ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x32,%ecx
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1c8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1b8(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rbp
  mov    %rsp,%rdx
  movabs $0x200000004,%rax
  mov    %rax,0x8(%rsp)
  mov    %rbp,%rdi
  movabs $0x500000003,%rax
  mov    %rax,0x10(%rsp)
  xor    %eax,%eax
  rep stos %rax,%es:(%rdi)
  movl   $0x2,0x4(%rsp)
  lea    0x18(%rsp),%rcx
  mov    $0x3,%eax
  jmp    L27
L28:
  movslq (%rdx),%rax
L27:
  add    $0x4,%rdx
  addl   $0x1,0x20(%rsp,%rax,4)
  cmp    %rcx,%rdx
  jne    L28
  xor    %r12d,%r12d
  lea    D6(%rip),%r13
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    0x0(%rbp),%ebx
  test   %ebx,%ebx
  jle    L22
  nopw   0x0(%rax,%rax,1)
L44:
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  mov    %ebx,0x0(%rbp)
  jne    L44
L22:
  add    $0x1,%r12d
  add    $0x4,%rbp
  cmp    $0x64,%r12d
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x1b8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x1c8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D6
type_infered: string

tool: D6:  %d

assistant: ```c
int main() {
  int a[6] = {3, 2, 4, 2, 3, 5};
  int b[100];
  int i;
  for (i = 0; i < 100; i++) {
    b[i] = 0;
  }
  for (i = 0; i < 6; i++) {
    b[a[i]]++;
  }
  for (i = 0; i < 100; i++) {
    while (b[i] > 0) {
      printf(" %d", i);
      b[i]--;
    }
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x32,%ecx
  lea    D10(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x1a8,%rsp
  movdqa D12(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  rep stos %rax,%es:(%rdi)
  movups %xmm0,0x8(%rsp)
  nopl   0x0(%rax)
L19:
  mov    0x0(%rbp),%ebx
  test   %ebx,%ebx
  jle    L31
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  mov    %ebx,0x0(%rbp)
  jne    L3
L31:
  add    $0x1,%r12d
  add    $0x4,%rbp
  cmp    $0x64,%r12d
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x1a8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D10
type_infered: string
name: unpack_data
arguments: tag: D12
type_infered: byte[16]

tool: D10:  %d

tool: {"D12": [0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int a[6] = {3, 2, 4, 2, 3, 5};
  int b[100];
  int i;
  for (i = 0; i < 100; i++) {
    b[i] = 0;
  }
  for (i = 0; i < 6; i++) {
    b[a[i]]++;
  }
  for (i = 0; i < 100; i++) {
    while (b[i] > 0) {
      printf(" %d", i);
      b[i]--;
    }
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0xc350,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int main() {
  long salary;
  long *salary_ptr;
  salary = 50000;
  salary_ptr = &salary;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  long salary;
  long *salary_ptr;
  salary = 50000;
  salary_ptr = &salary;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  long salary;
  long *salary_ptr;
  salary = 50000;
  salary_ptr = &salary;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  long salary;
  long *salary_ptr;
  salary = 50000;
  salary_ptr = &salary;
  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    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x1f55,%edi
  call   <server@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  server(8021);
  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
  sub    $0x8,%rsp
  mov    $0x1f55,%edi
  call   <server@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  server(8021);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1f55,%edi
  call   <server@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  server(8021);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1f55,%edi
  call   <server@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  server(8021);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x61,-0x5(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L1:
  mov    -0x4(%rbp),%eax
  add    $0x61,%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L4:
  cmpl   $0x19,-0x4(%rbp)
  jle    L1
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_guessed: string

tool: D9: this char is %c
 

assistant: ```c
int main() {
  char letter = 'a';
  for (int i = 0; i < 26; i++) {
    printf("this char is %c\n ", ('a' + i));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x61,%ebx
  lea    D0(%rip),%rbp
L7:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x7b,%ebx
  jne    L7
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
datatype: string

tool: D0: this char is %c
 

assistant: ```c
int main() {
  char letter = 'a';
  for (int i = 0; i < 26; i++) {
    printf("this char is %c\n ", ('a' + i));
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  mov    $0x61,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7b,%ebx
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type: string

tool: D1: this char is %c
 

assistant: ```c
int main() {
  char letter = 'a';
  for (int i = 0; i < 26; i++) {
    printf("this char is %c\n ", ('a' + i));
  }
  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   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  mov    $0x61,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7b,%ebx
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_infered: string

tool: D0: this char is %c
 

assistant: ```c
int main() {
  char letter = 'a';
  for (int i = 0; i < 26; i++) {
    printf("this char is %c\n ", ('a' + i));
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x3f60a4d,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  nop
  mov    -0x10(%rbp),%rax
  lea    -0x4409719c(%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  unsigned int copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] < local1) {
        copy11 = *((unsigned short *)(&state[0UL]) + 0);
        *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
        *((unsigned short *)(&state[0UL]) + 1) = copy11;
      } else {
        state[0UL] += state[0UL];
      }
      if (state[0UL] > local1) {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        copy12 = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = copy12;
      }
      local1++;
    }
    output[0UL] = (state[0UL] - 1029146479UL) + 0xfffffffff94e15d3UL;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  sub    $0x47ff7be9,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  unsigned int copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] < local1) {
        copy11 = *((unsigned short *)(&state[0UL]) + 0);
        *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
        *((unsigned short *)(&state[0UL]) + 1) = copy11;
      } else {
        state[0UL] += state[0UL];
      }
      if (state[0UL] > local1) {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        copy12 = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = copy12;
      }
      local1++;
    }
    output[0UL] = (state[0UL] - 1029146479UL) + 0xfffffffff94e15d3UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  sub    $0x47ff7be9,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  unsigned int copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] < local1) {
        copy11 = *((unsigned short *)(&state[0UL]) + 0);
        *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
        *((unsigned short *)(&state[0UL]) + 1) = copy11;
      } else {
        state[0UL] += state[0UL];
      }
      if (state[0UL] > local1) {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        copy12 = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = copy12;
      }
      local1++;
    }
    output[0UL] = (state[0UL] - 1029146479UL) + 0xfffffffff94e15d3UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  sub    $0x47ff7be9,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  unsigned int copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] < local1) {
        copy11 = *((unsigned short *)(&state[0UL]) + 0);
        *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
        *((unsigned short *)(&state[0UL]) + 1) = copy11;
      } else {
        state[0UL] += state[0UL];
      }
      if (state[0UL] > local1) {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        copy12 = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = copy12;
      }
      local1++;
    }
    output[0UL] = (state[0UL] - 1029146479UL) + 0xfffffffff94e15d3UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L37:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L37
  sub    $0xc70,%rsp
  mov    %edi,-0x9c64(%rbp)
  mov    %rsi,-0x9c70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x9c58(%rbp),%rax
  mov    %rax,%rsi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x9c54(%rbp),%rax
  mov    %rax,%rsi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x9c54(%rbp),%ecx
  mov    -0x9c58(%rbp),%eax
  lea    -0x9c50(%rbp),%rdx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <maker_matriz@plt>
  mov    -0x9c54(%rbp),%ecx
  mov    -0x9c58(%rbp),%eax
  lea    -0x9c50(%rbp),%rdx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <print@plt>
  mov    -0x9c54(%rbp),%ecx
  mov    -0x9c58(%rbp),%eax
  lea    -0x9c50(%rbp),%rdx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <repeat@plt>
  test   %rax,%rax
  je     L8
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L7
L8:
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L7:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D10
type_guessed: string
name: unpack_data
arguments: label: D59
type_guessed: string
name: unpack_data
arguments: label: D37
type_guessed: string
name: unpack_data
arguments: label: D52
type_guessed: string
name: unpack_data
arguments: label: D6
type_guessed: string
name: unpack_data
arguments: label: D71
type_guessed: string

tool: D10: Informe o numero de linhas da matriz:


tool: D59: %d

tool: D37: Informe o numero de colunas da matriz:


tool: D52: 


tool: D6: Existem elementos repetidos.


tool: D71: Nao existem elementos repetidos.


assistant: ```c
int main(int argc, const char *argv[]) {
  int matriz[100][100], m, n;
  printf("Informe o numero de linhas da matriz:\n");
  scanf("%d", &m);
  printf("Informe o numero de colunas da matriz:\n");
  scanf("%d", &n);
  maker_matriz(m, n, matriz);
  print(m, n, matriz);
  if (repeat(m, n, matriz)) {
    printf("\n");
    printf("Existem elementos repetidos.\n");
  } else {
    printf("\n");
    printf("Nao existem elementos repetidos.\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    -0x9000(%rsp),%r11
L16:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L16
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdx
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  call   <maker_matriz@plt>
  mov    %rbx,%rdx
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  call   <print@plt>
  mov    %rbx,%rdx
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  call   <repeat@plt>
  test   %rax,%rax
  je     L14
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D31(%rip),%rdi
  call   <printf@plt>
L21:
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x9c60,%rsp
  pop    %rbx
  ret
L14:
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L21
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D36
type_infered: string
name: extract_symbolic_data
arguments: label: D7
type_infered: string
name: extract_symbolic_data
arguments: label: D8
type_infered: string
name: extract_symbolic_data
arguments: label: D26
type_infered: string
name: extract_symbolic_data
arguments: label: D31
type_infered: string
name: extract_symbolic_data
arguments: label: D10
type_infered: string

tool: D36: Informe o numero de linhas da matriz:


tool: D7: %d

tool: D8: Informe o numero de colunas da matriz:


tool: D26: 


tool: D31: Existem elementos repetidos.


tool: D10: Nao existem elementos repetidos.


assistant: ```c
int main(int argc, const char *argv[]) {
  int matriz[100][100], m, n;
  printf("Informe o numero de linhas da matriz:\n");
  scanf("%d", &m);
  printf("Informe o numero de colunas da matriz:\n");
  scanf("%d", &n);
  maker_matriz(m, n, matriz);
  print(m, n, matriz);
  if (repeat(m, n, matriz)) {
    printf("\n");
    printf("Existem elementos repetidos.\n");
  } else {
    printf("\n");
    printf("Nao existem elementos repetidos.\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    -0x9000(%rsp),%r11
L32:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L32
  sub    $0xc60,%rsp
  lea    D19(%rip),%r12
  lea    D32(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D49(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  mov    %r12,%rdx
  call   <maker_matriz@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  mov    %r12,%rdx
  call   <print@plt>
  mov    0x8(%rsp),%edi
  mov    0xc(%rsp),%esi
  mov    %r12,%rdx
  call   <repeat@plt>
  lea    D10(%rip),%rdi
  test   %rax,%rax
  je     L10
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
L35:
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x9c60,%rsp
  xor    %eax,%eax
  pop    %r12
  ret
L10:
  call   <printf@plt>
  lea    D40(%rip),%rdi
  call   <printf@plt>
  jmp    L35
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
type_guessed: string
name: read_data
arguments: data_label: D32
type_guessed: string
name: read_data
arguments: data_label: D49
type_guessed: string
name: read_data
arguments: data_label: D10
type_guessed: string
name: read_data
arguments: data_label: D14
type_guessed: string
name: read_data
arguments: data_label: D40
type_guessed: string

tool: D19: %d

tool: D32: Informe o numero de linhas da matriz:


tool: D49: Informe o numero de colunas da matriz:


tool: D10: 


tool: D14: Existem elementos repetidos.


tool: D40: Nao existem elementos repetidos.


assistant: ```c
int main(int argc, const char *argv[]) {
  int matriz[100][100], m, n;
  printf("Informe o numero de linhas da matriz:\n");
  scanf("%d", &m);
  printf("Informe o numero de colunas da matriz:\n");
  scanf("%d", &n);
  maker_matriz(m, n, matriz);
  print(m, n, matriz);
  if (repeat(m, n, matriz)) {
    printf("\n");
    printf("Existem elementos repetidos.\n");
  } else {
    printf("\n");
    printf("Nao existem elementos repetidos.\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  lea    -0x9000(%rsp),%r11
L24:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L24
  sub    $0xc60,%rsp
  lea    D44(%rip),%r12
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  mov    %r12,%rdx
  call   <maker_matriz@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  mov    %r12,%rdx
  call   <print@plt>
  mov    0x8(%rsp),%edi
  mov    0xc(%rsp),%esi
  mov    %r12,%rdx
  call   <repeat@plt>
  lea    D49(%rip),%rdi
  test   %rax,%rax
  je     L39
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
L18:
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x9c60,%rsp
  xor    %eax,%eax
  pop    %r12
  ret
L39:
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L18
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D44
type_infered: string
name: parse_data
arguments: tag: D13
type_infered: string
name: parse_data
arguments: tag: D28
type_infered: string
name: parse_data
arguments: tag: D49
type_infered: string
name: parse_data
arguments: tag: D48
type_infered: string
name: parse_data
arguments: tag: D15
type_infered: string

tool: D44: %d

tool: D13: Informe o numero de linhas da matriz:


tool: D28: Informe o numero de colunas da matriz:


tool: D49: 


tool: D48: Existem elementos repetidos.


tool: D15: Nao existem elementos repetidos.


assistant: ```c
int main(int argc, const char *argv[]) {
  int matriz[100][100], m, n;
  printf("Informe o numero de linhas da matriz:\n");
  scanf("%d", &m);
  printf("Informe o numero de colunas da matriz:\n");
  scanf("%d", &n);
  maker_matriz(m, n, matriz);
  print(m, n, matriz);
  if (repeat(m, n, matriz)) {
    printf("\n");
    printf("Existem elementos repetidos.\n");
  } else {
    printf("\n");
    printf("Nao existem elementos repetidos.\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<tail>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L29
L32:
  subq   $0x1,-0x8(%rbp)
L29:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L6
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jne    L32
L6:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L9
  addq   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
const char *tail(const char *foo) {
  const char *tmp = foo + strlen(foo);
  while (*tmp != '/' && tmp != foo)
    tmp--;
  if (*tmp == '/')
    tmp++;
  return tmp;
}
```