ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<sort_test>:
  endbr64
  push   %r14
  xor    %edi,%edi
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0xfa0(%rsp),%r12
  call   <time@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %eax,%edi
  call   <srand@plt>
  nopl   0x0(%rax)
L1:
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <rand@plt>
  mov    %r13,%rdi
  movslq %eax,%rsi
  cltd
  imul   $0x10624dd3,%rsi,%rsi
  sar    $0x26,%rsi
  sub    %edx,%esi
  imul   $0x3e8,%esi,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x3e7,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <quickSort@plt>
  nop
L23:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L23
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D2
type_guessed: string

tool: D2: %d 

assistant: ```c
void sort_test() {
  int i = 0;
  int data[1000];
  srand((unsigned)time(NULL));
  for (i = 0; i < 1000; i++) {
    data[i] = rand() % 1000;
    printf("%d ", data[i]);
  }
  printf("\n");
  printf("\n");
  printf("\n");
  printf("\n");
  quickSort(data, 0, 1000 - 1);
  for (i = 0; i < 1000; i++) {
    printf("%d ", data[i]);
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<init_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void init_array(int *a, int size) {
  int i;
  for (i = 0; i < size; ++i) {
    *(a + i) = i;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<init_array>:
  endbr64
  test   %esi,%esi
  jle    L7
  mov    %esi,%esi
  mov    $0x0,%eax
L13:
  mov    %eax,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L13
L7:
  ret
```

assistant: ```c
void init_array(int *a, int size) {
  int i;
  for (i = 0; i < size; ++i) {
    *(a + i) = i;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<init_array>:
  endbr64
  test   %esi,%esi
  jle    L13
  movslq %esi,%rsi
  xor    %eax,%eax
  nopl   (%rax)
L18:
  mov    %eax,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L18
L13:
  ret
```

assistant: ```c
void init_array(int *a, int size) {
  int i;
  for (i = 0; i < size; ++i) {
    *(a + i) = i;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<init_array>:
  endbr64
  test   %esi,%esi
  jle    L44
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L41
  mov    %esi,%edx
  movdqa D12(%rip),%xmm0
  movdqa D6(%rip),%xmm2
  mov    %rdi,%rax
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L68:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L68
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%sil
  je     L51
L7:
  movslq %eax,%rdx
  mov    %eax,(%rdi,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%esi
  jle    L44
  add    $0x2,%eax
  mov    %edx,0x4(%rdi,%rcx,1)
  cmp    %eax,%esi
  jle    L44
  mov    %eax,0x8(%rdi,%rcx,1)
L44:
  ret
  nopl   (%rax)
L51:
  ret
L41:
  xor    %eax,%eax
  jmp    L7
```

tool_calls: name: disassemble_data
arguments: label: D12
data_type: byte[16]
name: disassemble_data
arguments: label: D6
data_type: byte[16]

tool: {"D12": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D6": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
void init_array(int *a, int size) {
  int i;
  for (i = 0; i < size; ++i) {
    *(a + i) = i;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  add    $0x6a,%eax
  xor    $0xb,%eax
  mov    %al,-0x9(%rbp)
  movb   $0x0,-0xa(%rbp)
  nop
  movzbl -0x9(%rbp),%eax
  lea    -0x2b(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned char)11;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] |= (state[local1] & (unsigned char)31) << 4UL;
      state[0UL] -= state[local1];
      local1++;
    }
    output[0UL] = (state[0UL] + 237763669UL) + (unsigned char)128;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  add    $0x6a,%eax
  xor    $0xb,%eax
  sub    $0x2b,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned char)11;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] |= (state[local1] & (unsigned char)31) << 4UL;
      state[0UL] -= state[local1];
      local1++;
    }
    output[0UL] = (state[0UL] + 237763669UL) + (unsigned char)128;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  add    $0x6a,%eax
  xor    $0xb,%eax
  sub    $0x2b,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned char)11;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] |= (state[local1] & (unsigned char)31) << 4UL;
      state[0UL] -= state[local1];
      local1++;
    }
    output[0UL] = (state[0UL] + 237763669UL) + (unsigned char)128;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  add    $0x6a,%eax
  xor    $0xb,%eax
  sub    $0x2b,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ (unsigned char)11;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] |= (state[local1] & (unsigned char)31) << 4UL;
      state[0UL] -= state[local1];
      local1++;
    }
    output[0UL] = (state[0UL] + 237763669UL) + (unsigned char)128;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Extract_Plane>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    $0x0,%ebx
  jmp    L38
L35:
  mov    $0x0,%r12d
  jmp    L28
L13:
  mov    -0x18(%rbp),%rax
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %rax,%rdi
  call   <XGetPixel@plt>
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%ecx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %rax,%rdi
  call   <XPutPixel@plt>
  add    $0x1,%r12d
L28:
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  cmp    %eax,%r12d
  jl     L13
  add    $0x1,%ebx
L38:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%ebx
  jl     L35
  nop
  nop
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void Extract_Plane(XImage *in_image, XImage *out_image, int plane) {
  register int x, y;
  for (y = 0; y < in_image->height; y++)
    for (x = 0; x < in_image->width; x++)
      XPutPixel(out_image, x, y, (XGetPixel(in_image, x, y) >> plane) & 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Extract_Plane>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%r14
  mov    %edx,%r13d
  mov    $0x0,%ebp
  cmpl   $0x0,(%rdi)
  jg     L19
L22:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L46:
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <XGetPixel@plt>
  mov    %r13d,%ecx
  sar    %cl,%eax
  and    $0x1,%eax
  mov    %eax,%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <XPutPixel@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%r12)
  jg     L46
L17:
  add    $0x1,%ebp
  cmp    %ebp,(%r12)
  jle    L22
L19:
  mov    $0x0,%ebx
  cmpl   $0x0,0x4(%r12)
  jg     L46
  jmp    L17
```

assistant: ```c
void Extract_Plane(XImage *in_image, XImage *out_image, int plane) {
  register int x, y;
  for (y = 0; y < in_image->height; y++)
    for (x = 0; x < in_image->width; x++)
      XPutPixel(out_image, x, y, (XGetPixel(in_image, x, y) >> plane) & 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Extract_Plane>:
  endbr64
  mov    (%rdi),%ecx
  test   %ecx,%ecx
  jle    L20
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L11:
  mov    0x4(%r14),%eax
  xor    %r13d,%r13d
  test   %eax,%eax
  jle    L24
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r13d,%esi
  mov    %ebx,%edx
  mov    %r14,%rdi
  call   <XGetPixel@plt>
  mov    %ebp,%ecx
  mov    %r13d,%esi
  mov    %ebx,%edx
  sar    %cl,%eax
  mov    %r12,%rdi
  add    $0x1,%r13d
  and    $0x1,%eax
  mov    %eax,%ecx
  call   <XPutPixel@plt>
  cmp    %r13d,0x4(%r14)
  jg     L19
L24:
  add    $0x1,%ebx
  cmp    %ebx,(%r14)
  jg     L11
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  ret
```

assistant: ```c
void Extract_Plane(XImage *in_image, XImage *out_image, int plane) {
  register int x, y;
  for (y = 0; y < in_image->height; y++)
    for (x = 0; x < in_image->width; x++)
      XPutPixel(out_image, x, y, (XGetPixel(in_image, x, y) >> plane) & 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Extract_Plane>:
  endbr64
  mov    (%rdi),%eax
  test   %eax,%eax
  jle    L19
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    0x4(%rdi),%eax
  xor    %ebx,%ebx
L27:
  test   %eax,%eax
  jle    L1
  xor    %r13d,%r13d
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %r13d,%esi
  mov    %ebx,%edx
  mov    %r14,%rdi
  call   <XGetPixel@plt>
  mov    %r12d,%ecx
  mov    %r13d,%esi
  mov    %ebx,%edx
  sar    %cl,%eax
  mov    %rbp,%rdi
  add    $0x1,%r13d
  and    $0x1,%eax
  mov    %eax,%ecx
  call   <XPutPixel@plt>
  mov    0x4(%r14),%eax
  cmp    %r13d,%eax
  jg     L32
  add    $0x1,%ebx
  cmp    %ebx,(%r14)
  jg     L27
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  ret
```

assistant: ```c
void Extract_Plane(XImage *in_image, XImage *out_image, int plane) {
  register int x, y;
  for (y = 0; y < in_image->height; y++)
    for (x = 0; x < in_image->width; x++)
      XPutPixel(out_image, x, y, (XGetPixel(in_image, x, y) >> plane) & 1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<netUDPPrint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%eax
  sar    $0x8,%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%eax
  sar    $0x10,%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%eax
  shr    $0x18,%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <ntohs@plt>
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D33
data_type: string
name: unpack_data
arguments: tag: D13
data_type: string
name: unpack_data
arguments: tag: D49
data_type: string

tool: D33: %d.

tool: D13: %d

tool: D49:  port %d


assistant: ```c
void netUDPPrint(struct sockaddr_in *from) {
  int i, j;
  printf("%d.", (from->sin_addr.s_addr) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 8) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 16) & 0xFF);
  printf("%d", (from->sin_addr.s_addr >> 24) & 0xFF);
  printf(" port %d\n", ntohs(from->sin_port));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<netUDPPrint>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movzbl 0x4(%rdi),%esi
  lea    D15(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x5(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x6(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x7(%rbx),%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%edi
  call   <ntohs@plt>
  mov    %eax,%esi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D17
type_infered: string

tool: D15: %d.

tool: D19: %d

tool: D17:  port %d


assistant: ```c
void netUDPPrint(struct sockaddr_in *from) {
  int i, j;
  printf("%d.", (from->sin_addr.s_addr) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 8) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 16) & 0xFF);
  printf("%d", (from->sin_addr.s_addr >> 24) & 0xFF);
  printf(" port %d\n", ntohs(from->sin_port));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<netUDPPrint>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movzbl 0x4(%rdi),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x5(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x6(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x7(%rbx),%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%edi
  call   <ntohs@plt>
  add    $0x8,%rsp
  lea    D2(%rip),%rdi
  pop    %rbx
  mov    %eax,%esi
  pop    %rbp
  jmp    1060 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D18
type_infered: string
name: unpack_data
arguments: tag: D8
type_infered: string
name: unpack_data
arguments: tag: D2
type_infered: string

tool: D18: %d.

tool: D8: %d

tool: D2:  port %d


assistant: ```c
void netUDPPrint(struct sockaddr_in *from) {
  int i, j;
  printf("%d.", (from->sin_addr.s_addr) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 8) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 16) & 0xFF);
  printf("%d", (from->sin_addr.s_addr >> 24) & 0xFF);
  printf(" port %d\n", ntohs(from->sin_port));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<netUDPPrint>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movzbl 0x4(%rdi),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x5(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x6(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movzbl 0x7(%rbx),%esi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%edi
  call   <ntohs@plt>
  add    $0x8,%rsp
  lea    D27(%rip),%rdi
  pop    %rbx
  mov    %eax,%esi
  pop    %rbp
  jmp    1060 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
data_type: string
name: unpack_data
arguments: data_label: D22
data_type: string
name: unpack_data
arguments: data_label: D27
data_type: string

tool: D5: %d.

tool: D22: %d

tool: D27:  port %d


assistant: ```c
void netUDPPrint(struct sockaddr_in *from) {
  int i, j;
  printf("%d.", (from->sin_addr.s_addr) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 8) & 0xFF);
  printf("%d.", (from->sin_addr.s_addr >> 16) & 0xFF);
  printf("%d", (from->sin_addr.s_addr >> 24) & 0xFF);
  printf(" port %d\n", ntohs(from->sin_port));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<lwc_lcase_memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  jmp    L14
L18:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <dolower@plt>
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L14:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  test   %rax,%rax
  jne    L18
  nop
  nop
  leave
  ret
```

assistant: ```c
void lwc_lcase_memcpy(char *target, const char *source, size_t n) {
  while (n--) {
    *target++ = dolower(*source++);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<lwc_lcase_memcpy>:
  endbr64
  test   %rdx,%rdx
  je     L18
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  lea    (%rdi,%rdx,1),%r12
L8:
  add    $0x1,%rbp
  movsbl -0x1(%rbp),%edi
  call   <dolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  cmp    %rbx,%r12
  jne    L8
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  ret
```

assistant: ```c
void lwc_lcase_memcpy(char *target, const char *source, size_t n) {
  while (n--) {
    *target++ = dolower(*source++);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lwc_lcase_memcpy>:
  endbr64
  test   %rdx,%rdx
  je     L17
  push   %r12
  lea    (%rdi,%rdx,1),%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L15:
  movsbl 0x0(%rbp),%edi
  add    $0x1,%rbx
  add    $0x1,%rbp
  call   <dolower@plt>
  mov    %al,-0x1(%rbx)
  cmp    %r12,%rbx
  jne    L15
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L17:
  ret
```

assistant: ```c
void lwc_lcase_memcpy(char *target, const char *source, size_t n) {
  while (n--) {
    *target++ = dolower(*source++);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<lwc_lcase_memcpy>:
  endbr64
  test   %rdx,%rdx
  je     L4
  push   %r12
  lea    (%rdi,%rdx,1),%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L2:
  movsbl 0x0(%rbp),%edi
  add    $0x1,%rbx
  add    $0x1,%rbp
  call   <dolower@plt>
  mov    %al,-0x1(%rbx)
  cmp    %r12,%rbx
  jne    L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L4:
  ret
```

assistant: ```c
void lwc_lcase_memcpy(char *target, const char *source, size_t n) {
  while (n--) {
    *target++ = dolower(*source++);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<int_sin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L10
  mov    $0x3243f,%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L10:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x517ccc83,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x11,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x6487e,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x4b65d,-0x14(%rbp)
  jle    L3
  subl   $0x6487e,-0x14(%rbp)
L3:
  cmpl   $0x1921e,-0x14(%rbp)
  jle    L26
  mov    $0x3243f,%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L26:
  mov    -0x14(%rbp),%eax
  mov    $0x3,%esi
  mov    %eax,%edi
  call   <int_pow@plt>
  movslq %eax,%rdx
  imul   $0x2aaaaaab,%rdx,%rdx
  mov    %rdx,%rcx
  shr    $0x20,%rcx
  sar    $0x1f,%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  lea    (%rdx,%rax,1),%ebx
  mov    -0x14(%rbp),%eax
  mov    $0x5,%esi
  mov    %eax,%edi
  call   <int_pow@plt>
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  add    %eax,%ebx
  mov    -0x14(%rbp),%eax
  mov    $0x7,%esi
  mov    %eax,%edi
  call   <int_pow@plt>
  movslq %eax,%rdx
  imul   $0xffffffffd00d00d1,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0xc,%edx
  sar    $0x1f,%eax
  sub    %edx,%eax
  add    %ebx,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int64_t int_sin(int64_t a) {
  if (a < 0)
    a = 205887 - a;
  a %= 2 * 205887;
  if (a >= 205887 * 3 / 2)
    a -= 2 * 205887;
  if (a >= 205887 / 2)
    a = 205887 - a;
  return a - int_pow(a, 3) / 6 + int_pow(a, 5) / 120 - int_pow(a, 7) / 5040;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<int_sin>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x3243f,%eax
  sub    %edi,%eax
  test   %edi,%edi
  cmovs  %eax,%edi
  movslq %edi,%rbx
  imul   $0x517ccc83,%rbx,%rbx
  sar    $0x31,%rbx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebx
  imul   $0x6487e,%ebx,%eax
  sub    %eax,%edi
  mov    %edi,%ebx
  cmp    $0x4b65d,%edi
  jle    L4
  sub    $0x6487e,%ebx
L13:
  mov    $0x3,%esi
  mov    %ebx,%edi
  call   <int_pow@plt>
  movslq %eax,%rdx
  imul   $0x2aaaaaab,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  sub    %edx,%eax
  lea    (%rax,%rbx,1),%r12d
  mov    $0x5,%esi
  mov    %ebx,%edi
  call   <int_pow@plt>
  movslq %eax,%rbp
  imul   $0xffffffff88888889,%rbp,%rbp
  shr    $0x20,%rbp
  add    %eax,%ebp
  sar    $0x6,%ebp
  sar    $0x1f,%eax
  sub    %eax,%ebp
  add    %r12d,%ebp
  mov    $0x7,%esi
  mov    %ebx,%edi
  call   <int_pow@plt>
  movslq %eax,%rdx
  imul   $0xffffffffd00d00d1,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0xc,%edx
  sar    $0x1f,%eax
  sub    %edx,%eax
  add    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  cmp    $0x1921e,%edi
  jle    L13
  mov    $0x3243f,%eax
  sub    %edi,%eax
  mov    %eax,%ebx
  jmp    L13
```

assistant: ```c
int64_t int_sin(int64_t a) {
  if (a < 0)
    a = 205887 - a;
  a %= 2 * 205887;
  if (a >= 205887 * 3 / 2)
    a -= 2 * 205887;
  if (a >= 205887 / 2)
    a = 205887 - a;
  return a - int_pow(a, 3) / 6 + int_pow(a, 5) / 120 - int_pow(a, 7) / 5040;
}
```