ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<nextcwedge>:
  endbr64
  cmp    $0xb,%edi
  ja     1040 <nextcwedge.cold>
  lea    D1(%rip),%rdx
  mov    %edi,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  xor    %eax,%eax
  cmp    $0x3,%esi
  setne  %al
  lea    0x1(%rax,%rax,4),%eax
  ret
  nopl   (%rax)
  xor    %eax,%eax
  cmp    $0x3,%esi
  sete   %al
  lea    0x3(%rax,%rax,1),%eax
  ret
  nopl   (%rax)
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xfffffffb,%eax
  add    $0x8,%eax
  ret
  nopl   0x0(%rax)
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xfffffff7,%eax
  add    $0xb,%eax
  ret
  nopl   0x0(%rax)
  neg    %esi
  sbb    %eax,%eax
  and    $0xa,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xfffffff8,%eax
  add    $0x9,%eax
  ret
  nopl   0x0(%rax)
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  lea    0x6(%rax,%rax,2),%eax
  ret
  nopl   (%rax)
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  lea    0x7(%rax,%rax,2),%eax
  ret
  nopl   (%rax)
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  lea    0x5(%rax,%rax,2),%eax
  ret
  nopl   (%rax)
  cmp    $0x1,%esi
  mov    $0x4,%edx
  mov    $0xb,%eax
  cmove  %edx,%eax
  ret
  nopl   0x0(%rax)
  xor    %eax,%eax
  cmp    $0x2,%esi
  sete   %al
  lea    0x2(%rax,%rax,1),%eax
  ret
  nopl   (%rax)
  cmp    $0x2,%esi
  mov    $0x0,%edx
  mov    $0x7,%eax
  cmove  %edx,%eax
  ret
```

assistant: ```c
int nextcwedge(edge, face)
int edge, face;
{
  switch (edge) {
  case 0:
    return (face == 0) ? 3 : 8;
  case 1:
    return (face == 0) ? 2 : 11;
  case 2:
    return (face == 0) ? 0 : 10;
  case 3:
    return (face == 0) ? 1 : 9;
  case 4:
    return (face == 1) ? 6 : 9;
  case 5:
    return (face == 1) ? 7 : 10;
  case 6:
    return (face == 1) ? 5 : 8;
  case 7:
    return (face == 1) ? 4 : 11;
  case 8:
    return (face == 2) ? 4 : 2;
  case 9:
    return (face == 2) ? 0 : 7;
  case 10:
    return (face == 3) ? 1 : 6;
  case 11:
    return (face == 3) ? 5 : 3;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nextcwedge>:
  endbr64
  cmp    $0xb,%edi
  ja     1040 <nextcwedge.cold>
  lea    D5(%rip),%rdx
  mov    %edi,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  xor    %eax,%eax
  cmp    $0x3,%esi
  setne  %al
  lea    0x1(%rax,%rax,4),%eax
  ret
  nopl   (%rax)
  xor    %eax,%eax
  cmp    $0x3,%esi
  sete   %al
  lea    0x3(%rax,%rax,1),%eax
  ret
  nopl   (%rax)
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xfffffffb,%eax
  add    $0x8,%eax
  ret
  nopl   0x0(%rax)
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xfffffff7,%eax
  add    $0xb,%eax
  ret
  nopl   0x0(%rax)
  neg    %esi
  sbb    %eax,%eax
  and    $0xa,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xfffffff8,%eax
  add    $0x9,%eax
  ret
  nopl   0x0(%rax)
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  lea    0x6(%rax,%rax,2),%eax
  ret
  nopl   (%rax)
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  lea    0x7(%rax,%rax,2),%eax
  ret
  nopl   (%rax)
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  lea    0x5(%rax,%rax,2),%eax
  ret
  nopl   (%rax)
  cmp    $0x1,%esi
  mov    $0x4,%edx
  mov    $0xb,%eax
  cmove  %edx,%eax
  ret
  nopl   0x0(%rax)
  xor    %eax,%eax
  cmp    $0x2,%esi
  sete   %al
  lea    0x2(%rax,%rax,1),%eax
  ret
  nopl   (%rax)
  cmp    $0x2,%esi
  mov    $0x0,%edx
  mov    $0x7,%eax
  cmove  %edx,%eax
  ret
```

assistant: ```c
int nextcwedge(edge, face)
int edge, face;
{
  switch (edge) {
  case 0:
    return (face == 0) ? 3 : 8;
  case 1:
    return (face == 0) ? 2 : 11;
  case 2:
    return (face == 0) ? 0 : 10;
  case 3:
    return (face == 0) ? 1 : 9;
  case 4:
    return (face == 1) ? 6 : 9;
  case 5:
    return (face == 1) ? 7 : 10;
  case 6:
    return (face == 1) ? 5 : 8;
  case 7:
    return (face == 1) ? 4 : 11;
  case 8:
    return (face == 2) ? 4 : 2;
  case 9:
    return (face == 2) ? 0 : 7;
  case 10:
    return (face == 3) ? 1 : 6;
  case 11:
    return (face == 3) ? 5 : 3;
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %esi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  add    $0xf,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L122:
  cmp    %rcx,%rsp
  je     L108
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L122
L108:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L112
  orq    $0x0,-0x8(%rsp,%rax,1)
L112:
  mov    %rsp,%rsi
  test   $0x1,%dl
  je     L134
  mov    %edx,%ecx
  shr    $0x1f,%ecx
  add    %edx,%ecx
  sar    %ecx
  mov    $0x0,%eax
  movslq %ecx,%r9
  add    %rdi,%r9
  cmp    $0x1,%edx
  jle    L61
L45:
  movzbl 0x1(%r9,%rax,1),%r8d
  mov    %r8b,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L45
L61:
  lea    0x1(%rcx),%eax
  cmp    %eax,%edx
  jle    L6
  cltq
  movslq %ecx,%r8
  mov    %rdi,%r9
  sub    %r8,%r9
L133:
  movzbl -0x1(%r9,%rax,1),%r8d
  mov    %r8b,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L133
L6:
  movslq %ecx,%rcx
  movzbl (%rdi,%rcx,1),%eax
  mov    %al,(%rsi,%rcx,1)
L121:
  movslq %edx,%rdx
  movb   $0x0,(%rsi,%rdx,1)
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  leave
  ret
L134:
  mov    %edx,%ecx
  shr    $0x1f,%ecx
  add    %edx,%ecx
  sar    %ecx
  cmp    $0x1,%edx
  jle    L14
  mov    $0x0,%eax
  movslq %ecx,%r9
  add    %rdi,%r9
L16:
  movzbl (%r9,%rax,1),%r8d
  mov    %r8b,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L16
L14:
  cmp    %ecx,%edx
  jle    L121
  movslq %ecx,%rcx
  mov    %rcx,%rax
  sub    %rcx,%rdi
L39:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L39
  jmp    L121
L8:
  call   <__stack_chk_fail@plt>
```

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

tool: D3: %s

assistant: ```c
void swap(char c[], int r) {
  char t[r];
  if (r % 2 == 0) {
    for (int i = 0; i < r / 2; i++) {
      t[i] = c[i + r / 2];
    }
    for (int j = r / 2; j < r; j++) {
      t[j] = c[j - r / 2];
    }
  } else {
    for (int i = 0; i < r / 2; i++) {
      t[i] = c[i + r / 2 + 1];
    }
    for (int j = r / 2 + 1; j < r; j++) {
      t[j] = c[j - r / 2 - 1];
    }
    t[r / 2] = c[r / 2];
  }
  t[r] = '\0';
  printf("%s", t);
}
```
user: What is the c source code of the assembly code below?

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  movslq %esi,%r12
  push   %rbx
  mov    %r12,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0xf(%r12),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L38
L1:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L1
L38:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L106
L86:
  mov    %ebx,%r13d
  mov    %rsp,%r15
  shr    $0x1f,%r13d
  add    %ebx,%r13d
  sar    %r13d
  test   $0x1,%bl
  je     L50
  cmp    $0x1,%ebx
  jle    L11
  lea    0x1(%r13),%ecx
  lea    -0x1(%r13),%edx
  mov    %rsp,%rdi
  movslq %ecx,%rsi
  add    $0x1,%rdx
  mov    %ecx,-0x44(%rbp)
  add    %r14,%rsi
  call   <memcpy@plt>
  mov    -0x44(%rbp),%ecx
L74:
  cmp    %ecx,%ebx
  jle    L63
  lea    -0x2(%rbx),%edx
  movslq %ecx,%rdi
  mov    %r14,%rsi
  mov    %edx,%ebx
  add    %r15,%rdi
  sub    %r13d,%ebx
  lea    0x1(%rbx),%rdx
  call   <memcpy@plt>
L63:
  movslq %r13d,%r13
  movzbl (%r14,%r13,1),%eax
  mov    %al,(%r15,%r13,1)
L49:
  mov    %r15,%rsi
  lea    D4(%rip),%rdi
  movb   $0x0,(%r15,%r12,1)
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L114
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nop
L50:
  cmp    $0x1,%ebx
  jle    L40
  lea    -0x1(%r13),%edx
  movslq %r13d,%rsi
  mov    %rsp,%rdi
  add    $0x1,%rdx
  add    %r14,%rsi
  call   <memcpy@plt>
L40:
  cmp    %ebx,%r13d
  jge    L49
  lea    -0x1(%rbx),%edx
  movslq %r13d,%rdi
  mov    %r14,%rsi
  mov    %edx,%ebx
  add    %r15,%rdi
  sub    %r13d,%ebx
  lea    0x1(%rbx),%rdx
  call   <memcpy@plt>
  jmp    L49
  nopl   0x0(%rax)
L106:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L86
  nopl   0x0(%rax,%rax,1)
L11:
  lea    0x1(%r13),%ecx
  jmp    L74
L114:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type: string

tool: D4: %s

assistant: ```c
void swap(char c[], int r) {
  char t[r];
  if (r % 2 == 0) {
    for (int i = 0; i < r / 2; i++) {
      t[i] = c[i + r / 2];
    }
    for (int j = r / 2; j < r; j++) {
      t[j] = c[j - r / 2];
    }
  } else {
    for (int i = 0; i < r / 2; i++) {
      t[i] = c[i + r / 2 + 1];
    }
    for (int j = r / 2 + 1; j < r; j++) {
      t[j] = c[j - r / 2 - 1];
    }
    t[r / 2] = c[r / 2];
  }
  t[r] = '\0';
  printf("%s", t);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  movslq %esi,%r12
  push   %rbx
  mov    %r12,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0xf(%r12),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L39
L15:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L15
L39:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L104
L19:
  mov    %ebx,%r13d
  mov    %rsp,%r15
  shr    $0x1f,%r13d
  add    %ebx,%r13d
  sar    %r13d
  test   $0x1,%bl
  je     L41
  cmp    $0x1,%ebx
  jle    L103
  lea    0x1(%r13),%ecx
  lea    -0x1(%r13),%edx
  mov    %rsp,%rdi
  movslq %ecx,%rsi
  add    $0x1,%rdx
  mov    %ecx,-0x44(%rbp)
  add    %r14,%rsi
  call   <memcpy@plt>
  mov    -0x44(%rbp),%ecx
L10:
  cmp    %ecx,%ebx
  jle    L31
  lea    -0x2(%rbx),%edx
  movslq %ecx,%rdi
  mov    %r14,%rsi
  mov    %edx,%ebx
  add    %r15,%rdi
  sub    %r13d,%ebx
  lea    0x1(%rbx),%rdx
  call   <memcpy@plt>
L31:
  movslq %r13d,%r13
  movzbl (%r14,%r13,1),%eax
  mov    %al,(%r15,%r13,1)
L114:
  mov    %r15,%rsi
  lea    D7(%rip),%rdi
  movb   $0x0,(%r15,%r12,1)
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nop
L41:
  cmp    $0x1,%ebx
  jle    L95
  lea    -0x1(%r13),%edx
  movslq %r13d,%rsi
  mov    %rsp,%rdi
  add    $0x1,%rdx
  add    %r14,%rsi
  call   <memcpy@plt>
L95:
  cmp    %r13d,%ebx
  jle    L114
  lea    -0x1(%rbx),%edx
  movslq %r13d,%rdi
  mov    %r14,%rsi
  mov    %edx,%ebx
  add    %r15,%rdi
  sub    %r13d,%ebx
  lea    0x1(%rbx),%rdx
  call   <memcpy@plt>
  jmp    L114
  nopl   0x0(%rax)
L104:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L103:
  lea    0x1(%r13),%ecx
  jmp    L10
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type: string

tool: D7: %s

assistant: ```c
void swap(char c[], int r) {
  char t[r];
  if (r % 2 == 0) {
    for (int i = 0; i < r / 2; i++) {
      t[i] = c[i + r / 2];
    }
    for (int j = r / 2; j < r; j++) {
      t[j] = c[j - r / 2];
    }
  } else {
    for (int i = 0; i < r / 2; i++) {
      t[i] = c[i + r / 2 + 1];
    }
    for (int j = r / 2 + 1; j < r; j++) {
      t[j] = c[j - r / 2 - 1];
    }
    t[r / 2] = c[r / 2];
  }
  t[r] = '\0';
  printf("%s", t);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L16
L39:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <PinSetINPUTS@plt>
  mov    $0x0,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <UpdateShiftCount@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L0
L46:
  movl   $0x0,-0x14(%rbp)
  jmp    L78
L23:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <PinSetINPUTS@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <UpdateShiftData@plt>
  mov    $0x0,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <PinSetINPUTS@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <UpdateShiftData@plt>
  mov    $0x0,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x0,%eax
  call   <PinGetOUTPUTS@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <GetAnswer@plt>
  cmp    %rax,%rbx
  je     L2
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L68
L2:
  addl   $0x1,-0x14(%rbp)
L78:
  cmpl   $0xff,-0x14(%rbp)
  jle    L23
  addl   $0x1,-0x18(%rbp)
L0:
  cmpl   $0xff,-0x18(%rbp)
  jle    L46
  addl   $0x1,-0x1c(%rbp)
L16:
  cmpl   $0x7,-0x1c(%rbp)
  jle    L39
  mov    $0x0,%eax
L68:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type_guessed: string

tool: D8: Mismatch


assistant: ```c
int main(int argc, char **argv) {
  int a, b, c;
  for (a = 0; a < 8; a++) {
    PinSetINPUTS(a);
    PinSetLATCH_SHIFT(0);
    PinSetLATCH_SHIFT(1);
    UpdateShiftCount(a);
    for (b = 0; b < 256; b++) {
      for (c = 0; c < 256; c++) {
        PinSetINPUTS(b);
        UpdateShiftData(b);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        PinSetINPUTS(c);
        UpdateShiftData(c);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        if (PinGetOUTPUTS() != GetAnswer()) {
          printf("Mismatch\n");
          return -1;
        }
      }
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%r13d
L22:
  mov    %r13d,%edi
  call   <PinSetINPUTS@plt>
  mov    $0x0,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    %r13d,%edi
  call   <UpdateShiftCount@plt>
  mov    $0x0,%r12d
L9:
  mov    $0x0,%ebx
L4:
  mov    %r12d,%edi
  call   <PinSetINPUTS@plt>
  mov    %r12d,%edi
  call   <UpdateShiftData@plt>
  mov    $0x0,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    %ebx,%edi
  call   <PinSetINPUTS@plt>
  mov    %ebx,%edi
  call   <UpdateShiftData@plt>
  mov    $0x0,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x0,%eax
  call   <PinGetOUTPUTS@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  call   <GetAnswer@plt>
  cmp    %rax,%rbp
  jne    L19
  add    $0x1,%ebx
  cmp    $0x100,%ebx
  jne    L4
  add    $0x1,%r12d
  cmp    $0x100,%r12d
  jne    L9
  add    $0x1,%r13d
  cmp    $0x8,%r13d
  jne    L22
  mov    $0x0,%eax
  jmp    L5
L19:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D1
type: string

tool: D1: Mismatch


assistant: ```c
int main(int argc, char **argv) {
  int a, b, c;
  for (a = 0; a < 8; a++) {
    PinSetINPUTS(a);
    PinSetLATCH_SHIFT(0);
    PinSetLATCH_SHIFT(1);
    UpdateShiftCount(a);
    for (b = 0; b < 256; b++) {
      for (c = 0; c < 256; c++) {
        PinSetINPUTS(b);
        UpdateShiftData(b);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        PinSetINPUTS(c);
        UpdateShiftData(c);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        if (PinGetOUTPUTS() != GetAnswer()) {
          printf("Mismatch\n");
          return -1;
        }
      }
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
L23:
  mov    %r13d,%edi
  xor    %r12d,%r12d
  call   <PinSetINPUTS@plt>
  xor    %edi,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    %r13d,%edi
  call   <UpdateShiftCount@plt>
L10:
  xor    %ebx,%ebx
  jmp    L5
  nopl   0x0(%rax)
L24:
  add    $0x1,%ebx
  cmp    $0x100,%ebx
  je     L59
L5:
  mov    %r12d,%edi
  call   <PinSetINPUTS@plt>
  mov    %r12d,%edi
  call   <UpdateShiftData@plt>
  xor    %edi,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    %ebx,%edi
  call   <PinSetINPUTS@plt>
  mov    %ebx,%edi
  call   <UpdateShiftData@plt>
  xor    %edi,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  xor    %eax,%eax
  call   <PinGetOUTPUTS@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <GetAnswer@plt>
  cmp    %rax,%rbp
  je     L24
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L59:
  add    $0x1,%r12d
  cmp    $0x100,%r12d
  jne    L10
  add    $0x1,%r13d
  cmp    $0x8,%r13d
  jne    L23
  xor    %eax,%eax
  jmp    L19
```

tool_calls: name: unpack_data
arguments: label: D1
type_infered: string

tool: D1: Mismatch


assistant: ```c
int main(int argc, char **argv) {
  int a, b, c;
  for (a = 0; a < 8; a++) {
    PinSetINPUTS(a);
    PinSetLATCH_SHIFT(0);
    PinSetLATCH_SHIFT(1);
    UpdateShiftCount(a);
    for (b = 0; b < 256; b++) {
      for (c = 0; c < 256; c++) {
        PinSetINPUTS(b);
        UpdateShiftData(b);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        PinSetINPUTS(c);
        UpdateShiftData(c);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        if (PinGetOUTPUTS() != GetAnswer()) {
          printf("Mismatch\n");
          return -1;
        }
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
L2:
  mov    %r13d,%edi
  xor    %r12d,%r12d
  call   <PinSetINPUTS@plt>
  xor    %edi,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_SHIFT@plt>
  mov    %r13d,%edi
  call   <UpdateShiftCount@plt>
L6:
  xor    %ebx,%ebx
  jmp    L4
  nopl   0x0(%rax)
L3:
  add    $0x1,%ebx
  cmp    $0x100,%ebx
  je     L13
L4:
  mov    %r12d,%edi
  call   <PinSetINPUTS@plt>
  mov    %r12d,%edi
  call   <UpdateShiftData@plt>
  xor    %edi,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    %ebx,%edi
  call   <PinSetINPUTS@plt>
  mov    %ebx,%edi
  call   <UpdateShiftData@plt>
  xor    %edi,%edi
  call   <PinSetLATCH_DATA@plt>
  mov    $0x1,%edi
  call   <PinSetLATCH_DATA@plt>
  xor    %eax,%eax
  call   <PinGetOUTPUTS@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <GetAnswer@plt>
  cmp    %rax,%rbp
  je     L3
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
L25:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  add    $0x1,%r12d
  cmp    $0x100,%r12d
  jne    L6
  add    $0x1,%r13d
  cmp    $0x8,%r13d
  jne    L2
  xor    %eax,%eax
  jmp    L25
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_infered: string

tool: D2: Mismatch


assistant: ```c
int main(int argc, char **argv) {
  int a, b, c;
  for (a = 0; a < 8; a++) {
    PinSetINPUTS(a);
    PinSetLATCH_SHIFT(0);
    PinSetLATCH_SHIFT(1);
    UpdateShiftCount(a);
    for (b = 0; b < 256; b++) {
      for (c = 0; c < 256; c++) {
        PinSetINPUTS(b);
        UpdateShiftData(b);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        PinSetINPUTS(c);
        UpdateShiftData(c);
        PinSetLATCH_DATA(0);
        PinSetLATCH_DATA(1);
        if (PinGetOUTPUTS() != GetAnswer()) {
          printf("Mismatch\n");
          return -1;
        }
      }
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L2:
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L2
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type: string
name: retrieve_data
arguments: data_tag: D7
type: string

tool: D3: i:%d


tool: D7: haha

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 1;
  printf("i:%d\n", i);
  do {
    printf("haha\n");
    continue;
  } while (--i);
  printf("i:%d\n", i);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%esi
  lea    D18(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: label: D18
type_guessed: string
name: retrieve_data
arguments: label: D14
type_guessed: string

tool: D18: i:%d


tool: D14: haha

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 1;
  printf("i:%d\n", i);
  do {
    printf("haha\n");
    continue;
  } while (--i);
  printf("i:%d\n", i);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rbp
  mov    $0x1,%esi
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type: string
name: extract_symbolic_data
arguments: tag: D0
type: string

tool: D14: i:%d


tool: D0: haha

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 1;
  printf("i:%d\n", i);
  do {
    printf("haha\n");
    continue;
  } while (--i);
  printf("i:%d\n", i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  mov    $0x1,%esi
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
datatype: string
name: parse_data
arguments: data_tag: D0
datatype: string

tool: D2: i:%d


tool: D0: haha

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 1;
  printf("i:%d\n", i);
  do {
    printf("haha\n");
    continue;
  } while (--i);
  printf("i:%d\n", i);
  return 0;
}
```