ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2,-0xd8(%rbp)
  movl   $0x2,-0xd0(%rbp)
  movl   $0x3,-0xcc(%rbp)
  movl   $0x5,-0xe0(%rbp)
  jmp    L80
L41:
  movl   $0x1,-0xd4(%rbp)
  movl   $0x1,-0xdc(%rbp)
  jmp    L59
L21:
  mov    -0xdc(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%ecx
  mov    -0xe0(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  jne    L67
  movl   $0x0,-0xd4(%rbp)
L67:
  addl   $0x1,-0xdc(%rbp)
L59:
  cmpl   $0x0,-0xd4(%rbp)
  je     L26
  mov    -0xdc(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%esi
  mov    -0xe0(%rbp),%eax
  cltd
  idiv   %esi
  mov    %eax,%edx
  mov    -0xdc(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jge    L21
L26:
  cmpl   $0x0,-0xd4(%rbp)
  je     L32
  mov    -0xd8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xd8(%rbp)
  cltq
  mov    -0xe0(%rbp),%edx
  mov    %edx,-0xd0(%rbp,%rax,4)
L32:
  addl   $0x2,-0xe0(%rbp)
L80:
  cmpl   $0x31,-0xe0(%rbp)
  jle    L41
  movl   $0x0,-0xdc(%rbp)
  jmp    L71
L48:
  mov    -0xdc(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xdc(%rbp)
L71:
  mov    -0xdc(%rbp),%eax
  cmp    -0xd8(%rbp),%eax
  jl     L48
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D5
type: string

tool: D5: %i 

assistant: ```c
int main(void) {
  int p, i, primes[50], primeIndex = 2;
  bool isPrime;
  primes[0] = 2;
  primes[1] = 3;
  for (p = 5; p < 50; p += 2) {
    for (isPrime = true, i = 1; isPrime && p / primes[i] >= primes[i]; ++i)
      if (!(p % primes[i]))
        isPrime = false;
    if (isPrime)
      primes[primeIndex++] = p;
  }
  for (i = 0; i < primeIndex; ++i)
    printf("%i ", primes[i]);
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movl   $0x2,(%rsp)
  movl   $0x3,0x4(%rsp)
  mov    $0x2,%r8d
  mov    $0x5,%esi
L14:
  lea    0x4(%rsp),%rdi
L29:
  mov    (%rdi),%ecx
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    %eax,%ecx
  jg     L26
  add    $0x4,%rdi
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L29
L41:
  add    $0x2,%esi
  cmp    $0x33,%esi
  jne    L14
  test   %r8d,%r8d
  jle    L18
  mov    %rsp,%rbx
  lea    -0x1(%r8),%eax
  lea    0x4(%rsp,%rax,4),%r12
  lea    D2(%rip),%rbp
L43:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L43
L18:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  mov    $0x0,%eax
  add    $0xd0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  movslq %r8d,%rax
  mov    %esi,(%rsp,%rax,4)
  lea    0x1(%r8),%r8d
  jmp    L41
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
type_guessed: string

tool: D2: %i 

assistant: ```c
int main(void) {
  int p, i, primes[50], primeIndex = 2;
  bool isPrime;
  primes[0] = 2;
  primes[1] = 3;
  for (p = 5; p < 50; p += 2) {
    for (isPrime = true, i = 1; isPrime && p / primes[i] >= primes[i]; ++i)
      if (!(p % primes[i]))
        isPrime = false;
    if (isPrime)
      primes[primeIndex++] = p;
  }
  for (i = 0; i < primeIndex; ++i)
    printf("%i ", primes[i]);
  printf("\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x2,%r8d
  mov    $0x5,%esi
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movl   $0x3,0x4(%rsp)
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L89:
  add    $0x2,%esi
  cmp    $0x33,%esi
  je     L54
L23:
  lea    0x8(%rsp),%rdi
  mov    $0x3,%ecx
  cmp    $0x8,%esi
  jle    L6
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L89
  mov    (%rdi),%ecx
  mov    %esi,%eax
  add    $0x4,%rdi
  cltd
  idiv   %ecx
  cmp    %eax,%ecx
  jle    L41
L6:
  movslq %r8d,%rax
  add    $0x1,%r8d
  mov    %esi,(%rsp,%rax,4)
  add    $0x2,%esi
  cmp    $0x33,%esi
  jne    L23
L54:
  lea    0x4(%rsp),%rbx
  lea    -0x1(%r8),%eax
  mov    $0x2,%esi
  lea    (%rbx,%rax,4),%r12
  lea    D5(%rip),%rbp
  jmp    L69
  nopl   (%rax)
L48:
  mov    (%rbx),%esi
  add    $0x4,%rbx
L69:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L48
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0xd0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: string

tool: D5: %i 

assistant: ```c
int main(void) {
  int p, i, primes[50], primeIndex = 2;
  bool isPrime;
  primes[0] = 2;
  primes[1] = 3;
  for (p = 5; p < 50; p += 2) {
    for (isPrime = true, i = 1; isPrime && p / primes[i] >= primes[i]; ++i)
      if (!(p % primes[i]))
        isPrime = false;
    if (isPrime)
      primes[primeIndex++] = p;
  }
  for (i = 0; i < primeIndex; ++i)
    printf("%i ", primes[i]);
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x2,%r8d
  mov    $0x5,%esi
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    D18(%rip),%rax
  mov    %rax,(%rsp)
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L38:
  add    $0x2,%esi
  cmp    $0x33,%esi
  je     L10
L33:
  lea    0x8(%rsp),%rdi
  mov    $0x3,%ecx
  cmp    $0x8,%esi
  jle    L29
  nopw   0x0(%rax,%rax,1)
L72:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L38
  mov    (%rdi),%ecx
  mov    %esi,%eax
  add    $0x4,%rdi
  cltd
  idiv   %ecx
  cmp    %eax,%ecx
  jle    L72
L29:
  movslq %r8d,%rax
  add    $0x1,%r8d
  mov    %esi,(%rsp,%rax,4)
  add    $0x2,%esi
  cmp    $0x33,%esi
  jne    L33
L10:
  lea    -0x1(%r8),%eax
  mov    %rsp,%rbx
  lea    D12(%rip),%rbp
  lea    0x4(%rsp,%rax,4),%r12
  nopl   (%rax)
L44:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L44
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0xd0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D18
type_infered: byte[8]
name: read_data
arguments: tag: D12
type_infered: string

tool: {"D18": [0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: D12: %i 

assistant: ```c
int main(void) {
  int p, i, primes[50], primeIndex = 2;
  bool isPrime;
  primes[0] = 2;
  primes[1] = 3;
  for (p = 5; p < 50; p += 2) {
    for (isPrime = true, i = 1; isPrime && p / primes[i] >= primes[i]; ++i)
      if (!(p % primes[i]))
        isPrime = false;
    if (isPrime)
      primes[primeIndex++] = p;
  }
  for (i = 0; i < primeIndex; ++i)
    printf("%i ", primes[i]);
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<check>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L50
L48:
  movl   $0x0,-0x8(%rbp)
  jmp    L36
L0:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  jne    L29
  movl   $0x1,-0x4(%rbp)
L29:
  addl   $0x1,-0x8(%rbp)
L36:
  cmpl   $0x8,-0x8(%rbp)
  jle    L0
  addl   $0x1,-0xc(%rbp)
L50:
  cmpl   $0x8,-0xc(%rbp)
  jg     L6
  cmpl   $0x1,-0x4(%rbp)
  jne    L48
L6:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int check(int S[9][9]) {
  int i, j, find_0 = 0;
  for (i = 0; i < 9 && find_0 != 1; i++) {
    for (j = 0; j < 9; j++) {
      if (S[i][j] == 0) {
        find_0 = 1;
      }
    }
  }
  return find_0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<check>:
  endbr64
  mov    $0x0,%ecx
  mov    $0x0,%r8d
  mov    $0x1,%esi
  jmp    L23
L56:
  cmp    $0x8,%eax
  jg     L17
  mov    %esi,%ecx
L10:
  add    $0x4,%rdx
  add    $0x1,%eax
L18:
  cmpl   $0x0,(%rdx)
  je     L56
  cmp    $0x8,%eax
  jle    L10
  add    $0x1,%r8d
  add    $0x24,%rdi
  cmp    $0x8,%r8d
  jg     L3
  test   %ecx,%ecx
  jne    L3
L23:
  mov    %rdi,%rdx
  mov    $0x1,%eax
  jmp    L18
L17:
  mov    $0x1,%ecx
L3:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int check(int S[9][9]) {
  int i, j, find_0 = 0;
  for (i = 0; i < 9 && find_0 != 1; i++) {
    for (j = 0; j < 9; j++) {
      if (S[i][j] == 0) {
        find_0 = 1;
      }
    }
  }
  return find_0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<check>:
  endbr64
  xor    %edx,%edx
  cs nopw 0x0(%rax,%rax,1)
L23:
  mov    $0x1,%eax
  xor    %r8d,%r8d
  jmp    L62
  nopw   0x0(%rax,%rax,1)
L38:
  cmp    $0x9,%rax
  je     L19
L65:
  add    $0x1,%rax
L62:
  mov    -0x4(%rdi,%rax,4),%ecx
  test   %ecx,%ecx
  jne    L38
  mov    $0x1,%r8d
  cmp    $0x9,%rax
  jne    L65
L66:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  add    $0x1,%edx
  add    $0x24,%rdi
  cmp    $0x8,%edx
  jg     L66
  test   %r8d,%r8d
  je     L23
  jmp    L66
```

assistant: ```c
int check(int S[9][9]) {
  int i, j, find_0 = 0;
  for (i = 0; i < 9 && find_0 != 1; i++) {
    for (j = 0; j < 9; j++) {
      if (S[i][j] == 0) {
        find_0 = 1;
      }
    }
  }
  return find_0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<check>:
  endbr64
  xor    %ecx,%ecx
  mov    $0x1,%edx
  nopl   0x0(%rax,%rax,1)
L33:
  mov    0x4(%rdi),%r10d
  mov    $0x1,%eax
  test   %r10d,%r10d
  je     L20
  mov    (%rdi),%r9d
  xor    %eax,%eax
  test   %r9d,%r9d
  sete   %al
L20:
  mov    0x8(%rdi),%esi
  mov    0xc(%rdi),%r8d
  mov    0x10(%rdi),%r9d
  mov    0x14(%rdi),%r10d
  test   %esi,%esi
  mov    0x18(%rdi),%r11d
  mov    0x1c(%rdi),%esi
  cmove  %edx,%eax
  test   %r8d,%r8d
  mov    0x20(%rdi),%r8d
  cmove  %edx,%eax
  test   %r9d,%r9d
  cmove  %edx,%eax
  test   %r10d,%r10d
  cmove  %edx,%eax
  test   %r11d,%r11d
  cmove  %edx,%eax
  test   %esi,%esi
  cmove  %edx,%eax
  test   %r8d,%r8d
  jne    L19
  mov    $0x1,%eax
L12:
  ret
L19:
  add    $0x1,%ecx
  add    $0x24,%rdi
  cmp    $0x8,%ecx
  jg     L12
  test   $0x1,%al
  je     L33
  ret
```

assistant: ```c
int check(int S[9][9]) {
  int i, j, find_0 = 0;
  for (i = 0; i < 9 && find_0 != 1; i++) {
    for (j = 0; j < 9; j++) {
      if (S[i][j] == 0) {
        find_0 = 1;
      }
    }
  }
  return find_0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<errata_6>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xff0130,%eax
  mov    (%rax),%eax
  cmp    $0x9,%eax
  jne    L27
  mov    $0xff0134,%eax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jne    L49
  mov    $0x1,%eax
  jmp    L12
L49:
  mov    $0xff0134,%eax
  mov    (%rax),%eax
  cmp    $0x2,%eax
  jne    L27
  mov    $0x1,%eax
  jmp    L12
L27:
  mov    $0x0,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
bool errata_6() {
  if (*(uint32_t *)0x00FF0130 == 0x9ul) {
    if (*(uint32_t *)0x00FF0134 == 0x01ul) {
      return true;
    }
    if (*(uint32_t *)0x00FF0134 == 0x02ul) {
      return true;
    }
  }
  return false;
}
```
user: Translate the below assembly code into C programming language:

```asm
<errata_6>:
  endbr64
  mov    $0x0,%eax
  cmpl   $0x9,0xff0130
  je     L13
L4:
  ret
L13:
  mov    0xff0134,%eax
  sub    $0x1,%eax
  cmp    $0x1,%eax
  setbe  %al
  movzbl %al,%eax
  jmp    L4
```

assistant: ```c
bool errata_6() {
  if (*(uint32_t *)0x00FF0130 == 0x9ul) {
    if (*(uint32_t *)0x00FF0134 == 0x01ul) {
      return true;
    }
    if (*(uint32_t *)0x00FF0134 == 0x02ul) {
      return true;
    }
  }
  return false;
}
```
user: Translate the below assembly code into C programming language:

```asm
<errata_6>:
  endbr64
  xor    %eax,%eax
  cmpl   $0x9,0xff0130
  je     L1
  ret
  nopl   0x0(%rax)
L1:
  mov    0xff0134,%eax
  sub    $0x1,%eax
  cmp    $0x1,%eax
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool errata_6() {
  if (*(uint32_t *)0x00FF0130 == 0x9ul) {
    if (*(uint32_t *)0x00FF0134 == 0x01ul) {
      return true;
    }
    if (*(uint32_t *)0x00FF0134 == 0x02ul) {
      return true;
    }
  }
  return false;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<errata_6>:
  endbr64
  xor    %eax,%eax
  cmpl   $0x9,0xff0130
  je     L8
  ret
  nopl   0x0(%rax)
L8:
  mov    0xff0134,%eax
  sub    $0x1,%eax
  cmp    $0x1,%eax
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool errata_6() {
  if (*(uint32_t *)0x00FF0130 == 0x9ul) {
    if (*(uint32_t *)0x00FF0134 == 0x01ul) {
      return true;
    }
    if (*(uint32_t *)0x00FF0134 == 0x02ul) {
      return true;
    }
  }
  return false;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<lwan_tpl_str_is_empty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <__builtin_expect@plt>
  test   %rax,%rax
  je     L16
  mov    $0x1,%eax
  jmp    L26
L16:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L14
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
L14:
  mov    $0x1,%eax
  jmp    L26
L9:
  mov    $0x0,%eax
L26:
  leave
  ret
```

assistant: ```c
bool lwan_tpl_str_is_empty(void *ptr) {
  if (__builtin_expect(((!ptr)), (0)))
    return true;
  const char *str = *(const char **)ptr;
  return !str || *str == '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<lwan_tpl_str_is_empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  sete   %dil
  movzbl %dil,%edi
  mov    $0x0,%esi
  call   <__builtin_expect@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L0
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L0
  cmpb   $0x0,(%rdx)
  sete   %al
  movzbl %al,%eax
L0:
  pop    %rbx
  ret
```

assistant: ```c
bool lwan_tpl_str_is_empty(void *ptr) {
  if (__builtin_expect(((!ptr)), (0)))
    return true;
  const char *str = *(const char **)ptr;
  return !str || *str == '\0';
}
```
user: What is the c source code of the assembly code below?

```asm
<lwan_tpl_str_is_empty>:
  endbr64
  test   %rdi,%rdi
  push   %rbx
  mov    %rdi,%rbx
  sete   %dil
  xor    %esi,%esi
  movzbl %dil,%edi
  call   <__builtin_expect@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  jne    L17
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L17
  xor    %eax,%eax
  cmpb   $0x0,(%rdx)
  sete   %al
L17:
  pop    %rbx
  ret
```

assistant: ```c
bool lwan_tpl_str_is_empty(void *ptr) {
  if (__builtin_expect(((!ptr)), (0)))
    return true;
  const char *str = *(const char **)ptr;
  return !str || *str == '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<lwan_tpl_str_is_empty>:
  endbr64
  test   %rdi,%rdi
  push   %rbx
  mov    %rdi,%rbx
  sete   %dil
  xor    %esi,%esi
  movzbl %dil,%edi
  call   <__builtin_expect@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  jne    L3
  mov    (%rbx),%rdx
  test   %rdx,%rdx
  je     L3
  xor    %eax,%eax
  cmpb   $0x0,(%rdx)
  sete   %al
L3:
  pop    %rbx
  ret
```

assistant: ```c
bool lwan_tpl_str_is_empty(void *ptr) {
  if (__builtin_expect(((!ptr)), (0)))
    return true;
  const char *str = *(const char **)ptr;
  return !str || *str == '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calcRoot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  neg    %eax
  pxor   %xmm3,%xmm3
  cvtsi2sd %eax,%xmm3
  movsd  %xmm3,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  imul   %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  imul   %ecx,%eax
  lea    0x0(,%rax,4),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pxor   %xmm4,%xmm4
  cvtsi2sd %eax,%xmm4
  movq   %xmm4,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  addsd  -0x18(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  add    %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  cvtss2sd %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  mov    -0x10(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  neg    %eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movsd  %xmm2,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  imul   %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  imul   %ecx,%eax
  lea    0x0(,%rax,4),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pxor   %xmm5,%xmm5
  cvtsi2sd %eax,%xmm5
  movq   %xmm5,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  add    %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  cvtss2sd %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void calcRoot(int *a, float *roots) {
  *roots = (-(*(a + 1)) + (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
  *(roots + 1) = (-(*(a + 1)) - (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
}
```
user: Translate the below assembly code into C programming language:

```asm
<calcRoot>:
  endbr64
  push   %r14
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    0x4(%rdi),%eax
  mov    %eax,%edx
  neg    %edx
  pxor   %xmm2,%xmm2
  cvtsi2sd %edx,%xmm2
  movsd  %xmm2,0x8(%rsp)
  imul   %eax,%eax
  mov    (%rdi),%edx
  imul   0x8(%rdi),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L21
  sqrtsd %xmm0,%xmm0
L15:
  addsd  0x8(%rsp),%xmm0
  mov    (%rbx),%eax
  add    %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  cvtss2sd %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0x0(%rbp)
  mov    0x4(%rbx),%eax
  mov    %eax,%edx
  neg    %edx
  pxor   %xmm4,%xmm4
  cvtsi2sd %edx,%xmm4
  movq   %xmm4,%r14
  imul   %eax,%eax
  mov    (%rbx),%edx
  imul   0x8(%rbx),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  pxor   %xmm0,%xmm0
  ucomisd %xmm1,%xmm0
  ja     L16
  sqrtsd %xmm1,%xmm1
L37:
  movq   %r14,%xmm0
  subsd  %xmm1,%xmm0
  mov    (%rbx),%eax
  add    %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  cvtss2sd %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0x4(%rbp)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r14
  ret
L21:
  call   <sqrt@plt>
  jmp    L15
L16:
  movapd %xmm1,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  jmp    L37
```

assistant: ```c
void calcRoot(int *a, float *roots) {
  *roots = (-(*(a + 1)) + (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
  *(roots + 1) = (-(*(a + 1)) - (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calcRoot>:
  endbr64
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  pxor   %xmm5,%xmm5
  sub    $0x28,%rsp
  mov    0x4(%rdi),%eax
  mov    0x8(%rdi),%ecx
  mov    %eax,%edx
  imul   %eax,%eax
  neg    %edx
  cvtsi2sd %edx,%xmm1
  mov    (%rdi),%edx
  imul   %edx,%ecx
  shl    $0x2,%ecx
  sub    %ecx,%eax
  cvtsi2sd %eax,%xmm0
  ucomisd %xmm0,%xmm5
  ja     L1
  sqrtsd %xmm0,%xmm0
  pxor   %xmm3,%xmm3
  movapd %xmm1,%xmm2
  add    %edx,%edx
  cvtsi2ss %edx,%xmm3
  addsd  %xmm0,%xmm2
  cvtss2sd %xmm3,%xmm3
  divsd  %xmm3,%xmm2
  cvtsd2ss %xmm2,%xmm2
  movss  %xmm2,(%rsi)
L37:
  subsd  %xmm0,%xmm1
  divsd  %xmm3,%xmm1
  cvtsd2ss %xmm1,%xmm1
  movss  %xmm1,0x4(%rsi)
  add    $0x28,%rsp
  ret
L1:
  mov    %rsi,0x18(%rsp)
  mov    %rdi,0x10(%rsp)
  movsd  %xmm1,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x10(%rsp),%rdi
  pxor   %xmm2,%xmm2
  movsd  0x8(%rsp),%xmm1
  pxor   %xmm3,%xmm3
  movapd %xmm0,%xmm4
  pxor   %xmm0,%xmm0
  mov    0x18(%rsp),%rsi
  mov    0x4(%rdi),%eax
  mov    (%rdi),%edx
  addsd  %xmm4,%xmm1
  pxor   %xmm6,%xmm6
  mov    %eax,%ecx
  imul   %eax,%eax
  neg    %ecx
  cvtsi2sd %ecx,%xmm2
  mov    0x8(%rdi),%ecx
  imul   %edx,%ecx
  add    %edx,%edx
  cvtsi2ss %edx,%xmm3
  shl    $0x2,%ecx
  sub    %ecx,%eax
  cvtss2sd %xmm3,%xmm3
  divsd  %xmm3,%xmm1
  cvtsi2sd %eax,%xmm0
  ucomisd %xmm0,%xmm6
  cvtsd2ss %xmm1,%xmm1
  movss  %xmm1,(%rsi)
  jbe    L34
  mov    %rsi,0x18(%rsp)
  mov    %rdi,0x10(%rsp)
  movsd  %xmm2,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x10(%rsp),%rdi
  pxor   %xmm3,%xmm3
  movsd  0x8(%rsp),%xmm2
  mov    0x18(%rsp),%rsi
  mov    (%rdi),%eax
  movapd %xmm2,%xmm1
  add    %eax,%eax
  cvtsi2ss %eax,%xmm3
  cvtss2sd %xmm3,%xmm3
  jmp    L37
L34:
  sqrtsd %xmm0,%xmm0
  movapd %xmm2,%xmm1
  jmp    L37
```

assistant: ```c
void calcRoot(int *a, float *roots) {
  *roots = (-(*(a + 1)) + (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
  *(roots + 1) = (-(*(a + 1)) - (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
}
```
user: What is the c source code of the assembly code below?

```asm
<calcRoot>:
  endbr64
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  pxor   %xmm5,%xmm5
  sub    $0x28,%rsp
  mov    0x4(%rdi),%eax
  mov    0x8(%rdi),%ecx
  mov    %eax,%edx
  imul   %eax,%eax
  neg    %edx
  cvtsi2sd %edx,%xmm1
  mov    (%rdi),%edx
  imul   %edx,%ecx
  shl    $0x2,%ecx
  sub    %ecx,%eax
  cvtsi2sd %eax,%xmm0
  ucomisd %xmm0,%xmm5
  ja     L27
  sqrtsd %xmm0,%xmm0
  pxor   %xmm3,%xmm3
  movapd %xmm1,%xmm2
  add    %edx,%edx
  cvtsi2ss %edx,%xmm3
  addsd  %xmm0,%xmm2
  cvtss2sd %xmm3,%xmm3
  divsd  %xmm3,%xmm2
  cvtsd2ss %xmm2,%xmm2
  movss  %xmm2,(%rsi)
L0:
  subsd  %xmm0,%xmm1
  divsd  %xmm3,%xmm1
  cvtsd2ss %xmm1,%xmm1
  movss  %xmm1,0x4(%rsi)
  add    $0x28,%rsp
  ret
L27:
  mov    %rsi,0x18(%rsp)
  mov    %rdi,0x10(%rsp)
  movsd  %xmm1,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x10(%rsp),%rdi
  pxor   %xmm2,%xmm2
  movsd  0x8(%rsp),%xmm1
  pxor   %xmm3,%xmm3
  movapd %xmm0,%xmm4
  pxor   %xmm0,%xmm0
  mov    0x18(%rsp),%rsi
  mov    0x4(%rdi),%eax
  mov    (%rdi),%edx
  addsd  %xmm4,%xmm1
  pxor   %xmm6,%xmm6
  mov    %eax,%ecx
  imul   %eax,%eax
  neg    %ecx
  cvtsi2sd %ecx,%xmm2
  mov    0x8(%rdi),%ecx
  imul   %edx,%ecx
  add    %edx,%edx
  cvtsi2ss %edx,%xmm3
  shl    $0x2,%ecx
  sub    %ecx,%eax
  cvtss2sd %xmm3,%xmm3
  divsd  %xmm3,%xmm1
  cvtsi2sd %eax,%xmm0
  ucomisd %xmm0,%xmm6
  cvtsd2ss %xmm1,%xmm1
  movss  %xmm1,(%rsi)
  jbe    L5
  mov    %rsi,0x18(%rsp)
  mov    %rdi,0x10(%rsp)
  movsd  %xmm2,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x10(%rsp),%rdi
  pxor   %xmm3,%xmm3
  movsd  0x8(%rsp),%xmm2
  mov    0x18(%rsp),%rsi
  mov    (%rdi),%eax
  movapd %xmm2,%xmm1
  add    %eax,%eax
  cvtsi2ss %eax,%xmm3
  cvtss2sd %xmm3,%xmm3
  jmp    L0
L5:
  sqrtsd %xmm0,%xmm0
  movapd %xmm2,%xmm1
  jmp    L0
```

assistant: ```c
void calcRoot(int *a, float *roots) {
  *roots = (-(*(a + 1)) + (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
  *(roots + 1) = (-(*(a + 1)) - (sqrt(*(a + 1) * *(a + 1) - 4 * *(a) * *(a + 2)))) / (float)(2 * *(a));
}
```