ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L1
  lea    0x10(%rsp),%r13
  lea    D1(%rip),%r12
  movabs $0x86377f7332675bb7,%rbx
  lea    D19(%rip),%rbp
L2:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rax
  lea    0x1(%rax),%rsi
  imul   %rax,%rsi
  add    %rax,%rax
  mov    %rsi,%rdx
  sub    %rax,%rdx
  mov    %rdx,%rdi
  shr    $0x3f,%rdi
  add    %rdx,%rdi
  mov    %rdi,%rcx
  sar    %rcx
  mov    %rsi,%rax
  imul   %rbx
  add    %rsi,%rdx
  sar    $0x13,%rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rsi
  mov    %rcx,%rax
  imul   %rbx
  add    %rcx,%rdx
  sar    $0x13,%rdx
  sar    $0x3f,%rdi
  sub    %rdi,%rdx
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rcx
  add    %rcx,%rsi
  mov    %rsi,%rax
  imul   %rbx
  add    %rsi,%rdx
  sar    $0x13,%rdx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L2
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
type: string
name: unpack_data
arguments: data_label: D19
type: string

tool: D1: %lld

tool: D19: %lld


assistant: ```c
int main() {
  long long test, num, i, j, k;
  scanf("%lld", &test);
  while (test--) {
    scanf("%lld", &num);
    i = num * (num + 1);
    i = i % 1000007;
    j = ((num - 1) * (num)) / 2;
    j = j % 1000007;
    printf("%lld\n", (i + j) % 1000007);
  }
  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
  lea    D3(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L26
  lea    0x10(%rsp),%r13
  lea    D1(%rip),%r12
  movabs $0x86377f7332675bb7,%rbx
  nop
L3:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rax
  lea    0x1(%rax),%rsi
  imul   %rax,%rsi
  add    %rax,%rax
  mov    %rsi,%rdx
  sub    %rax,%rdx
  mov    %rsi,%rax
  mov    %rdx,%rdi
  shr    $0x3f,%rdi
  add    %rdx,%rdi
  imul   %rbx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  mov    %rdi,%rcx
  sar    $0x3f,%rdi
  sar    %rcx
  add    %rsi,%rdx
  sar    $0x13,%rdx
  sub    %rax,%rdx
  mov    %rcx,%rax
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rsi
  imul   %rbx
  add    %rcx,%rdx
  sar    $0x13,%rdx
  sub    %rdi,%rdx
  mov    %r12,%rdi
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rcx
  add    %rcx,%rsi
  mov    %rsi,%rax
  imul   %rbx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  add    %rsi,%rdx
  sar    $0x13,%rdx
  sub    %rax,%rdx
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rsi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L3
L26:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
type: string
name: disassemble_data
arguments: label: D1
type: string

tool: D3: %lld

tool: D1: %lld


assistant: ```c
int main() {
  long long test, num, i, j, k;
  scanf("%lld", &test);
  while (test--) {
    scanf("%lld", &num);
    i = num * (num + 1);
    i = i % 1000007;
    j = ((num - 1) * (num)) / 2;
    j = j % 1000007;
    printf("%lld\n", (i + j) % 1000007);
  }
  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
  lea    D16(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L7
  lea    0x10(%rsp),%r13
  lea    D15(%rip),%r12
  movabs $0x86377f7332675bb7,%rbx
  nop
L9:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rax
  lea    0x1(%rax),%rsi
  imul   %rax,%rsi
  add    %rax,%rax
  mov    %rsi,%rdx
  sub    %rax,%rdx
  mov    %rsi,%rax
  mov    %rdx,%rdi
  shr    $0x3f,%rdi
  add    %rdx,%rdi
  imul   %rbx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  mov    %rdi,%rcx
  sar    $0x3f,%rdi
  sar    %rcx
  add    %rsi,%rdx
  sar    $0x13,%rdx
  sub    %rax,%rdx
  mov    %rcx,%rax
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rsi
  imul   %rbx
  add    %rcx,%rdx
  sar    $0x13,%rdx
  sub    %rdi,%rdx
  mov    %r12,%rdi
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rcx
  add    %rcx,%rsi
  mov    %rsi,%rax
  imul   %rbx
  mov    %rsi,%rax
  sar    $0x3f,%rax
  add    %rsi,%rdx
  sar    $0x13,%rdx
  sub    %rax,%rdx
  imul   $0xf4247,%rdx,%rdx
  sub    %rdx,%rsi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L9
L7:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D16
datatype: string
name: disassemble_data
arguments: label: D15
datatype: string

tool: D16: %lld

tool: D15: %lld


assistant: ```c
int main() {
  long long test, num, i, j, k;
  scanf("%lld", &test);
  while (test--) {
    scanf("%lld", &num);
    i = num * (num + 1);
    i = i % 1000007;
    j = ((num - 1) * (num)) / 2;
    j = j % 1000007;
    printf("%lld\n", (i + j) % 1000007);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__put_unaligned_2_le>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,%rax
  mov    %rax,%rdx
  lea    0x4(%rdx),%rax
  mov    -0x4(%rbp),%ecx
  mov    %ecx,(%rdx)
  mov    %rax,%rdx
  lea    0x4(%rdx),%rax
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void __put_unaligned_2_le(__u32 __v, register __u8 *__p) {
  *__p++ = __v;
  *__p++ = __v >> 8;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__put_unaligned_2_le>:
  endbr64
  mov    %edi,(%rsi)
  sar    $0x8,%edi
  mov    %edi,0x4(%rsi)
  ret
```

assistant: ```c
void __put_unaligned_2_le(__u32 __v, register __u8 *__p) {
  *__p++ = __v;
  *__p++ = __v >> 8;
}
```
user: Translate the below assembly code into C programming language:

```asm
<__put_unaligned_2_le>:
  endbr64
  mov    %edi,%eax
  movd   %edi,%xmm0
  sar    $0x8,%eax
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%rsi)
  ret
```

assistant: ```c
void __put_unaligned_2_le(__u32 __v, register __u8 *__p) {
  *__p++ = __v;
  *__p++ = __v >> 8;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mystrcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L16
L12:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *mystrcpy(char *dest, char *src) {
  char *d = dest;
  char *s = src;
  while (*s != '\0') {
    *d = *s;
    d++;
    s++;
  }
  *d = '\0';
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mystrcpy>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L2
  mov    %rdi,%rcx
L14:
  mov    %dl,(%rcx)
  add    $0x1,%rcx
  add    $0x1,%rsi
  movzbl (%rsi),%edx
  test   %dl,%dl
  jne    L14
L1:
  movb   $0x0,(%rcx)
  ret
L2:
  mov    %rdi,%rcx
  jmp    L1
```

assistant: ```c
char *mystrcpy(char *dest, char *src) {
  char *d = dest;
  char *s = src;
  while (*s != '\0') {
    *d = *s;
    d++;
    s++;
  }
  *d = '\0';
  return dest;
}
```
user: What is the c source code of the assembly code below?

```asm
<mystrcpy>:
  endbr64
  movzbl (%rsi),%edx
  mov    %rdi,%rax
  mov    %rdi,%rcx
  test   %dl,%dl
  je     L2
  nopl   0x0(%rax)
L6:
  add    $0x1,%rsi
  mov    %dl,(%rcx)
  add    $0x1,%rcx
  movzbl (%rsi),%edx
  test   %dl,%dl
  jne    L6
L2:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
char *mystrcpy(char *dest, char *src) {
  char *d = dest;
  char *s = src;
  while (*s != '\0') {
    *d = *s;
    d++;
    s++;
  }
  *d = '\0';
  return dest;
}
```
user: What is the c source code of the assembly code below?

```asm
<mystrcpy>:
  endbr64
  movzbl (%rsi),%edx
  mov    %rdi,%rax
  mov    %rdi,%rcx
  test   %dl,%dl
  je     L9
  nopl   0x0(%rax)
L4:
  add    $0x1,%rsi
  mov    %dl,(%rcx)
  add    $0x1,%rcx
  movzbl (%rsi),%edx
  test   %dl,%dl
  jne    L4
L9:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
char *mystrcpy(char *dest, char *src) {
  char *d = dest;
  char *s = src;
  while (*s != '\0') {
    *d = *s;
    d++;
    s++;
  }
  *d = '\0';
  return dest;
}
```
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    $0x10,%rsp
  lea    D2(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L10
L13:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addq   $0x1,-0x10(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  add    $0xe,%rax
  cmp    %rax,-0x10(%rbp)
  jb     L13
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type: string

tool: D2: Hello, world!


assistant: ```c
int main() {
  char *s = "Hello, world!\n";
  char *p;
  for (p = s; p < s + 14; p++)
    putchar(*p);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D0(%rip),%rbx
  lea    0xe(%rbx),%rbp
L2:
  movsbl (%rbx),%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L2
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
type_infered: string

tool: D0: Hello, world!


assistant: ```c
int main() {
  char *s = "Hello, world!\n";
  char *p;
  for (p = s; p < s + 14; p++)
    putchar(*p);
  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
  mov    $0x48,%edi
  push   %rbx
  lea    D2(%rip),%rbx
  lea    0xe(%rbx),%rbp
  sub    $0x8,%rsp
  jmp    L10
  nopl   0x0(%rax)
L9:
  movsbl (%rbx),%edi
L10:
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
datatype: string

tool: D2: Hello, world!


assistant: ```c
int main() {
  char *s = "Hello, world!\n";
  char *p;
  for (p = s; p < s + 14; p++)
    putchar(*p);
  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   %rbp
  push   %rbx
  lea    D9(%rip),%rbx
  lea    0xe(%rbx),%rbp
  sub    $0x8,%rsp
  nopl   (%rax)
L2:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %rbp,%rbx
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D9: Hello, world!


assistant: ```c
int main() {
  char *s = "Hello, world!\n";
  char *p;
  for (p = s; p < s + 14; p++)
    putchar(*p);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<update_flags>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %edx,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <engine_update_flags@plt>
  nop
  nop
  leave
  ret
```

assistant: ```c
int update_flags(unsigned char *buf) {
  unsigned long flagmap;
  flagmap = (unsigned long)((buf[3] << 8) | buf[4]);
  engine_update_flags(0, flagmap, buf[2]);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<update_flags>:
  endbr64
  sub    $0x8,%rsp
  movzbl 0x2(%rdi),%edx
  movzbl 0x3(%rdi),%esi
  shl    $0x8,%esi
  movzbl 0x4(%rdi),%eax
  or     %eax,%esi
  movslq %esi,%rsi
  mov    $0x0,%edi
  call   <engine_update_flags@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int update_flags(unsigned char *buf) {
  unsigned long flagmap;
  flagmap = (unsigned long)((buf[3] << 8) | buf[4]);
  engine_update_flags(0, flagmap, buf[2]);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<update_flags>:
  endbr64
  movzwl 0x3(%rdi),%esi
  movzbl 0x2(%rdi),%edx
  xor    %edi,%edi
  rol    $0x8,%si
  movzwl %si,%esi
  jmp    1050 <engine_update_flags@plt>
```

assistant: ```c
int update_flags(unsigned char *buf) {
  unsigned long flagmap;
  flagmap = (unsigned long)((buf[3] << 8) | buf[4]);
  engine_update_flags(0, flagmap, buf[2]);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<update_flags>:
  endbr64
  movzwl 0x3(%rdi),%esi
  movzbl 0x2(%rdi),%edx
  xor    %edi,%edi
  rol    $0x8,%si
  movzwl %si,%esi
  jmp    1050 <engine_update_flags@plt>
```

assistant: ```c
int update_flags(unsigned char *buf) {
  unsigned long flagmap;
  flagmap = (unsigned long)((buf[3] << 8) | buf[4]);
  engine_update_flags(0, flagmap, buf[2]);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x78(%rbp)
  lea    -0x80(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L80
L140:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x78(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x74(%rbp)
  movl   $0x0,-0x7c(%rbp)
  jmp    L117
L120:
  mov    -0x7c(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L22
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D12(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D12(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  addl   $0x6,-0x78(%rbp)
  jmp    L22
  addl   $0x2,-0x78(%rbp)
  jmp    L22
  addl   $0x5,-0x78(%rbp)
  jmp    L22
  addl   $0x5,-0x78(%rbp)
  jmp    L22
  addl   $0x4,-0x78(%rbp)
  jmp    L22
  addl   $0x5,-0x78(%rbp)
  jmp    L22
  addl   $0x6,-0x78(%rbp)
  jmp    L22
  addl   $0x3,-0x78(%rbp)
  jmp    L22
  addl   $0x7,-0x78(%rbp)
  jmp    L22
  addl   $0x6,-0x78(%rbp)
  nop
L22:
  addl   $0x1,-0x7c(%rbp)
L117:
  mov    -0x7c(%rbp),%eax
  cmp    -0x74(%rbp),%eax
  jle    L120
  mov    -0x78(%rbp),%eax
  mov    %eax,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x80(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x80(%rbp)
L80:
  mov    -0x80(%rbp),%eax
  test   %eax,%eax
  jg     L140
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L105
  call   <__stack_chk_fail@plt>
L105:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D43
datatype: string
name: disassemble_data
arguments: data_tag: D17
datatype: string
name: disassemble_data
arguments: data_tag: D34
datatype: string

tool: D43:  %d

tool: D17:  %s

tool: D34: %d leds


assistant: ```c
int main() {
  char V[100];
  int i, N, tam, j, qtd_l = 0;
  scanf(" %d", &N);
  while (N > 0) {
    scanf(" %s", V);
    qtd_l = 0;
    tam = strlen(V);
    j = 0;
    while (j <= tam) {
      switch (V[j]) {
      case '0':
        qtd_l += 6;
        break;
      case '1':
        qtd_l += 2;
        break;
      case '2':
        qtd_l += 5;
        break;
      case '3':
        qtd_l += 5;
        break;
      case '4':
        qtd_l += 4;
        break;
      case '5':
        qtd_l += 5;
        break;
      case '6':
        qtd_l += 6;
        break;
      case '7':
        qtd_l += 3;
        break;
      case '8':
        qtd_l += 7;
        break;
      case '9':
        qtd_l += 6;
        break;
      }
      j++;
    }
    printf("%d leds\n", qtd_l);
    N--;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L26
  lea    D13(%rip),%r12
  lea    D11(%rip),%rbx
  lea    D17(%rip),%rbp
  jmp    L159
  add    $0x6,%esi
L149:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  je     L66
L76:
  movzbl (%rdx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L149
  movzbl %al,%eax
  movslq (%rbx,%rax,4),%rax
  add    %rbx,%rax
  notrack jmp *%rax
  add    $0x2,%esi
  jmp    L149
  add    $0x5,%esi
  jmp    L149
  add    $0x5,%esi
  jmp    L149
  add    $0x4,%esi
  jmp    L149
  add    $0x5,%esi
  jmp    L149
  add    $0x6,%esi
  jmp    L149
  add    $0x3,%esi
  jmp    L149
  add    $0x7,%esi
  jmp    L149
  add    $0x6,%esi
  jmp    L149
L141:
  mov    $0x0,%esi
L66:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0xc(%rsp)
  test   %eax,%eax
  jle    L26
L159:
  lea    0x10(%rsp),%r13
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  js     L141
  mov    %r13,%rdx
  mov    %eax,%eax
  lea    0x11(%rsp,%rax,1),%rcx
  mov    $0x0,%esi
  jmp    L76
L26:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L127
  mov    $0x0,%eax
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L127:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
type_infered: string
name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D17
type_infered: string

tool: D25:  %d

tool: D13:  %s

tool: D17: %d leds


assistant: ```c
int main() {
  char V[100];
  int i, N, tam, j, qtd_l = 0;
  scanf(" %d", &N);
  while (N > 0) {
    scanf(" %s", V);
    qtd_l = 0;
    tam = strlen(V);
    j = 0;
    while (j <= tam) {
      switch (V[j]) {
      case '0':
        qtd_l += 6;
        break;
      case '1':
        qtd_l += 2;
        break;
      case '2':
        qtd_l += 5;
        break;
      case '3':
        qtd_l += 5;
        break;
      case '4':
        qtd_l += 4;
        break;
      case '5':
        qtd_l += 5;
        break;
      case '6':
        qtd_l += 6;
        break;
      case '7':
        qtd_l += 3;
        break;
      case '8':
        qtd_l += 7;
        break;
      case '9':
        qtd_l += 6;
        break;
      }
      j++;
    }
    printf("%d leds\n", qtd_l);
    N--;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L86
  lea    0x10(%rsp),%rbx
  lea    D18(%rip),%r12
  lea    D38(%rip),%rbp
  lea    D0(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L41:
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  js     L58
  cltq
  mov    %rbx,%rdx
  xor    %esi,%esi
  lea    0x11(%rsp,%rax,1),%rcx
  nopl   0x0(%rax)
L59:
  movzbl (%rdx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L76
  movzbl %al,%eax
  movslq 0x0(%r13,%rax,4),%rax
  add    %r13,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  add    $0x5,%esi
L76:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L59
L98:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0xc(%rsp)
  test   %eax,%eax
  jg     L41
L86:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
  add    $0x6,%esi
  jmp    L76
  nopl   (%rax)
  add    $0x7,%esi
  jmp    L76
  nopl   (%rax)
  add    $0x2,%esi
  jmp    L76
  nopl   (%rax)
  add    $0x3,%esi
  jmp    L76
  nopl   (%rax)
  add    $0x4,%esi
  jmp    L76
  nopl   (%rax)
L58:
  xor    %esi,%esi
  jmp    L98
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D10
type_infered: string
name: read_data
arguments: data_tag: D18
type_infered: string
name: read_data
arguments: data_tag: D38
type_infered: string

tool: D10:  %d

tool: D18:  %s

tool: D38: %d leds


assistant: ```c
int main() {
  char V[100];
  int i, N, tam, j, qtd_l = 0;
  scanf(" %d", &N);
  while (N > 0) {
    scanf(" %s", V);
    qtd_l = 0;
    tam = strlen(V);
    j = 0;
    while (j <= tam) {
      switch (V[j]) {
      case '0':
        qtd_l += 6;
        break;
      case '1':
        qtd_l += 2;
        break;
      case '2':
        qtd_l += 5;
        break;
      case '3':
        qtd_l += 5;
        break;
      case '4':
        qtd_l += 4;
        break;
      case '5':
        qtd_l += 5;
        break;
      case '6':
        qtd_l += 6;
        break;
      case '7':
        qtd_l += 3;
        break;
      case '8':
        qtd_l += 7;
        break;
      case '9':
        qtd_l += 6;
        break;
      }
      j++;
    }
    printf("%d leds\n", qtd_l);
    N--;
  }
  return 0;
}
```