ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<ustime>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ustime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ustime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rotateR_fun_uint16_t>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cltd
  shr    $0x1c,%edx
  add    %edx,%eax
  and    $0xf,%eax
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L8
  mov    -0x4(%rbp),%eax
  jmp    L14
L8:
  mov    $0x10,%eax
  sub    -0x8(%rbp),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    -0x8(%rbp),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  or     %esi,%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
uint16_t rotateR_fun_uint16_t(uint16_t x, int32_t i) {
  if ((i %= 16) == 0)
    return x;
  return (x << (16 - i)) | (x >> i);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rotateR_fun_uint16_t>:
  endbr64
  mov    %edi,%eax
  test   $0xf,%sil
  je     L8
  mov    %esi,%eax
  sar    $0x1f,%eax
  shr    $0x1c,%eax
  lea    (%rsi,%rax,1),%edx
  and    $0xf,%edx
  sub    %edx,%eax
  lea    0x10(%rax),%ecx
  mov    %edi,%eax
  shl    %cl,%eax
  and    $0xf,%esi
  mov    %esi,%ecx
  sar    %cl,%edi
  or     %edi,%eax
L8:
  ret
```

assistant: ```c
uint16_t rotateR_fun_uint16_t(uint16_t x, int32_t i) {
  if ((i %= 16) == 0)
    return x;
  return (x << (16 - i)) | (x >> i);
}
```
user: What is the c source code of the assembly code below:

```asm
<rotateR_fun_uint16_t>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edi
  and    $0xf,%edi
  je     L2
  mov    %esi,%edx
  sar    $0x1f,%edx
  shr    $0x1c,%edx
  add    %edx,%esi
  and    $0xf,%esi
  sub    %esi,%edx
  lea    0x10(%rdx),%ecx
  mov    %eax,%edx
  shl    %cl,%edx
  mov    %edi,%ecx
  sar    %cl,%eax
  or     %edx,%eax
L2:
  ret
```

assistant: ```c
uint16_t rotateR_fun_uint16_t(uint16_t x, int32_t i) {
  if ((i %= 16) == 0)
    return x;
  return (x << (16 - i)) | (x >> i);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rotateR_fun_uint16_t>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edi
  and    $0xf,%edi
  je     L2
  mov    %esi,%edx
  sar    $0x1f,%edx
  shr    $0x1c,%edx
  add    %edx,%esi
  and    $0xf,%esi
  sub    %esi,%edx
  lea    0x10(%rdx),%ecx
  mov    %eax,%edx
  shl    %cl,%edx
  mov    %edi,%ecx
  sar    %cl,%eax
  or     %edx,%eax
L2:
  ret
```

assistant: ```c
uint16_t rotateR_fun_uint16_t(uint16_t x, int32_t i) {
  if ((i %= 16) == 0)
    return x;
  return (x << (16 - i)) | (x >> i);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<init_extensions6>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <libip6t_ah_init@plt>
  mov    $0x0,%eax
  call   <libip6t_dst_init@plt>
  mov    $0x0,%eax
  call   <libip6t_eui64_init@plt>
  mov    $0x0,%eax
  call   <libip6t_frag_init@plt>
  mov    $0x0,%eax
  call   <libip6t_hbh_init@plt>
  mov    $0x0,%eax
  call   <libip6t_hl_init@plt>
  mov    $0x0,%eax
  call   <libip6t_HL_init@plt>
  mov    $0x0,%eax
  call   <libip6t_icmp6_init@plt>
  mov    $0x0,%eax
  call   <libip6t_ipv6header_init@plt>
  mov    $0x0,%eax
  call   <libip6t_LOG_init@plt>
  mov    $0x0,%eax
  call   <libip6t_mh_init@plt>
  mov    $0x0,%eax
  call   <libip6t_REJECT_init@plt>
  mov    $0x0,%eax
  call   <libip6t_rt_init@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void init_extensions6(void) {
  libip6t_ah_init();
  libip6t_dst_init();
  libip6t_eui64_init();
  libip6t_frag_init();
  libip6t_hbh_init();
  libip6t_hl_init();
  libip6t_HL_init();
  libip6t_icmp6_init();
  libip6t_ipv6header_init();
  libip6t_LOG_init();
  libip6t_mh_init();
  libip6t_REJECT_init();
  libip6t_rt_init();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<init_extensions6>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <libip6t_ah_init@plt>
  mov    $0x0,%eax
  call   <libip6t_dst_init@plt>
  mov    $0x0,%eax
  call   <libip6t_eui64_init@plt>
  mov    $0x0,%eax
  call   <libip6t_frag_init@plt>
  mov    $0x0,%eax
  call   <libip6t_hbh_init@plt>
  mov    $0x0,%eax
  call   <libip6t_hl_init@plt>
  mov    $0x0,%eax
  call   <libip6t_HL_init@plt>
  mov    $0x0,%eax
  call   <libip6t_icmp6_init@plt>
  mov    $0x0,%eax
  call   <libip6t_ipv6header_init@plt>
  mov    $0x0,%eax
  call   <libip6t_LOG_init@plt>
  mov    $0x0,%eax
  call   <libip6t_mh_init@plt>
  mov    $0x0,%eax
  call   <libip6t_REJECT_init@plt>
  mov    $0x0,%eax
  call   <libip6t_rt_init@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void init_extensions6(void) {
  libip6t_ah_init();
  libip6t_dst_init();
  libip6t_eui64_init();
  libip6t_frag_init();
  libip6t_hbh_init();
  libip6t_hl_init();
  libip6t_HL_init();
  libip6t_icmp6_init();
  libip6t_ipv6header_init();
  libip6t_LOG_init();
  libip6t_mh_init();
  libip6t_REJECT_init();
  libip6t_rt_init();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<init_extensions6>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <libip6t_ah_init@plt>
  xor    %eax,%eax
  call   <libip6t_dst_init@plt>
  xor    %eax,%eax
  call   <libip6t_eui64_init@plt>
  xor    %eax,%eax
  call   <libip6t_frag_init@plt>
  xor    %eax,%eax
  call   <libip6t_hbh_init@plt>
  xor    %eax,%eax
  call   <libip6t_hl_init@plt>
  xor    %eax,%eax
  call   <libip6t_HL_init@plt>
  xor    %eax,%eax
  call   <libip6t_icmp6_init@plt>
  xor    %eax,%eax
  call   <libip6t_ipv6header_init@plt>
  xor    %eax,%eax
  call   <libip6t_LOG_init@plt>
  xor    %eax,%eax
  call   <libip6t_mh_init@plt>
  xor    %eax,%eax
  call   <libip6t_REJECT_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1170 <libip6t_rt_init@plt>
```

assistant: ```c
void init_extensions6(void) {
  libip6t_ah_init();
  libip6t_dst_init();
  libip6t_eui64_init();
  libip6t_frag_init();
  libip6t_hbh_init();
  libip6t_hl_init();
  libip6t_HL_init();
  libip6t_icmp6_init();
  libip6t_ipv6header_init();
  libip6t_LOG_init();
  libip6t_mh_init();
  libip6t_REJECT_init();
  libip6t_rt_init();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<init_extensions6>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <libip6t_ah_init@plt>
  xor    %eax,%eax
  call   <libip6t_dst_init@plt>
  xor    %eax,%eax
  call   <libip6t_eui64_init@plt>
  xor    %eax,%eax
  call   <libip6t_frag_init@plt>
  xor    %eax,%eax
  call   <libip6t_hbh_init@plt>
  xor    %eax,%eax
  call   <libip6t_hl_init@plt>
  xor    %eax,%eax
  call   <libip6t_HL_init@plt>
  xor    %eax,%eax
  call   <libip6t_icmp6_init@plt>
  xor    %eax,%eax
  call   <libip6t_ipv6header_init@plt>
  xor    %eax,%eax
  call   <libip6t_LOG_init@plt>
  xor    %eax,%eax
  call   <libip6t_mh_init@plt>
  xor    %eax,%eax
  call   <libip6t_REJECT_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1170 <libip6t_rt_init@plt>
```

assistant: ```c
void init_extensions6(void) {
  libip6t_ah_init();
  libip6t_dst_init();
  libip6t_eui64_init();
  libip6t_frag_init();
  libip6t_hbh_init();
  libip6t_hl_init();
  libip6t_HL_init();
  libip6t_icmp6_init();
  libip6t_ipv6header_init();
  libip6t_LOG_init();
  libip6t_mh_init();
  libip6t_REJECT_init();
  libip6t_rt_init();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<set_contenido_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <set_contenido@plt>
  mov    -0x8(%rbp),%eax
  sar    $0x8,%eax
  movzbl %al,%eax
  mov    -0x4(%rbp),%edx
  add    $0x1,%edx
  mov    %eax,%esi
  mov    %edx,%edi
  call   <set_contenido@plt>
  mov    -0x8(%rbp),%eax
  sar    $0x10,%eax
  movzbl %al,%eax
  mov    -0x4(%rbp),%edx
  add    $0x2,%edx
  mov    %eax,%esi
  mov    %edx,%edi
  call   <set_contenido@plt>
  mov    -0x8(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    $0x3,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <set_contenido@plt>
  nop
  leave
  ret
```

assistant: ```c
void set_contenido_mem(uint32_t direccion, uint32_t data) {
  set_contenido(direccion, (data & 0xFF));
  set_contenido(direccion + 1, (data >> 8) & 0xFF);
  set_contenido(direccion + 2, (data >> 16) & 0xFF);
  set_contenido(direccion + 3, (data >> 24) & 0xFF);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<set_contenido_mem>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  movzbl %sil,%esi
  call   <set_contenido@plt>
  mov    %ebp,%eax
  movzbl %ah,%esi
  lea    0x1(%rbx),%edi
  call   <set_contenido@plt>
  mov    %ebp,%esi
  sar    $0x10,%esi
  movzbl %sil,%esi
  lea    0x2(%rbx),%edi
  call   <set_contenido@plt>
  mov    %ebp,%esi
  shr    $0x18,%esi
  lea    0x3(%rbx),%edi
  call   <set_contenido@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void set_contenido_mem(uint32_t direccion, uint32_t data) {
  set_contenido(direccion, (data & 0xFF));
  set_contenido(direccion + 1, (data >> 8) & 0xFF);
  set_contenido(direccion + 2, (data >> 16) & 0xFF);
  set_contenido(direccion + 3, (data >> 24) & 0xFF);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<set_contenido_mem>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  movzbl %sil,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <set_contenido@plt>
  mov    %ebp,%eax
  lea    0x1(%rbx),%edi
  movzbl %ah,%esi
  call   <set_contenido@plt>
  mov    %ebp,%esi
  lea    0x2(%rbx),%edi
  sar    $0x10,%esi
  movzbl %sil,%esi
  call   <set_contenido@plt>
  mov    %ebp,%esi
  add    $0x8,%rsp
  lea    0x3(%rbx),%edi
  shr    $0x18,%esi
  pop    %rbx
  pop    %rbp
  jmp    1050 <set_contenido@plt>
```

assistant: ```c
void set_contenido_mem(uint32_t direccion, uint32_t data) {
  set_contenido(direccion, (data & 0xFF));
  set_contenido(direccion + 1, (data >> 8) & 0xFF);
  set_contenido(direccion + 2, (data >> 16) & 0xFF);
  set_contenido(direccion + 3, (data >> 24) & 0xFF);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<set_contenido_mem>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  movzbl %sil,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <set_contenido@plt>
  mov    %ebp,%eax
  lea    0x1(%rbx),%edi
  movzbl %ah,%esi
  call   <set_contenido@plt>
  mov    %ebp,%esi
  lea    0x2(%rbx),%edi
  sar    $0x10,%esi
  movzbl %sil,%esi
  call   <set_contenido@plt>
  mov    %ebp,%esi
  add    $0x8,%rsp
  lea    0x3(%rbx),%edi
  shr    $0x18,%esi
  pop    %rbx
  pop    %rbp
  jmp    1050 <set_contenido@plt>
```

assistant: ```c
void set_contenido_mem(uint32_t direccion, uint32_t data) {
  set_contenido(direccion, (data & 0xFF));
  set_contenido(direccion + 1, (data >> 8) & 0xFF);
  set_contenido(direccion + 2, (data >> 16) & 0xFF);
  set_contenido(direccion + 3, (data >> 24) & 0xFF);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dump2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L10:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0xf,-0x4(%rbp)
  jle    L10
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type: string

tool: D2: %d:%d 

assistant: ```c
void dump2(void *p, void *q) {
  int i;
  for (i = 0; i < 16; i++) {
    printf("%d:%d ", ((unsigned char *)p)[i], ((unsigned char *)q)[i]);
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dump2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  lea    D5(%rip),%r13
L0:
  movzbl 0x0(%rbp,%rbx,1),%edx
  movzbl (%r12,%rbx,1),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x10,%rbx
  jne    L0
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
type_guessed: string

tool: D5: %d:%d 

assistant: ```c
void dump2(void *p, void *q) {
  int i;
  for (i = 0; i < 16; i++) {
    printf("%d:%d ", ((unsigned char *)p)[i], ((unsigned char *)q)[i]);
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dump2>:
  endbr64
  push   %r13
  lea    D4(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   (%rax)
L1:
  movzbl 0x0(%rbp,%rbx,1),%edx
  movzbl (%r12,%rbx,1),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x10,%rbx
  jne    L1
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
datatype: string

tool: D4: %d:%d 

assistant: ```c
void dump2(void *p, void *q) {
  int i;
  for (i = 0; i < 16; i++) {
    printf("%d:%d ", ((unsigned char *)p)[i], ((unsigned char *)q)[i]);
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dump2>:
  endbr64
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   (%rax)
L9:
  movzbl 0x0(%rbp,%rbx,1),%edx
  movzbl (%r12,%rbx,1),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x10,%rbx
  jne    L9
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: tag: D2
type: string

tool: D2: %d:%d 

assistant: ```c
void dump2(void *p, void *q) {
  int i;
  for (i = 0; i < 16; i++) {
    printf("%d:%d ", ((unsigned char *)p)[i], ((unsigned char *)q)[i]);
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<luinv>:
  endbr64
  test   %edi,%edi
  jle    L33
  push   %rbp
  push   %rbx
  mov    %rsi,%r11
  mov    %rdx,%rbx
  movsd  (%rdx),%xmm1
  mov    $0x1,%r10d
  sub    %edi,%r10d
  lea    0x8(%rcx),%rsi
  lea    -0x1(%rdi),%ebp
  shl    $0x3,%rbp
  movslq %edi,%r8
  shl    $0x3,%r8
  mov    $0x0,%r9d
  jmp    L53
L33:
  ret
L16:
  add    $0x8,%r9
  movsd  (%rbx,%r9,1),%xmm1
  lea    (%r10,%rdi,1),%eax
  test   %eax,%eax
  jle    L3
  lea    (%r11,%r9,1),%rdx
  mov    %rcx,%rax
L57:
  movsd  (%rdx),%xmm0
  mulsd  (%rax),%xmm0
  subsd  %xmm0,%xmm1
  add    %r8,%rdx
  add    $0x8,%rax
  cmp    %rsi,%rax
  jne    L57
L3:
  add    $0x1,%r10d
  add    $0x8,%rsi
L53:
  movsd  %xmm1,-0x8(%rsi)
  cmp    %rbp,%r9
  jne    L16
  lea    -0x1(%rdi),%r8d
  movslq %r8d,%r8
  mov    %edi,%r10d
  mov    %edi,%ebx
  not    %ebx
  movslq %ebx,%rbx
  shl    $0x3,%rbx
  movslq %edi,%rax
  mov    %edi,%edx
  imul   %edi,%edx
  movslq %edx,%rdx
  add    %r8,%rdx
  lea    (%r11,%rdx,8),%r9
  lea    0x0(,%rax,8),%rsi
  neg    %rax
  lea    0x0(,%rax,8),%r11
L4:
  movsd  (%rcx,%r8,8),%xmm1
  cmp    %r10d,%edi
  jle    L31
  movslq %r10d,%rax
  mov    %r9,%rdx
L19:
  movsd  (%rdx),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  subsd  %xmm0,%xmm1
  add    $0x1,%rax
  add    %rsi,%rdx
  cmp    %eax,%edi
  jg     L19
L31:
  mulsd  (%r9,%r11,1),%xmm1
  movsd  %xmm1,(%rcx,%r8,8)
  sub    $0x1,%r8
  sub    $0x1,%r10d
  add    %rbx,%r9
  test   %r8d,%r8d
  jns    L4
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void luinv(int n, double *a, double *x, double *y) {
  int i, j, bsA, bsB;
  double sum;
  for (i = 0; i < n; i++) {
    sum = x[i];
    bsA = i - n;
    for (j = 0; j < i; j++) {
      bsA += n;
      sum -= a[bsA] * y[j];
    }
    y[i] = sum;
  }
  bsB = i * n;
  for (i = n - 1; i >= 0; i--) {
    sum = y[i];
    bsB -= n;
    bsA = i + bsB;
    for (j = i + 1; j < n; j++) {
      bsA += n;
      sum -= a[bsA] * y[j];
    }
    y[i] = sum * a[bsB + i];
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<luinv>:
  endbr64
  push   %r13
  lea    -0x1(%rdi),%r10d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  test   %edi,%edi
  jle    L95
  movsd  (%rdx),%xmm1
  mov    %r10d,%r13d
  mov    $0x1,%ebp
  movslq %edi,%r8
  shl    $0x3,%r13
  xor    %r11d,%r11d
  lea    0x8(%rcx),%rsi
  mov    %rdx,%r12
  sub    %edi,%ebp
  lea    0x0(,%r8,8),%r9
  movsd  %xmm1,-0x8(%rsi)
  cmp    %r13,%r11
  je     L32
  nopw   0x0(%rax,%rax,1)
L12:
  add    $0x8,%r11
  lea    0x0(%rbp,%rdi,1),%eax
  movsd  (%r12,%r11,1),%xmm1
  test   %eax,%eax
  jle    L66
  lea    (%rbx,%r11,1),%rdx
  mov    %rcx,%rax
  nopl   0x0(%rax)
L10:
  movsd  (%rdx),%xmm0
  mulsd  (%rax),%xmm0
  add    $0x8,%rax
  add    %r9,%rdx
  subsd  %xmm0,%xmm1
  cmp    %rsi,%rax
  jne    L10
L66:
  add    $0x8,%rsi
  add    $0x1,%ebp
  movsd  %xmm1,-0x8(%rsi)
  cmp    %r13,%r11
  jne    L12
L32:
  mov    %edi,%eax
  imul   %edi,%eax
L34:
  mov    %edi,%ebp
  cltq
  lea    0x0(,%r8,8),%rsi
  movslq %r10d,%r9
  not    %ebp
  lea    -0x1(%rax,%r8,1),%rax
  mov    %rsi,%r8
  mov    %edi,%r11d
  movslq %ebp,%rbp
  lea    (%rbx,%rax,8),%r10
  neg    %r8
  shl    $0x3,%rbp
  nopl   0x0(%rax,%rax,1)
L97:
  movsd  (%rcx,%r9,8),%xmm1
  cmp    %r11d,%edi
  jle    L43
  movslq %r11d,%rax
  mov    %r10,%rdx
  nopl   0x0(%rax)
L98:
  movsd  (%rdx),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  add    $0x1,%rax
  add    %rsi,%rdx
  subsd  %xmm0,%xmm1
  cmp    %eax,%edi
  jg     L98
L43:
  mulsd  (%r10,%r8,1),%xmm1
  sub    $0x1,%r11d
  add    %rbp,%r10
  movsd  %xmm1,(%rcx,%r9,8)
  sub    $0x1,%r9
  test   %r9d,%r9d
  jns    L97
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L95:
  test   %r10d,%r10d
  js     L47
  xor    %eax,%eax
  movslq %edi,%r8
  jmp    L34
```

assistant: ```c
void luinv(int n, double *a, double *x, double *y) {
  int i, j, bsA, bsB;
  double sum;
  for (i = 0; i < n; i++) {
    sum = x[i];
    bsA = i - n;
    for (j = 0; j < i; j++) {
      bsA += n;
      sum -= a[bsA] * y[j];
    }
    y[i] = sum;
  }
  bsB = i * n;
  for (i = n - 1; i >= 0; i--) {
    sum = y[i];
    bsB -= n;
    bsA = i + bsB;
    for (j = i + 1; j < n; j++) {
      bsA += n;
      sum -= a[bsA] * y[j];
    }
    y[i] = sum * a[bsB + i];
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<luinv>:
  endbr64
  push   %r12
  mov    %edi,%r8d
  lea    -0x1(%rdi),%r11d
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rcx,%rsi
  test   %edi,%edi
  jle    L41
  movsd  (%rdx),%xmm0
  mov    %rdx,%r12
  mov    $0x1,%r9d
  movsd  %xmm0,(%rcx)
  cmp    $0x1,%edi
  je     L2
  movslq %r8d,%r9
  lea    0x8(%rsi),%rdi
  lea    0x8(%rbx),%rbp
  mov    $0x1,%edx
  lea    0x0(,%r9,8),%r10
  nopl   0x0(%rax,%rax,1)
L51:
  movsd  (%r12,%rdx,8),%xmm1
  mov    %rsi,%rax
  mov    %rbp,%rcx
  nopl   0x0(%rax)
L10:
  movsd  (%rcx),%xmm0
  mulsd  (%rax),%xmm0
  add    $0x8,%rax
  add    %r10,%rcx
  subsd  %xmm0,%xmm1
  cmp    %rax,%rdi
  jne    L10
  add    $0x1,%rdx
  movsd  %xmm1,(%rdi)
  add    $0x8,%rbp
  add    $0x8,%rdi
  cmp    %rdx,%r9
  jne    L51
L2:
  mov    %r8d,%eax
  imul   %r8d,%eax
L68:
  mov    %r8d,%ebp
  cltq
  lea    0x0(,%r9,8),%rcx
  movslq %r11d,%rdi
  not    %ebp
  lea    -0x1(%rax,%r9,1),%rax
  mov    %rcx,%r9
  mov    %r8d,%r11d
  movslq %ebp,%rbp
  lea    (%rbx,%rax,8),%r10
  neg    %r9
  shl    $0x3,%rbp
  nop
L43:
  movsd  (%rsi,%rdi,8),%xmm1
  cmp    %r11d,%r8d
  jle    L69
  movslq %r11d,%rax
  mov    %r10,%rdx
L16:
  movsd  (%rdx),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  add    $0x1,%rax
  add    %rcx,%rdx
  subsd  %xmm0,%xmm1
  cmp    %eax,%r8d
  jg     L16
L69:
  mulsd  (%r10,%r9,1),%xmm1
  sub    $0x1,%r11d
  add    %rbp,%r10
  movsd  %xmm1,(%rsi,%rdi,8)
  sub    $0x1,%rdi
  test   %edi,%edi
  jns    L43
L19:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L41:
  test   %r11d,%r11d
  js     L19
  xor    %eax,%eax
  movslq %edi,%r9
  jmp    L68
```

assistant: ```c
void luinv(int n, double *a, double *x, double *y) {
  int i, j, bsA, bsB;
  double sum;
  for (i = 0; i < n; i++) {
    sum = x[i];
    bsA = i - n;
    for (j = 0; j < i; j++) {
      bsA += n;
      sum -= a[bsA] * y[j];
    }
    y[i] = sum;
  }
  bsB = i * n;
  for (i = n - 1; i >= 0; i--) {
    sum = y[i];
    bsB -= n;
    bsA = i + bsB;
    for (j = i + 1; j < n; j++) {
      bsA += n;
      sum -= a[bsA] * y[j];
    }
    y[i] = sum * a[bsB + i];
  }
}
```