ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rbx
L22:
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  mov    0xc(%rsp),%edx
  test   %eax,%eax
  js     L22
  cmp    %edx,%eax
  jg     L22
  mov    0x10(%rsp),%esi
  mov    %edx,%ecx
  or     %esi,%ecx
  js     L22
  sub    %eax,%edx
  mov    %edx,0xc(%rsp)
  imul   %eax,%esi
  sub    %eax,%edx
  js     L35
  mov    %esi,%r9d
  neg    %r9d
  add    %r9d,%r9d
  imul   $0x62,%esi,%edi
L10:
  mov    %edx,%r10d
  movslq %edi,%rcx
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x25,%rcx
  mov    %edi,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%ecx
  add    %ecx,%esi
  add    %r9d,%edi
  sub    %eax,%edx
  jns    L10
  mov    %r10d,0xc(%rsp)
L35:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D11
datatype: string

tool: D11: %d

assistant: ```c
int main() {
  int n, m, k, s = 0;
  do {
    scanf("%d", &n);
    scanf("%d", &m);
    scanf("%d", &k);
  } while (k > n || k < 0 || m < 0 || n < 0);
  n = n - k;
  s = k * m;
  for (int i = 2; n - k >= 0; i += 2) {
    n = n - k;
    s = s + k * m * (100 - i) / 100;
  }
  printf("%d", s);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    0x14(%rsp),%rbx
  nopl   0x0(%rax)
L51:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%edx
  mov    0xc(%rsp),%eax
  cmp    %eax,%edx
  jg     L51
  test   %edx,%edx
  js     L51
  mov    0x10(%rsp),%esi
  mov    %eax,%ecx
  or     %esi,%ecx
  js     L51
  sub    %edx,%eax
  imul   %edx,%esi
  mov    %eax,0xc(%rsp)
  sub    %edx,%eax
  js     L10
  mov    %esi,%r9d
  imul   $0x62,%esi,%edi
  neg    %r9d
  add    %r9d,%r9d
  nopl   0x0(%rax)
L1:
  movslq %edi,%rcx
  mov    %edi,%r8d
  mov    %eax,%r10d
  add    %r9d,%edi
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x1f,%r8d
  sar    $0x25,%rcx
  sub    %r8d,%ecx
  add    %ecx,%esi
  sub    %edx,%eax
  jns    L1
  mov    %r10d,0xc(%rsp)
L10:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D9
data_type: string

tool: D9: %d

assistant: ```c
int main() {
  int n, m, k, s = 0;
  do {
    scanf("%d", &n);
    scanf("%d", &m);
    scanf("%d", &k);
  } while (k > n || k < 0 || m < 0 || n < 0);
  n = n - k;
  s = k * m;
  for (int i = 2; n - k >= 0; i += 2) {
    n = n - k;
    s = s + k * m * (100 - i) / 100;
  }
  printf("%d", s);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    0x14(%rsp),%rbx
  nopl   0x0(%rax)
L3:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%edx
  mov    0xc(%rsp),%eax
  cmp    %eax,%edx
  jg     L3
  test   %edx,%edx
  js     L3
  mov    0x10(%rsp),%esi
  mov    %eax,%ecx
  or     %esi,%ecx
  js     L3
  sub    %edx,%eax
  imul   %edx,%esi
  mov    %eax,0xc(%rsp)
  sub    %edx,%eax
  js     L7
  mov    %esi,%r9d
  imul   $0x62,%esi,%edi
  neg    %r9d
  add    %r9d,%r9d
  nopl   0x0(%rax)
L24:
  movslq %edi,%rcx
  mov    %edi,%r8d
  mov    %eax,%r10d
  add    %r9d,%edi
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x1f,%r8d
  sar    $0x25,%rcx
  sub    %r8d,%ecx
  add    %ecx,%esi
  sub    %edx,%eax
  jns    L24
  mov    %r10d,0xc(%rsp)
L7:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D3
type: string

tool: D3: %d

assistant: ```c
int main() {
  int n, m, k, s = 0;
  do {
    scanf("%d", &n);
    scanf("%d", &m);
    scanf("%d", &k);
  } while (k > n || k < 0 || m < 0 || n < 0);
  n = n - k;
  s = k * m;
  for (int i = 2; n - k >= 0; i += 2) {
    n = n - k;
    s = s + k * m * (100 - i) / 100;
  }
  printf("%d", s);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rop3_216_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movzbl -0x1c(%rbp),%eax
  xor    -0x20(%rbp),%al
  mov    %al,-0x3(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x20(%rbp),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_216_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S ^ T;
  stk2 = *D & stk3;
  stk1 = T ^ stk2;
  *D = stk1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rop3_216_0_0>:
  endbr64
  xor    %edx,%esi
  and    (%rdi),%sil
  xor    %edx,%esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_216_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S ^ T;
  stk2 = *D & stk3;
  stk1 = T ^ stk2;
  *D = stk1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rop3_216_0_0>:
  endbr64
  xor    %edx,%esi
  and    (%rdi),%sil
  xor    %edx,%esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_216_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S ^ T;
  stk2 = *D & stk3;
  stk1 = T ^ stk2;
  *D = stk1;
}
```
user: What is the c source code of the assembly code below?

```asm
<rop3_216_0_0>:
  endbr64
  xor    %edx,%esi
  and    (%rdi),%sil
  xor    %edx,%esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_216_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S ^ T;
  stk2 = *D & stk3;
  stk1 = T ^ stk2;
  *D = stk1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<least_power_of_2_ge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L2
L10:
  mov    -0x8(%rbp),%rax
  xor    %rax,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%rax
  neg    %rax
  and    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jne    L10
  mov    -0x18(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  je     L21
  shlq   -0x10(%rbp)
L21:
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t least_power_of_2_ge(size_t size) {
  size_t remaining_bits, lowbit;
  remaining_bits = size;
  while ((lowbit = (remaining_bits & -remaining_bits)) != remaining_bits)
    remaining_bits ^= lowbit;
  if (size != remaining_bits)
    remaining_bits <<= 1;
  return remaining_bits;
}
```
user: What is the c source code of the assembly code below:

```asm
<least_power_of_2_ge>:
  endbr64
  mov    %rdi,%rdx
  neg    %rdx
  and    %rdi,%rdx
  cmp    %rdx,%rdi
  je     L3
  mov    %rdi,%rax
L0:
  xor    %rdx,%rax
  mov    %rax,%rdx
  neg    %rdx
  and    %rax,%rdx
  cmp    %rax,%rdx
  jne    L0
  lea    (%rax,%rax,1),%rdx
  cmp    %rax,%rdi
  cmovne %rdx,%rax
  ret
L3:
  mov    %rdx,%rax
  ret
```

assistant: ```c
size_t least_power_of_2_ge(size_t size) {
  size_t remaining_bits, lowbit;
  remaining_bits = size;
  while ((lowbit = (remaining_bits & -remaining_bits)) != remaining_bits)
    remaining_bits ^= lowbit;
  if (size != remaining_bits)
    remaining_bits <<= 1;
  return remaining_bits;
}
```
user: What is the c source code of the assembly code below:

```asm
<least_power_of_2_ge>:
  endbr64
  mov    %rdi,%rdx
  mov    %rdi,%rax
  neg    %rdx
  and    %rdi,%rdx
  cmp    %rdx,%rdi
  je     L0
  nopl   (%rax)
L19:
  xor    %rdx,%rax
  mov    %rax,%rdx
  neg    %rdx
  and    %rax,%rdx
  cmp    %rax,%rdx
  jne    L19
  lea    (%rax,%rax,1),%rdx
  cmp    %rax,%rdi
  cmovne %rdx,%rax
  ret
  nopl   (%rax)
L0:
  ret
```

assistant: ```c
size_t least_power_of_2_ge(size_t size) {
  size_t remaining_bits, lowbit;
  remaining_bits = size;
  while ((lowbit = (remaining_bits & -remaining_bits)) != remaining_bits)
    remaining_bits ^= lowbit;
  if (size != remaining_bits)
    remaining_bits <<= 1;
  return remaining_bits;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<least_power_of_2_ge>:
  endbr64
  mov    %rdi,%rdx
  mov    %rdi,%rax
  neg    %rdx
  and    %rdi,%rdx
  cmp    %rdx,%rdi
  je     L12
  nopl   (%rax)
L10:
  xor    %rdx,%rax
  mov    %rax,%rdx
  neg    %rdx
  and    %rax,%rdx
  cmp    %rax,%rdx
  jne    L10
  lea    (%rax,%rax,1),%rdx
  cmp    %rax,%rdi
  cmovne %rdx,%rax
  ret
  nopl   (%rax)
L12:
  ret
```

assistant: ```c
size_t least_power_of_2_ge(size_t size) {
  size_t remaining_bits, lowbit;
  remaining_bits = size;
  while ((lowbit = (remaining_bits & -remaining_bits)) != remaining_bits)
    remaining_bits ^= lowbit;
  if (size != remaining_bits)
    remaining_bits <<= 1;
  return remaining_bits;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<addressToWord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void addressToWord(char *dest, char *src) {
  int addr_trans = (int)src;
  dest[3] = (char)(addr_trans & 0x000000FF);
  dest[2] = (char)((addr_trans & 0x0000FF00) >> 8);
  dest[1] = (char)((addr_trans & 0x00FF0000) >> 16);
  dest[0] = (char)((addr_trans & 0xFF000000) >> 24);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<addressToWord>:
  endbr64
  mov    %rsi,%rax
  mov    %sil,0x3(%rdi)
  movzbl %ah,%ecx
  mov    %cl,0x2(%rdi)
  mov    %esi,%edx
  sar    $0x10,%edx
  mov    %dl,0x1(%rdi)
  shr    $0x18,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void addressToWord(char *dest, char *src) {
  int addr_trans = (int)src;
  dest[3] = (char)(addr_trans & 0x000000FF);
  dest[2] = (char)((addr_trans & 0x0000FF00) >> 8);
  dest[1] = (char)((addr_trans & 0x00FF0000) >> 16);
  dest[0] = (char)((addr_trans & 0xFF000000) >> 24);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<addressToWord>:
  endbr64
  bswap  %esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void addressToWord(char *dest, char *src) {
  int addr_trans = (int)src;
  dest[3] = (char)(addr_trans & 0x000000FF);
  dest[2] = (char)((addr_trans & 0x0000FF00) >> 8);
  dest[1] = (char)((addr_trans & 0x00FF0000) >> 16);
  dest[0] = (char)((addr_trans & 0xFF000000) >> 24);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<addressToWord>:
  endbr64
  bswap  %esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void addressToWord(char *dest, char *src) {
  int addr_trans = (int)src;
  dest[3] = (char)(addr_trans & 0x000000FF);
  dest[2] = (char)((addr_trans & 0x0000FF00) >> 8);
  dest[1] = (char)((addr_trans & 0x00FF0000) >> 16);
  dest[0] = (char)((addr_trans & 0xFF000000) >> 24);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ft_sqrt_d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L7
  pxor   %xmm0,%xmm0
  jmp    L21
L7:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <ft_sqrt@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%esi
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  add    %esi,%eax
  imul   %eax,%ecx
  mov    %ecx,%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  add    %edx,%eax
  mov    %eax,%eax
  test   %rax,%rax
  js     L36
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L41
L36:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L41:
  movsd  %xmm0,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  imul   -0xc(%rbp),%eax
  shl    $0x2,%eax
  mov    %eax,%eax
  test   %rax,%rax
  js     L46
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L58
L46:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L58:
  movsd  -0x8(%rbp),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
L21:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double ft_sqrt_d(unsigned int n) {
  double res;
  unsigned int s;
  if (!n)
    return (0);
  s = ft_sqrt(n);
  res = (s * s) * ((s * s) + 6 * n) + (n * n);
  res /= (4 * s) * ((s * s) + n);
  return (res);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_sqrt_d>:
  endbr64
  pxor   %xmm0,%xmm0
  test   %edi,%edi
  jne    L2
  ret
L2:
  push   %rbx
  mov    %edi,%ebx
  call   <ft_sqrt@plt>
  mov    %eax,%edx
  imul   %eax,%edx
  lea    (%rbx,%rbx,2),%ecx
  lea    (%rdx,%rcx,2),%ecx
  imul   %edx,%ecx
  mov    %ebx,%esi
  imul   %ebx,%esi
  lea    (%rcx,%rsi,1),%ecx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rcx,%xmm0
  add    %edx,%ebx
  imul   %eax,%ebx
  shl    $0x2,%ebx
  mov    %ebx,%ebx
  pxor   %xmm1,%xmm1
  cvtsi2sd %rbx,%xmm1
  divsd  %xmm1,%xmm0
  pop    %rbx
  ret
```

assistant: ```c
double ft_sqrt_d(unsigned int n) {
  double res;
  unsigned int s;
  if (!n)
    return (0);
  s = ft_sqrt(n);
  res = (s * s) * ((s * s) + 6 * n) + (n * n);
  res /= (4 * s) * ((s * s) + n);
  return (res);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_sqrt_d>:
  endbr64
  test   %edi,%edi
  jne    L8
  pxor   %xmm0,%xmm0
  ret
  nopl   (%rax)
L8:
  push   %rbx
  mov    %edi,%ebx
  call   <ft_sqrt@plt>
  lea    (%rbx,%rbx,2),%ecx
  mov    %ebx,%esi
  pxor   %xmm0,%xmm0
  mov    %eax,%edx
  imul   %ebx,%esi
  pxor   %xmm1,%xmm1
  imul   %eax,%edx
  lea    (%rdx,%rcx,2),%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  imul   %eax,%ebx
  add    %esi,%ecx
  shl    $0x2,%ebx
  mov    %ecx,%ecx
  cvtsi2sd %rbx,%xmm1
  pop    %rbx
  cvtsi2sd %rcx,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double ft_sqrt_d(unsigned int n) {
  double res;
  unsigned int s;
  if (!n)
    return (0);
  s = ft_sqrt(n);
  res = (s * s) * ((s * s) + 6 * n) + (n * n);
  res /= (4 * s) * ((s * s) + n);
  return (res);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_sqrt_d>:
  endbr64
  test   %edi,%edi
  jne    L1
  pxor   %xmm0,%xmm0
  ret
  nopl   (%rax)
L1:
  push   %rbx
  mov    %edi,%ebx
  call   <ft_sqrt@plt>
  lea    (%rbx,%rbx,2),%ecx
  mov    %ebx,%esi
  pxor   %xmm0,%xmm0
  mov    %eax,%edx
  imul   %ebx,%esi
  pxor   %xmm1,%xmm1
  imul   %eax,%edx
  lea    (%rdx,%rcx,2),%ecx
  add    %edx,%ebx
  imul   %edx,%ecx
  imul   %eax,%ebx
  add    %esi,%ecx
  shl    $0x2,%ebx
  mov    %ecx,%ecx
  cvtsi2sd %rbx,%xmm1
  pop    %rbx
  cvtsi2sd %rcx,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double ft_sqrt_d(unsigned int n) {
  double res;
  unsigned int s;
  if (!n)
    return (0);
  s = ft_sqrt(n);
  res = (s * s) * ((s * s) + 6 * n) + (n * n);
  res /= (4 * s) * ((s * s) + n);
  return (res);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<calculate_double_hash_key>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  movq   $0x1,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L42
L12:
  fildll -0x20(%rbp)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   <is_prime@plt>
  add    $0x10,%rsp
  cmp    $0x1,%eax
  jne    L23
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L34
L23:
  subq   $0x1,-0x20(%rbp)
L42:
  cmpq   $0x0,-0x20(%rbp)
  jg     L12
L34:
  movq   $0x0,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x21(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L1
L51:
  movsbq -0x21(%rbp),%rax
  add    %rax,-0x10(%rbp)
  addq   $0x1,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x21(%rbp)
L1:
  cmpb   $0x0,-0x21(%rbp)
  jne    L51
  mov    -0x10(%rbp),%rax
  cqto
  idivq  -0x18(%rbp)
  mov    -0x18(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long int calculate_double_hash_key(wchar_t *str, long int hash_table_length) {
  long int i, R, second_hash_key, str_sum;
  wchar_t c;
  R = 1;
  for (i = hash_table_length - 1; i > 0; i--) {
    if (is_prime((long double)i) == 1) {
      R = i;
      break;
    }
  }
  i = 0;
  c = str[i];
  str_sum = 0;
  while (c != '\0') {
    str_sum += (long int)c;
    i++;
    c = str[i];
  }
  second_hash_key = R - (str_sum % R);
  return second_hash_key;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<calculate_double_hash_key>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%rbx
  test   %rbx,%rbx
  jle    L46
L65:
  mov    %rbx,0x8(%rsp)
  fildll 0x8(%rsp)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   <is_prime@plt>
  add    $0x10,%rsp
  cmp    $0x1,%eax
  je     L33
  sub    $0x1,%rbx
  jne    L65
  mov    $0x1,%ebx
L33:
  movzbl 0x0(%rbp),%edx
  test   %dl,%dl
  je     L68
  lea    0x1(%rbp),%rdi
  mov    $0x0,%eax
L37:
  movsbq %dl,%rdx
  add    %rdx,%rax
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%edx
  test   %dl,%dl
  jne    L37
L27:
  cqto
  idiv   %rbx
  mov    %rbx,%rax
  sub    %rdx,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L46:
  mov    $0x1,%ebx
  jmp    L33
L68:
  mov    $0x0,%eax
  jmp    L27
```

assistant: ```c
long int calculate_double_hash_key(wchar_t *str, long int hash_table_length) {
  long int i, R, second_hash_key, str_sum;
  wchar_t c;
  R = 1;
  for (i = hash_table_length - 1; i > 0; i--) {
    if (is_prime((long double)i) == 1) {
      R = i;
      break;
    }
  }
  i = 0;
  c = str[i];
  str_sum = 0;
  while (c != '\0') {
    str_sum += (long int)c;
    i++;
    c = str[i];
  }
  second_hash_key = R - (str_sum % R);
  return second_hash_key;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<calculate_double_hash_key>:
  endbr64
  push   %r12
  lea    -0x1(%rsi),%r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  test   %r12,%r12
  jg     L53
  jmp    L43
  nopl   0x0(%rax)
L39:
  sub    $0x1,%r12
  je     L43
L53:
  mov    %r12,0x8(%rsp)
  fildll 0x8(%rsp)
  sub    $0x10,%rsp
  fstpt  (%rsp)
  call   <is_prime@plt>
  pop    %rdx
  pop    %rcx
  cmp    $0x1,%eax
  jne    L39
L14:
  movsbq (%rbx),%rdx
  test   %dl,%dl
  je     L1
  lea    0x1(%rbx),%rdi
  xor    %eax,%eax
L40:
  add    $0x1,%rdi
  add    %rdx,%rax
  movsbq -0x1(%rdi),%rdx
  test   %dl,%dl
  jne    L40
  cqto
  idiv   %r12
  sub    %rdx,%r12
L1:
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  mov    $0x1,%r12d
  jmp    L14
```

assistant: ```c
long int calculate_double_hash_key(wchar_t *str, long int hash_table_length) {
  long int i, R, second_hash_key, str_sum;
  wchar_t c;
  R = 1;
  for (i = hash_table_length - 1; i > 0; i--) {
    if (is_prime((long double)i) == 1) {
      R = i;
      break;
    }
  }
  i = 0;
  c = str[i];
  str_sum = 0;
  while (c != '\0') {
    str_sum += (long int)c;
    i++;
    c = str[i];
  }
  second_hash_key = R - (str_sum % R);
  return second_hash_key;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calculate_double_hash_key>:
  endbr64
  push   %r12
  lea    -0x1(%rsi),%r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  test   %r12,%r12
  jg     L49
  jmp    L5
  nopl   0x0(%rax)
L39:
  sub    $0x1,%r12
  je     L5
L49:
  mov    %r12,0x8(%rsp)
  fildll 0x8(%rsp)
  sub    $0x10,%rsp
  fstpt  (%rsp)
  call   <is_prime@plt>
  pop    %rdx
  pop    %rcx
  cmp    $0x1,%eax
  jne    L39
L67:
  movsbq (%rbx),%rdx
  test   %dl,%dl
  je     L52
  lea    0x1(%rbx),%rdi
  xor    %eax,%eax
L56:
  add    $0x1,%rdi
  add    %rdx,%rax
  movsbq -0x1(%rdi),%rdx
  test   %dl,%dl
  jne    L56
  cqto
  idiv   %r12
  sub    %rdx,%r12
L52:
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    $0x1,%r12d
  jmp    L67
```

assistant: ```c
long int calculate_double_hash_key(wchar_t *str, long int hash_table_length) {
  long int i, R, second_hash_key, str_sum;
  wchar_t c;
  R = 1;
  for (i = hash_table_length - 1; i > 0; i--) {
    if (is_prime((long double)i) == 1) {
      R = i;
      break;
    }
  }
  i = 0;
  c = str[i];
  str_sum = 0;
  while (c != '\0') {
    str_sum += (long int)c;
    i++;
    c = str[i];
  }
  second_hash_key = R - (str_sum % R);
  return second_hash_key;
}
```