ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D10
type_guessed: string
name: extract_symbolic_data
arguments: tag: D0
type_guessed: string

tool: D10: begin

tool: D0: %d


assistant: ```c
int main() {
  printf("begin\n");
  pid_t pid = getpid();
  printf("%d\n", pid);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_nanbits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x2,%eax
  sub    -0x2c(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  movl   $0x7ff00000,-0x10(%rbp,%rax,4)
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%edx
  mov    -0x2c(%rbp),%eax
  cltq
  mov    %edx,-0x18(%rbp,%rax,4)
  mov    $0x1,%eax
  sub    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  movl   $0x0,-0x10(%rbp,%rdx,4)
  mov    $0x1,%edx
  mov    %edx,%ecx
  sub    -0x2c(%rbp),%ecx
  cltq
  mov    -0x10(%rbp,%rax,4),%edx
  movslq %ecx,%rax
  mov    %edx,-0x18(%rbp,%rax,4)
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x1c(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void get_nanbits(unsigned int *b, int k) {
  union {
    double d;
    unsigned int z[2];
  } u, u1, u2;
  k = 2 - k;
  u1.z[k] = u2.z[k] = 0x7ff00000;
  u1.z[1 - k] = u2.z[1 - k] = 0;
  u.d = u1.d - u2.d;
  b[0] = u.z[0];
  b[1] = u.z[1];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_nanbits>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x2,%eax
  sub    %esi,%eax
  cltq
  movl   $0x7ff00000,0x10(%rsp,%rax,4)
  movl   $0x7ff00000,0x8(%rsp,%rax,4)
  sub    $0x1,%esi
  movslq %esi,%rsi
  movl   $0x0,0x10(%rsp,%rsi,4)
  movl   $0x0,0x8(%rsp,%rsi,4)
  movsd  0x8(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  movq   %xmm0,%rax
  movd   %xmm0,(%rdi)
  shr    $0x20,%rax
  mov    %eax,0x4(%rdi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_nanbits(unsigned int *b, int k) {
  union {
    double d;
    unsigned int z[2];
  } u, u1, u2;
  k = 2 - k;
  u1.z[k] = u2.z[k] = 0x7ff00000;
  u1.z[1 - k] = u2.z[1 - k] = 0;
  u.d = u1.d - u2.d;
  b[0] = u.z[0];
  b[1] = u.z[1];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_nanbits>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    $0x2,%eax
  sub    %esi,%eax
  sub    $0x1,%esi
  cltq
  movslq %esi,%rsi
  movl   $0x7ff00000,0x10(%rsp,%rax,4)
  movl   $0x7ff00000,0x8(%rsp,%rax,4)
  movl   $0x0,0x10(%rsp,%rsi,4)
  movl   $0x0,0x8(%rsp,%rsi,4)
  movsd  0x8(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  movq   %xmm0,%rax
  movd   %xmm0,(%rdi)
  shr    $0x20,%rax
  mov    %eax,0x4(%rdi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_nanbits(unsigned int *b, int k) {
  union {
    double d;
    unsigned int z[2];
  } u, u1, u2;
  k = 2 - k;
  u1.z[k] = u2.z[k] = 0x7ff00000;
  u1.z[1 - k] = u2.z[1 - k] = 0;
  u.d = u1.d - u2.d;
  b[0] = u.z[0];
  b[1] = u.z[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_nanbits>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    $0x2,%eax
  sub    %esi,%eax
  sub    $0x1,%esi
  cltq
  movslq %esi,%rsi
  movl   $0x7ff00000,0x10(%rsp,%rax,4)
  movl   $0x7ff00000,0x8(%rsp,%rax,4)
  movl   $0x0,0x10(%rsp,%rsi,4)
  movl   $0x0,0x8(%rsp,%rsi,4)
  movsd  0x8(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  movq   %xmm0,(%rdi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_nanbits(unsigned int *b, int k) {
  union {
    double d;
    unsigned int z[2];
  } u, u1, u2;
  k = 2 - k;
  u1.z[k] = u2.z[k] = 0x7ff00000;
  u1.z[1 - k] = u2.z[1 - k] = 0;
  u.d = u1.d - u2.d;
  b[0] = u.z[0];
  b[1] = u.z[1];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ReferenceDct>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  lea    0x200(%rsp),%rdx
L49:
  add    $0x8,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sdl (%rdi),%xmm0
  movsd  %xmm0,-0x8(%rax)
  add    $0x4,%rdi
  cmp    %rdx,%rax
  jne    L49
  lea    0x200(%rsp),%r12
  mov    %rsp,%rbp
  mov    %r12,%r13
L62:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <DoubleReferenceDct1D@plt>
  add    $0x40,%r12
  add    $0x40,%rbp
  cmp    %r13,%rbp
  jne    L62
  mov    %rsp,%rbp
  lea    0x200(%rsp),%r12
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <DoubleTransposeMatrix@plt>
  mov    %r12,%r13
L61:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <DoubleReferenceDct1D@plt>
  add    $0x40,%r12
  add    $0x40,%rbp
  cmp    %r13,%rbp
  jne    L61
  mov    %rsp,%rsi
  lea    0x200(%rsp),%rdi
  call   <DoubleTransposeMatrix@plt>
  mov    $0x0,%eax
  pxor   %xmm1,%xmm1
  movsd  D0(%rip),%xmm3
  movapd %xmm3,%xmm2
  jmp    L10
L67:
  addsd  %xmm2,%xmm0
  cvttsd2si %xmm0,%edx
L7:
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    $0x100,%rax
  je     L15
L10:
  movsd  (%rsp,%rax,2),%xmm0
  comisd %xmm1,%xmm0
  ja     L67
  subsd  %xmm3,%xmm0
  cvttsd2si %xmm0,%edx
  jmp    L7
L15:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L70:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
data_type: f64

tool: D0: 0.5

assistant: ```c
void ReferenceDct(int *matrix, int *newmatrix) {
  static char RoutineName[] = "ReferenceDct";
  int *mptr;
  double *sptr, *dptr;
  double sourcematrix[64], destmatrix[64];
  for (sptr = sourcematrix, mptr = matrix; mptr < matrix + 64; mptr++) {
    *(sptr++) = (double)*mptr;
  }
  for (dptr = destmatrix, sptr = sourcematrix; sptr < sourcematrix + 64; sptr += 8) {
    DoubleReferenceDct1D(sptr, dptr);
    dptr += 8;
  }
  DoubleTransposeMatrix(destmatrix, sourcematrix);
  for (dptr = destmatrix, sptr = sourcematrix; sptr < sourcematrix + 64; sptr += 8) {
    DoubleReferenceDct1D(sptr, dptr);
    dptr += 8;
  }
  DoubleTransposeMatrix(destmatrix, sourcematrix);
  for (sptr = sourcematrix, mptr = newmatrix; mptr < newmatrix + 64; sptr++) {
    *(mptr++) = (int)(*sptr > 0 ? (*(sptr) + 0.5) : (*(sptr)-0.5));
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ReferenceDct>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x200(%rsp),%r12
  mov    %rbx,%rax
  nopw   0x0(%rax,%rax,1)
L25:
  pxor   %xmm0,%xmm0
  add    $0x8,%rax
  add    $0x4,%rdi
  cvtsi2sdl -0x4(%rdi),%xmm0
  movsd  %xmm0,-0x8(%rax)
  cmp    %r12,%rax
  jne    L25
  mov    %r12,%r14
  mov    %rbx,%r13
  nopl   0x0(%rax)
L74:
  mov    %r14,%rsi
  mov    %r13,%rdi
  add    $0x40,%r13
  add    $0x40,%r14
  call   <DoubleReferenceDct1D@plt>
  cmp    %r12,%r13
  jne    L74
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    %r12,%r14
  mov    %rbx,%r13
  call   <DoubleTransposeMatrix@plt>
  nopl   0x0(%rax)
L79:
  mov    %r14,%rsi
  mov    %r13,%rdi
  add    $0x40,%r13
  add    $0x40,%r14
  call   <DoubleReferenceDct1D@plt>
  cmp    %r12,%r13
  jne    L79
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <DoubleTransposeMatrix@plt>
  movsd  D6(%rip),%xmm2
  mov    %rbx,%rax
  pxor   %xmm1,%xmm1
  jmp    L70
  nopl   0x0(%rax)
L63:
  addsd  %xmm2,%xmm0
  add    $0x4,%rbp
  add    $0x8,%rax
  cvttsd2si %xmm0,%edx
  mov    %edx,-0x4(%rbp)
  cmp    %r12,%rax
  je     L1
L70:
  movsd  (%rax),%xmm0
  comisd %xmm1,%xmm0
  ja     L63
  subsd  %xmm2,%xmm0
  add    $0x4,%rbp
  add    $0x8,%rax
  cvttsd2si %xmm0,%edx
  mov    %edx,-0x4(%rbp)
  cmp    %r12,%rax
  jne    L70
L1:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
datatype: f64

tool: D6: 0.5

assistant: ```c
void ReferenceDct(int *matrix, int *newmatrix) {
  static char RoutineName[] = "ReferenceDct";
  int *mptr;
  double *sptr, *dptr;
  double sourcematrix[64], destmatrix[64];
  for (sptr = sourcematrix, mptr = matrix; mptr < matrix + 64; mptr++) {
    *(sptr++) = (double)*mptr;
  }
  for (dptr = destmatrix, sptr = sourcematrix; sptr < sourcematrix + 64; sptr += 8) {
    DoubleReferenceDct1D(sptr, dptr);
    dptr += 8;
  }
  DoubleTransposeMatrix(destmatrix, sourcematrix);
  for (dptr = destmatrix, sptr = sourcematrix; sptr < sourcematrix + 64; sptr += 8) {
    DoubleReferenceDct1D(sptr, dptr);
    dptr += 8;
  }
  DoubleTransposeMatrix(destmatrix, sourcematrix);
  for (sptr = sourcematrix, mptr = newmatrix; mptr < newmatrix + 64; sptr++) {
    *(mptr++) = (int)(*sptr > 0 ? (*(sptr) + 0.5) : (*(sptr)-0.5));
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_palidrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x410,%rsp
  mov    %edi,-0x404(%rbp)
  mov    %esi,-0x408(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3fc(%rbp)
  jmp    L38
L26:
  mov    -0x404(%rbp),%eax
  cltd
  idivl  -0x408(%rbp)
  mov    %edx,-0x3f4(%rbp)
  mov    -0x404(%rbp),%eax
  cltd
  idivl  -0x408(%rbp)
  mov    %eax,-0x404(%rbp)
  mov    -0x3f4(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x3fc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3fc(%rbp)
  mov    %ecx,%edx
  cltq
  mov    %dl,-0x3f0(%rbp,%rax,1)
L38:
  mov    -0x404(%rbp),%eax
  cltd
  idivl  -0x408(%rbp)
  test   %eax,%eax
  jg     L26
  mov    -0x404(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0x3fc(%rbp),%eax
  cltq
  mov    %dl,-0x3f0(%rbp,%rax,1)
  mov    -0x3fc(%rbp),%eax
  add    $0x1,%eax
  cltq
  movb   $0x0,-0x3f0(%rbp,%rax,1)
  movl   $0x0,-0x3fc(%rbp)
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x3f8(%rbp)
  jmp    L67
L73:
  addl   $0x1,-0x3fc(%rbp)
  subl   $0x1,-0x3f8(%rbp)
L67:
  mov    -0x3fc(%rbp),%eax
  cmp    -0x3f8(%rbp),%eax
  jge    L4
  mov    -0x3fc(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%edx
  mov    -0x3f8(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L73
L4:
  mov    -0x3fc(%rbp),%eax
  cmp    -0x3f8(%rbp),%eax
  jge    L72
  mov    $0x0,%eax
  jmp    L53
L72:
  mov    $0x1,%eax
L53:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

assistant: ```c
int is_palidrome(int num, int base) {
  char li[1000];
  int mod;
  int i = 0;
  while (num / base > 0) {
    mod = num % base;
    num /= base;
    li[i++] = '0' + mod;
  }
  li[i] = '0' + num;
  li[i + 1] = '\0';
  i = 0;
  int j = strlen(li) - 1;
  while (i < j && li[i] == li[j]) {
    ++i;
    --j;
  }
  if (i < j)
    return 0;
  else
    return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<is_palidrome>:
  endbr64
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %eax,%eax
  jle    L22
  mov    %esi,%ecx
  mov    %eax,%esi
  mov    $0x1,%r8d
  lea    -0x1(%rsp),%r10
  jmp    L33
L11:
  mov    %eax,%esi
L33:
  mov    %r8d,%r9d
  mov    %edi,%eax
  cltd
  idiv   %ecx
  add    $0x30,%edx
  mov    %dl,(%r10,%r8,1)
  mov    %esi,%eax
  cltd
  idiv   %ecx
  add    $0x1,%r8
  mov    %esi,%edi
  test   %eax,%eax
  jg     L11
L58:
  movslq %r9d,%rax
  add    $0x30,%esi
  mov    %sil,(%rsp,%rax,1)
  add    $0x1,%r9d
  movslq %r9d,%r9
  movb   $0x0,(%rsp,%r9,1)
  mov    %rsp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  test   %edx,%edx
  jle    L23
  mov    %rsp,%rcx
  cltq
  add    %rcx,%rax
  mov    $0x0,%esi
L83:
  movzbl -0x1(%rax),%edi
  cmp    %dil,(%rcx)
  jne    L89
  add    $0x1,%esi
  sub    $0x1,%edx
  add    $0x1,%rcx
  sub    $0x1,%rax
  cmp    %edx,%esi
  jl     L83
L89:
  cmp    %edx,%esi
  setge  %al
  movzbl %al,%eax
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x3f8,%rsp
  ret
L22:
  mov    %edi,%esi
  mov    $0x0,%r9d
  jmp    L58
L23:
  mov    $0x0,%esi
  jmp    L89
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_palidrome(int num, int base) {
  char li[1000];
  int mod;
  int i = 0;
  while (num / base > 0) {
    mod = num % base;
    num /= base;
    li[i++] = '0' + mod;
  }
  li[i] = '0' + num;
  li[i + 1] = '\0';
  i = 0;
  int j = strlen(li) - 1;
  while (i < j && li[i] == li[j]) {
    ++i;
    --j;
  }
  if (i < j)
    return 0;
  else
    return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<is_palidrome>:
  endbr64
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %eax,%eax
  jle    L20
  mov    %eax,%ecx
  mov    $0x1,%r8d
  lea    -0x1(%rsp),%r10
  jmp    L61
  nopl   0x0(%rax)
L41:
  mov    %eax,%ecx
L61:
  mov    %edi,%eax
  movslq %r8d,%r9
  mov    %ecx,%edi
  cltd
  idiv   %esi
  mov    %ecx,%eax
  add    $0x30,%edx
  mov    %dl,(%r10,%r8,1)
  cltd
  add    $0x1,%r8
  idiv   %esi
  test   %eax,%eax
  jg     L41
  lea    0x1(%r9),%eax
L66:
  mov    %rsp,%rbx
  add    $0x30,%ecx
  cltq
  mov    %rbx,%rdi
  mov    %cl,(%rsp,%r9,1)
  movb   $0x0,(%rsp,%rax,1)
  call   <strlen@plt>
  sub    $0x1,%eax
  test   %eax,%eax
  jle    L14
  cltq
  xor    %edx,%edx
  jmp    L37
  nopl   0x0(%rax)
L54:
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %edx,%eax
  jle    L14
L37:
  movzbl (%rbx,%rax,1),%esi
  cmp    %sil,(%rbx,%rdx,1)
  je     L54
  xor    %eax,%eax
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L14:
  mov    $0x1,%eax
L15:
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L55
  add    $0x3f0,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax)
L20:
  mov    %edi,%ecx
  mov    $0x1,%eax
  xor    %r9d,%r9d
  jmp    L66
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_palidrome(int num, int base) {
  char li[1000];
  int mod;
  int i = 0;
  while (num / base > 0) {
    mod = num % base;
    num /= base;
    li[i++] = '0' + mod;
  }
  li[i] = '0' + num;
  li[i + 1] = '\0';
  i = 0;
  int j = strlen(li) - 1;
  while (i < j && li[i] == li[j]) {
    ++i;
    --j;
  }
  if (i < j)
    return 0;
  else
    return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_palidrome>:
  endbr64
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %eax,%eax
  jle    L73
  mov    %eax,%ecx
  mov    $0x1,%r8d
  lea    -0x1(%rsp),%r10
  jmp    L81
  nopl   0x0(%rax)
L16:
  mov    %eax,%ecx
L81:
  mov    %edi,%eax
  movslq %r8d,%r9
  mov    %ecx,%edi
  cltd
  idiv   %esi
  mov    %ecx,%eax
  add    $0x30,%edx
  mov    %dl,(%r10,%r8,1)
  cltd
  add    $0x1,%r8
  idiv   %esi
  test   %eax,%eax
  jg     L16
  lea    0x1(%r9),%eax
L42:
  mov    %rsp,%rbx
  add    $0x30,%ecx
  cltq
  mov    %rbx,%rdi
  mov    %cl,(%rsp,%r9,1)
  movb   $0x0,(%rsp,%rax,1)
  call   <strlen@plt>
  sub    $0x1,%eax
  test   %eax,%eax
  jle    L93
  cltq
  xor    %edx,%edx
  jmp    L2
  nopl   0x0(%rax)
L92:
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %edx,%eax
  jle    L93
L2:
  movzbl (%rbx,%rax,1),%esi
  cmp    %sil,(%rbx,%rdx,1)
  je     L92
  xor    %eax,%eax
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L93:
  mov    $0x1,%eax
L30:
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L64
  add    $0x3f0,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax)
L73:
  mov    %edi,%ecx
  mov    $0x1,%eax
  xor    %r9d,%r9d
  jmp    L42
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_palidrome(int num, int base) {
  char li[1000];
  int mod;
  int i = 0;
  while (num / base > 0) {
    mod = num % base;
    num /= base;
    li[i++] = '0' + mod;
  }
  li[i] = '0' + num;
  li[i + 1] = '\0';
  i = 0;
  int j = strlen(li) - 1;
  while (i < j && li[i] == li[j]) {
    ++i;
    --j;
  }
  if (i < j)
    return 0;
  else
    return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<PECIAdvCmdSend>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %ecx,%eax
  mov    %r8d,%edi
  mov    %r9,-0x20(%rbp)
  mov    0x10(%rbp),%ecx
  mov    %sil,-0xc(%rbp)
  mov    %dl,-0x10(%rbp)
  mov    %al,-0x14(%rbp)
  mov    %edi,%eax
  mov    %al,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x24(%rbp)
  cmpq   $0x400b0000,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <DEBUG_ASSERT@plt>
  nop
L0:
  mov    -0x8(%rbp),%rax
  add    $0x114,%rax
  mov    (%rax),%rax
  mov    $0xffffffff,%edx
  cmp    %rdx,%rax
  je     L0
  mov    -0x8(%rbp),%rax
  add    $0x104,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x108,%rax
  mov    %rax,%rdx
  mov    0x18(%rbp),%rax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x10c,%rax
  mov    %rax,%rdx
  mov    0x20(%rbp),%rax
  mov    %rax,(%rdx)
  movzbl -0x10(%rbp),%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  shl    $0x10,%eax
  or     %eax,%edx
  movzbl -0x14(%rbp),%eax
  shl    $0x8,%eax
  or     %eax,%edx
  movzbl -0x18(%rbp),%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x100,%rax
  mov    %rax,%rcx
  movslq %edx,%rax
  mov    %rax,(%rcx)
  mov    -0x8(%rbp),%rax
  add    $0x110,%rax
  mov    %rax,%rdx
  movzbl -0xc(%rbp),%eax
  mov    %rax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void PECIAdvCmdSend(unsigned long ulBase, unsigned char ucCmd, unsigned char ucHidRe, unsigned char ucDomain, unsigned char ucProcAdd, unsigned long ulArg, unsigned char ucSize, unsigned long ulData0, unsigned long ulData1) {
  DEBUG_ASSERT(ulBase == 0x400B0000);
  while ((*((volatile unsigned long *)(ulBase + 0x00000114))) == 0xFFFFFFFF) {
  }
  (*((volatile unsigned long *)(ulBase + 0x00000104))) = ulArg;
  (*((volatile unsigned long *)(ulBase + 0x00000108))) = ulData0;
  (*((volatile unsigned long *)(ulBase + 0x0000010C))) = ulData1;
  (*((volatile unsigned long *)(ulBase + 0x00000100))) = (ucHidRe << 24) | (ucSize << 16) | (ucDomain << 8) | (ucProcAdd << 0);
  (*((volatile unsigned long *)(ulBase + 0x00000110))) = ucCmd;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<PECIAdvCmdSend>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %esi,0xc(%rsp)
  mov    %edx,%r12d
  mov    %ecx,%ebp
  mov    %r8d,%r14d
  mov    %r9,%r15
  mov    0x50(%rsp),%r13d
  cmp    $0x400b0000,%rdi
  sete   %dil
  movzbl %dil,%edi
  call   <DEBUG_ASSERT@plt>
  mov    $0xffffffff,%edx
L3:
  mov    0x114(%rbx),%rax
  cmp    %rdx,%rax
  je     L3
  mov    %r15,0x104(%rbx)
  mov    0x58(%rsp),%rax
  mov    %rax,0x108(%rbx)
  mov    0x60(%rsp),%rax
  mov    %rax,0x10c(%rbx)
  movzbl %bpl,%ebp
  shl    $0x8,%ebp
  movzbl %r14b,%r14d
  or     %r14d,%ebp
  movzbl %r13b,%r13d
  shl    $0x10,%r13d
  or     %ebp,%r13d
  shl    $0x18,%r12d
  or     %r13d,%r12d
  movslq %r12d,%r12
  mov    %r12,0x100(%rbx)
  movzbl 0xc(%rsp),%eax
  mov    %rax,0x110(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void PECIAdvCmdSend(unsigned long ulBase, unsigned char ucCmd, unsigned char ucHidRe, unsigned char ucDomain, unsigned char ucProcAdd, unsigned long ulArg, unsigned char ucSize, unsigned long ulData0, unsigned long ulData1) {
  DEBUG_ASSERT(ulBase == 0x400B0000);
  while ((*((volatile unsigned long *)(ulBase + 0x00000114))) == 0xFFFFFFFF) {
  }
  (*((volatile unsigned long *)(ulBase + 0x00000104))) = ulArg;
  (*((volatile unsigned long *)(ulBase + 0x00000108))) = ulData0;
  (*((volatile unsigned long *)(ulBase + 0x0000010C))) = ulData1;
  (*((volatile unsigned long *)(ulBase + 0x00000100))) = (ucHidRe << 24) | (ucSize << 16) | (ucDomain << 8) | (ucProcAdd << 0);
  (*((volatile unsigned long *)(ulBase + 0x00000110))) = ucCmd;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<PECIAdvCmdSend>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  xor    %edi,%edi
  push   %r14
  mov    %r8d,%r14d
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x18,%rsp
  cmp    $0x400b0000,%r15
  sete   %dil
  mov    %r9,0x8(%rsp)
  mov    0x50(%rsp),%r13d
  call   <DEBUG_ASSERT@plt>
  mov    0x8(%rsp),%r9
  mov    $0xffffffff,%edx
  lea    0x114(%r15),%rcx
  nopl   0x0(%rax)
L3:
  mov    (%rcx),%rax
  cmp    %rdx,%rax
  je     L3
  movzbl %r14b,%r14d
  movzbl %r13b,%r13d
  shl    $0x18,%r12d
  movzbl %bpl,%ebp
  mov    0x58(%rsp),%rax
  mov    %r9,0x104(%r15)
  shl    $0x10,%r13d
  mov    %rax,0x108(%r15)
  mov    0x60(%rsp),%rax
  mov    %rax,0x10c(%r15)
  movzbl %bl,%eax
  shl    $0x8,%eax
  or     %r14d,%eax
  or     %r13d,%eax
  or     %r12d,%eax
  cltq
  mov    %rax,0x100(%r15)
  mov    %rbp,0x110(%r15)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void PECIAdvCmdSend(unsigned long ulBase, unsigned char ucCmd, unsigned char ucHidRe, unsigned char ucDomain, unsigned char ucProcAdd, unsigned long ulArg, unsigned char ucSize, unsigned long ulData0, unsigned long ulData1) {
  DEBUG_ASSERT(ulBase == 0x400B0000);
  while ((*((volatile unsigned long *)(ulBase + 0x00000114))) == 0xFFFFFFFF) {
  }
  (*((volatile unsigned long *)(ulBase + 0x00000104))) = ulArg;
  (*((volatile unsigned long *)(ulBase + 0x00000108))) = ulData0;
  (*((volatile unsigned long *)(ulBase + 0x0000010C))) = ulData1;
  (*((volatile unsigned long *)(ulBase + 0x00000100))) = (ucHidRe << 24) | (ucSize << 16) | (ucDomain << 8) | (ucProcAdd << 0);
  (*((volatile unsigned long *)(ulBase + 0x00000110))) = ucCmd;
}
```