %ecx,%eax
  mov    %eax,%ebx
  neg    %ebx
  cmovs  %eax,%ebx
  movslq %ebx,%rbx
  shl    $0x4,%rbx
  add    %rbp,%rbx
  mov    (%rbx,%r12,4),%eax
  test   %eax,%eax
  jne    L7
  xor    %eax,%eax
  call   <rand@plt>
  mov    %eax,%r8d
  neg    %eax
  cmovs  %r8d,%eax
  and    $0x1,%eax
  lea    0x2(%rax,%rax,1),%eax
  mov    %eax,(%rbx,%r12,4)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void add_num(int num[4][4]) {
  int i = abs(rand() % 4);
  int j = abs(rand() % 4);
  while (num[i][j] != 0) {
    i = abs(rand() % 4);
    j = abs(rand() % 4);
  }
  num[i][j] = (abs(rand()) % 2 + 1) * 2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  add    $0x136ff696,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L41
L33:
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x1a(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  add    $0x6,%rax
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  shl    $0x3,%rdx
  add    %rcx,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  lea    0x6(%rax),%rdx
  movzwl -0x1a(%rbp),%eax
  mov    %ax,(%rdx)
  addq   $0x2,-0x18(%rbp)
L41:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L33
  movq   $0x0,-0x18(%rbp)
  jmp    L45
L46:
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L45:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L46
  mov    -0x10(%rbp),%rax
  lea    0x2b2feb28(%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] + 724560680UL;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%rax
  add    $0x136ff696,%rax
  mov    %rax,(%rsp)
  mov    0x8(%rdi),%rdx
  test   %rdx,%rdx
  je     L16
  mov    %rsp,%rax
  lea    -0x8(,%rdx,8),%rdi
  and    $0xfffffffffffffff0,%rdi
  lea    0x10(%rsp),%rcx
  add    %rcx,%rdi
L28:
  movzwl (%rax),%edx
  movzwl 0x6(%rax),%ecx
  mov    %cx,(%rax)
  mov    %dx,0x6(%rax)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L28
  shlq   (%rsp)
L16:
  mov    (%rsp),%rax
  add    $0x2b2feb28,%rax
  mov    %rax,(%rsi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] + 724560680UL;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rdx
  lea    0x136ff696(%rax),%rcx
  test   %rdx,%rdx
  je     L17
  xor    %eax,%eax
  nopl   0x0(%rax)
L11:
  add    $0x2,%rax
  cmp    %rax,%rdx
  ja     L11
  add    %rcx,%rcx
L17:
  add    $0x2b2feb28,%rcx
  mov    %rcx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] + 724560680UL;
  }
}
```