ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ipfw_append>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  xor    %edx,%edx
  push   %r12
  mov    %rsi,%r12
  mov    $0x10,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <add_command@plt>
  mov    0x4(%rsp),%edx
  xor    %r8d,%r8d
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <ipfw_init@plt>
  mov    D9(%rip),%rax
  movq   $0x0,(%rax)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ipfw_append(int args, char **arglist, const char *chain) {
  unsigned int command = 0;
  add_command(&command, 0x0010U, 0x0000U);
  ipfw_init(args, arglist, command, chain, 0);
  optind = 0;
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_invalid_clause>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L7:
  addl   $0x1,-0x4(%rbp)
L2:
  cmpl   $0xf,-0x4(%rbp)
  jle    L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_invalid_clause() {
  int i;
#pragma omp target teams distribute foo bar
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test_invalid_clause>:
  endbr64
  ret
```

assistant: ```c
void test_invalid_clause() {
  int i;
#pragma omp target teams distribute foo bar
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_invalid_clause>:
  endbr64
  ret
```

assistant: ```c
void test_invalid_clause() {
  int i;
#pragma omp target teams distribute foo bar
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test_invalid_clause>:
  endbr64
  ret
```

assistant: ```c
void test_invalid_clause() {
  int i;
#pragma omp target teams distribute foo bar
  for (i = 0; i < 16; ++i)
    ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L14:
  mov    -0x20(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L5:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L7
  mov    -0x20(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L14
L7:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  const char *e = s + n;
  char *ret = d;
  while (*s && s < e)
    *(d++) = *(s++);
  *d = '\0';
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  add    %rsi,%rdx
  movzbl (%rsi),%r8d
  cmp    %rdx,%rsi
  jae    L1
  test   %r8b,%r8b
  je     L1
  mov    %rdi,%rcx
L41:
  add    $0x1,%rsi
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  movzbl (%rsi),%r8d
  cmp    %rsi,%rdx
  jbe    L27
  test   %r8b,%r8b
  jne    L41
L27:
  movb   $0x0,(%rcx)
  ret
L1:
  mov    %rax,%rcx
  jmp    L27
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  const char *e = s + n;
  char *ret = d;
  while (*s && s < e)
    *(d++) = *(s++);
  *d = '\0';
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strncpy>:
  endbr64
  add    %rsi,%rdx
  movzbl (%rsi),%r8d
  mov    %rdi,%rax
  mov    %rdi,%rcx
  cmp    %rdx,%rsi
  jb     L25
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L23:
  add    $0x1,%rsi
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  movzbl (%rsi),%r8d
  cmp    %rsi,%rdx
  jbe    L4
L25:
  test   %r8b,%r8b
  jne    L23
L4:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  const char *e = s + n;
  char *ret = d;
  while (*s && s < e)
    *(d++) = *(s++);
  *d = '\0';
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strncpy>:
  endbr64
  add    %rsi,%rdx
  movzbl (%rsi),%r8d
  mov    %rdi,%rax
  mov    %rdi,%rcx
  cmp    %rdx,%rsi
  jb     L36
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L14:
  add    $0x1,%rsi
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  movzbl (%rsi),%r8d
  cmp    %rsi,%rdx
  jbe    L25
L36:
  test   %r8b,%r8b
  jne    L14
L25:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  const char *e = s + n;
  char *ret = d;
  while (*s && s < e)
    *(d++) = *(s++);
  *d = '\0';
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x1a4(%rbp)
  movl   $0x0,-0x1ac(%rbp)
  jmp    L58
L98:
  mov    -0x1ac(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x1ac(%rbp)
L58:
  mov    -0x1ac(%rbp),%eax
  cmp    -0x1a4(%rbp),%eax
  jl     L98
  movl   $0x0,-0x1ac(%rbp)
  jmp    L0
L156:
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  cmp    $0x1,%eax
  je     L2
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L2
  mov    -0x1ac(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1a8(%rbp)
  jmp    L16
L18:
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    -0x1ac(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x1a0(%rbp,%rdx,4),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  jne    L102
  mov    -0x1a8(%rbp),%eax
  cltq
  movl   $0x0,-0x1a0(%rbp,%rax,4)
L102:
  addl   $0x1,-0x1a8(%rbp)
L16:
  mov    -0x1a8(%rbp),%eax
  cmp    -0x1a4(%rbp),%eax
  jl     L18
  jmp    L59
L2:
  nop
L59:
  addl   $0x1,-0x1ac(%rbp)
L0:
  mov    -0x1ac(%rbp),%eax
  cmp    -0x1a4(%rbp),%eax
  jl     L156
  movl   $0x0,-0x1ac(%rbp)
  jmp    L62
L126:
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L38
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  cmp    $0x1,%eax
  je     L38
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L73
L38:
  nop
L73:
  addl   $0x1,-0x1ac(%rbp)
L62:
  mov    -0x1ac(%rbp),%eax
  cmp    -0x1a4(%rbp),%eax
  jl     L126
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L84
  call   <__stack_chk_fail@plt>
L84:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type_guessed: string

tool: D1: %5d 

assistant: ```c
void main() {
  int prime[100];
  int i, j, amount = 100;
  for (i = 0; i < amount; i++) {
    prime[i] = i + 1;
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 1 || prime[i] == 0)
      continue;
    for (j = i + 1; j < amount; j++) {
      if (prime[j] % prime[i] == 0) {
        prime[j] = 0;
      }
    }
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 0 || prime[i] == 1)
      continue;
    else
      printf("%5d ", prime[i]);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1a0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    $0x1,%eax
L53:
  mov    %eax,-0x4(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x65,%rax
  jne    L53
  mov    %rsp,%rbx
  mov    %rbx,%r9
  mov    $0x1,%r10d
  mov    $0x0,%r11d
  mov    $0x62,%r12d
  lea    0x4(%rbx),%rbp
  jmp    L130
L38:
  mov    %eax,0x4(%rdi)
  add    $0x4,%rcx
  cmp    %r8,%rcx
  je     L45
L122:
  mov    %rcx,%rdi
  mov    0x4(%rcx),%eax
  cltd
  idivl  (%rsi)
  mov    %edx,%eax
  test   %edx,%edx
  je     L38
  mov    0x4(%rcx),%eax
  jmp    L38
L100:
  cmp    $0x63,%r10d
  jg     L7
L45:
  add    $0x1,%r11
  add    $0x4,%r9
  add    $0x1,%r10d
L130:
  mov    %r9,%rsi
  cmpl   $0x1,(%r9)
  jbe    L100
  cmp    $0x63,%r10d
  jg     L7
  mov    %r12d,%eax
  sub    %r11d,%eax
  add    %r11,%rax
  lea    0x0(%rbp,%rax,4),%r8
  mov    %r9,%rcx
  jmp    L122
L7:
  lea    0x190(%rbx),%rbp
  lea    D3(%rip),%r12
  jmp    L105
L67:
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  je     L83
L105:
  mov    (%rbx),%esi
  cmp    $0x1,%esi
  jbe    L67
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L67
L83:
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x1a0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
data_type: string

tool: D3: %5d 

assistant: ```c
void main() {
  int prime[100];
  int i, j, amount = 100;
  for (i = 0; i < amount; i++) {
    prime[i] = i + 1;
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 1 || prime[i] == 0)
      continue;
    for (j = i + 1; j < amount; j++) {
      if (prime[j] % prime[i] == 0) {
        prime[j] = 0;
      }
    }
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 0 || prime[i] == 1)
      continue;
    else
      printf("%5d ", prime[i]);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1a0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  mov    $0x1,%eax
  mov    %rsp,%r11
  nopl   0x0(%rax,%rax,1)
L56:
  mov    %eax,-0x4(%r11,%rax,4)
  add    $0x1,%rax
  cmp    $0x65,%rax
  jne    L56
  mov    %r11,%r8
  mov    %r11,%rbx
  mov    $0x1,%r10d
  mov    $0x63,%ebp
  mov    (%r8),%edi
  cmp    $0x1,%edi
  jbe    L96
  nopl   0x0(%rax,%rax,1)
L7:
  cmp    $0x64,%r10
  je     L11
  mov    %ebp,%edx
  mov    %r8,%rcx
  sub    %r10d,%edx
  lea    -0x1(%r10,%rdx,1),%rax
  lea    (%r11,%rax,4),%r9
  jmp    L73
  nopl   0x0(%rax)
L20:
  mov    (%r8),%edi
  add    $0x4,%rcx
L73:
  mov    0x4(%rcx),%esi
  mov    %esi,%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  cmovne %esi,%edx
  mov    %edx,0x4(%rcx)
  cmp    %r9,%rcx
  jne    L20
L33:
  mov    0x4(%r8),%edi
  add    $0x4,%r8
  add    $0x1,%r10
  cmp    $0x1,%edi
  ja     L7
L96:
  cmp    $0x64,%r10
  jne    L33
  nopl   0x0(%rax,%rax,1)
L11:
  lea    0x190(%r11),%rbp
  lea    D9(%rip),%r12
  cs nopw 0x0(%rax,%rax,1)
L97:
  mov    (%rbx),%esi
  cmp    $0x1,%esi
  jbe    L84
  mov    %r12,%rdi
  call   <printf@plt>
L84:
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L97
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L91
  add    $0x1a0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L91:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string

tool: D9: %5d 

assistant: ```c
void main() {
  int prime[100];
  int i, j, amount = 100;
  for (i = 0; i < amount; i++) {
    prime[i] = i + 1;
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 1 || prime[i] == 0)
      continue;
    for (j = i + 1; j < amount; j++) {
      if (prime[j] % prime[i] == 0) {
        prime[j] = 0;
      }
    }
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 0 || prime[i] == 1)
      continue;
    else
      printf("%5d ", prime[i]);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1a0,%rsp
  movdqa D7(%rip),%xmm1
  movdqa D30(%rip),%xmm3
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movdqa D28(%rip),%xmm2
  lea    0x190(%rsp),%rbx
  mov    %rbp,%rax
  nopl   0x0(%rax,%rax,1)
L54:
  movdqa %xmm1,%xmm0
  add    $0x10,%rax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  movaps %xmm0,-0x10(%rax)
  cmp    %rbx,%rax
  jne    L54
  mov    %rbp,%r8
  mov    $0x1,%r10d
  mov    $0x63,%r11d
  mov    (%r8),%edi
  cmp    $0x1,%edi
  jbe    L12
L21:
  cmp    $0x64,%r10
  je     L61
  mov    %r11d,%edx
  mov    %r8,%rcx
  sub    %r10d,%edx
  lea    -0x1(%r10,%rdx,1),%rax
  lea    0x0(%rbp,%rax,4),%r9
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L40:
  movl   $0x0,0x4(%rcx)
  cmp    %r9,%rcx
  je     L42
L45:
  mov    (%r8),%edi
  add    $0x4,%rcx
L44:
  mov    0x4(%rcx),%esi
  mov    %esi,%eax
  cltd
  idiv   %edi
  test   %edx,%edx
  je     L40
  mov    %esi,0x4(%rcx)
  cmp    %r9,%rcx
  jne    L45
L42:
  mov    0x4(%r8),%edi
  add    $0x4,%r8
  add    $0x1,%r10
  cmp    $0x1,%edi
  ja     L21
L12:
  cmp    $0x64,%r10
  jne    L42
  xchg   %ax,%ax
L61:
  lea    D11(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L46:
  mov    0x0(%rbp),%esi
  cmp    $0x1,%esi
  jbe    L102
  mov    %r12,%rdi
  call   <printf@plt>
L102:
  add    $0x4,%rbp
  cmp    %rbx,%rbp
  jne    L46
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L101
  add    $0x1a0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L101:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
type_infered: byte[16]
name: retrieve_data
arguments: label: D30
type_infered: byte[16]
name: retrieve_data
arguments: label: D28
type_infered: byte[16]
name: retrieve_data
arguments: label: D11
type_infered: string

tool: {"D7": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D30": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D28": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D11: %5d 

assistant: ```c
void main() {
  int prime[100];
  int i, j, amount = 100;
  for (i = 0; i < amount; i++) {
    prime[i] = i + 1;
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 1 || prime[i] == 0)
      continue;
    for (j = i + 1; j < amount; j++) {
      if (prime[j] % prime[i] == 0) {
        prime[j] = 0;
      }
    }
  }
  for (i = 0; i < amount; i++) {
    if (prime[i] == 0 || prime[i] == 1)
      continue;
    else
      printf("%5d ", prime[i]);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<copyvert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L13
L9:
  mov    -0x10(%rbp),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
L13:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  test   %eax,%eax
  jne    L9
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void copyvert(float *dst, float *src, int n) {
  while (n--)
    *dst++ = *src++;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<copyvert>:
  endbr64
  test   %edx,%edx
  je     L13
  mov    %edx,%edx
  mov    $0x0,%eax
L16:
  movss  (%rsi,%rax,4),%xmm0
  movss  %xmm0,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L16
L13:
  ret
```

assistant: ```c
void copyvert(float *dst, float *src, int n) {
  while (n--)
    *dst++ = *src++;
}
```
user: What is the c source code of the assembly code below?

```asm
<copyvert>:
  endbr64
  test   %edx,%edx
  je     L9
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L12:
  movss  (%rsi,%rax,4),%xmm0
  movss  %xmm0,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L12
L9:
  ret
```

assistant: ```c
void copyvert(float *dst, float *src, int n) {
  while (n--)
    *dst++ = *src++;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<copyvert>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%r8
  lea    -0x1(%rdx),%edi
  test   %edx,%edx
  je     L11
  lea    0x4(%rsi),%rsi
  mov    %rcx,%rax
  sub    %rsi,%rax
  cmp    $0x8,%rax
  jbe    L73
  cmp    $0x2,%edi
  jbe    L73
  mov    %edx,%esi
  xor    %eax,%eax
  shr    $0x2,%esi
  shl    $0x4,%rsi
  nopl   (%rax)
L52:
  movups (%r8,%rax,1),%xmm1
  movups %xmm1,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L52
  mov    %edx,%esi
  and    $0xfffffffc,%esi
  mov    %esi,%eax
  sub    %esi,%edi
  shl    $0x2,%rax
  add    %rax,%rcx
  add    %r8,%rax
  cmp    %esi,%edx
  je     L11
  movss  (%rax),%xmm0
  movss  %xmm0,(%rcx)
  test   %edi,%edi
  je     L11
  movss  0x4(%rax),%xmm0
  movss  %xmm0,0x4(%rcx)
  cmp    $0x1,%edi
  je     L11
  movss  0x8(%rax),%xmm0
  movss  %xmm0,0x8(%rcx)
  ret
  nopw   0x0(%rax,%rax,1)
L73:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L59:
  movss  (%r8,%rax,4),%xmm0
  movss  %xmm0,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L59
L11:
  ret
```

assistant: ```c
void copyvert(float *dst, float *src, int n) {
  while (n--)
    *dst++ = *src++;
}
```
user: Converte this assembly snippet into its equivalent C code:

```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
  xor    $0x29be1cdb,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L24
L1:
  lea    -0x10(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x1a(%rbp)
  lea    -0x10(%rbp),%rdx
  add    $0x4,%rdx
  lea    -0x10(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rdx),%edx
  mov    %dx,(%rax)
  lea    -0x10(%rbp),%rax
  add    $0x4,%rax
  movzwl -0x1a(%rbp),%edx
  mov    %dx,(%rax)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  mov    %al,-0x1b(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  add    $0x2,%rax
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  shl    $0x3,%rdx
  add    %rcx,%rdx
  add    $0x5,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  lea    0x2(%rax),%rdx
  movzbl -0x1b(%rbp),%eax
  mov    %al,(%rdx)
  addq   $0x2,-0x18(%rbp)
L24:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L1
  mov    -0x10(%rbp),%rax
  and    $0x10201,%eax
  mov    %rax,%rdx
  mov    $0x9fbcf9fc,%eax
  or     %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  char copy12;
  {
    state[0UL] = input[0UL] ^ 700325083UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
      *((unsigned short *)(&state[0UL]) + 2) = copy11;
      copy12 = *((char *)(&state[local1]) + 5);
      *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 2);
      *((char *)(&state[local1]) + 2) = copy12;
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] & 498160537UL) | 2679962108UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%rax
  xor    $0x29be1cdb,%rax
  mov    %rax,(%rsp)
  mov    0x8(%rdi),%rdx
  test   %rdx,%rdx
  je     L11
  lea    0x2(%rsp),%rax
  lea    -0x8(,%rdx,8),%rdx
  and    $0xfffffffffffffff0,%rdx
  lea    0x12(%rsp,%rdx,1),%rsi
L15:
  movzwl 0x2(%rsp),%edx
  movzwl 0x4(%rsp),%ecx
  mov    %cx,0x2(%rsp)
  mov    %dx,0x4(%rsp)
  movzbl 0x3(%rax),%edx
  movzbl (%rax),%ecx
  mov    %cl,0x3(%rax)
  mov    %dl,(%rax)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L15
L11:
  mov    (%rsp),%rax
  and    $0x10201,%eax
  mov    $0x9fbcf9fc,%edx
  or     %rdx,%rax
  mov    %rax,(%r8)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  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;
  char copy12;
  {
    state[0UL] = input[0UL] ^ 700325083UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
      *((unsigned short *)(&state[0UL]) + 2) = copy11;
      copy12 = *((char *)(&state[local1]) + 5);
      *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 2);
      *((char *)(&state[local1]) + 2) = copy12;
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] & 498160537UL) | 2679962108UL;
  }
}
```
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    0x8(%rdi),%rdx
  mov    %rsi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%rax
  xor    $0x29be1cdb,%rax
  mov    %rax,(%rsp)
  test   %rdx,%rdx
  je     L15
  lea    -0x8(,%rdx,8),%rdx
  lea    0x2(%rsp),%rax
  and    $0xfffffffffffffff0,%rdx
  lea    0x12(%rsp,%rdx,1),%rsi
  nopl   (%rax)
L25:
  movzwl 0x2(%rsp),%edx
  movzwl 0x4(%rsp),%ecx
  add    $0x10,%rax
  mov    %cx,0x2(%rsp)
  mov    %dx,0x4(%rsp)
  movzbl -0xd(%rax),%edx
  movzbl -0x10(%rax),%ecx
  mov    %dl,-0x10(%rax)
  mov    %cl,-0xd(%rax)
  cmp    %rsi,%rax
  jne    L25
  mov    (%rsp),%rax
L15:
  and    $0x10201,%eax
  mov    $0x9fbcf9fc,%edx
  or     %rdx,%rax
  mov    %rax,(%r8)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  ret
L17:
  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;
  char copy12;
  {
    state[0UL] = input[0UL] ^ 700325083UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
      *((unsigned short *)(&state[0UL]) + 2) = copy11;
      copy12 = *((char *)(&state[local1]) + 5);
      *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 2);
      *((char *)(&state[local1]) + 2) = copy12;
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] & 498160537UL) | 2679962108UL;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    0x8(%rdi),%rdx
  mov    %rsi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%rax
  xor    $0x29be1cdb,%rax
  mov    %rax,(%rsp)
  test   %rdx,%rdx
  je     L0
  lea    -0x8(,%rdx,8),%rdx
  lea    0x2(%rsp),%rax
  and    $0xfffffffffffffff0,%rdx
  lea    0x12(%rsp,%rdx,1),%rsi
  nopl   (%rax)
L21:
  movzwl 0x2(%rsp),%edx
  movzwl 0x4(%rsp),%ecx
  add    $0x10,%rax
  mov    %cx,0x2(%rsp)
  mov    %dx,0x4(%rsp)
  movzbl -0xd(%rax),%edx
  movzbl -0x10(%rax),%ecx
  mov    %dl,-0x10(%rax)
  mov    %cl,-0xd(%rax)
  cmp    %rsi,%rax
  jne    L21
  mov    (%rsp),%rax
L0:
  and    $0x10201,%eax
  mov    $0x9fbcf9fc,%edx
  or     %rdx,%rax
  mov    %rax,(%r8)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x18,%rsp
  ret
L18:
  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;
  char copy12;
  {
    state[0UL] = input[0UL] ^ 700325083UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 2);
      *((unsigned short *)(&state[0UL]) + 2) = copy11;
      copy12 = *((char *)(&state[local1]) + 5);
      *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 2);
      *((char *)(&state[local1]) + 2) = copy12;
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] & 498160537UL) | 2679962108UL;
  }
}
```