ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L9:
  mov    %rbx,%rdi
  call   <xingenc_read_stat@plt>
  test   %rax,%rax
  jne    L8
  movsd  (%rsp),%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
L8:
  mov    D18(%rip),%rax
  movq   %rax,%xmm0
  call   <usleep@plt>
  jmp    L9
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
data_type: string
name: extract_symbolic_data
arguments: label: D18
data_type: f32[2]

tool: D7: [P %f]


tool: {"D18": [0.0,9.9073486328125]}

assistant: ```c
int main(int argc, char **argv) {
  double current = 0;
  while (1) {
    if (xingenc_read_stat(&current) == 0)
      printf("[P %f]\n", current);
    usleep(0.5 * 1000000);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rc6_generateKeySchedule>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,-0x78(%rbp)
  mov    %esi,-0x7c(%rbp)
  mov    %rdx,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    -0x60(%rbp),%rax
  mov    %rax,-0x68(%rbp)
  mov    -0x7c(%rbp),%edx
  mov    -0x78(%rbp),%rcx
  mov    -0x68(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x88(%rbp),%rax
  movl   $0xb7e15163,(%rax)
  mov    $0x1,%ebx
  jmp    L37
L43:
  lea    -0x1(%rbx),%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    %ebx,%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0x88(%rbp),%rax
  add    %rcx,%rax
  sub    $0x61c88647,%edx
  mov    %edx,(%rax)
  add    $0x1,%ebx
L37:
  cmp    $0x2b,%ebx
  jbe    L43
  mov    $0x0,%r14d
  mov    %r14d,%ebx
  mov    %ebx,%r12d
  mov    %r12d,%r15d
  mov    $0x84,%esi
  shrl   $0x2,-0x7c(%rbp)
  mov    $0x1,%r13d
  jmp    L1
L33:
  mov    %ebx,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %r15d,%eax
  lea    (%r12,%rax,1),%edx
  mov    %ebx,%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0x88(%rbp),%rax
  add    %rcx,%rax
  rol    $0x3,%edx
  mov    %edx,(%rax)
  mov    (%rax),%r15d
  mov    %r14d,%eax
  mov    -0x60(%rbp,%rax,4),%eax
  add    %r15d,%eax
  lea    (%r12,%rax,1),%edx
  lea    (%r15,%r12,1),%eax
  and    $0x1f,%eax
  mov    %eax,%ecx
  rol    %cl,%edx
  mov    %edx,%eax
  mov    %r14d,%edx
  mov    %eax,-0x60(%rbp,%rdx,4)
  mov    %r14d,%eax
  mov    -0x60(%rbp,%rax,4),%r12d
  lea    0x1(%rbx),%eax
  mov    %eax,%ecx
  mov    $0xba2e8ba3,%edx
  imul   %rcx,%rdx
  shr    $0x20,%rdx
  mov    %edx,%ebx
  shr    $0x5,%ebx
  imul   $0x2c,%ebx,%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  lea    0x1(%r14),%eax
  mov    $0x0,%edx
  divl   -0x7c(%rbp)
  mov    %edx,%r14d
  add    $0x1,%r13d
L1:
  mov    %esi,%eax
  cmp    %eax,%r13d
  jbe    L33
  nop
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  add    $0x68,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void rc6_generateKeySchedule(unsigned char *initKey, unsigned int keyLength, unsigned int S[]) {
  unsigned int L[8];
  register unsigned int A, B, i, j, s, v;
  unsigned char *bPtr = (unsigned char *)L;
  memcpy(bPtr, initKey, keyLength);
  S[0] = 0xB7E15163;
  for (i = 1; i < 44; i++)
    S[i] = S[i - 1] + 0x9E3779B9;
  A = B = i = j = 0;
  v = 132;
  keyLength >>= 2;
  for (s = 1; s <= v; s++) {
    A = S[i] = (((S[i] + A + B) << ((3) & 0x1F)) | ((S[i] + A + B) >> (32 - ((3) & 0x1F))));
    B = L[j] = (((L[j] + A + B) << ((A + B) & 0x1F)) | ((L[j] + A + B) >> (32 - ((A + B) & 0x1F))));
    i = (i + 1) % 44;
    j = (j + 1) % keyLength;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rc6_generateKeySchedule>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %esi,%ebp
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %esi,%edx
  mov    %rdi,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0xb7e15163,(%rbx)
  mov    %rbx,%rax
  lea    0xac(%rbx),%rcx
L1:
  mov    (%rax),%edx
  sub    $0x61c88647,%edx
  mov    %edx,0x4(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L1
  mov    %ebp,%r9d
  shr    $0x2,%r9d
  mov    $0x84,%r8d
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%ecx
  mov    $0x0,%edi
  mov    $0xba2e8ba3,%r10d
L29:
  mov    %esi,%eax
  lea    (%rbx,%rax,4),%r11
  mov    %ecx,%eax
  add    (%r11),%eax
  add    %edi,%eax
  rol    $0x3,%eax
  mov    %eax,%edi
  mov    %eax,(%r11)
  mov    %edx,%r11d
  mov    %ecx,%eax
  add    (%rsp,%r11,4),%eax
  add    %edi,%eax
  add    %edi,%ecx
  rol    %cl,%eax
  mov    %eax,%ecx
  mov    %eax,(%rsp,%r11,4)
  lea    0x1(%rsi),%eax
  mov    %eax,%esi
  imul   %r10,%rsi
  shr    $0x25,%rsi
  imul   $0x2c,%esi,%r11d
  sub    %r11d,%eax
  mov    %eax,%esi
  lea    0x1(%rdx),%eax
  mov    $0x0,%edx
  div    %r9d
  sub    $0x1,%r8d
  jne    L29
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void rc6_generateKeySchedule(unsigned char *initKey, unsigned int keyLength, unsigned int S[]) {
  unsigned int L[8];
  register unsigned int A, B, i, j, s, v;
  unsigned char *bPtr = (unsigned char *)L;
  memcpy(bPtr, initKey, keyLength);
  S[0] = 0xB7E15163;
  for (i = 1; i < 44; i++)
    S[i] = S[i - 1] + 0x9E3779B9;
  A = B = i = j = 0;
  v = 132;
  keyLength >>= 2;
  for (s = 1; s <= v; s++) {
    A = S[i] = (((S[i] + A + B) << ((3) & 0x1F)) | ((S[i] + A + B) >> (32 - ((3) & 0x1F))));
    B = L[j] = (((L[j] + A + B) << ((A + B) & 0x1F)) | ((L[j] + A + B) >> (32 - ((A + B) & 0x1F))));
    i = (i + 1) % 44;
    j = (j + 1) % keyLength;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rc6_generateKeySchedule>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  mov    %esi,%edx
  mov    %r8,%rsi
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memcpy@plt>
  movl   $0xb7e15163,(%rbx)
  lea    0x4(%rbx),%rax
  lea    0xac(%rbx),%rcx
  mov    $0xb7e15163,%edx
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L24:
  add    $0x4,%rax
L9:
  sub    $0x61c88647,%edx
  mov    %edx,(%rax)
  cmp    %rax,%rcx
  jne    L24
  mov    %ebp,%r9d
  mov    $0x84,%r8d
  xor    %edx,%edx
  xor    %esi,%esi
  shr    $0x2,%r9d
  xor    %ecx,%ecx
  xor    %edi,%edi
  mov    $0xba2e8ba3,%r10d
  nopl   0x0(%rax)
L19:
  mov    %esi,%eax
  add    $0x1,%esi
  lea    (%rbx,%rax,4),%r11
  mov    (%r11),%eax
  add    %ecx,%eax
  add    %edi,%eax
  rol    $0x3,%eax
  mov    %eax,(%r11)
  mov    %edx,%r11d
  mov    %eax,%edi
  mov    (%rsp,%r11,4),%eax
  add    %ecx,%eax
  add    %edi,%ecx
  add    %edi,%eax
  rol    %cl,%eax
  mov    %eax,(%rsp,%r11,4)
  mov    %eax,%ecx
  mov    %rsi,%rax
  imul   %r10,%rsi
  shr    $0x25,%rsi
  imul   $0x2c,%esi,%r11d
  sub    %r11d,%eax
  mov    %eax,%esi
  lea    0x1(%rdx),%eax
  xor    %edx,%edx
  div    %r9d
  sub    $0x1,%r8d
  jne    L19
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void rc6_generateKeySchedule(unsigned char *initKey, unsigned int keyLength, unsigned int S[]) {
  unsigned int L[8];
  register unsigned int A, B, i, j, s, v;
  unsigned char *bPtr = (unsigned char *)L;
  memcpy(bPtr, initKey, keyLength);
  S[0] = 0xB7E15163;
  for (i = 1; i < 44; i++)
    S[i] = S[i - 1] + 0x9E3779B9;
  A = B = i = j = 0;
  v = 132;
  keyLength >>= 2;
  for (s = 1; s <= v; s++) {
    A = S[i] = (((S[i] + A + B) << ((3) & 0x1F)) | ((S[i] + A + B) >> (32 - ((3) & 0x1F))));
    B = L[j] = (((L[j] + A + B) << ((A + B) & 0x1F)) | ((L[j] + A + B) >> (32 - ((A + B) & 0x1F))));
    i = (i + 1) % 44;
    j = (j + 1) % keyLength;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rc6_generateKeySchedule>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  mov    %esi,%edx
  mov    %r8,%rsi
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memcpy@plt>
  movl   $0xb7e15163,(%rbx)
  lea    0x4(%rbx),%rax
  lea    0xb0(%rbx),%rcx
  mov    $0xb7e15163,%edx
  nopl   (%rax)
L6:
  sub    $0x61c88647,%edx
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rax,%rcx
  jne    L6
  mov    %ebp,%r9d
  mov    $0x84,%r8d
  xor    %edx,%edx
  xor    %esi,%esi
  shr    $0x2,%r9d
  xor    %ecx,%ecx
  xor    %edi,%edi
  mov    $0xba2e8ba3,%r10d
  nopl   (%rax)
L24:
  mov    %esi,%eax
  add    $0x1,%esi
  lea    (%rbx,%rax,4),%r11
  mov    (%r11),%eax
  add    %ecx,%eax
  add    %edi,%eax
  rol    $0x3,%eax
  mov    %eax,(%r11)
  mov    %edx,%r11d
  mov    %eax,%edi
  mov    (%rsp,%r11,4),%eax
  add    %ecx,%eax
  add    %edi,%ecx
  add    %edi,%eax
  rol    %cl,%eax
  mov    %eax,(%rsp,%r11,4)
  mov    %eax,%ecx
  mov    %rsi,%rax
  imul   %r10,%rsi
  shr    $0x25,%rsi
  imul   $0x2c,%esi,%r11d
  sub    %r11d,%eax
  mov    %eax,%esi
  lea    0x1(%rdx),%eax
  xor    %edx,%edx
  div    %r9d
  sub    $0x1,%r8d
  jne    L24
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void rc6_generateKeySchedule(unsigned char *initKey, unsigned int keyLength, unsigned int S[]) {
  unsigned int L[8];
  register unsigned int A, B, i, j, s, v;
  unsigned char *bPtr = (unsigned char *)L;
  memcpy(bPtr, initKey, keyLength);
  S[0] = 0xB7E15163;
  for (i = 1; i < 44; i++)
    S[i] = S[i - 1] + 0x9E3779B9;
  A = B = i = j = 0;
  v = 132;
  keyLength >>= 2;
  for (s = 1; s <= v; s++) {
    A = S[i] = (((S[i] + A + B) << ((3) & 0x1F)) | ((S[i] + A + B) >> (32 - ((3) & 0x1F))));
    B = L[j] = (((L[j] + A + B) << ((A + B) & 0x1F)) | ((L[j] + A + B) >> (32 - ((A + B) & 0x1F))));
    i = (i + 1) % 44;
    j = (j + 1) % keyLength;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ipow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L17
L16:
  mov    -0x4(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L17:
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  test   %eax,%eax
  jne    L16
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned ipow(unsigned b, unsigned e) {
  unsigned r = 1;
  while (e-- > 0)
    r *= b;
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ipow>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  je     L19
  mov    $0x1,%edx
L14:
  imul   %edi,%edx
  sub    $0x1,%eax
  cmp    $0xffffffff,%eax
  jne    L14
L20:
  mov    %edx,%eax
  ret
L19:
  mov    $0x1,%edx
  jmp    L20
```

assistant: ```c
unsigned ipow(unsigned b, unsigned e) {
  unsigned r = 1;
  while (e-- > 0)
    r *= b;
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ipow>:
  endbr64
  lea    -0x1(%rsi),%eax
  mov    $0x1,%r8d
  test   %esi,%esi
  je     L10
  nopl   0x0(%rax)
L14:
  imul   %edi,%r8d
  sub    $0x1,%eax
  jae    L14
L10:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned ipow(unsigned b, unsigned e) {
  unsigned r = 1;
  while (e-- > 0)
    r *= b;
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ipow>:
  endbr64
  lea    -0x1(%rsi),%eax
  mov    $0x1,%r8d
  test   %esi,%esi
  je     L5
  nopl   0x0(%rax)
L7:
  imul   %edi,%r8d
  sub    $0x1,%eax
  jae    L7
L5:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned ipow(unsigned b, unsigned e) {
  unsigned r = 1;
  while (e-- > 0)
    r *= b;
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<step2>:
  endbr64
  mov    %rdx,%rsi
  mov    %r8,%rdx
  mov    0x8(%rsp),%r8
  mov    0x10(%rsp),%rdi
  mov    $0x0,%r9d
  jmp    L39
L29:
  mov    %r9d,%eax
  and    $0x7,%eax
  mov    %r9d,%ecx
  and    $0xfffffff8,%ecx
  movslq (%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    (%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x4(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x42(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x8(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x84(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0xc(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0xc6(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x10(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x108(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x14(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x14a(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x18(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x18c(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x1c(%rdi),%r10
  mov    (%rdx,%r10,4),%r10d
  lea    0x1ce(%rcx,%rax,1),%eax
  cltq
  mov    %r10d,(%r8,%rax,4)
  add    $0x1,%r9d
  add    $0x20,%rsi
  cmp    $0x40,%r9d
  je     L23
L39:
  mov    $0x0,%eax
L33:
  mov    (%rsi,%rax,1),%ecx
  mov    %ecx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    $0x20,%rax
  jne    L33
  jmp    L29
L23:
  ret
```

assistant: ```c
void step2(int work_x[], int work_y[], int DATA_x[], int DATA_y[], int data_x[], int data_y[], int smem[], int reversed[], float sin_64[448], float cos_64[448], float sin_512[448], float cos_512[448]) {
  int tid, hi, lo, i, j, stride;
  stride = 64;
outer:
  for (tid = 0; tid < 64; tid++) {
  load:
    for (i = 0; i < 8; i++) {
      data_x[i] = DATA_x[tid * 8 + i];
    }
    hi = tid >> 3;
    lo = tid & 7;
    {
      smem[0 * 66 + hi * 8 + lo] = data_x[reversed[0]];
      smem[1 * 66 + hi * 8 + lo] = data_x[reversed[1]];
      smem[2 * 66 + hi * 8 + lo] = data_x[reversed[2]];
      smem[3 * 66 + hi * 8 + lo] = data_x[reversed[3]];
      smem[4 * 66 + hi * 8 + lo] = data_x[reversed[4]];
      smem[5 * 66 + hi * 8 + lo] = data_x[reversed[5]];
      smem[6 * 66 + hi * 8 + lo] = data_x[reversed[6]];
      smem[7 * 66 + hi * 8 + lo] = data_x[reversed[7]];
    };
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<step2>:
  endbr64
  mov    %rdx,%rsi
  mov    0x10(%rsp),%rdi
  mov    %r8,%rdx
  mov    0x8(%rsp),%r8
  xor    %r9d,%r9d
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L15:
  mov    (%rsi,%rax,1),%ecx
  mov    %ecx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    $0x20,%rax
  jne    L15
  movslq (%rdi),%r10
  mov    %r9d,%eax
  mov    %r9d,%ecx
  add    $0x1,%r9d
  and    $0x7,%eax
  and    $0xfffffff8,%ecx
  add    $0x20,%rsi
  mov    (%rdx,%r10,4),%r11d
  lea    (%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x4(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x42(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x8(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x84(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0xc(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0xc6(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x10(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x108(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x14(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x14a(%rcx,%rax,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x18(%rdi),%r10
  mov    (%rdx,%r10,4),%r11d
  lea    0x18c(%rcx,%rax,1),%r10d
  lea    0x1ce(%rcx,%rax,1),%eax
  movslq %r10d,%r10
  cltq
  mov    %r11d,(%r8,%r10,4)
  movslq 0x1c(%rdi),%r10
  mov    (%rdx,%r10,4),%r10d
  mov    %r10d,(%r8,%rax,4)
  cmp    $0x40,%r9d
  jne    L17
  ret
```

assistant: ```c
void step2(int work_x[], int work_y[], int DATA_x[], int DATA_y[], int data_x[], int data_y[], int smem[], int reversed[], float sin_64[448], float cos_64[448], float sin_512[448], float cos_512[448]) {
  int tid, hi, lo, i, j, stride;
  stride = 64;
outer:
  for (tid = 0; tid < 64; tid++) {
  load:
    for (i = 0; i < 8; i++) {
      data_x[i] = DATA_x[tid * 8 + i];
    }
    hi = tid >> 3;
    lo = tid & 7;
    {
      smem[0 * 66 + hi * 8 + lo] = data_x[reversed[0]];
      smem[1 * 66 + hi * 8 + lo] = data_x[reversed[1]];
      smem[2 * 66 + hi * 8 + lo] = data_x[reversed[2]];
      smem[3 * 66 + hi * 8 + lo] = data_x[reversed[3]];
      smem[4 * 66 + hi * 8 + lo] = data_x[reversed[4]];
      smem[5 * 66 + hi * 8 + lo] = data_x[reversed[5]];
      smem[6 * 66 + hi * 8 + lo] = data_x[reversed[6]];
      smem[7 * 66 + hi * 8 + lo] = data_x[reversed[7]];
    };
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<step2>:
  endbr64
  mov    %r8,%rax
  mov    0x10(%rsp),%rdi
  mov    0x8(%rsp),%r8
  xor    %r9d,%r9d
  nopl   0x0(%rax)
L8:
  mov    (%rdx),%ecx
  mov    %r9d,%esi
  add    $0x20,%rdx
  and    $0xfffffff8,%esi
  mov    %ecx,(%rax)
  mov    -0x1c(%rdx),%ecx
  mov    %ecx,0x4(%rax)
  mov    -0x18(%rdx),%ecx
  mov    %ecx,0x8(%rax)
  mov    -0x14(%rdx),%ecx
  mov    %ecx,0xc(%rax)
  mov    -0x10(%rdx),%ecx
  mov    %ecx,0x10(%rax)
  mov    -0xc(%rdx),%ecx
  mov    %ecx,0x14(%rax)
  mov    -0x8(%rdx),%ecx
  mov    %ecx,0x18(%rax)
  mov    -0x4(%rdx),%ecx
  mov    %ecx,0x1c(%rax)
  mov    %r9d,%ecx
  movslq (%rdi),%r10
  add    $0x1,%r9d
  and    $0x7,%ecx
  mov    (%rax,%r10,4),%r11d
  lea    (%rsi,%rcx,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x4(%rdi),%r10
  mov    (%rax,%r10,4),%r11d
  lea    0x42(%rsi,%rcx,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x8(%rdi),%r10
  mov    (%rax,%r10,4),%r11d
  lea    0x84(%rsi,%rcx,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0xc(%rdi),%r10
  mov    (%rax,%r10,4),%r11d
  lea    0xc6(%rsi,%rcx,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x10(%rdi),%r10
  mov    (%rax,%r10,4),%r11d
  lea    0x108(%rsi,%rcx,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x14(%rdi),%r10
  mov    (%rax,%r10,4),%r11d
  lea    0x14a(%rsi,%rcx,1),%r10d
  movslq %r10d,%r10
  mov    %r11d,(%r8,%r10,4)
  movslq 0x18(%rdi),%r10
  mov    (%rax,%r10,4),%r11d
  lea    0x18c(%rsi,%rcx,1),%r10d
  lea    0x1ce(%rsi,%rcx,1),%ecx
  movslq %r10d,%r10
  movslq %ecx,%rcx
  mov    %r11d,(%r8,%r10,4)
  movslq 0x1c(%rdi),%r10
  mov    (%rax,%r10,4),%r10d
  mov    %r10d,(%r8,%rcx,4)
  cmp    $0x40,%r9d
  jne    L8
  ret
```

assistant: ```c
void step2(int work_x[], int work_y[], int DATA_x[], int DATA_y[], int data_x[], int data_y[], int smem[], int reversed[], float sin_64[448], float cos_64[448], float sin_512[448], float cos_512[448]) {
  int tid, hi, lo, i, j, stride;
  stride = 64;
outer:
  for (tid = 0; tid < 64; tid++) {
  load:
    for (i = 0; i < 8; i++) {
      data_x[i] = DATA_x[tid * 8 + i];
    }
    hi = tid >> 3;
    lo = tid & 7;
    {
      smem[0 * 66 + hi * 8 + lo] = data_x[reversed[0]];
      smem[1 * 66 + hi * 8 + lo] = data_x[reversed[1]];
      smem[2 * 66 + hi * 8 + lo] = data_x[reversed[2]];
      smem[3 * 66 + hi * 8 + lo] = data_x[reversed[3]];
      smem[4 * 66 + hi * 8 + lo] = data_x[reversed[4]];
      smem[5 * 66 + hi * 8 + lo] = data_x[reversed[5]];
      smem[6 * 66 + hi * 8 + lo] = data_x[reversed[6]];
      smem[7 * 66 + hi * 8 + lo] = data_x[reversed[7]];
    };
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <hotel_stack@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  hotel_stack();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <hotel_stack@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  hotel_stack();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <hotel_stack@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  hotel_stack();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <hotel_stack@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  hotel_stack();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D63(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D63
type_guessed: string
name: retrieve_data
arguments: data_label: D7
type_guessed: string
name: retrieve_data
arguments: data_label: D42
type_guessed: string
name: retrieve_data
arguments: data_label: D34
type_guessed: string
name: retrieve_data
arguments: data_label: D4
type_guessed: string
name: retrieve_data
arguments: data_label: D36
type_guessed: string
name: retrieve_data
arguments: data_label: D57
type_guessed: string
name: retrieve_data
arguments: data_label: D70
type_guessed: string

tool: D63: usage: iopp -h|--help


tool: D7: usage: iopp [-ci] [-k|-m] [delay [count]]


tool: D42:             -c, --command display full command line


tool: D34:             -h, --help display help


tool: D4:             -i, --idle hides idle processes


tool: D36:             -k, --kilobytes display data in kilobytes


tool: D57:             -m, --megabytes display data in megabytes


tool: D70:             -u, --human-readable display data in kilo-, mega-, or giga-bytes


assistant: ```c
void usage() {
  printf("usage: iopp -h|--help\n");
  printf("usage: iopp [-ci] [-k|-m] [delay [count]]\n");
  printf("            -c, --command display full command line\n");
  printf("            -h, --help display help\n");
  printf("            -i, --idle hides idle processes\n");
  printf("            -k, --kilobytes display data in kilobytes\n");
  printf("            -m, --megabytes display data in megabytes\n");
  printf("            -u, --human-readable display data in kilo-, mega-, or giga-bytes\n");
}
```