ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dup2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  nop
L6:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x3f,%edi
  call   <__syscall@plt>
  mov    %eax,-0x4(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  cmp    %eax,-0x4(%rbp)
  je     L6
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <__syscall_ret@plt>
  leave
  ret
```

assistant: ```c
int dup2(int old, int new) {
  int r;
  while ((r = __syscall(63, old, new)) == -EBUSY)
    ;
  return __syscall_ret(r);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<dup2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
L1:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    $0x3f,%edi
  call   <__syscall@plt>
  mov    D8(%rip),%rdx
  mov    %eax,%ecx
  add    (%rdx),%ecx
  je     L1
  mov    %eax,%edi
  call   <__syscall_ret@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int dup2(int old, int new) {
  int r;
  while ((r = __syscall(63, old, new)) == -EBUSY)
    ;
  return __syscall_ret(r);
}
```
user: What is the c source code of the assembly code below?

```asm
<dup2>:
  endbr64
  push   %r12
  mov    D5(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  nopl   0x0(%rax,%rax,1)
L6:
  mov    $0x3f,%edi
  mov    %ebx,%edx
  mov    %ebp,%esi
  call   <__syscall@plt>
  mov    %eax,%edi
  add    (%r12),%eax
  je     L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <__syscall_ret@plt>
```

assistant: ```c
int dup2(int old, int new) {
  int r;
  while ((r = __syscall(63, old, new)) == -EBUSY)
    ;
  return __syscall_ret(r);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dup2>:
  endbr64
  push   %r12
  mov    D1(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0x3f,%edi
  mov    %ebx,%edx
  mov    %ebp,%esi
  call   <__syscall@plt>
  mov    %eax,%edi
  add    (%r12),%eax
  je     L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <__syscall_ret@plt>
```

assistant: ```c
int dup2(int old, int new) {
  int r;
  while ((r = __syscall(63, old, new)) == -EBUSY)
    ;
  return __syscall_ret(r);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<build_partition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  cmpl   $0x0,-0x18(%rbp)
  jne    L29
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  movl   $0x0,(%rax)
  jmp    L38
L29:
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movl   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movl   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  movl   $0xff,(%rax)
L38:
  mov    -0x8(%rbp),%rax
  lea    0x10(%rax),%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
  cmpl   $0x0,-0x18(%rbp)
  jne    L27
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1c,%rax
  movl   $0x0,(%rax)
  jmp    L26
L27:
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  movl   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  movl   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1c,%rax
  movl   $0xff,(%rax)
L26:
  mov    -0x8(%rbp),%rax
  lea    0x20(%rax),%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x24,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x28,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x10,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x2c,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x18,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x30(%rax),%rdx
  mov    -0x18(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x34,%rax
  mov    -0x18(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x38,%rax
  mov    -0x18(%rbp),%edx
  sar    $0x10,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x3c,%rax
  mov    -0x18(%rbp),%edx
  sar    $0x18,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void build_partition(uint8_t *buf, int boot, int type, uint32_t start_block, uint32_t num_blocks) {
  buf[0] = boot;
  if (num_blocks == 0) {
    buf[1] = 0;
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[1] = 0xff;
    buf[2] = 0xff;
    buf[3] = 0xff;
  }
  buf[4] = type;
  if (num_blocks == 0) {
    buf[5] = 0;
    buf[6] = 0;
    buf[7] = 0;
  } else {
    buf[5] = 0xff;
    buf[6] = 0xff;
    buf[7] = 0xff;
  }
  buf[8] = start_block;
  buf[9] = start_block >> 8;
  buf[10] = start_block >> 16;
  buf[11] = start_block >> 24;
  buf[12] = num_blocks;
  buf[13] = num_blocks >> 8;
  buf[14] = num_blocks >> 16;
  buf[15] = num_blocks >> 24;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<build_partition>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%edi
  mov    %edx,%esi
  test   %r8d,%r8d
  mov    $0xff,%edx
  cmove  %r8d,%edx
  mov    %edi,(%rax)
  mov    %edx,0x4(%rax)
  mov    %edx,0x8(%rax)
  mov    %edx,0xc(%rax)
  mov    %esi,0x10(%rax)
  mov    %edx,0x14(%rax)
  mov    %edx,0x18(%rax)
  mov    %edx,0x1c(%rax)
  mov    %ecx,0x20(%rax)
  mov    %ecx,%edx
  sar    $0x8,%edx
  mov    %edx,0x24(%rax)
  mov    %ecx,%edx
  sar    $0x10,%edx
  mov    %edx,0x28(%rax)
  sar    $0x18,%ecx
  mov    %ecx,0x2c(%rax)
  mov    %r8d,0x30(%rax)
  mov    %r8d,%edx
  sar    $0x8,%edx
  mov    %edx,0x34(%rax)
  mov    %r8d,%edx
  sar    $0x10,%edx
  mov    %edx,0x38(%rax)
  sar    $0x18,%r8d
  mov    %r8d,0x3c(%rax)
  ret
```

assistant: ```c
void build_partition(uint8_t *buf, int boot, int type, uint32_t start_block, uint32_t num_blocks) {
  buf[0] = boot;
  if (num_blocks == 0) {
    buf[1] = 0;
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[1] = 0xff;
    buf[2] = 0xff;
    buf[3] = 0xff;
  }
  buf[4] = type;
  if (num_blocks == 0) {
    buf[5] = 0;
    buf[6] = 0;
    buf[7] = 0;
  } else {
    buf[5] = 0xff;
    buf[6] = 0xff;
    buf[7] = 0xff;
  }
  buf[8] = start_block;
  buf[9] = start_block >> 8;
  buf[10] = start_block >> 16;
  buf[11] = start_block >> 24;
  buf[12] = num_blocks;
  buf[13] = num_blocks >> 8;
  buf[14] = num_blocks >> 16;
  buf[15] = num_blocks >> 24;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<build_partition>:
  endbr64
  mov    %rdi,%rax
  mov    %r8d,%edi
  mov    %edx,%r9d
  neg    %edi
  mov    %ecx,0x20(%rax)
  sbb    %edx,%edx
  mov    %r8d,0x30(%rax)
  movzbl %dl,%edx
  mov    %esi,(%rax)
  mov    %edx,0x4(%rax)
  mov    %edx,0x8(%rax)
  mov    %edx,0xc(%rax)
  mov    %edx,0x14(%rax)
  mov    %edx,0x18(%rax)
  mov    %edx,0x1c(%rax)
  mov    %ecx,%edx
  sar    $0x8,%edx
  mov    %r9d,0x10(%rax)
  mov    %edx,0x24(%rax)
  mov    %ecx,%edx
  sar    $0x18,%ecx
  sar    $0x10,%edx
  mov    %ecx,0x2c(%rax)
  mov    %edx,0x28(%rax)
  mov    %r8d,%edx
  sar    $0x8,%edx
  mov    %edx,0x34(%rax)
  mov    %r8d,%edx
  sar    $0x18,%r8d
  sar    $0x10,%edx
  mov    %r8d,0x3c(%rax)
  mov    %edx,0x38(%rax)
  ret
```

assistant: ```c
void build_partition(uint8_t *buf, int boot, int type, uint32_t start_block, uint32_t num_blocks) {
  buf[0] = boot;
  if (num_blocks == 0) {
    buf[1] = 0;
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[1] = 0xff;
    buf[2] = 0xff;
    buf[3] = 0xff;
  }
  buf[4] = type;
  if (num_blocks == 0) {
    buf[5] = 0;
    buf[6] = 0;
    buf[7] = 0;
  } else {
    buf[5] = 0xff;
    buf[6] = 0xff;
    buf[7] = 0xff;
  }
  buf[8] = start_block;
  buf[9] = start_block >> 8;
  buf[10] = start_block >> 16;
  buf[11] = start_block >> 24;
  buf[12] = num_blocks;
  buf[13] = num_blocks >> 8;
  buf[14] = num_blocks >> 16;
  buf[15] = num_blocks >> 24;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<build_partition>:
  endbr64
  mov    %ecx,%eax
  mov    %r8d,%ecx
  movd   %esi,%xmm0
  neg    %ecx
  mov    %eax,%esi
  sbb    %ecx,%ecx
  sar    $0x10,%esi
  movzbl %cl,%ecx
  movd   %ecx,%xmm2
  movd   %ecx,%xmm1
  mov    %eax,%ecx
  punpckldq %xmm1,%xmm1
  punpckldq %xmm2,%xmm0
  sar    $0x18,%ecx
  punpcklqdq %xmm1,%xmm0
  movd   %ecx,%xmm5
  movups %xmm0,(%rdi)
  movd   %edx,%xmm0
  mov    %eax,%edx
  punpckldq %xmm2,%xmm0
  sar    $0x8,%edx
  punpcklqdq %xmm1,%xmm0
  movd   %edx,%xmm6
  movd   %esi,%xmm1
  mov    %r8d,%edx
  movups %xmm0,0x10(%rdi)
  movd   %eax,%xmm0
  punpckldq %xmm5,%xmm1
  mov    %r8d,%eax
  sar    $0x8,%eax
  punpckldq %xmm6,%xmm0
  mov    %r8d,%esi
  sar    $0x18,%edx
  punpcklqdq %xmm1,%xmm0
  sar    $0x10,%esi
  movd   %edx,%xmm7
  movd   %eax,%xmm2
  movups %xmm0,0x20(%rdi)
  movd   %esi,%xmm1
  movd   %r8d,%xmm0
  punpckldq %xmm7,%xmm1
  punpckldq %xmm2,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,0x30(%rdi)
  ret
```

assistant: ```c
void build_partition(uint8_t *buf, int boot, int type, uint32_t start_block, uint32_t num_blocks) {
  buf[0] = boot;
  if (num_blocks == 0) {
    buf[1] = 0;
    buf[2] = 0;
    buf[3] = 0;
  } else {
    buf[1] = 0xff;
    buf[2] = 0xff;
    buf[3] = 0xff;
  }
  buf[4] = type;
  if (num_blocks == 0) {
    buf[5] = 0;
    buf[6] = 0;
    buf[7] = 0;
  } else {
    buf[5] = 0xff;
    buf[6] = 0xff;
    buf[7] = 0xff;
  }
  buf[8] = start_block;
  buf[9] = start_block >> 8;
  buf[10] = start_block >> 16;
  buf[11] = start_block >> 24;
  buf[12] = num_blocks;
  buf[13] = num_blocks >> 8;
  buf[14] = num_blocks >> 16;
  buf[15] = num_blocks >> 24;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<combination>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %rdx,-0x98(%rbp)
  mov    %ecx,-0x9c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rax
  mov    $0x64,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x9c(%rbp),%edx
  mov    -0x90(%rbp),%rcx
  mov    -0x88(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <reflection@plt>
  movl   $0x0,-0x74(%rbp)
  jmp    L32
L18:
  mov    -0x9c(%rbp),%edx
  lea    -0x70(%rbp),%rcx
  mov    -0x90(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <rotate_90@plt>
  mov    -0x9c(%rbp),%edx
  mov    -0x98(%rbp),%rcx
  lea    -0x70(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <compare@plt>
  test   %eax,%eax
  jne    L20
  mov    $0x0,%eax
  jmp    L19
L20:
  lea    -0x70(%rbp),%rcx
  mov    -0x90(%rbp),%rax
  mov    $0x64,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  addl   $0x1,-0x74(%rbp)
L32:
  cmpl   $0x2,-0x74(%rbp)
  jle    L18
  mov    $0x1,%eax
L19:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

assistant: ```c
int combination(char before[][10], char trans[][10], char target[][10], int n) {
  char tmp[10][10];
  memset(tmp, 0, sizeof(tmp));
  reflection(before, trans, n);
  int i;
  for (i = 0; i < 3; ++i) {
    rotate_90(trans, tmp, n);
    if (!compare(tmp, target, n))
      return 0;
    memcpy(trans, tmp, sizeof(tmp));
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<combination>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%r14
  mov    %ecx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x64,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <reflection@plt>
  mov    $0x3,%r13d
  mov    %rsp,%r12
L19:
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <rotate_90@plt>
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <compare@plt>
  test   %eax,%eax
  je     L3
  mov    $0x64,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  sub    $0x1,%r13d
  jne    L19
  mov    $0x1,%eax
L3:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int combination(char before[][10], char trans[][10], char target[][10], int n) {
  char tmp[10][10];
  memset(tmp, 0, sizeof(tmp));
  reflection(before, trans, n);
  int i;
  for (i = 0; i < 3; ++i) {
    rotate_90(trans, tmp, n);
    if (!compare(tmp, target, n))
      return 0;
    memcpy(trans, tmp, sizeof(tmp));
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<combination>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  mov    $0x64,%edx
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %r13,%rdi
  mov    %ebp,%edx
  mov    %r12,%rsi
  call   <reflection@plt>
  mov    $0x3,%r13d
L1:
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <rotate_90@plt>
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <compare@plt>
  test   %eax,%eax
  je     L22
  mov    $0x64,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  sub    $0x1,%r13d
  jne    L1
  mov    $0x1,%eax
L22:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int combination(char before[][10], char trans[][10], char target[][10], int n) {
  char tmp[10][10];
  memset(tmp, 0, sizeof(tmp));
  reflection(before, trans, n);
  int i;
  for (i = 0; i < 3; ++i) {
    rotate_90(trans, tmp, n);
    if (!compare(tmp, target, n))
      return 0;
    memcpy(trans, tmp, sizeof(tmp));
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<combination>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  mov    $0x64,%edx
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %r13,%rdi
  mov    %ebp,%edx
  mov    %r12,%rsi
  call   <reflection@plt>
  mov    $0x3,%r13d
L3:
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <rotate_90@plt>
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <compare@plt>
  test   %eax,%eax
  je     L9
  mov    $0x64,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  sub    $0x1,%r13d
  jne    L3
  mov    $0x1,%eax
L9:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int combination(char before[][10], char trans[][10], char target[][10], int n) {
  char tmp[10][10];
  memset(tmp, 0, sizeof(tmp));
  reflection(before, trans, n);
  int i;
  for (i = 0; i < 3; ++i) {
    rotate_90(trans, tmp, n);
    if (!compare(tmp, target, n))
      return 0;
    memcpy(trans, tmp, sizeof(tmp));
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L9
L2:
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <doubleRound@plt>
  addl   $0x1,-0x24(%rbp)
L9:
  cmpl   $0x14,-0x24(%rbp)
  jle    L2
  nop
  nop
  leave
  ret
```

assistant: ```c
void Encrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  uint32_t *rk = (uint32_t *)roundKeys;
  uint32_t *rightPtr = (uint32_t *)block;
  uint32_t *leftPtr = (rightPtr + 1);
  uint32_t rk2i;
  uint32_t rk2ip1;
  for (i = 0; i < 42 / 2; ++i) {
    rk2i = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i]);
    rk2ip1 = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i + 1]);
    doubleRound(leftPtr, rightPtr, rk2i, rk2ip1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<Encrypt>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  lea    0x4(%rdi),%r14
  mov    %rsi,%rbx
  lea    0xa8(%rsi),%r12
L7:
  mov    (%rbx),%edi
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    %eax,%ebp
  mov    0x4(%rbx),%edi
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    %eax,%ecx
  mov    %ebp,%edx
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <doubleRound@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void Encrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  uint32_t *rk = (uint32_t *)roundKeys;
  uint32_t *rightPtr = (uint32_t *)block;
  uint32_t *leftPtr = (rightPtr + 1);
  uint32_t rk2i;
  uint32_t rk2ip1;
  for (i = 0; i < 42 / 2; ++i) {
    rk2i = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i]);
    rk2ip1 = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i + 1]);
    doubleRound(leftPtr, rightPtr, rk2i, rk2ip1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Encrypt>:
  endbr64
  push   %r14
  lea    0xa8(%rsi),%r14
  push   %r13
  lea    0x4(%rdi),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   (%rax)
L8:
  mov    (%rbx),%edi
  add    $0x8,%rbx
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    -0x4(%rbx),%edi
  mov    %eax,%r12d
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %eax,%ecx
  call   <doubleRound@plt>
  cmp    %r14,%rbx
  jne    L8
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void Encrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  uint32_t *rk = (uint32_t *)roundKeys;
  uint32_t *rightPtr = (uint32_t *)block;
  uint32_t *leftPtr = (rightPtr + 1);
  uint32_t rk2i;
  uint32_t rk2ip1;
  for (i = 0; i < 42 / 2; ++i) {
    rk2i = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i]);
    rk2ip1 = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i + 1]);
    doubleRound(leftPtr, rightPtr, rk2i, rk2ip1);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<Encrypt>:
  endbr64
  push   %r14
  lea    0xa8(%rsi),%r14
  push   %r13
  lea    0x4(%rdi),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   (%rax)
L0:
  mov    (%rbx),%edi
  add    $0x8,%rbx
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    -0x4(%rbx),%edi
  mov    %eax,%r12d
  call   <READ_ROUND_KEY_DOUBLE_WORD@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %eax,%ecx
  call   <doubleRound@plt>
  cmp    %rbx,%r14
  jne    L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void Encrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i;
  uint32_t *rk = (uint32_t *)roundKeys;
  uint32_t *rightPtr = (uint32_t *)block;
  uint32_t *leftPtr = (rightPtr + 1);
  uint32_t rk2i;
  uint32_t rk2ip1;
  for (i = 0; i < 42 / 2; ++i) {
    rk2i = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i]);
    rk2ip1 = READ_ROUND_KEY_DOUBLE_WORD(rk[2 * i + 1]);
    doubleRound(leftPtr, rightPtr, rk2i, rk2ip1);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getNWSInterval>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L12
  mov    $0xe10,%eax
  jmp    L33
L12:
  mov    -0x8(%rbp),%rax
  lea    D25(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L18
  mov    $0x384,%eax
  jmp    L33
L18:
  mov    -0x8(%rbp),%rax
  lea    D28(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L54
  mov    $0x384,%eax
  jmp    L33
L54:
  mov    $0x0,%eax
L33:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type: string
name: extract_symbolic_data
arguments: data_tag: D25
type: string
name: extract_symbolic_data
arguments: data_tag: D28
type: string

tool: D3: HPCP

tool: D25: QPCP

tool: D28: QGAG

assistant: ```c
int getNWSInterval(char *elemType) {
  if (strcmp(elemType, "HPCP") == 0)
    return 3600;
  else if (strcmp(elemType, "QPCP") == 0)
    return 900;
  else if (strcmp(elemType, "QGAG") == 0)
    return 900;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getNWSInterval>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D12(%rip),%rsi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0xe10,%eax
  test   %rdx,%rdx
  je     L5
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x384,%eax
  test   %rdx,%rdx
  je     L5
  lea    D13(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x384,%eax
L5:
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D12
data_type: string
name: unpack_data
arguments: label: D0
data_type: string
name: unpack_data
arguments: label: D13
data_type: string

tool: D12: HPCP

tool: D0: QPCP

tool: D13: QGAG

assistant: ```c
int getNWSInterval(char *elemType) {
  if (strcmp(elemType, "HPCP") == 0)
    return 3600;
  else if (strcmp(elemType, "QPCP") == 0)
    return 900;
  else if (strcmp(elemType, "QGAG") == 0)
    return 900;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getNWSInterval>:
  endbr64
  push   %rbp
  lea    D17(%rip),%rsi
  mov    %rdi,%rbp
  call   <strcmp@plt>
  mov    %rax,%r8
  mov    $0xe10,%eax
  test   %r8,%r8
  je     L17
  lea    D11(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L23
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L23
  xor    %eax,%eax
L17:
  pop    %rbp
  ret
  nopl   (%rax)
L23:
  mov    $0x384,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D17
data_type: string
name: parse_data
arguments: label: D11
data_type: string
name: parse_data
arguments: label: D9
data_type: string

tool: D17: HPCP

tool: D11: QPCP

tool: D9: QGAG

assistant: ```c
int getNWSInterval(char *elemType) {
  if (strcmp(elemType, "HPCP") == 0)
    return 3600;
  else if (strcmp(elemType, "QPCP") == 0)
    return 900;
  else if (strcmp(elemType, "QGAG") == 0)
    return 900;
  else
    return 0;
}
```