ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<chug>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  je     L6
  lea    D86(%rip),%rbp
  mov    %edi,%ebx
  mov    %edi,%esi
  mov    %rbp,%rdx
  lea    D109(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%esi
  lea    D97(%rip),%rdi
  xor    %eax,%eax
  mov    %rbp,%rdx
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
  sub    $0x1,%ebx
  je     L12
  mov    %rsp,%r12
  lea    D75(%rip),%rdx
  mov    %ebx,%ecx
  mov    $0x8,%esi
  mov    %r12,%rdi
  call   <snprintf@plt>
  cmp    $0x1,%ebx
  lea    D127(%rip),%rax
  cmovne %rbp,%rax
  mov    %rax,%rdx
L13:
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  lea    D127(%rip),%r12
  mov    $0x1,%esi
  lea    D109(%rip),%rdi
  xor    %eax,%eax
  mov    %r12,%rdx
  call   <printf@plt>
  mov    %r12,%rdx
  mov    $0x1,%esi
  xor    %eax,%eax
  lea    D97(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
L12:
  mov    %rsp,%r12
  lea    D37(%rip),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  lea    D86(%rip),%rdx
  jmp    L13
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D86
data_type: string
name: parse_data
arguments: data_tag: D109
data_type: string
name: parse_data
arguments: data_tag: D97
data_type: string
name: parse_data
arguments: data_tag: D16
data_type: string
name: parse_data
arguments: data_tag: D75
data_type: string
name: parse_data
arguments: data_tag: D127
data_type: string
name: parse_data
arguments: data_tag: D4
data_type: string
name: parse_data
arguments: data_tag: D37
data_type: string

tool: D86: s

tool: D109: %d bottle%s of beer on the wall,


tool: D97: %d bottle%s of beeeeer . . . ,


tool: D16: Take one down, pass it around,

tool: D75: %d

tool: 

tool: D4: %s bottle%s of beer on the wall.


tool: D37: No more

assistant: ```c
void chug(int beers) {
  char howmany[8], *s;
  s = beers != 1 ? "s" : "";
  printf("%d bottle%s of beer on the wall,\n", beers, s);
  printf("%d bottle%s of beeeeer . . . ,\n", beers, s);
  printf("Take one down, pass it around,\n");
  if (--beers)
    snprintf(howmany, sizeof(howmany), "%d", beers);
  else
    strcpy(howmany, "No more");
  s = beers != 1 ? "s" : "";
  printf("%s bottle%s of beer on the wall.\n", howmany, s);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hsiehHash>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  test   %esi,%esi
  setg   %al
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  and    %eax,%edi
  call   <assert@plt>
  mov    %ebx,%r9d
  and    $0x3,%r9d
  mov    %ebx,%r8d
  sar    $0x2,%r8d
  test   %r8d,%r8d
  jle    L62
  mov    %r8d,%eax
  lea    0x0(%rbp,%rax,4),%rdi
  mov    %rbp,%rdx
L69:
  movzwl (%rdx),%ecx
  add    %ebx,%ecx
  movzwl 0x2(%rdx),%eax
  shl    $0xb,%eax
  mov    %ecx,%esi
  shl    $0x10,%esi
  xor    %esi,%eax
  xor    %ecx,%eax
  add    $0x4,%rdx
  mov    %eax,%ebx
  shr    $0xb,%ebx
  add    %eax,%ebx
  cmp    %rdi,%rdx
  jne    L69
  mov    %r8d,%r8d
  lea    0x0(%rbp,%r8,4),%rbp
L62:
  cmp    $0x2,%r9d
  je     L66
  cmp    $0x3,%r9d
  je     L16
  cmp    $0x1,%r9d
  je     L21
L68:
  lea    0x0(,%rbx,8),%edx
  xor    %ebx,%edx
  mov    %edx,%eax
  shr    $0x5,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shl    $0x4,%edx
  xor    %eax,%edx
  mov    %edx,%eax
  shr    $0x11,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shl    $0x19,%edx
  xor    %eax,%edx
  mov    %edx,%eax
  shr    $0x6,%eax
  add    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  movzwl 0x0(%rbp),%edx
  add    %ebx,%edx
  movsbl 0x2(%rbp),%eax
  shl    $0x12,%eax
  mov    %edx,%ecx
  shl    $0x10,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  mov    %eax,%ebx
  shr    $0xb,%ebx
  add    %eax,%ebx
  jmp    L68
L66:
  movzwl 0x0(%rbp),%edx
  add    %ebx,%edx
  mov    %edx,%eax
  shl    $0xb,%eax
  xor    %edx,%eax
  mov    %eax,%ebx
  shr    $0x11,%ebx
  add    %eax,%ebx
  jmp    L68
L21:
  movsbl 0x0(%rbp),%edx
  add    %ebx,%edx
  mov    %edx,%eax
  shl    $0xa,%eax
  xor    %edx,%eax
  mov    %eax,%ebx
  shr    %ebx
  add    %eax,%ebx
  jmp    L68
```

assistant: ```c
unsigned int hsiehHash(const char *data, int len) {
  unsigned int hash = len, tmp;
  int rem;
  assert(len > 0 && data != NULL);
  rem = len & 3;
  len >>= 2;
  for (; len > 0; len--) {
    hash += (*((const unsigned short *)(data)));
    tmp = ((*((const unsigned short *)(data + 2))) << 11) ^ hash;
    hash = (hash << 16) ^ tmp;
    data += 2 * sizeof(unsigned short);
    hash += hash >> 11;
  }
  switch (rem) {
  case 3:
    hash += (*((const unsigned short *)(data)));
    hash ^= hash << 16;
    hash ^= data[sizeof(unsigned short)] << 18;
    hash += hash >> 11;
    break;
  case 2:
    hash += (*((const unsigned short *)(data)));
    hash ^= hash << 11;
    hash += hash >> 17;
    break;
  case 1:
    hash += *data;
    hash ^= hash << 10;
    hash += hash >> 1;
  }
  hash ^= hash << 3;
  hash += hash >> 5;
  hash ^= hash << 4;
  hash += hash >> 17;
  hash ^= hash << 25;
  hash += hash >> 6;
  return hash;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hsiehHash>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  setg   %al
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  and    %eax,%edi
  call   <assert@plt>
  mov    %ebp,%eax
  mov    %ebp,%edi
  sar    $0x2,%eax
  and    $0x3,%edi
  test   %eax,%eax
  jle    L15
  cltq
  lea    (%rbx,%rax,4),%rsi
  xchg   %ax,%ax
L32:
  movzwl (%rbx),%edx
  movzwl 0x2(%rbx),%eax
  add    $0x4,%rbx
  add    %ebp,%edx
  shl    $0xb,%eax
  mov    %edx,%ecx
  shl    $0x10,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    $0xb,%ebp
  add    %eax,%ebp
  cmp    %rsi,%rbx
  jne    L32
L34:
  cmp    $0x2,%edi
  je     L44
  cmp    $0x3,%edi
  je     L1
  cmp    $0x1,%edi
  je     L53
L76:
  lea    0x0(,%rbp,8),%edx
  add    $0x8,%rsp
  xor    %ebp,%edx
  pop    %rbx
  pop    %rbp
  mov    %edx,%eax
  shr    $0x5,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shl    $0x4,%edx
  xor    %eax,%edx
  mov    %edx,%eax
  shr    $0x11,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shl    $0x19,%edx
  xor    %eax,%edx
  mov    %edx,%eax
  shr    $0x6,%eax
  add    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L44:
  movzwl (%rsi),%edx
  add    %ebp,%edx
  mov    %edx,%eax
  shl    $0xb,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    $0x11,%ebp
  add    %eax,%ebp
  jmp    L76
  nopl   (%rax)
L53:
  movsbl (%rsi),%edx
  add    %ebp,%edx
  mov    %edx,%eax
  shl    $0xa,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    %ebp
  add    %eax,%ebp
  jmp    L76
  nopl   0x0(%rax)
L1:
  movzwl (%rsi),%edx
  movsbl 0x2(%rsi),%eax
  add    %ebp,%edx
  shl    $0x12,%eax
  mov    %edx,%ecx
  shl    $0x10,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    $0xb,%ebp
  add    %eax,%ebp
  jmp    L76
  nopl   0x0(%rax)
L15:
  mov    %rbx,%rsi
  jmp    L34
```

assistant: ```c
unsigned int hsiehHash(const char *data, int len) {
  unsigned int hash = len, tmp;
  int rem;
  assert(len > 0 && data != NULL);
  rem = len & 3;
  len >>= 2;
  for (; len > 0; len--) {
    hash += (*((const unsigned short *)(data)));
    tmp = ((*((const unsigned short *)(data + 2))) << 11) ^ hash;
    hash = (hash << 16) ^ tmp;
    data += 2 * sizeof(unsigned short);
    hash += hash >> 11;
  }
  switch (rem) {
  case 3:
    hash += (*((const unsigned short *)(data)));
    hash ^= hash << 16;
    hash ^= data[sizeof(unsigned short)] << 18;
    hash += hash >> 11;
    break;
  case 2:
    hash += (*((const unsigned short *)(data)));
    hash ^= hash << 11;
    hash += hash >> 17;
    break;
  case 1:
    hash += *data;
    hash ^= hash << 10;
    hash += hash >> 1;
  }
  hash ^= hash << 3;
  hash += hash >> 5;
  hash ^= hash << 4;
  hash += hash >> 17;
  hash ^= hash << 25;
  hash += hash >> 6;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<hsiehHash>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  setg   %al
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  and    %eax,%edi
  call   <assert@plt>
  mov    %ebp,%eax
  mov    %ebp,%edi
  sar    $0x2,%eax
  and    $0x3,%edi
  test   %eax,%eax
  jle    L59
  cltq
  lea    (%rbx,%rax,4),%rsi
  xchg   %ax,%ax
L74:
  movzwl (%rbx),%edx
  movzwl 0x2(%rbx),%eax
  add    $0x4,%rbx
  add    %ebp,%edx
  shl    $0xb,%eax
  mov    %edx,%ecx
  shl    $0x10,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    $0xb,%ebp
  add    %eax,%ebp
  cmp    %rsi,%rbx
  jne    L74
L56:
  cmp    $0x2,%edi
  je     L2
  cmp    $0x3,%edi
  je     L40
  cmp    $0x1,%edi
  je     L50
L79:
  lea    0x0(,%rbp,8),%edx
  add    $0x8,%rsp
  xor    %ebp,%edx
  pop    %rbx
  pop    %rbp
  mov    %edx,%eax
  shr    $0x5,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shl    $0x4,%edx
  xor    %eax,%edx
  mov    %edx,%eax
  shr    $0x11,%eax
  add    %edx,%eax
  mov    %eax,%edx
  shl    $0x19,%edx
  xor    %eax,%edx
  mov    %edx,%eax
  shr    $0x6,%eax
  add    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  movzwl (%rsi),%edx
  add    %ebp,%edx
  mov    %edx,%eax
  shl    $0xb,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    $0x11,%ebp
  add    %eax,%ebp
  jmp    L79
  nopl   (%rax)
L50:
  movsbl (%rsi),%edx
  add    %ebp,%edx
  mov    %edx,%eax
  shl    $0xa,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    %ebp
  add    %eax,%ebp
  jmp    L79
  nopl   0x0(%rax)
L40:
  movzwl (%rsi),%edx
  movsbl 0x2(%rsi),%eax
  add    %ebp,%edx
  shl    $0x12,%eax
  mov    %edx,%ecx
  shl    $0x10,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  mov    %eax,%ebp
  shr    $0xb,%ebp
  add    %eax,%ebp
  jmp    L79
  nopl   0x0(%rax)
L59:
  mov    %rbx,%rsi
  jmp    L56
```

assistant: ```c
unsigned int hsiehHash(const char *data, int len) {
  unsigned int hash = len, tmp;
  int rem;
  assert(len > 0 && data != NULL);
  rem = len & 3;
  len >>= 2;
  for (; len > 0; len--) {
    hash += (*((const unsigned short *)(data)));
    tmp = ((*((const unsigned short *)(data + 2))) << 11) ^ hash;
    hash = (hash << 16) ^ tmp;
    data += 2 * sizeof(unsigned short);
    hash += hash >> 11;
  }
  switch (rem) {
  case 3:
    hash += (*((const unsigned short *)(data)));
    hash ^= hash << 16;
    hash ^= data[sizeof(unsigned short)] << 18;
    hash += hash >> 11;
    break;
  case 2:
    hash += (*((const unsigned short *)(data)));
    hash ^= hash << 11;
    hash += hash >> 17;
    break;
  case 1:
    hash += *data;
    hash ^= hash << 10;
    hash += hash >> 1;
  }
  hash ^= hash << 3;
  hash += hash >> 5;
  hash ^= hash << 4;
  hash += hash >> 17;
  hash ^= hash << 25;
  hash += hash >> 6;
  return hash;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jiecheng>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movsd  D8(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L7
L1:
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  movsd  -0x8(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  subl   $0x1,-0xc(%rbp)
L7:
  cmpl   $0x0,-0xc(%rbp)
  jg     L1
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
type_infered: f64

tool: D8: 1

assistant: ```c
double jiecheng(int n) {
  int i = n;
  double outcome = 1;
  for (; i >= 1; i--) {
    outcome = outcome * i;
  }
  return outcome;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<jiecheng>:
  endbr64
  test   %edi,%edi
  jle    L9
  movsd  D4(%rip),%xmm0
L2:
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  mulsd  %xmm1,%xmm0
  sub    $0x1,%edi
  jne    L2
  ret
L9:
  movsd  D4(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D4
data_type: f64

tool: D4: 1

assistant: ```c
double jiecheng(int n) {
  int i = n;
  double outcome = 1;
  for (; i >= 1; i--) {
    outcome = outcome * i;
  }
  return outcome;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<jiecheng>:
  endbr64
  movsd  D5(%rip),%xmm0
  test   %edi,%edi
  jle    L13
L18:
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  mulsd  %xmm1,%xmm0
  sub    $0x1,%edi
  jne    L18
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  ret
```

tool_calls: name: parse_data
arguments: label: D5
type_infered: f64

tool: D5: 1

assistant: ```c
double jiecheng(int n) {
  int i = n;
  double outcome = 1;
  for (; i >= 1; i--) {
    outcome = outcome * i;
  }
  return outcome;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jiecheng>:
  endbr64
  movsd  D5(%rip),%xmm0
  test   %edi,%edi
  jle    L19
L1:
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  mulsd  %xmm1,%xmm0
  sub    $0x1,%edi
  jne    L1
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
type_guessed: f64

tool: D5: 1

assistant: ```c
double jiecheng(int n) {
  int i = n;
  double outcome = 1;
  for (; i >= 1; i--) {
    outcome = outcome * i;
  }
  return outcome;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nnodup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L6
L11:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  shl    $0x2,%rax
  lea    -0x104(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jne    L25
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
L25:
  addl   $0x1,-0x10(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x10(%rbp)
  jl     L11
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *nnodup(char *input, int *dict) {
  char *output = malloc(strlen(input) + 1);
  int i, outpt = 0;
  for (i = 0; i < strlen(input); i++)
    if (dict[input[i] - 'A'] == 1)
      output[outpt++] = input[i];
  output[outpt] = '\0';
  return output;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<nnodup>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r13
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  jmp    L30
L20:
  add    $0x1,%rbx
L30:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L19
  movzbl 0x0(%rbp,%rbx,1),%eax
  movsbq %al,%rdx
  cmpl   $0x1,-0x104(%r13,%rdx,4)
  jne    L20
  movslq %r12d,%rdx
  mov    %al,(%r14,%rdx,1)
  lea    0x1(%r12),%r12d
  jmp    L20
L19:
  movslq %r12d,%r12
  movb   $0x0,(%r14,%r12,1)
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *nnodup(char *input, int *dict) {
  char *output = malloc(strlen(input) + 1);
  int i, outpt = 0;
  for (i = 0; i < strlen(input); i++)
    if (dict[input[i] - 'A'] == 1)
      output[outpt++] = input[i];
  output[outpt] = '\0';
  return output;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<nnodup>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L0:
  movsbq 0x0(%rbp,%rbx,1),%rdx
  cmpl   $0x1,-0x104(%r14,%rdx,4)
  jne    L10
  mov    %dl,(%rcx)
  add    $0x1,%r15d
  movslq %r15d,%r13
L10:
  add    $0x1,%rbx
L3:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%r12,%r13,1),%rcx
  cmp    %ebx,%eax
  jg     L0
  movb   $0x0,(%rcx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *nnodup(char *input, int *dict) {
  char *output = malloc(strlen(input) + 1);
  int i, outpt = 0;
  for (i = 0; i < strlen(input); i++)
    if (dict[input[i] - 'A'] == 1)
      output[outpt++] = input[i];
  output[outpt] = '\0';
  return output;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<nnodup>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L3:
  movsbq 0x0(%rbp,%rbx,1),%rdx
  cmpl   $0x1,-0x104(%r14,%rdx,4)
  jne    L12
  mov    %dl,(%rcx)
  add    $0x1,%r15d
  movslq %r15d,%r13
L12:
  add    $0x1,%rbx
L25:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%r12,%r13,1),%rcx
  cmp    %ebx,%eax
  jg     L3
  movb   $0x0,(%rcx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *nnodup(char *input, int *dict) {
  char *output = malloc(strlen(input) + 1);
  int i, outpt = 0;
  for (i = 0; i < strlen(input); i++)
    if (dict[input[i] - 'A'] == 1)
      output[outpt++] = input[i];
  output[outpt] = '\0';
  return output;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_10_pixel>:
  endbr64
  mov    %edx,%eax
  shr    $0x3,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %eax,%eax
  imul   %esi,%ecx
  add    %rcx,%rax
  lea    (%rdi,%rax,4),%rcx
  and    $0x7,%edx
  lea    D5(%rip),%rsi
  movslq (%rsi,%rdx,4),%rax
  add    %rsi,%rax
  notrack jmp *%rax
  mov    0x4(%rcx),%eax
  shl    $0x2,%eax
  and    $0x3fc,%eax
  mov    (%rcx),%edx
  sar    $0x6,%edx
  or     %edx,%eax
  ret
  mov    (%rcx),%eax
  shl    $0x4,%eax
  and    $0x3f0,%eax
  mov    0xc(%rcx),%edx
  sar    $0x4,%edx
  or     %edx,%eax
  ret
  mov    0xc(%rcx),%eax
  shl    $0x6,%eax
  and    $0x3c0,%eax
  mov    0x8(%rcx),%edx
  sar    $0x2,%edx
  or     %edx,%eax
  ret
  mov    0x8(%rcx),%eax
  shl    $0x8,%eax
  and    $0x300,%eax
  or     0x14(%rcx),%eax
  ret
  mov    0x10(%rcx),%eax
  shl    $0x2,%eax
  and    $0x3fc,%eax
  mov    0x1c(%rcx),%edx
  sar    $0x6,%edx
  or     %edx,%eax
  ret
  mov    0x1c(%rcx),%eax
  shl    $0x4,%eax
  and    $0x3f0,%eax
  mov    0x18(%rcx),%edx
  sar    $0x4,%edx
  or     %edx,%eax
  ret
  mov    0x18(%rcx),%eax
  shl    $0x6,%eax
  and    $0x3c0,%eax
  mov    0x24(%rcx),%edx
  sar    $0x2,%edx
  or     %edx,%eax
  ret
  mov    0x24(%rcx),%eax
  shl    $0x8,%eax
  and    $0x300,%eax
  or     0x20(%rcx),%eax
  ret
```

assistant: ```c
unsigned get_10_pixel(uint8_t *p, unsigned row_bytes, unsigned x, unsigned y) {
  uint8_t *addr = p + y * row_bytes + (x >> 3) * 10;
  switch (x & 7) {
  case 0:
    return ((0x3fc & (((unsigned short)addr[1]) << 2)) | (addr[0] >> 6));
  case 1:
    return ((0x3f0 & (((unsigned short)addr[0]) << 4)) | (addr[3] >> 4));
  case 2:
    return ((0x3c0 & (((unsigned short)addr[3]) << 6)) | (addr[2] >> 2));
  case 3:
    return ((0x300 & (((unsigned short)addr[2]) << 8)) | (addr[5]));
  case 4:
    return ((0x3fc & (((unsigned short)addr[4]) << 2)) | (addr[7] >> 6));
  case 5:
    return ((0x3f0 & (((unsigned short)addr[7]) << 4)) | (addr[6] >> 4));
  case 6:
    return ((0x3c0 & (((unsigned short)addr[6]) << 6)) | (addr[9] >> 2));
  case 7:
    return ((0x300 & (((unsigned short)addr[9]) << 8)) | (addr[8]));
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_10_pixel>:
  endbr64
  mov    %edx,%eax
  imul   %esi,%ecx
  lea    D0(%rip),%rsi
  and    $0x7,%edx
  shr    $0x3,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  add    %rcx,%rax
  lea    (%rdi,%rax,4),%rcx
  movslq (%rsi,%rdx,4),%rax
  add    %rsi,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
  mov    0x24(%rcx),%eax
  shl    $0x8,%eax
  and    $0x300,%eax
  or     0x20(%rcx),%eax
  ret
  nop
  mov    (%rcx),%eax
  mov    0xc(%rcx),%edx
  shl    $0x4,%eax
  sar    $0x4,%edx
  and    $0x3f0,%eax
  or     %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    0xc(%rcx),%eax
  mov    0x8(%rcx),%edx
  shl    $0x6,%eax
  sar    $0x2,%edx
  and    $0x3c0,%eax
  or     %edx,%eax
  ret
  nopl   0x0(%rax)
  mov    0x8(%rcx),%eax
  shl    $0x8,%eax
  and    $0x300,%eax
  or     0x14(%rcx),%eax
  ret
  nop
  mov    0x10(%rcx),%eax
  mov    0x1c(%rcx),%edx
  shl    $0x2,%eax
  sar    $0x6,%edx
  and    $0x3fc,%eax
  or     %edx,%eax
  ret
  nopl   0x0(%rax)
  mov    0x1c(%rcx),%eax
  mov    0x18(%rcx),%edx
  shl    $0x4,%eax
  sar    $0x4,%edx
  and    $0x3f0,%eax
  or     %edx,%eax
  ret
  nopl   0x0(%rax)
  mov    0x18(%rcx),%eax
  mov    0x24(%rcx),%edx
  shl    $0x6,%eax
  sar    $0x2,%edx
  and    $0x3c0,%eax
  or     %edx,%eax
  ret
  mov    0x4(%rcx),%eax
  mov    (%rcx),%edx
  shl    $0x2,%eax
  sar    $0x6,%edx
  and    $0x3fc,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned get_10_pixel(uint8_t *p, unsigned row_bytes, unsigned x, unsigned y) {
  uint8_t *addr = p + y * row_bytes + (x >> 3) * 10;
  switch (x & 7) {
  case 0:
    return ((0x3fc & (((unsigned short)addr[1]) << 2)) | (addr[0] >> 6));
  case 1:
    return ((0x3f0 & (((unsigned short)addr[0]) << 4)) | (addr[3] >> 4));
  case 2:
    return ((0x3c0 & (((unsigned short)addr[3]) << 6)) | (addr[2] >> 2));
  case 3:
    return ((0x300 & (((unsigned short)addr[2]) << 8)) | (addr[5]));
  case 4:
    return ((0x3fc & (((unsigned short)addr[4]) << 2)) | (addr[7] >> 6));
  case 5:
    return ((0x3f0 & (((unsigned short)addr[7]) << 4)) | (addr[6] >> 4));
  case 6:
    return ((0x3c0 & (((unsigned short)addr[6]) << 6)) | (addr[9] >> 2));
  case 7:
    return ((0x300 & (((unsigned short)addr[9]) << 8)) | (addr[8]));
  }
  return 0;
}
```