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

```asm
<c>:
  endbr64
  mov    %rsi,%r9
  lea    -0x1(%rdi),%esi
  mov    %rdx,%r10
  mov    %rcx,%r11
  movslq %esi,%rsi
  cmp    %rsi,%r9
  jle    L47
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L39:
  mov    (%r10,%rsi,8),%rax
  lea    0x1(%rdi),%edx
  mov    %edi,%ecx
  sub    %eax,%edx
  cmpb   $0x0,(%rax)
  je     L31
L4:
  lea    (%rdx,%rax,1),%ecx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L4
L31:
  lea    0x1(%rcx),%edi
  cmp    %ecx,%r8d
  jl     L35
  add    $0x1,%rsi
  cmp    %r9,%rsi
  jne    L39
L44:
  xor    %eax,%eax
  cmp    %r11,%r9
  je     L41
  sub    %edi,%r8d
  lea    0x1(%r8),%eax
  imul   %eax,%eax
  ret
L35:
  mov    $0xffffffff,%eax
L41:
  ret
L47:
  cmp    $0xffffffff,%r8d
  jl     L35
  xor    %edi,%edi
  jmp    L44
```

assistant: ```c
int c(long i, long j, char **words, long words_no, int w) {
  int k, wij = 0, x;
  for (k = i - 1; k < j; k++) {
    x = 0;
    while (words[k][x] != '\0') {
      wij++;
      x++;
    }
    wij++;
    if (wij - 1 > w)
      break;
  }
  if (wij - 1 > w)
    return -1;
  if (j == words_no)
    return 0;
  return (w - wij + 1) * (w - wij + 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<c>:
  endbr64
  mov    %rsi,%r9
  lea    -0x1(%rdi),%esi
  mov    %rdx,%r10
  mov    %rcx,%r11
  movslq %esi,%rsi
  cmp    %rsi,%r9
  jle    L30
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L68:
  mov    (%r10,%rsi,8),%rax
  lea    0x1(%rdi),%edx
  mov    %edi,%ecx
  sub    %eax,%edx
  cmpb   $0x0,(%rax)
  je     L56
L21:
  lea    (%rdx,%rax,1),%ecx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L21
L56:
  lea    0x1(%rcx),%edi
  cmp    %ecx,%r8d
  jl     L15
  add    $0x1,%rsi
  cmp    %rsi,%r9
  jne    L68
L50:
  xor    %eax,%eax
  cmp    %r11,%r9
  je     L62
  sub    %edi,%r8d
  lea    0x1(%r8),%eax
  imul   %eax,%eax
  ret
L15:
  mov    $0xffffffff,%eax
L62:
  ret
L30:
  cmp    $0xffffffff,%r8d
  jl     L15
  xor    %edi,%edi
  jmp    L50
```

assistant: ```c
int c(long i, long j, char **words, long words_no, int w) {
  int k, wij = 0, x;
  for (k = i - 1; k < j; k++) {
    x = 0;
    while (words[k][x] != '\0') {
      wij++;
      x++;
    }
    wij++;
    if (wij - 1 > w)
      break;
  }
  if (wij - 1 > w)
    return -1;
  if (j == words_no)
    return 0;
  return (w - wij + 1) * (w - wij + 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L33
L31:
  movl   $0x0,-0x10(%rbp)
  jmp    L66
L67:
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L66:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L67
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L15:
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x18(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0xc(%rbp),%eax
  sub    %edx,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L15
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L33:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L31
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L44
  call   <__stack_chk_fail@plt>
L44:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D10
type_guessed: string
name: disassemble_data
arguments: data_label: D20
type_guessed: string
name: disassemble_data
arguments: data_label: D55
type_guessed: string
name: disassemble_data
arguments: data_label: D35
type_guessed: string
name: disassemble_data
arguments: data_label: D0
type_guessed: string

tool: D10: Wuwedi n: 

tool: D20: %d

tool: D55: 


tool: D35:  

tool: D0: .

assistant: ```c
int main() {
  int n, i, k, h, d;
  printf("Wuwedi n: ");
  scanf("%d", &n);
  h = 2 * n - 1;
  printf("\n");
  for (i = 0; i < n; i++) {
    for (d = 0; d < i; d++)
      printf(" ");
    for (k = 0; k < h - 2 * i; k++) {
      printf(".");
    }
    printf("\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D46(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax,%rax,1),%r14d
  lea    D24(%rip),%rdi
  call   <printf@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L68
  mov    $0x0,%ebp
  lea    D6(%rip),%r13
  lea    D10(%rip),%r12
  jmp    L32
L27:
  mov    $0x0,%ebx
L69:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L69
L3:
  sub    $0x2,%r14d
  mov    %r15d,%ebp
L32:
  mov    %r14d,%r15d
  test   %r14d,%r14d
  jle    L75
  mov    $0x0,%ebx
L71:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r15d
  jne    L71
L75:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0x1(%rbp),%r15d
  cmp    %r15d,0x4(%rsp)
  jle    L68
  test   %r15d,%r15d
  jg     L27
  jmp    L3
L68:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D46
type: string
name: disassemble_data
arguments: data_tag: D4
type: string
name: disassemble_data
arguments: data_tag: D24
type: string
name: disassemble_data
arguments: data_tag: D6
type: string
name: disassemble_data
arguments: data_tag: D10
type: string

tool: D46: Wuwedi n: 

tool: D4: %d

tool: D24: 


tool: D6: .

tool: D10:  

assistant: ```c
int main() {
  int n, i, k, h, d;
  printf("Wuwedi n: ");
  scanf("%d", &n);
  h = 2 * n - 1;
  printf("\n");
  for (i = 0; i < n; i++) {
    for (d = 0; d < i; d++)
      printf(" ");
    for (k = 0; k < h - 2 * i; k++) {
      printf(".");
    }
    printf("\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  lea    D5(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D43(%rip),%rdi
  lea    -0x1(%rax,%rax,1),%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L42
  xor    %r12d,%r12d
  lea    D45(%rip),%r14
  lea    D13(%rip),%r13
  nopl   0x0(%rax)
L40:
  test   %ebp,%ebp
  jle    L21
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L48:
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L48
L21:
  lea    D43(%rip),%rdi
  lea    0x1(%r12),%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jle    L42
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L56:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r15d,%eax
  add    $0x1,%r15d
  cmp    %eax,%r12d
  jne    L56
  sub    $0x2,%ebp
  mov    %ebx,%r12d
  jmp    L40
  nopl   0x0(%rax)
L42:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_infered: string
name: unpack_data
arguments: data_label: D28
type_infered: string
name: unpack_data
arguments: data_label: D43
type_infered: string
name: unpack_data
arguments: data_label: D45
type_infered: string
name: unpack_data
arguments: data_label: D13
type_infered: string

tool: D5: Wuwedi n: 

tool: D28: %d

tool: D43: 


tool: D45: .

tool: D13:  

assistant: ```c
int main() {
  int n, i, k, h, d;
  printf("Wuwedi n: ");
  scanf("%d", &n);
  h = 2 * n - 1;
  printf("\n");
  for (i = 0; i < n; i++) {
    for (d = 0; d < i; d++)
      printf(" ");
    for (k = 0; k < h - 2 * i; k++) {
      printf(".");
    }
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D25(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D57(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D45(%rip),%rdi
  lea    -0x1(%rax,%rax,1),%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L19
  xor    %r12d,%r12d
  lea    D1(%rip),%r14
  lea    D49(%rip),%r13
  nopl   0x0(%rax)
L53:
  test   %ebp,%ebp
  jle    L41
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L9
L41:
  lea    D45(%rip),%rdi
  lea    0x1(%r12),%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jle    L19
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L38:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r15d,%eax
  add    $0x1,%r15d
  cmp    %eax,%r12d
  jne    L38
  sub    $0x2,%ebp
  mov    %ebx,%r12d
  jmp    L53
  nopl   0x0(%rax)
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D25
data_type: string
name: extract_symbolic_data
arguments: tag: D57
data_type: string
name: extract_symbolic_data
arguments: tag: D45
data_type: string
name: extract_symbolic_data
arguments: tag: D1
data_type: string
name: extract_symbolic_data
arguments: tag: D49
data_type: string

tool: D25: Wuwedi n: 

tool: D57: %d

tool: D45: 


tool: D1: .

tool: D49:  

assistant: ```c
int main() {
  int n, i, k, h, d;
  printf("Wuwedi n: ");
  scanf("%d", &n);
  h = 2 * n - 1;
  printf("\n");
  for (i = 0; i < n; i++) {
    for (d = 0; d < i; d++)
      printf(" ");
    for (k = 0; k < h - 2 * i; k++) {
      printf(".");
    }
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  mov    %rsi,%rdx
  movzbl (%rdi),%eax
  mov    %eax,%ecx
  xor    $0xffffffdb,%ecx
  test   $0x10,%al
  jne    L65
  mov    %ecx,%esi
  shr    $0x2,%sil
  test   $0x4,%cl
  je     L27
  movzbl %cl,%eax
  test   $0x1,%cl
  je     L89
  shr    $0x3,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  movzbl %al,%esi
  shr    $0x4,%al
  and    $0x6,%eax
  mov    %eax,%ecx
  or     $0x1,%ecx
  sar    %cl,%esi
  mov    %esi,%eax
  jmp    L86
L89:
  shr    %cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  shr    %cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  jmp    L86
L27:
  test   $0x2,%cl
  je     L51
  movzbl %cl,%eax
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  mov    %eax,%ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  jmp    L86
L51:
  movzbl %cl,%eax
  and    $0x6,%esi
  mov    %esi,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  shr    $0x4,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  jmp    L86
L65:
  mov    %ecx,%esi
  shr    %sil
  test   $0x2,%cl
  je     L24
  movzbl %cl,%eax
  and    $0x6,%esi
  mov    %esi,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  mov    %eax,%ecx
  shr    $0x2,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
L86:
  shr    $0x3,%al
  xor    $0xffffffa4,%eax
  mov    %al,(%rdx)
  ret
L24:
  movzbl %cl,%eax
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  movzbl %al,%eax
  shr    $0x4,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  jmp    L86
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = input[0UL] ^ (unsigned char)219;
    if ((state[0UL] >> (unsigned char)4) & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)2) & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] <<= ((state[0UL] >> (unsigned char)3) & (unsigned char)7) | 1UL;
          state[0UL] >>= ((state[0UL] >> (unsigned char)4) & (unsigned char)7) | 1UL;
        } else {
          state[0UL] <<= ((state[0UL] >> (unsigned char)1) & (unsigned char)7) | 1UL;
          state[0UL] <<= ((state[0UL] >> (unsigned char)1) & (unsigned char)7) | 1UL;
        }
      } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
        state[0UL] >>= (state[0UL] & (unsigned char)7) | 1UL;
        state[0UL] <<= (state[0UL] & (unsigned char)7) | 1UL;
      } else {
        state[0UL] <<= ((state[0UL] >> (unsigned char)2) & (unsigned char)7) | 1UL;
        state[0UL] <<= ((state[0UL] >> (unsigned char)4) & (unsigned char)7) | 1UL;
      }
    } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
      state[0UL] >>= ((state[0UL] >> (unsigned char)1) & (unsigned char)7) | 1UL;
      state[0UL] >>= ((state[0UL] >> (unsigned char)2) & (unsigned char)7) | 1UL;
    } else {
      state[0UL] <<= (state[0UL] & (unsigned char)7) | 1UL;
      state[0UL] >>= ((state[0UL] >> (unsigned char)4) & (unsigned char)7) | 1UL;
    }
    output[0UL] = (state[0UL] >> (unsigned char)3) ^ (unsigned char)164;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rsi,%rdx
  xor    $0xffffffdb,%ecx
  movzbl %cl,%eax
  mov    %ecx,%esi
  test   $0x10,%cl
  je     L44
  shr    $0x2,%sil
  test   $0x4,%cl
  je     L26
  test   $0x1,%cl
  je     L24
  shr    $0x3,%cl
L9:
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  movzbl %al,%eax
  shr    $0x4,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  shr    $0x3,%al
  xor    $0xffffffa4,%eax
  mov    %al,(%rdx)
  ret
  nopl   0x0(%rax)
L44:
  shr    %sil
  test   $0x2,%cl
  je     L9
  and    $0x6,%esi
  mov    %esi,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  mov    %eax,%ecx
  shr    $0x2,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  shr    $0x3,%al
  xor    $0xffffffa4,%eax
  mov    %al,(%rdx)
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  test   $0x2,%cl
  je     L18
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%eax
  mov    %eax,%ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  shr    $0x3,%al
  xor    $0xffffffa4,%eax
  mov    %al,(%rdx)
  ret
L24:
  shr    %cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  shr    %cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  shr    $0x3,%al
  xor    $0xffffffa4,%eax
  mov    %al,(%rdx)
  ret
  nop
L18:
  and    $0x6,%esi
  mov    %esi,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  mov    %eax,%ecx
  shr    $0x4,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%eax
  shr    $0x3,%al
  xor    $0xffffffa4,%eax
  mov    %al,(%rdx)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = input[0UL] ^ (unsigned char)219;
    if ((state[0UL] >> (unsigned char)4) & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)2) & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] <<= ((state[0UL] >> (unsigned char)3) & (unsigned char)7) | 1UL;
          state[0UL] >>= ((state[0UL] >> (unsigned char)4) & (unsigned char)7) | 1UL;
        } else {
          state[0UL] <<= ((state[0UL] >> (unsigned char)1) & (unsigned char)7) | 1UL;
          state[0UL] <<= ((state[0UL] >> (unsigned char)1) & (unsigned char)7) | 1UL;
        }
      } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
        state[0UL] >>= (state[0UL] & (unsigned char)7) | 1UL;
        state[0UL] <<= (state[0UL] & (unsigned char)7) | 1UL;
      } else {
        state[0UL] <<= ((state[0UL] >> (unsigned char)2) & (unsigned char)7) | 1UL;
        state[0UL] <<= ((state[0UL] >> (unsigned char)4) & (unsigned char)7) | 1UL;
      }
    } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
      state[0UL] >>= ((state[0UL] >> (unsigned char)1) & (unsigned char)7) | 1UL;
      state[0UL] >>= ((state[0UL] >> (unsigned char)2) & (unsigned char)7) | 1UL;
    } else {
      state[0UL] <<= (state[0UL] & (unsigned char)7) | 1UL;
      state[0UL] >>= ((state[0UL] >> (unsigned char)4) & (unsigned char)7) | 1UL;
    }
    output[0UL] = (state[0UL] >> (unsigned char)3) ^ (unsigned char)164;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<Grad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movss  %xmm0,-0x18(%rbp)
  movss  %xmm1,-0x1c(%rbp)
  movss  %xmm2,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0xf,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x7,-0xc(%rbp)
  jg     L33
  movss  -0x18(%rbp),%xmm0
  jmp    L86
L33:
  movss  -0x1c(%rbp),%xmm0
L86:
  movss  %xmm0,-0x8(%rbp)
  cmpl   $0x3,-0xc(%rbp)
  jle    L68
  cmpl   $0xc,-0xc(%rbp)
  je     L52
  cmpl   $0xe,-0xc(%rbp)
  jne    L79
L52:
  movss  -0x18(%rbp),%xmm0
  jmp    L80
L79:
  movss  -0x20(%rbp),%xmm0
  jmp    L80
L68:
  movss  -0x1c(%rbp),%xmm0
L80:
  movss  %xmm0,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L53
  movss  -0x8(%rbp),%xmm0
  movss  D15(%rip),%xmm1
  xorps  %xmm0,%xmm1
  jmp    L55
L53:
  movss  -0x8(%rbp),%xmm1
L55:
  mov    -0xc(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L97
  movss  -0x4(%rbp),%xmm0
  movss  D15(%rip),%xmm2
  xorps  %xmm2,%xmm0
  jmp    L25
L97:
  movss  -0x4(%rbp),%xmm0
L25:
  addss  %xmm1,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
type: byte[4]

tool: {"D15": [0x00,0x00,0x00,0x80]}

assistant: ```c
float Grad(int hash, float x, float y, float z) {
  int h = hash & 15;
  float u = h < 8 ? x : y, v = h < 4 ? y : h == 12 || h == 14 ? x : z;
  return ((h & 1) == 0 ? u : -u) + ((h & 2) == 0 ? v : -v);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<Grad>:
  endbr64
  movaps %xmm0,%xmm3
  test   $0x8,%dil
  je     L42
  movaps %xmm1,%xmm3
L42:
  test   $0xc,%dil
  je     L27
  mov    %edi,%eax
  and    $0xd,%eax
  movaps %xmm0,%xmm1
  cmp    $0xc,%eax
  je     L27
  movaps %xmm2,%xmm1
L27:
  test   $0x1,%dil
  je     L39
  xorps  D14(%rip),%xmm3
L39:
  test   $0x2,%dil
  je     L43
  xorps  D14(%rip),%xmm1
L43:
  addss  %xmm1,%xmm3
  movaps %xmm3,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D14
data_type: byte[16]

tool: {"D14": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float Grad(int hash, float x, float y, float z) {
  int h = hash & 15;
  float u = h < 8 ? x : y, v = h < 4 ? y : h == 12 || h == 14 ? x : z;
  return ((h & 1) == 0 ? u : -u) + ((h & 2) == 0 ? v : -v);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Grad>:
  endbr64
  movaps %xmm0,%xmm3
  test   $0x8,%dil
  je     L48
  movaps %xmm1,%xmm3
L48:
  test   $0xc,%dil
  je     L49
  mov    %edi,%eax
  movaps %xmm0,%xmm1
  and    $0xd,%eax
  cmp    $0xc,%eax
  jne    L21
L49:
  test   $0x1,%dil
  je     L17
  xorps  D17(%rip),%xmm3
L17:
  and    $0x2,%edi
  je     L19
  xorps  D17(%rip),%xmm1
L19:
  addss  %xmm1,%xmm3
  movaps %xmm3,%xmm0
  ret
  nopl   0x0(%rax)
L21:
  movaps %xmm2,%xmm1
  jmp    L49
```

tool_calls: name: parse_data
arguments: data_label: D17
type_guessed: byte[16]

tool: {"D17": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float Grad(int hash, float x, float y, float z) {
  int h = hash & 15;
  float u = h < 8 ? x : y, v = h < 4 ? y : h == 12 || h == 14 ? x : z;
  return ((h & 1) == 0 ? u : -u) + ((h & 2) == 0 ? v : -v);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Grad>:
  endbr64
  movaps %xmm0,%xmm3
  test   $0x8,%dil
  je     L17
  movaps %xmm1,%xmm3
L17:
  test   $0xc,%dil
  je     L8
  mov    %edi,%eax
  movaps %xmm0,%xmm1
  and    $0xd,%eax
  cmp    $0xc,%eax
  jne    L36
L8:
  test   $0x1,%dil
  je     L0
  xorps  D17(%rip),%xmm3
L0:
  and    $0x2,%edi
  je     L58
  xorps  D17(%rip),%xmm1
L58:
  addss  %xmm1,%xmm3
  movaps %xmm3,%xmm0
  ret
  nopl   0x0(%rax)
L36:
  movaps %xmm2,%xmm1
  jmp    L8
```

tool_calls: name: extract_symbolic_data
arguments: label: D17
datatype: byte[16]

tool: {"D17": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float Grad(int hash, float x, float y, float z) {
  int h = hash & 15;
  float u = h < 8 ? x : y, v = h < 4 ? y : h == 12 || h == 14 ? x : z;
  return ((h & 1) == 0 ? u : -u) + ((h & 2) == 0 ? v : -v);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<talloc_array_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x14(%rbp),%ecx
  mov    $0x10000000,%eax
  mov    $0x0,%edx
  divq   -0x10(%rbp)
  cmp    %rax,%rcx
  jb     L9
  mov    $0x0,%eax
  jmp    L17
L9:
  mov    -0x14(%rbp),%eax
  imul   -0x10(%rbp),%rax
  mov    %rax,%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <talloc_named_const@plt>
L17:
  leave
  ret
```

assistant: ```c
void *talloc_array_size(const void *ctx, size_t el_size, unsigned count, const char *name) {
  if (count >= 0x10000000 / el_size) {
    return NULL;
  }
  return talloc_named_const(ctx, el_size * count, name);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<talloc_array_size>:
  endbr64
  mov    %rdi,%r8
  mov    %edx,%edi
  mov    $0x10000000,%eax
  mov    $0x0,%edx
  div    %rsi
  mov    $0x0,%edx
  cmp    %rax,%rdi
  jae    L9
  sub    $0x8,%rsp
  imul   %rdi,%rsi
  mov    %rcx,%rdx
  mov    %r8,%rdi
  call   <talloc_named_const@plt>
  add    $0x8,%rsp
  ret
L9:
  mov    %rdx,%rax
  ret
```

assistant: ```c
void *talloc_array_size(const void *ctx, size_t el_size, unsigned count, const char *name) {
  if (count >= 0x10000000 / el_size) {
    return NULL;
  }
  return talloc_named_const(ctx, el_size * count, name);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<talloc_array_size>:
  endbr64
  mov    %edx,%r8d
  mov    $0x10000000,%eax
  xor    %edx,%edx
  div    %rsi
  cmp    %rax,%r8
  jae    L6
  imul   %r8,%rsi
  mov    %rcx,%rdx
  jmp    1050 <talloc_named_const@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *talloc_array_size(const void *ctx, size_t el_size, unsigned count, const char *name) {
  if (count >= 0x10000000 / el_size) {
    return NULL;
  }
  return talloc_named_const(ctx, el_size * count, name);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<talloc_array_size>:
  endbr64
  mov    %edx,%r8d
  mov    $0x10000000,%eax
  xor    %edx,%edx
  div    %rsi
  cmp    %rax,%r8
  jae    L6
  imul   %r8,%rsi
  mov    %rcx,%rdx
  jmp    1050 <talloc_named_const@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *talloc_array_size(const void *ctx, size_t el_size, unsigned count, const char *name) {
  if (count >= 0x10000000 / el_size) {
    return NULL;
  }
  return talloc_named_const(ctx, el_size * count, name);
}
```