ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<bubble_sort>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L30
  lea    -0x4(%rdi,%rsi,4),%r8
  xor    %r9d,%r9d
  nopw   0x0(%rax,%rax,1)
L31:
  mov    %rdi,%rdx
  nopl   0x0(%rax,%rax,1)
L37:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%ecx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L39
  movq   %xmm1,(%rdx)
L39:
  add    $0x4,%rdx
  cmp    %r8,%rdx
  jne    L37
  add    $0x1,%r9
  cmp    %r9,%rsi
  jne    L31
L30:
  ret
```

assistant: ```c
void bubble_sort(int *array, size_t n) {
  for (size_t i = 0; i < n; ++i) {
    for (size_t j = 0; j < n - 1; ++j) {
      if (array[j] > array[j + 1]) {
        int tmp = array[j];
        array[j] = array[j + 1];
        array[j + 1] = tmp;
      }
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<my_strncmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L89
L20:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L47
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  test   %eax,%eax
  jle    L110
  mov    $0x1,%eax
  jmp    L32
L110:
  mov    $0xffffffff,%eax
  jmp    L32
L47:
  nop
  addl   $0x1,-0x4(%rbp)
L89:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jge    L136
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jge    L136
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L20
L136:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  je     L130
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jne    L22
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jne    L22
L130:
  mov    $0x0,%eax
  jmp    L32
L22:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jne    L127
  mov    $0xffffffff,%eax
  jmp    L32
L127:
  mov    $0x1,%eax
L32:
  leave
  ret
```

assistant: ```c
int my_strncmp(char *firstArgument, char *secondArgument, int n) {
  int i;
  for (i = 0; i < n && i < strlen(firstArgument) && i < strlen(secondArgument); i++) {
    if (firstArgument[i] - secondArgument[i] == 0) {
      continue;
    } else {
      return (firstArgument[i] - secondArgument[i] > 0) ? 1 : -1;
    }
  }
  if (i == n || (i == strlen(firstArgument) && i == strlen(secondArgument))) {
    return 0;
  }
  if (i == strlen(firstArgument)) {
    return -1;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_strncmp>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %edx,%r13d
  mov    %edx,%r15d
  mov    $0x0,%ebx
  mov    $0x0,%r14d
  test   %edx,%edx
  jle    L8
L38:
  mov    %ebx,0xc(%rsp)
  mov    %ebx,%r14d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L8
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L8
  movzbl 0x0(%rbp,%rbx,1),%edx
  movzbl (%r12,%rbx,1),%eax
  cmp    %al,%dl
  je     L39
  movsbl %dl,%edx
  movsbl %al,%eax
  cmp    %eax,%edx
  setg   %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
L14:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  mov    0xc(%rsp),%r14d
  add    $0x1,%r14d
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L38
L8:
  mov    $0x0,%eax
  cmp    %r14d,%r13d
  je     L14
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  je     L9
L92:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  setne  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
  jmp    L14
L9:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    %r14d,%edx
  jne    L92
  jmp    L14
```

assistant: ```c
int my_strncmp(char *firstArgument, char *secondArgument, int n) {
  int i;
  for (i = 0; i < n && i < strlen(firstArgument) && i < strlen(secondArgument); i++) {
    if (firstArgument[i] - secondArgument[i] == 0) {
      continue;
    } else {
      return (firstArgument[i] - secondArgument[i] > 0) ? 1 : -1;
    }
  }
  if (i == n || (i == strlen(firstArgument) && i == strlen(secondArgument))) {
    return 0;
  }
  if (i == strlen(firstArgument)) {
    return -1;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<my_strncmp>:
  endbr64
  push   %r15
  movslq %edx,%r15
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %r15,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %r15d,%r15d
  jle    L80
  xor    %ebx,%ebx
L63:
  mov    %rbp,%rdi
  mov    %ebx,%r14d
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L25
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L25
  movzbl 0x0(%rbp,%rbx,1),%eax
  movzbl (%r12,%rbx,1),%edx
  cmp    %dl,%al
  je     L98
  cmp    %al,%dl
  jge    L77
L3:
  mov    $0x1,%eax
L116:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L98:
  add    $0x1,%rbx
  add    $0x1,%r14d
  cmp    %rbx,%r15
  jne    L63
L80:
  cmp    %r13d,%r14d
  jne    L25
L37:
  xor    %eax,%eax
  jmp    L116
  cs nopw 0x0(%rax,%rax,1)
L25:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  je     L72
L26:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  jne    L3
L77:
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L72:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  jne    L26
  jmp    L37
```

assistant: ```c
int my_strncmp(char *firstArgument, char *secondArgument, int n) {
  int i;
  for (i = 0; i < n && i < strlen(firstArgument) && i < strlen(secondArgument); i++) {
    if (firstArgument[i] - secondArgument[i] == 0) {
      continue;
    } else {
      return (firstArgument[i] - secondArgument[i] > 0) ? 1 : -1;
    }
  }
  if (i == n || (i == strlen(firstArgument) && i == strlen(secondArgument))) {
    return 0;
  }
  if (i == strlen(firstArgument)) {
    return -1;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<my_strncmp>:
  endbr64
  push   %r15
  movslq %edx,%r15
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %r15,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %r15d,%r15d
  jle    L52
  xor    %ebx,%ebx
L84:
  mov    %rbp,%rdi
  mov    %ebx,%r14d
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L72
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L72
  movzbl 0x0(%rbp,%rbx,1),%eax
  movzbl (%r12,%rbx,1),%edx
  cmp    %dl,%al
  je     L112
  cmp    %al,%dl
  jge    L85
L33:
  mov    $0x1,%eax
L69:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L112:
  add    $0x1,%rbx
  add    $0x1,%r14d
  cmp    %rbx,%r15
  jne    L84
L52:
  cmp    %r13d,%r14d
  jne    L72
L2:
  xor    %eax,%eax
  jmp    L69
  cs nopw 0x0(%rax,%rax,1)
L72:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  je     L90
L47:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  jne    L33
L85:
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L90:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  jne    L47
  jmp    L2
```

assistant: ```c
int my_strncmp(char *firstArgument, char *secondArgument, int n) {
  int i;
  for (i = 0; i < n && i < strlen(firstArgument) && i < strlen(secondArgument); i++) {
    if (firstArgument[i] - secondArgument[i] == 0) {
      continue;
    } else {
      return (firstArgument[i] - secondArgument[i] > 0) ? 1 : -1;
    }
  }
  if (i == n || (i == strlen(firstArgument) && i == strlen(secondArgument))) {
    return 0;
  }
  if (i == strlen(firstArgument)) {
    return -1;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ifc_disable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <ifc_init@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ifc_down@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <ifc_set_addr@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L28
L15:
  lea    -0x14(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ifc_get_addr@plt>
  test   %rax,%rax
  js     L41
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L52
  mov    -0x28(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <ifc_set_addr@plt>
L52:
  addl   $0x1,-0x10(%rbp)
L28:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L23
  cmpl   $0xfe,-0x10(%rbp)
  jbe    L15
  jmp    L23
L41:
  nop
L23:
  mov    $0x0,%eax
  call   <ifc_close@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int ifc_disable(const char *ifname) {
  unsigned addr, count;
  int result;
  ifc_init();
  result = ifc_down(ifname);
  ifc_set_addr(ifname, 0);
  for (count = 0, addr = 1; ((addr != 0) && (count < 255)); count++) {
    if (ifc_get_addr(ifname, &addr) < 0)
      break;
    if (addr)
      ifc_set_addr(ifname, 0);
  }
  ifc_close();
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ifc_disable>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <ifc_init@plt>
  mov    %rbp,%rdi
  call   <ifc_down@plt>
  mov    %eax,%r13d
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <ifc_set_addr@plt>
  movl   $0x1,0x4(%rsp)
  mov    $0x0,%ebx
  lea    0x4(%rsp),%r12
  jmp    L28
L57:
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <ifc_set_addr@plt>
  add    $0x1,%ebx
  cmpl   $0x0,0x4(%rsp)
  je     L48
  cmp    $0xfe,%ebx
  ja     L48
L28:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ifc_get_addr@plt>
  test   %rax,%rax
  js     L48
  cmpl   $0x0,0x4(%rsp)
  jne    L57
L48:
  mov    $0x0,%eax
  call   <ifc_close@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    %r13d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ifc_disable(const char *ifname) {
  unsigned addr, count;
  int result;
  ifc_init();
  result = ifc_down(ifname);
  ifc_set_addr(ifname, 0);
  for (count = 0, addr = 1; ((addr != 0) && (count < 255)); count++) {
    if (ifc_get_addr(ifname, &addr) < 0)
      break;
    if (addr)
      ifc_set_addr(ifname, 0);
  }
  ifc_close();
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ifc_disable>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  call   <ifc_init@plt>
  mov    %rbp,%rdi
  call   <ifc_down@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <ifc_set_addr@plt>
  movl   $0x1,0x4(%rsp)
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ifc_get_addr@plt>
  test   %rax,%rax
  js     L7
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jne    L14
L7:
  xor    %eax,%eax
  call   <ifc_close@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L40
  add    $0x18,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L14:
  xor    %esi,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <ifc_set_addr@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L7
  cmp    $0xfe,%ebx
  jbe    L11
  jmp    L7
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ifc_disable(const char *ifname) {
  unsigned addr, count;
  int result;
  ifc_init();
  result = ifc_down(ifname);
  ifc_set_addr(ifname, 0);
  for (count = 0, addr = 1; ((addr != 0) && (count < 255)); count++) {
    if (ifc_get_addr(ifname, &addr) < 0)
      break;
    if (addr)
      ifc_set_addr(ifname, 0);
  }
  ifc_close();
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ifc_disable>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  call   <ifc_init@plt>
  mov    %rbp,%rdi
  call   <ifc_down@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <ifc_set_addr@plt>
  movl   $0x1,0x4(%rsp)
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <ifc_get_addr@plt>
  test   %rax,%rax
  js     L13
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jne    L25
L13:
  xor    %eax,%eax
  call   <ifc_close@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x18,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L25:
  xor    %esi,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <ifc_set_addr@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L13
  cmp    $0xfe,%ebx
  jbe    L19
  jmp    L13
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ifc_disable(const char *ifname) {
  unsigned addr, count;
  int result;
  ifc_init();
  result = ifc_down(ifname);
  ifc_set_addr(ifname, 0);
  for (count = 0, addr = 1; ((addr != 0) && (count < 255)); count++) {
    if (ifc_get_addr(ifname, &addr) < 0)
      break;
    if (addr)
      ifc_set_addr(ifname, 0);
  }
  ifc_close();
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L10
L9:
  mov    -0x10(%rbp),%eax
  sub    $0x7a8,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  add    $0x2,%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%edi
  call   <solve@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L10:
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D18
type_guessed: string
name: parse_data
arguments: data_label: D9
type_guessed: string

tool: D18: %d


tool: D9: %d

assistant: ```c
int main() {
  int year;
  int ans;
  while (scanf("%d", &year), year) {
    ans = solve(1 << (2 + (year - 1960) / 10));
    printf("%d\n", ans);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  lea    D4(%rip),%rbx
  mov    $0x1,%r13d
  lea    D6(%rip),%r12
  jmp    L12
L15:
  sub    $0x7a8,%eax
  movslq %eax,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sar    $0x1f,%eax
  sub    %eax,%ecx
  add    $0x2,%ecx
  mov    %r13d,%edi
  shl    %cl,%edi
  call   <solve@plt>
  mov    %eax,%esi
  mov    %r12,%rdi
  call   <printf@plt>
L12:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L15
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D4
type: string
name: read_data
arguments: data_label: D6
type: string

tool: D4: %d

tool: D6: %d


assistant: ```c
int main() {
  int year;
  int ans;
  while (scanf("%d", &year), year) {
    ans = solve(1 << (2 + (year - 1960) / 10));
    printf("%d\n", ans);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  lea    D19(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D10(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  jmp    L4
  nopl   0x0(%rax)
L3:
  sub    $0x7a8,%eax
  mov    $0x1,%edi
  movslq %eax,%rcx
  sar    $0x1f,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sub    %eax,%ecx
  add    $0x2,%ecx
  shl    %cl,%edi
  call   <solve@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
L4:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L3
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D19
type: string
name: extract_symbolic_data
arguments: data_label: D10
type: string

tool: D19: %d


tool: D10: %d

assistant: ```c
int main() {
  int year;
  int ans;
  while (scanf("%d", &year), year) {
    ans = solve(1 << (2 + (year - 1960) / 10));
    printf("%d\n", ans);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D0(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  jmp    L5
  nopl   0x0(%rax)
L13:
  sub    $0x7a8,%eax
  mov    $0x1,%edi
  movslq %eax,%rcx
  sar    $0x1f,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sub    %eax,%ecx
  add    $0x2,%ecx
  shl    %cl,%edi
  call   <solve@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
L5:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L13
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D6
datatype: string
name: read_data
arguments: data_label: D0
datatype: string

tool: D6: %d


tool: D0: %d

assistant: ```c
int main() {
  int year;
  int ans;
  while (scanf("%d", &year), year) {
    ans = solve(1 << (2 + (year - 1960) / 10));
    printf("%d\n", ans);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D13(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int rename(const char *oldname, const char *newname) {
  errno = ENOENT;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rename>:
  endbr64
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    D0(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int rename(const char *oldname, const char *newname) {
  errno = ENOENT;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rename>:
  endbr64
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int rename(const char *oldname, const char *newname) {
  errno = ENOENT;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rename>:
  endbr64
  mov    D18(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int rename(const char *oldname, const char *newname) {
  errno = ENOENT;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f7>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <allocate_and_init@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L1
  mov    $0x0,%eax
  call   <do_stuff@plt>
L1:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int f7() {
  int rc = allocate_and_init();
  if (rc != 0)
    do_stuff();
  return rc;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<f7>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <allocate_and_init@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L16
L1:
  mov    %ebx,%eax
  pop    %rbx
  ret
L16:
  mov    $0x0,%eax
  call   <do_stuff@plt>
  jmp    L1
```

assistant: ```c
int f7() {
  int rc = allocate_and_init();
  if (rc != 0)
    do_stuff();
  return rc;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<f7>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <allocate_and_init@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L2
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  call   <do_stuff@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int f7() {
  int rc = allocate_and_init();
  if (rc != 0)
    do_stuff();
  return rc;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f7>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <allocate_and_init@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L5
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  call   <do_stuff@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int f7() {
  int rc = allocate_and_init();
  if (rc != 0)
    do_stuff();
  return rc;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1020101,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  movzwl %ax,%eax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %u


assistant: ```c
int main() {
  unsigned int a = 1 + (1 << 8) + (1 << 17) + (1 << 24);
  unsigned short *b = *((unsigned short *)(&a) + 1);
  printf("%u\n", b);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x102,%esi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type: string

tool: D0: %u


assistant: ```c
int main() {
  unsigned int a = 1 + (1 << 8) + (1 << 17) + (1 << 24);
  unsigned short *b = *((unsigned short *)(&a) + 1);
  printf("%u\n", b);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x102,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
datatype: string

tool: D6: %u


assistant: ```c
int main() {
  unsigned int a = 1 + (1 << 8) + (1 << 17) + (1 << 24);
  unsigned short *b = *((unsigned short *)(&a) + 1);
  printf("%u\n", b);
  return 0;
}
```