ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x10,%rsp
  movl   $0x1,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  movl   $0x3,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    -0x8(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%esi
  mov    -0x8(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type: string
name: disassemble_data
arguments: data_tag: D17
type: string

tool: D1: Numero 1: %d
Numero 2: %d
Numero 3: %d

Total   : %d



tool: D17: Fim de Programa

assistant: ```c
main() {
  int intNum1 = 1, intNum2 = 2, intNum3 = 3, intProduto = 0;
  intProduto = intNum1 * intNum2 * intNum3;
  printf("Numero 1: %d\nNumero 2: %d\nNumero 3: %d\n\nTotal   : %d\n\n", intNum1, intNum2, intNum3, intProduto);
  printf("Fim de Programa\n");
}
```
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    $0x6,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_guessed: string
name: parse_data
arguments: data_tag: D19
type_guessed: string

tool: D2: Numero 1: %d
Numero 2: %d
Numero 3: %d

Total   : %d



tool: D19: Fim de Programa

assistant: ```c
main() {
  int intNum1 = 1, intNum2 = 2, intNum3 = 3, intProduto = 0;
  intProduto = intNum1 * intNum2 * intNum3;
  printf("Numero 1: %d\nNumero 2: %d\nNumero 3: %d\n\nTotal   : %d\n\n", intNum1, intNum2, intNum3, intProduto);
  printf("Fim de Programa\n");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%ecx
  mov    $0x2,%edx
  xor    %eax,%eax
  mov    $0x6,%r8d
  mov    $0x1,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
datatype: string
name: read_data
arguments: tag: D5
datatype: string

tool: D4: Numero 1: %d
Numero 2: %d
Numero 3: %d

Total   : %d



tool: D5: Fim de Programa

assistant: ```c
main() {
  int intNum1 = 1, intNum2 = 2, intNum3 = 3, intProduto = 0;
  intProduto = intNum1 * intNum2 * intNum3;
  printf("Numero 1: %d\nNumero 2: %d\nNumero 3: %d\n\nTotal   : %d\n\n", intNum1, intNum2, intNum3, intProduto);
  printf("Fim de Programa\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%ecx
  mov    $0x2,%edx
  xor    %eax,%eax
  mov    $0x6,%r8d
  mov    $0x1,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D17(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_guessed: string
name: retrieve_data
arguments: data_tag: D17
type_guessed: string

tool: D7: Numero 1: %d
Numero 2: %d
Numero 3: %d

Total   : %d



tool: D17: Fim de Programa

assistant: ```c
main() {
  int intNum1 = 1, intNum2 = 2, intNum3 = 3, intProduto = 0;
  intProduto = intNum1 * intNum2 * intNum3;
  printf("Numero 1: %d\nNumero 2: %d\nNumero 3: %d\n\nTotal   : %d\n\n", intNum1, intNum2, intNum3, intProduto);
  printf("Fim de Programa\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L11
  mov    $0x0,%r12d
  lea    0x10(%rsp),%rbp
L83:
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L84
  lea    0x10(%rsp),%rdx
  lea    -0x1(%rax),%eax
  lea    0x11(%rsp,%rax,1),%rdi
L33:
  movzbl (%rdx),%eax
  movsbl %al,%ecx
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  lea    0x3(%rcx),%esi
  cmp    $0x1a,%al
  cmovb  %esi,%ecx
  mov    %cl,(%rdx)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L33
L84:
  mov    %ebx,%edx
  shr    $0x1f,%edx
  add    %ebx,%edx
  sar    %edx
  mov    %ebx,%eax
  and    $0x1,%eax
  cmp    $0x1,%eax
  sbb    $0xffffffff,%edx
  test   %edx,%edx
  jle    L28
  lea    0x10(%rsp),%rax
  lea    -0x1(%rdx),%edx
  lea    0x11(%rsp,%rdx,1),%rdx
L80:
  subb   $0x1,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L80
L28:
  sub    $0x1,%ebx
  js     L4
  movslq %ebx,%rbx
L77:
  movsbl (%rbx,%rbp,1),%edi
  call   <putchar@plt>
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jns    L77
L4:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r12d
  cmp    %r12d,0xc(%rsp)
  jg     L83
L11:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  call   <__stack_chk_fail@plt>
```

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

tool: D7: %d

assistant: ```c
int main() {
  int N, i, j, k, inverte, tam;
  char palavra[1001];
  scanf("%d", &N);
  getchar();
  for (i = 0; i < N; i++) {
    gets(palavra);
    inverte = strlen(palavra);
    tam = inverte;
    for (j = 0, inverte; j < tam; j++, inverte--) {
      k = palavra[j];
      if (palavra[j] >= 65 && palavra[j] <= 90 || palavra[j] >= 97 && palavra[j] <= 122) {
        k += 3;
      }
      palavra[j] = k;
    }
    inverte = tam / 2;
    if (tam % 2 != 0) {
      inverte++;
    }
    for (j = 0; j < inverte; j++) {
      k = palavra[j];
      k--;
      palavra[j] = k;
    }
    for (j = tam - 1; j >= 0; j--) {
      printf("%c", palavra[j]);
    }
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  lea    D7(%rip),%rdi
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  call   <scanf@plt>
  xor    %eax,%eax
  lea    0x11(%rsp),%r12
  call   <getchar@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L59
  nopl   0x0(%rax)
L70:
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L62
  lea    -0x1(%rax),%r8d
  mov    %rbx,%rcx
  mov    %r8,%r9
  add    %r12,%r8
  nopl   (%rax)
L22:
  movzbl (%rcx),%edx
  mov    %edx,%esi
  lea    0x3(%rdx),%edi
  and    $0xffffffdf,%esi
  sub    $0x41,%esi
  cmp    $0x1a,%sil
  cmovb  %edi,%edx
  add    $0x1,%rcx
  mov    %dl,-0x1(%rcx)
  cmp    %r8,%rcx
  jne    L22
L40:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  and    $0x1,%eax
  sar    %edx
  cmp    $0x1,%eax
  sbb    $0xffffffff,%edx
  test   %edx,%edx
  jle    L0
  sub    $0x1,%edx
  mov    %rbx,%rax
  add    %r12,%rdx
L55:
  subb   $0x1,(%rax)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L55
L0:
  movslq %r9d,%r13
  test   %r9d,%r9d
  js     L7
  nopl   0x0(%rax)
L36:
  movsbl (%rbx,%r13,1),%edi
  sub    $0x1,%r13
  call   <putchar@plt>
  test   %r13d,%r13d
  jns    L36
L7:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0xc(%rsp)
  jg     L70
L59:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x418,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L62:
  lea    -0x1(%rax),%r9d
  jmp    L40
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
datatype: string

tool: D7: %d

assistant: ```c
int main() {
  int N, i, j, k, inverte, tam;
  char palavra[1001];
  scanf("%d", &N);
  getchar();
  for (i = 0; i < N; i++) {
    gets(palavra);
    inverte = strlen(palavra);
    tam = inverte;
    for (j = 0, inverte; j < tam; j++, inverte--) {
      k = palavra[j];
      if (palavra[j] >= 65 && palavra[j] <= 90 || palavra[j] >= 97 && palavra[j] <= 122) {
        k += 3;
      }
      palavra[j] = k;
    }
    inverte = tam / 2;
    if (tam % 2 != 0) {
      inverte++;
    }
    for (j = 0; j < inverte; j++) {
      k = palavra[j];
      k--;
      palavra[j] = k;
    }
    for (j = tam - 1; j >= 0; j--) {
      printf("%c", palavra[j]);
    }
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_addressable_void_variable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <do_dbl_alloc@plt>
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <assertTrue@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <do_check_dbl@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
datatype: string

tool: D8: pv != NULL

assistant: ```c
void test_addressable_void_variable() {
  void *pv;
  do_dbl_alloc(&pv);
  assertTrue("pv != NULL", pv != NULL);
  ;
  do_check_dbl(pv);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<test_addressable_void_variable>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <do_dbl_alloc@plt>
  cmpq   $0x0,(%rsp)
  setne  %sil
  movzbl %sil,%esi
  lea    D7(%rip),%rdi
  call   <assertTrue@plt>
  mov    (%rsp),%rdi
  call   <do_check_dbl@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D7
type_infered: string

tool: D7: pv != NULL

assistant: ```c
void test_addressable_void_variable() {
  void *pv;
  do_dbl_alloc(&pv);
  assertTrue("pv != NULL", pv != NULL);
  ;
  do_check_dbl(pv);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_addressable_void_variable>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <do_dbl_alloc@plt>
  xor    %esi,%esi
  cmpq   $0x0,(%rsp)
  lea    D7(%rip),%rdi
  setne  %sil
  call   <assertTrue@plt>
  mov    (%rsp),%rdi
  call   <do_check_dbl@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_guessed: string

tool: D7: pv != NULL

assistant: ```c
void test_addressable_void_variable() {
  void *pv;
  do_dbl_alloc(&pv);
  assertTrue("pv != NULL", pv != NULL);
  ;
  do_check_dbl(pv);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test_addressable_void_variable>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <do_dbl_alloc@plt>
  xor    %esi,%esi
  cmpq   $0x0,(%rsp)
  lea    D4(%rip),%rdi
  setne  %sil
  call   <assertTrue@plt>
  mov    (%rsp),%rdi
  call   <do_check_dbl@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_infered: string

tool: D4: pv != NULL

assistant: ```c
void test_addressable_void_variable() {
  void *pv;
  do_dbl_alloc(&pv);
  assertTrue("pv != NULL", pv != NULL);
  ;
  do_check_dbl(pv);
}
```
user: What is the c source code of the assembly code below?

```asm
<enm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0xc8(%rbp),%rax
  mov    (%rax),%rax
  lea    -0xb0(%rbp),%rdx
  mov    $0x7c,%esi
  mov    %rax,%rdi
  call   <get_fields@plt>
  mov    %eax,-0xb4(%rbp)
  cmpl   $0x0,-0xb4(%rbp)
  jle    L14
  mov    -0xb0(%rbp),%rax
  jmp    L1
L14:
  lea    D9(%rip),%rax
L1:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
datatype: string

tool: D9: Unknown

assistant: ```c
char *enm(char *flds[20]) {
  char *sflds[20];
  int nsf = get_fields(flds[0], '|', sflds);
  return (nsf > 0 ? sflds[0] : "Unknown");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<enm>:
  endbr64
  sub    $0xb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    $0x7c,%esi
  mov    (%rdi),%rdi
  call   <get_fields@plt>
  test   %eax,%eax
  lea    D3(%rip),%rax
  cmovg  (%rsp),%rax
  mov    0xa8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0xb8,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
data_type: string

tool: D3: Unknown

assistant: ```c
char *enm(char *flds[20]) {
  char *sflds[20];
  int nsf = get_fields(flds[0], '|', sflds);
  return (nsf > 0 ? sflds[0] : "Unknown");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<enm>:
  endbr64
  sub    $0xb8,%rsp
  mov    (%rdi),%rdi
  mov    $0x7c,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0xa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  call   <get_fields@plt>
  mov    %eax,%r8d
  lea    D1(%rip),%rax
  test   %r8d,%r8d
  cmovg  (%rsp),%rax
  mov    0xa8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0xb8,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

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

tool: D1: Unknown

assistant: ```c
char *enm(char *flds[20]) {
  char *sflds[20];
  int nsf = get_fields(flds[0], '|', sflds);
  return (nsf > 0 ? sflds[0] : "Unknown");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<enm>:
  endbr64
  sub    $0xb8,%rsp
  mov    (%rdi),%rdi
  mov    $0x7c,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0xa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  call   <get_fields@plt>
  mov    %eax,%r8d
  lea    D9(%rip),%rax
  test   %r8d,%r8d
  cmovg  (%rsp),%rax
  mov    0xa8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0xb8,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D9
type: string

tool: D9: Unknown

assistant: ```c
char *enm(char *flds[20]) {
  char *sflds[20];
  int nsf = get_fields(flds[0], '|', sflds);
  return (nsf > 0 ? sflds[0] : "Unknown");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x150,%rsp
  mov    $0x28,%ecx
  mov    %fs:(%rcx),%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D14(%rip),%rsi
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  mov    %rsp,%rsi
  mov    $0x0,%edi
  call   <find_solution@plt>
  test   %rax,%rax
  je     L57
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D29(%rip),%r12
  jmp    L106
L81:
  movslq %ebp,%rax
  imul   $0x38e38e39,%rax,%rax
  sar    $0x21,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,8),%eax
  cmp    %eax,%ebp
  je     L39
L12:
  movslq %ebp,%rax
  imul   $0x4bda12f7,%rax,%rax
  sar    $0x23,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  lea    (%rax,%rax,8),%eax
  cmp    %eax,%ebp
  je     L19
L55:
  mov    (%rsp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x51,%rbx
  je     L24
L106:
  mov    %ebx,%ebp
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  cmp    %ebx,%eax
  jne    L81
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L81
L39:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L12
L19:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L55
L24:
  mov    $0xa,%edi
  call   <putchar@plt>
L95:
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L90
  mov    $0x0,%eax
  add    $0x150,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L57:
  lea    D24(%rip),%rdi
  call   <puts@plt>
  jmp    L95
L90:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D14
data_type: string
name: retrieve_data
arguments: data_tag: D26
data_type: string
name: retrieve_data
arguments: data_tag: D29
data_type: string
name: retrieve_data
arguments: data_tag: D24
data_type: string

tool: 

tool: D26: Solution Found!

tool: D29: %d 

tool: D24: No Solution

assistant: ```c
int main() {
  int s[] = {0, 0, 5, 1, 0, 9, 4, 7, 3, 0, 0, 9, 5, 0, 0, 8, 0, 6, 1, 4, 0, 8, 0, 0, 2, 0, 0, 4, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 6, 7, 0, 2, 5, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 2, 8, 5, 0, 2, 0, 0, 8, 6, 0, 0, 3, 9, 8, 2, 0, 7, 1, 0, 0};
  if (find_solution(0, s)) {
    printf("Solution Found!");
    int i;
    for (i = 0; i < 81; i++) {
      if (i % 3 == 0)
        printf(" ");
      if (i % 9 == 0)
        printf("\n");
      if (i % (3 * 9) == 0)
        printf("\n");
      printf("%d ", s[i]);
    }
    printf("\n");
  } else
    printf("No Solution\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x28,%ecx
  lea    D2(%rip),%rsi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rsp,%r12
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %r12,%rsi
  mov    %eax,(%rdi)
  xor    %edi,%edi
  call   <find_solution@plt>
  test   %rax,%rax
  je     L58
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    D0(%rip),%r13
  jmp    L31
  nopl   0x0(%rax)
L92:
  imul   $0x38e38e39,%ebp,%eax
  cmp    $0x1c71c71c,%eax
  jbe    L30
L3:
  imul   $0x684bda13,%ebp,%ebp
  cmp    $0x97b425e,%ebp
  jbe    L8
L35:
  mov    (%r12,%rbx,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x51,%rbx
  je     L102
L31:
  imul   $0xaaaaaaab,%ebx,%eax
  mov    %ebx,%ebp
  cmp    $0x55555555,%eax
  ja     L92
  mov    $0x20,%edi
  call   <putchar@plt>
  imul   $0x38e38e39,%ebp,%eax
  cmp    $0x1c71c71c,%eax
  ja     L3
L30:
  imul   $0x684bda13,%ebp,%ebp
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x97b425e,%ebp
  ja     L35
L8:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L35
L58:
  lea    D31(%rip),%rdi
  call   <puts@plt>
L46:
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x158,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L102:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L46
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D2
datatype: string
name: read_data
arguments: data_label: D11
datatype: string
name: read_data
arguments: data_label: D0
datatype: string
name: read_data
arguments: data_label: D31
datatype: string

tool: 

tool: D11: Solution Found!

tool: D0: %d 

tool: D31: No Solution

assistant: ```c
int main() {
  int s[] = {0, 0, 5, 1, 0, 9, 4, 7, 3, 0, 0, 9, 5, 0, 0, 8, 0, 6, 1, 4, 0, 8, 0, 0, 2, 0, 0, 4, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 6, 7, 0, 2, 5, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 2, 8, 5, 0, 2, 0, 0, 8, 6, 0, 0, 3, 9, 8, 2, 0, 7, 1, 0, 0};
  if (find_solution(0, s)) {
    printf("Solution Found!");
    int i;
    for (i = 0; i < 81; i++) {
      if (i % 3 == 0)
        printf(" ");
      if (i % 9 == 0)
        printf("\n");
      if (i % (3 * 9) == 0)
        printf("\n");
      printf("%d ", s[i]);
    }
    printf("\n");
  } else
    printf("No Solution\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x28,%ecx
  lea    D38(%rip),%rsi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rsp,%r12
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %r12,%rsi
  mov    %eax,(%rdi)
  xor    %edi,%edi
  call   <find_solution@plt>
  test   %rax,%rax
  je     L76
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    D28(%rip),%r13
  jmp    L31
  nopl   0x0(%rax)
L69:
  imul   $0x38e38e39,%ebp,%eax
  cmp    $0x1c71c71c,%eax
  jbe    L84
L44:
  imul   $0x684bda13,%ebp,%ebp
  cmp    $0x97b425e,%ebp
  jbe    L100
L60:
  mov    (%r12,%rbx,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x51,%rbx
  je     L86
L31:
  imul   $0xaaaaaaab,%ebx,%eax
  mov    %ebx,%ebp
  cmp    $0x55555555,%eax
  ja     L69
  mov    $0x20,%edi
  call   <putchar@plt>
  imul   $0x38e38e39,%ebp,%eax
  cmp    $0x1c71c71c,%eax
  ja     L44
L84:
  imul   $0x684bda13,%ebp,%ebp
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x97b425e,%ebp
  ja     L60
L100:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L60
L76:
  lea    D14(%rip),%rdi
  call   <puts@plt>
L42:
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x158,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L86:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L42
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D38
datatype: string
name: parse_data
arguments: tag: D24
datatype: string
name: parse_data
arguments: tag: D28
datatype: string
name: parse_data
arguments: tag: D14
datatype: string

tool: 

tool: D24: Solution Found!

tool: D28: %d 

tool: D14: No Solution

assistant: ```c
int main() {
  int s[] = {0, 0, 5, 1, 0, 9, 4, 7, 3, 0, 0, 9, 5, 0, 0, 8, 0, 6, 1, 4, 0, 8, 0, 0, 2, 0, 0, 4, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 6, 7, 0, 2, 5, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 2, 8, 5, 0, 2, 0, 0, 8, 6, 0, 0, 3, 9, 8, 2, 0, 7, 1, 0, 0};
  if (find_solution(0, s)) {
    printf("Solution Found!");
    int i;
    for (i = 0; i < 81; i++) {
      if (i % 3 == 0)
        printf(" ");
      if (i % 9 == 0)
        printf("\n");
      if (i % (3 * 9) == 0)
        printf("\n");
      printf("%d ", s[i]);
    }
    printf("\n");
  } else
    printf("No Solution\n");
}
```