ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inet6_opt_append>:
  endbr64
  push   %r12
  movd   %ecx,%xmm0
  movd   %r8d,%xmm2
  push   %rbp
  punpckldq %xmm2,%xmm0
  push   %rbx
  sub    $0x10,%rsp
  cmp    $0xff,%r8d
  jg     L110
  cmp    $0x1,%ecx
  jle    L110
  lea    -0x4(%r9),%eax
  mov    %esi,%r10d
  movslq %edx,%rsi
  and    $0xfffffffb,%eax
  je     L92
  lea    -0x1(%r9),%eax
  cmp    $0x1,%eax
  ja     L110
L92:
  cmp    %r9d,%r8d
  jl     L110
  lea    0x2(%rsi,%r8,1),%r12d
  mov    %r12d,%eax
  cltd
  idiv   %r9d
  movslq %edx,%rbx
  test   %ebx,%ebx
  je     L67
  sub    %ebx,%r9d
  movslq %r9d,%rbx
  add    %ebx,%r12d
L67:
  test   %r10d,%r10d
  je     L46
  cmp    %r12d,%r10d
  jl     L110
L46:
  test   %rdi,%rdi
  je     L96
  lea    (%rdi,%rsi,4),%rax
  cmp    $0x1,%ebx
  je     L16
  test   %ebx,%ebx
  jle    L62
  mov    D14(%rip),%rcx
  lea    -0x2(%rbx),%edx
  lea    0x8(%rax),%rbp
  xor    %esi,%esi
  movd   %edx,%xmm3
  mov    %rbp,%rdi
  movq   %xmm0,0x8(%rsp)
  movd   (%rcx),%xmm1
  punpckldq %xmm3,%xmm1
  movq   %xmm1,(%rax)
  call   <memset@plt>
  movq   0x8(%rsp),%xmm0
  lea    -0x8(%rbp,%rbx,4),%rax
  jmp    L62
  nopl   0x0(%rax,%rax,1)
L16:
  mov    D17(%rip),%rdx
  add    $0x4,%rax
  mov    (%rdx),%edx
  mov    %edx,-0x4(%rax)
L62:
  mov    0x30(%rsp),%rdx
  movq   %xmm0,(%rax)
  add    $0x8,%rax
  mov    %rax,(%rdx)
L96:
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L110:
  mov    $0xffffffff,%r12d
  jmp    L96
```

assistant: ```c
__attribute__((visibility("default"))) int inet6_opt_append(void *extbuf, socklen_t extlen, int offset, u_int8_t type, socklen_t len, u_int8_t align, void **databufp) {
  int currentlen = offset, padlen = 0;
  if (type < 2)
    return (-1);
  if (len > 255) {
    return (-1);
  }
  if (align != 1 && align != 2 && align != 4 && align != 8)
    return (-1);
  if (align > len)
    return (-1);
  currentlen += 2 + len;
  if (currentlen % align)
    padlen = align - (currentlen % align);
  currentlen += padlen;
  if (extlen && currentlen > extlen)
    return (-1);
  if (extbuf) {
    u_int8_t *optp = (u_int8_t *)extbuf + offset;
    if (padlen == 1) {
      *optp = IP6OPT_PAD1;
      optp++;
    } else if (padlen > 0) {
      *optp++ = IP6OPT_PADN;
      *optp++ = padlen - 2;
      memset(optp, 0, padlen - 2);
      optp += (padlen - 2);
    }
    *optp++ = type;
    *optp++ = len;
    *databufp = optp;
  }
  return (currentlen);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
L5:
  cmpl   $0x0,-0x4(%rbp)
  jne    L5
  nop
  pop    %rbp
  ret
```

assistant: ```c
test8(x) {
  {
    struct {
      int foo;
    } bar(charstar);
  }
  {
    do {
      int foo;
    } while (x);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test8>:
  endbr64
L0:
  test   %edi,%edi
  jne    L0
  ret
```

assistant: ```c
test8(x) {
  {
    struct {
      int foo;
    } bar(charstar);
  }
  {
    do {
      int foo;
    } while (x);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test8>:
  endbr64
  nopl   0x0(%rax)
L7:
  test   %edi,%edi
  jne    L7
  ret
```

assistant: ```c
test8(x) {
  {
    struct {
      int foo;
    } bar(charstar);
  }
  {
    do {
      int foo;
    } while (x);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<test8>:
  endbr64
  test   %edi,%edi
  je     L8
L13:
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L8:
  ret
```

assistant: ```c
test8(x) {
  {
    struct {
      int foo;
    } bar(charstar);
  }
  {
    do {
      int foo;
    } while (x);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <geteuid@plt>
  mov    %eax,-0x4(%rbp)
  nop
  leave
  ret
```

assistant: ```c
void test(void) {
  uid_t uid;
  uid = geteuid();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <geteuid@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test(void) {
  uid_t uid;
  uid = geteuid();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <geteuid@plt>
```

assistant: ```c
void test(void) {
  uid_t uid;
  uid = geteuid();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <geteuid@plt>
```

assistant: ```c
void test(void) {
  uid_t uid;
  uid = geteuid();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<do_seg_not_present>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
L6:
  jmp    L6
```

tool_calls: name: unpack_data
arguments: data_label: D19
data_type: string
name: unpack_data
arguments: data_label: D13
data_type: string

tool: D19: do_seg_not_present

tool: D13: Unsupport Now...[%s]


assistant: ```c
void do_seg_not_present(void *x) {
  do {
    kprintf("Unsupport Now...[%s]\n", __FUNCTION__);
    while (1)
      ;
  } while (0);
  ;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<do_seg_not_present>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rsi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
L8:
  jmp    L8
```

tool_calls: name: unpack_data
arguments: data_label: D5
type: string
name: unpack_data
arguments: data_label: D0
type: string

tool: D5: do_seg_not_present

tool: D0: Unsupport Now...[%s]


assistant: ```c
void do_seg_not_present(void *x) {
  do {
    kprintf("Unsupport Now...[%s]\n", __FUNCTION__);
    while (1)
      ;
  } while (0);
  ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<do_seg_not_present>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rsi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <kprintf@plt>
L6:
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D19
type_guessed: string

tool: D5: do_seg_not_present

tool: D19: Unsupport Now...[%s]


assistant: ```c
void do_seg_not_present(void *x) {
  do {
    kprintf("Unsupport Now...[%s]\n", __FUNCTION__);
    while (1)
      ;
  } while (0);
  ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<do_seg_not_present>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rsi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <kprintf@plt>
L5:
  jmp    L5
```

tool_calls: name: retrieve_data
arguments: label: D16
type: string
name: retrieve_data
arguments: label: D7
type: string

tool: D16: do_seg_not_present

tool: D7: Unsupport Now...[%s]


assistant: ```c
void do_seg_not_present(void *x) {
  do {
    kprintf("Unsupport Now...[%s]\n", __FUNCTION__);
    while (1)
      ;
  } while (0);
  ;
}
```
user: What is the c source code of the assembly code below:

```asm
<prim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0x32,%ecx
  rep stos %rax,%es:(%rdi)
  lea    D0(%rip),%rdi
  call   <puts@plt>
  cmp    $0x1,%r12d
  jle    L11
  lea    -0x1(%r12),%r15d
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  jmp    L92
L22:
  lea    0x1(%rax),%rdx
  cmp    %rax,%r8
  je     L0
  mov    %rdx,%rax
L121:
  mov    (%rdi,%rax,4),%edx
  cmp    %ecx,%edx
  jge    L22
  cmpl   $0x0,(%rsi)
  cmove  %eax,%ebx
  cmove  %r9d,%ebp
  cmove  %edx,%ecx
  jmp    L22
L0:
  add    $0x1,%r9d
  add    $0x4,%rsi
  add    $0x190,%r10
  cmp    %r9d,%r12d
  je     L71
L56:
  mov    %r10,%rdi
  mov    %r13,%rax
  jmp    L121
L117:
  mov    $0xffffffff,%ebx
  mov    $0xffffffff,%ebp
L71:
  movslq %ebp,%rax
  cmpl   $0x0,0x10(%rsp,%rax,4)
  je     L59
  movslq %ebx,%rax
  cmpl   $0x0,0x10(%rsp,%rax,4)
  jne    L65
L59:
  lea    0x1(%rbx),%edx
  lea    0x1(%rbp),%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movslq %ebp,%rbp
  lea    0x0(%rbp,%rbp,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  mov    0x8(%rsp),%rdi
  add    %rdi,%rax
  movslq %ebx,%rbx
  movl   $0x270f,(%rax,%rbx,4)
  lea    (%rbx,%rbx,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  add    %rdi,%rax
  movl   $0x270f,(%rax,%rbp,4)
  movl   $0x1,0x10(%rsp,%rbx,4)
L65:
  add    $0x1,%r14d
  cmp    %r15d,%r14d
  je     L11
L92:
  test   %r12d,%r12d
  jle    L117
  lea    0x10(%rsp),%rsi
  mov    0x8(%rsp),%r10
  mov    $0xffffffff,%ebx
  mov    $0xffffffff,%ebp
  mov    $0x270f,%ecx
  mov    $0x0,%r9d
  mov    %r15d,%r8d
  jmp    L56
L11:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  add    $0x1b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
data_type: string
name: unpack_data
arguments: data_label: D3
data_type: string

tool: D0: Minimum Spanning Tree : 

tool: D3: %d - %d


assistant: ```c
void prim(int adj[100][100], int N) {
  int visited[100] = {0};
  int i, j, count = 0;
  printf("Minimum Spanning Tree : \n");
  while (count < N - 1) {
    int min = 9999, u = -1, v = -1;
    for (i = 0; i < N; i++) {
      for (j = 0; j < N; j++) {
        if (adj[i][j] < min && !visited[i]) {
          min = adj[i][j];
          u = i;
          v = j;
        }
      }
    }
    if (visited[u] == 0 || visited[v] == 0) {
      printf("%d - %d\n", (u + 1), (v + 1));
      adj[u][v] = 9999;
      adj[v][u] = 9999;
      visited[v] = 1;
    }
    count++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<prim>:
  endbr64
  push   %r15
  mov    $0x32,%ecx
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  lea    -0x1(%r12),%r15d
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    %rdi,0x8(%rsp)
  rep stos %rax,%es:(%rdi)
  lea    D1(%rip),%rdi
  call   <puts@plt>
  test   %r15d,%r15d
  jle    L50
  xor    %r13d,%r13d
  nopl   (%rax)
L60:
  test   %r12d,%r12d
  jle    L108
  mov    0x8(%rsp),%r11
  mov    %r14,%rsi
  mov    $0xffffffff,%edi
  xor    %r10d,%r10d
  mov    $0xffffffff,%r8d
  mov    $0x270f,%ecx
  mov    %r15d,%r9d
  nopw   0x0(%rax,%rax,1)
L47:
  xor    %eax,%eax
  jmp    L62
  nopl   0x0(%rax)
L79:
  mov    %rdx,%rax
L62:
  mov    (%rsi,%rax,4),%edx
  cmp    %ecx,%edx
  jge    L9
  mov    (%r11),%ebx
  test   %ebx,%ebx
  cmove  %eax,%r8d
  cmove  %r10d,%edi
  cmove  %edx,%ecx
L9:
  lea    0x1(%rax),%rdx
  cmp    %rax,%r9
  jne    L79
  add    $0x1,%r10d
  add    $0x190,%rsi
  add    $0x4,%r11
  cmp    %r10d,%r12d
  jne    L47
L81:
  movslq %edi,%rbp
  movslq %r8d,%rbx
  mov    0x10(%rsp,%rbp,4),%edx
  test   %edx,%edx
  je     L3
  mov    0x10(%rsp,%rbx,4),%eax
  test   %eax,%eax
  jne    L46
L3:
  lea    0x1(%rdi),%esi
  lea    0x1(%r8),%edx
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    0x0(%rbp,%rbp,4),%rax
  movl   $0x1,0x10(%rsp,%rbx,4)
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  add    %r14,%rax
  movl   $0x270f,(%rax,%rbx,4)
  lea    (%rbx,%rbx,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  add    %r14,%rax
  movl   $0x270f,(%rax,%rbp,4)
L46:
  add    $0x1,%r13d
  cmp    %r13d,%r15d
  jne    L60
L50:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L89
  add    $0x1b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L108:
  mov    $0xffffffff,%r8d
  mov    $0xffffffff,%edi
  jmp    L81
L89:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D1
type: string
name: retrieve_data
arguments: data_label: D3
type: string

tool: D1: Minimum Spanning Tree : 

tool: D3: %d - %d


assistant: ```c
void prim(int adj[100][100], int N) {
  int visited[100] = {0};
  int i, j, count = 0;
  printf("Minimum Spanning Tree : \n");
  while (count < N - 1) {
    int min = 9999, u = -1, v = -1;
    for (i = 0; i < N; i++) {
      for (j = 0; j < N; j++) {
        if (adj[i][j] < min && !visited[i]) {
          min = adj[i][j];
          u = i;
          v = j;
        }
      }
    }
    if (visited[u] == 0 || visited[v] == 0) {
      printf("%d - %d\n", (u + 1), (v + 1));
      adj[u][v] = 9999;
      adj[v][u] = 9999;
      visited[v] = 1;
    }
    count++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<prim>:
  endbr64
  push   %r15
  mov    $0x32,%ecx
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  lea    -0x1(%r12),%r15d
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    %rdi,0x8(%rsp)
  rep stos %rax,%es:(%rdi)
  lea    D7(%rip),%rdi
  call   <puts@plt>
  test   %r15d,%r15d
  jle    L29
  xor    %r13d,%r13d
  nopl   (%rax)
L43:
  test   %r12d,%r12d
  jle    L88
  mov    0x8(%rsp),%r11
  mov    %r14,%rsi
  mov    $0xffffffff,%edi
  xor    %r10d,%r10d
  mov    $0xffffffff,%r8d
  mov    $0x270f,%ecx
  mov    %r15d,%r9d
  nopw   0x0(%rax,%rax,1)
L64:
  xor    %eax,%eax
  jmp    L38
  nopl   0x0(%rax)
L0:
  mov    %rdx,%rax
L38:
  mov    (%rsi,%rax,4),%edx
  cmp    %ecx,%edx
  jge    L95
  mov    (%r11),%ebx
  test   %ebx,%ebx
  cmove  %eax,%r8d
  cmove  %r10d,%edi
  cmove  %edx,%ecx
L95:
  lea    0x1(%rax),%rdx
  cmp    %r9,%rax
  jne    L0
  add    $0x1,%r10d
  add    $0x4,%r11
  add    $0x190,%rsi
  cmp    %r10d,%r12d
  jne    L64
L40:
  movslq %edi,%rbp
  movslq %r8d,%rbx
  mov    0x10(%rsp,%rbp,4),%edx
  test   %edx,%edx
  je     L19
  mov    0x10(%rsp,%rbx,4),%eax
  test   %eax,%eax
  jne    L3
L19:
  lea    0x1(%rdi),%esi
  lea    0x1(%r8),%edx
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    0x0(%rbp,%rbp,4),%rax
  movl   $0x1,0x10(%rsp,%rbx,4)
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  add    %r14,%rax
  movl   $0x270f,(%rax,%rbx,4)
  lea    (%rbx,%rbx,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  add    %r14,%rax
  movl   $0x270f,(%rax,%rbp,4)
L3:
  add    $0x1,%r13d
  cmp    %r13d,%r15d
  jne    L43
L29:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x1b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L88:
  mov    $0xffffffff,%r8d
  mov    $0xffffffff,%edi
  jmp    L40
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type: string
name: extract_symbolic_data
arguments: data_tag: D1
type: string

tool: D7: Minimum Spanning Tree : 

tool: D1: %d - %d


assistant: ```c
void prim(int adj[100][100], int N) {
  int visited[100] = {0};
  int i, j, count = 0;
  printf("Minimum Spanning Tree : \n");
  while (count < N - 1) {
    int min = 9999, u = -1, v = -1;
    for (i = 0; i < N; i++) {
      for (j = 0; j < N; j++) {
        if (adj[i][j] < min && !visited[i]) {
          min = adj[i][j];
          u = i;
          v = j;
        }
      }
    }
    if (visited[u] == 0 || visited[v] == 0) {
      printf("%d - %d\n", (u + 1), (v + 1));
      adj[u][v] = 9999;
      adj[v][u] = 9999;
      visited[v] = 1;
    }
    count++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<showTime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <difftime@plt>
  mov    %eax,%edx
  mov    -0x10(%rbp),%esi
  mov    -0xc(%rbp),%eax
  mov    %edx,%ecx
  lea    D5(%rip),%rdx
  mov    %eax,%edi
  call   <mvprintw@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
type_infered: string

tool: D5: time: %.0lf sec

assistant: ```c
void showTime(time_t startTime, time_t playTime, int Xtxt, int Ytxt) {
  mvprintw(Xtxt, Ytxt, "time: %.0lf sec", difftime(playTime, startTime));
  refresh();
}
```
user: What is the c source code of the assembly code below:

```asm
<showTime>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    %esi,%edi
  mov    %edx,%ebx
  mov    %ecx,%ebp
  mov    %eax,%esi
  call   <difftime@plt>
  mov    %eax,%ecx
  lea    D5(%rip),%rdx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <mvprintw@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
datatype: string

tool: D5: time: %.0lf sec

assistant: ```c
void showTime(time_t startTime, time_t playTime, int Xtxt, int Ytxt) {
  mvprintw(Xtxt, Ytxt, "time: %.0lf sec", difftime(playTime, startTime));
  refresh();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<showTime>:
  endbr64
  push   %r12
  mov    %edi,%r8d
  mov    %esi,%edi
  mov    %ecx,%r12d
  push   %rbp
  mov    %r8d,%esi
  mov    %edx,%ebp
  sub    $0x8,%rsp
  call   <difftime@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  lea    D5(%rip),%rdx
  mov    %eax,%ecx
  call   <mvprintw@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1080 <refresh@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type: string

tool: D5: time: %.0lf sec

assistant: ```c
void showTime(time_t startTime, time_t playTime, int Xtxt, int Ytxt) {
  mvprintw(Xtxt, Ytxt, "time: %.0lf sec", difftime(playTime, startTime));
  refresh();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<showTime>:
  endbr64
  push   %r12
  mov    %edi,%r8d
  mov    %esi,%edi
  mov    %ecx,%r12d
  push   %rbp
  mov    %r8d,%esi
  mov    %edx,%ebp
  sub    $0x8,%rsp
  call   <difftime@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  lea    D7(%rip),%rdx
  mov    %eax,%ecx
  call   <mvprintw@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1080 <refresh@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D7
datatype: string

tool: D7: time: %.0lf sec

assistant: ```c
void showTime(time_t startTime, time_t playTime, int Xtxt, int Ytxt) {
  mvprintw(Xtxt, Ytxt, "time: %.0lf sec", difftime(playTime, startTime));
  refresh();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<restore_state>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0xc(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x10(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void restore_state(vec, a, b, c, d, e) int vec[5];
int *a, *b, *c, *d, *e;
{
  *a = vec[0];
  *b = vec[1];
  *c = vec[2];
  *d = vec[3];
  *e = vec[4];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<restore_state>:
  endbr64
  mov    (%rdi),%eax
  mov    %eax,(%rsi)
  mov    0x4(%rdi),%eax
  mov    %eax,(%rdx)
  mov    0x8(%rdi),%eax
  mov    %eax,(%rcx)
  mov    0xc(%rdi),%eax
  mov    %eax,(%r8)
  mov    0x10(%rdi),%eax
  mov    %eax,(%r9)
  ret
```

assistant: ```c
void restore_state(vec, a, b, c, d, e) int vec[5];
int *a, *b, *c, *d, *e;
{
  *a = vec[0];
  *b = vec[1];
  *c = vec[2];
  *d = vec[3];
  *e = vec[4];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<restore_state>:
  endbr64
  mov    (%rdi),%eax
  mov    %eax,(%rsi)
  mov    0x4(%rdi),%eax
  mov    %eax,(%rdx)
  mov    0x8(%rdi),%eax
  mov    %eax,(%rcx)
  mov    0xc(%rdi),%eax
  mov    %eax,(%r8)
  mov    0x10(%rdi),%eax
  mov    %eax,(%r9)
  ret
```

assistant: ```c
void restore_state(vec, a, b, c, d, e) int vec[5];
int *a, *b, *c, *d, *e;
{
  *a = vec[0];
  *b = vec[1];
  *c = vec[2];
  *d = vec[3];
  *e = vec[4];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<restore_state>:
  endbr64
  mov    (%rdi),%eax
  mov    %eax,(%rsi)
  mov    0x4(%rdi),%eax
  mov    %eax,(%rdx)
  mov    0x8(%rdi),%eax
  mov    %eax,(%rcx)
  mov    0xc(%rdi),%eax
  mov    %eax,(%r8)
  mov    0x10(%rdi),%eax
  mov    %eax,(%r9)
  ret
```

assistant: ```c
void restore_state(vec, a, b, c, d, e) int vec[5];
int *a, *b, *c, *d, *e;
{
  *a = vec[0];
  *b = vec[1];
  *c = vec[2];
  *d = vec[3];
  *e = vec[4];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D28(%rip),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0xd8(%rbp)
  cmpq   $0x0,-0xd8(%rbp)
  jne    L18
  mov    $0xfffffffb,%edi
  call   <exit@plt>
L18:
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <fscanf@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: retrieve_data
arguments: tag: D28
datatype: string
name: retrieve_data
arguments: tag: D21
datatype: string
name: retrieve_data
arguments: tag: D13
datatype: string
name: retrieve_data
arguments: tag: D29
datatype: string

tool: D28: r

tool: D21: aviao.txt

tool: D13: Insira o nome do tripulante:

tool: D29: %s

assistant: ```c
int main() {
  FILE *man = fopen("aviao.txt", "r");
  if (man == NULL)
    exit(-5);
  int i[21];
  char nome[100];
  printf("Insira o nome do tripulante:");
  fscanf("%s", nome);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rsi
  lea    D25(%rip),%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L29
L4:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D15(%rip),%rdi
  call   <fscanf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x78,%rsp
  ret
L29:
  mov    $0xfffffffb,%edi
  call   <exit@plt>
  jmp    L4
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: string
name: unpack_data
arguments: data_tag: D25
datatype: string
name: unpack_data
arguments: data_tag: D8
datatype: string
name: unpack_data
arguments: data_tag: D15
datatype: string

tool: D0: r

tool: D25: aviao.txt

tool: D8: Insira o nome do tripulante:

tool: D15: %s

assistant: ```c
int main() {
  FILE *man = fopen("aviao.txt", "r");
  if (man == NULL)
    exit(-5);
  int i[21];
  char nome[100];
  printf("Insira o nome do tripulante:");
  fscanf("%s", nome);
  return 0;
}
```