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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x7,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  imul   $0x190,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L4
  cmpl   $0x0,-0x10(%rbp)
  jne    L2
L4:
  cmpl   $0x0,-0xc(%rbp)
  jne    L10
L2:
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L42
L10:
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L42:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D31
type_guessed: string
name: disassemble_data
arguments: data_tag: D38
type_guessed: string
name: disassemble_data
arguments: data_tag: D36
type_guessed: string
name: disassemble_data
arguments: data_tag: D4
type_guessed: string

tool: D31: Enter the year: 

tool: D38: %i

tool: D36: It's a leap year.


tool: D4: Nope, it's not a leap year.


assistant: ```c
int main(void) {
  int year, rem_4, rem_100, rem_400;
  printf("Enter the year: ");
  scanf("%i", &year);
  rem_4 = year % 4;
  rem_100 = year % 100;
  rem_400 = year % 400;
  if ((rem_4 == 0 && rem_100 != 0) || rem_400 == 0)
    printf("It's a leap year.\n");
  else
    printf("Nope, it's not a leap year.\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D37(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   $0x3,%dl
  sete   %cl
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  imul   $0x64,%eax,%eax
  cmp    %eax,%edx
  setne  %al
  test   %al,%cl
  jne    L17
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x27,%rax
  sub    %esi,%eax
  imul   $0x190,%eax,%eax
  cmp    %eax,%edx
  jne    L22
L17:
  lea    D28(%rip),%rdi
  call   <printf@plt>
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L22:
  lea    D26(%rip),%rdi
  call   <printf@plt>
  jmp    L6
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D37
datatype: string
name: unpack_data
arguments: label: D12
datatype: string
name: unpack_data
arguments: label: D28
datatype: string
name: unpack_data
arguments: label: D26
datatype: string

tool: D37: Enter the year: 

tool: D12: %i

tool: D28: It's a leap year.


tool: D26: Nope, it's not a leap year.


assistant: ```c
int main(void) {
  int year, rem_4, rem_100, rem_400;
  printf("Enter the year: ");
  scanf("%i", &year);
  rem_4 = year % 4;
  rem_100 = year % 100;
  rem_400 = year % 400;
  if ((rem_4 == 0 && rem_100 != 0) || rem_400 == 0)
    printf("It's a leap year.\n");
  else
    printf("Nope, it's not a leap year.\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
  sub    $0x18,%rsp
  lea    D23(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x3,%al
  sete   %cl
  imul   $0xc28f5c29,%eax,%eax
  add    $0x51eb850,%eax
  mov    %eax,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  seta   %dl
  test   %dl,%cl
  jne    L36
  ror    $0x4,%eax
  cmp    $0xa3d70a,%eax
  ja     L39
L36:
  lea    D27(%rip),%rdi
  call   <printf@plt>
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L39:
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D23
type: string
name: disassemble_data
arguments: data_label: D14
type: string
name: disassemble_data
arguments: data_label: D27
type: string
name: disassemble_data
arguments: data_label: D22
type: string

tool: D23: Enter the year: 

tool: D14: %i

tool: D27: It's a leap year.


tool: D22: Nope, it's not a leap year.


assistant: ```c
int main(void) {
  int year, rem_4, rem_100, rem_400;
  printf("Enter the year: ");
  scanf("%i", &year);
  rem_4 = year % 4;
  rem_100 = year % 100;
  rem_400 = year % 400;
  if ((rem_4 == 0 && rem_100 != 0) || rem_400 == 0)
    printf("It's a leap year.\n");
  else
    printf("Nope, it's not a leap year.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D34(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D36(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x3,%al
  sete   %cl
  imul   $0xc28f5c29,%eax,%eax
  add    $0x51eb850,%eax
  mov    %eax,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  seta   %dl
  test   %dl,%cl
  jne    L33
  ror    $0x4,%eax
  cmp    $0xa3d70a,%eax
  ja     L3
L33:
  lea    D3(%rip),%rdi
  call   <printf@plt>
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  jmp    L28
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D34
data_type: string
name: disassemble_data
arguments: data_tag: D36
data_type: string
name: disassemble_data
arguments: data_tag: D3
data_type: string
name: disassemble_data
arguments: data_tag: D2
data_type: string

tool: D34: Enter the year: 

tool: D36: %i

tool: D3: It's a leap year.


tool: D2: Nope, it's not a leap year.


assistant: ```c
int main(void) {
  int year, rem_4, rem_100, rem_400;
  printf("Enter the year: ");
  scanf("%i", &year);
  rem_4 = year % 4;
  rem_100 = year % 100;
  rem_400 = year % 400;
  if ((rem_4 == 0 && rem_100 != 0) || rem_400 == 0)
    printf("It's a leap year.\n");
  else
    printf("Nope, it's not a leap year.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gt1_name_context_hash_func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L18
L2:
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L18:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L2
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int gt1_name_context_hash_func(const char *string) {
  unsigned int result;
  int c;
  int i;
  result = 0;
  for (i = 0; (c = ((const unsigned char *)string)[i]) != '\0'; i++)
    result += (result << 3) + c;
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gt1_name_context_hash_func>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L4
  add    $0x1,%rdi
  mov    $0x0,%edx
L3:
  movzbl %al,%eax
  lea    (%rax,%rdx,8),%eax
  add    %eax,%edx
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L3
L7:
  mov    %edx,%eax
  ret
L4:
  mov    $0x0,%edx
  jmp    L7
```

assistant: ```c
unsigned int gt1_name_context_hash_func(const char *string) {
  unsigned int result;
  int c;
  int i;
  result = 0;
  for (i = 0; (c = ((const unsigned char *)string)[i]) != '\0'; i++)
    result += (result << 3) + c;
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<gt1_name_context_hash_func>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L16
  add    $0x1,%rdi
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L8:
  lea    (%rax,%r8,8),%eax
  add    $0x1,%rdi
  add    %eax,%r8d
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L8
  mov    %r8d,%eax
  ret
  nop
L16:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int gt1_name_context_hash_func(const char *string) {
  unsigned int result;
  int c;
  int i;
  result = 0;
  for (i = 0; (c = ((const unsigned char *)string)[i]) != '\0'; i++)
    result += (result << 3) + c;
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gt1_name_context_hash_func>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L3
  add    $0x1,%rdi
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L6:
  lea    (%rax,%r8,8),%eax
  add    $0x1,%rdi
  add    %eax,%r8d
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L6
  mov    %r8d,%eax
  ret
  nop
L3:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int gt1_name_context_hash_func(const char *string) {
  unsigned int result;
  int c;
  int i;
  result = 0;
  for (i = 0; (c = ((const unsigned char *)string)[i]) != '\0'; i++)
    result += (result << 3) + c;
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<variant_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x7,%eax
  test   %eax,%eax
  je     L7
  addl   $0x1,-0x4(%rbp)
L7:
  sarl   $0x7,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L7
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int variant_length(int proto_id, int value) {
  int n = 0;
  if (proto_id >> 7 != 0) {
    n += 1;
  }
  do {
    value >>= 7;
    n++;
  } while (value != 0);
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<variant_length>:
  endbr64
  shr    $0x7,%edi
  setne  %al
  movzbl %al,%eax
L4:
  add    $0x1,%eax
  sar    $0x7,%esi
  jne    L4
  ret
```

assistant: ```c
int variant_length(int proto_id, int value) {
  int n = 0;
  if (proto_id >> 7 != 0) {
    n += 1;
  }
  do {
    value >>= 7;
    n++;
  } while (value != 0);
  return n;
}
```
user: What is the c source code of the assembly code below:

```asm
<variant_length>:
  endbr64
  xor    %eax,%eax
  shr    $0x7,%edi
  setne  %al
  nopl   0x0(%rax)
L7:
  add    $0x1,%eax
  sar    $0x7,%esi
  jne    L7
  ret
```

assistant: ```c
int variant_length(int proto_id, int value) {
  int n = 0;
  if (proto_id >> 7 != 0) {
    n += 1;
  }
  do {
    value >>= 7;
    n++;
  } while (value != 0);
  return n;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<variant_length>:
  endbr64
  xor    %eax,%eax
  shr    $0x7,%edi
  setne  %al
  nopl   0x0(%rax)
L7:
  add    $0x1,%eax
  sar    $0x7,%esi
  jne    L7
  ret
```

assistant: ```c
int variant_length(int proto_id, int value) {
  int n = 0;
  if (proto_id >> 7 != 0) {
    n += 1;
  }
  do {
    value >>= 7;
    n++;
  } while (value != 0);
  return n;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<eatext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L35
  mov    $0x0,%eax
  jmp    L6
L35:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  jmp    L52
L18:
  movslq %ebx,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L52
  movslq %ebx,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L42
L52:
  mov    %ebx,%eax
  lea    -0x1(%rax),%ebx
  test   %eax,%eax
  jne    L18
L42:
  mov    -0x18(%rbp),%rax
L6:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *eatext(char *fname) {
  register int n;
  if (!fname)
    return NULL;
  n = strlen(fname);
  while (n--) {
    if (fname[n] == '.') {
      fname[n] = '\0';
      break;
    }
  }
  return fname;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<eatext>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L1
  call   <strlen@plt>
  movslq %eax,%rcx
  lea    -0x1(%rbx,%rcx,1),%rdx
  mov    %rdx,%rcx
  mov    %eax,%eax
  sub    %rax,%rcx
L3:
  cmp    %rcx,%rdx
  je     L1
  mov    %rdx,%rax
  sub    $0x1,%rdx
  cmpb   $0x2e,0x1(%rdx)
  jne    L3
  movb   $0x0,(%rax)
L1:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *eatext(char *fname) {
  register int n;
  if (!fname)
    return NULL;
  n = strlen(fname);
  while (n--) {
    if (fname[n] == '.') {
      fname[n] = '\0';
      break;
    }
  }
  return fname;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<eatext>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L0
  call   <strlen@plt>
  movslq %eax,%rcx
  mov    %eax,%eax
  lea    -0x1(%r12,%rcx,1),%rdx
  mov    %rdx,%rcx
  sub    %rax,%rcx
  jmp    L16
  nopl   (%rax)
L11:
  mov    %rdx,%rax
  sub    $0x1,%rdx
  cmpb   $0x2e,0x1(%rdx)
  je     L8
L16:
  cmp    %rcx,%rdx
  jne    L11
L0:
  mov    %r12,%rax
  pop    %r12
  ret
L8:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *eatext(char *fname) {
  register int n;
  if (!fname)
    return NULL;
  n = strlen(fname);
  while (n--) {
    if (fname[n] == '.') {
      fname[n] = '\0';
      break;
    }
  }
  return fname;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<eatext>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L33
  call   <strlen@plt>
  movslq %eax,%rcx
  mov    %eax,%eax
  lea    -0x1(%r12,%rcx,1),%rdx
  mov    %rdx,%rcx
  sub    %rax,%rcx
  jmp    L9
  nopl   (%rax)
L10:
  mov    %rdx,%rax
  sub    $0x1,%rdx
  cmpb   $0x2e,0x1(%rdx)
  je     L35
L9:
  cmp    %rcx,%rdx
  jne    L10
L33:
  mov    %r12,%rax
  pop    %r12
  ret
L35:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *eatext(char *fname) {
  register int n;
  if (!fname)
    return NULL;
  n = strlen(fname);
  while (n--) {
    if (fname[n] == '.') {
      fname[n] = '\0';
      break;
    }
  }
  return fname;
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L26
L29:
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L26:
  cmpl   $0xa,-0x10(%rbp)
  jle    L29
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D33
type_guessed: string
name: retrieve_data
arguments: label: D18
type_guessed: string
name: retrieve_data
arguments: label: D38
type_guessed: string
name: retrieve_data
arguments: label: D21
type_guessed: string

tool: D33: 
Tabuada de mult 

tool: D18: 
Entre com um numero para mult:

tool: D38: %d

tool: D21: 
 %d x %d = %d

assistant: ```c
int main() {
  int num1;
  int num2;
  int mult;
  printf("\nTabuada de mult \n");
  printf("\nEntre com um numero para mult:");
  scanf("%d", &num1);
  for (num2 = 1; num2 <= 10; num2++) {
    mult = num1 * num2;
    printf("\n %d x %d = %d", num1, num2, mult);
  }
  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
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <puts@plt>
  lea    D31(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    $0x1,%ebx
  lea    D17(%rip),%rbp
L5:
  mov    0x4(%rsp),%esi
  mov    %esi,%ecx
  imul   %ebx,%ecx
  mov    %ebx,%edx
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xb,%ebx
  jne    L5
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D28
type_infered: string
name: read_data
arguments: data_tag: D31
type_infered: string
name: read_data
arguments: data_tag: D20
type_infered: string
name: read_data
arguments: data_tag: D17
type_infered: string

tool: D28: 
Tabuada de mult 

tool: D31: 
Entre com um numero para mult:

tool: D20: %d

tool: D17: 
 %d x %d = %d

assistant: ```c
int main() {
  int num1;
  int num2;
  int mult;
  printf("\nTabuada de mult \n");
  printf("\nEntre com um numero para mult:");
  scanf("%d", &num1);
  for (num2 = 1; num2 <= 10; num2++) {
    mult = num1 * num2;
    printf("\n %d x %d = %d", num1, num2, mult);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D26(%rip),%rdi
  lea    D1(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf@plt>
  nopl   0x0(%rax)
L3:
  mov    0x4(%rsp),%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %esi,%ecx
  imul   %ebx,%ecx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xb,%ebx
  jne    L3
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D26
type_guessed: string
name: disassemble_data
arguments: tag: D1
type_guessed: string
name: disassemble_data
arguments: tag: D34
type_guessed: string
name: disassemble_data
arguments: tag: D38
type_guessed: string

tool: D26: 
Tabuada de mult 

tool: D1: 
 %d x %d = %d

tool: D34: 
Entre com um numero para mult:

tool: D38: %d

assistant: ```c
int main() {
  int num1;
  int num2;
  int mult;
  printf("\nTabuada de mult \n");
  printf("\nEntre com um numero para mult:");
  scanf("%d", &num1);
  for (num2 = 1; num2 <= 10; num2++) {
    mult = num1 * num2;
    printf("\n %d x %d = %d", num1, num2, mult);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rdi
  lea    D27(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  nopl   0x0(%rax)
L14:
  mov    0x4(%rsp),%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %esi,%ecx
  imul   %ebx,%ecx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xb,%ebx
  jne    L14
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
type_infered: string
name: extract_symbolic_data
arguments: label: D27
type_infered: string
name: extract_symbolic_data
arguments: label: D1
type_infered: string
name: extract_symbolic_data
arguments: label: D0
type_infered: string

tool: D11: 
Tabuada de mult 

tool: D27: 
 %d x %d = %d

tool: D1: 
Entre com um numero para mult:

tool: D0: %d

assistant: ```c
int main() {
  int num1;
  int num2;
  int mult;
  printf("\nTabuada de mult \n");
  printf("\nEntre com um numero para mult:");
  scanf("%d", &num1);
  for (num2 = 1; num2 <= 10; num2++) {
    mult = num1 * num2;
    printf("\n %d x %d = %d", num1, num2, mult);
  }
  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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  movl   $0x1,-0x14(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L8:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  -0x10(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L8
  mov    -0x10(%rbp),%eax
  movd   %eax,%xmm0
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  movss  -0x10(%rbp),%xmm0
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D24
data_type: string
name: read_data
arguments: label: D19
data_type: string
name: read_data
arguments: label: D14
data_type: string

tool: D24: %d

tool: D19: Sum: %.0f


tool: D14: Avg: %.2f


assistant: ```c
int main() {
  int n, i, m;
  float Avg, Sum;
  Sum = 0;
  i = 1;
  scanf("%d", &n);
  do {
    scanf("%d", &m);
    Sum = Sum + m;
    i++;
  } while (i <= n);
  printf("Sum: %.0f\n", Sum);
  Avg = Sum / n;
  printf("Avg: %.2f\n", Avg);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  pxor   %xmm3,%xmm3
  movss  %xmm3,0xc(%rsp)
  mov    $0x1,%ebx
  lea    D31(%rip),%rbp
L6:
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ssl 0x14(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  add    $0x1,%ebx
  cmp    %ebx,0x10(%rsp)
  jge    L6
  lea    D1(%rip),%rdi
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtsi2ssl 0x10(%rsp),%xmm1
  movss  0xc(%rsp),%xmm0
  divss  %xmm1,%xmm0
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D31
datatype: string
name: retrieve_data
arguments: data_label: D1
datatype: string
name: retrieve_data
arguments: data_label: D24
datatype: string

tool: D31: %d

tool: D1: Sum: %.0f


tool: D24: Avg: %.2f


assistant: ```c
int main() {
  int n, i, m;
  float Avg, Sum;
  Sum = 0;
  i = 1;
  scanf("%d", &n);
  do {
    scanf("%d", &m);
    Sum = Sum + m;
    i++;
  } while (i <= n);
  printf("Sum: %.0f\n", Sum);
  Avg = Sum / n;
  printf("Avg: %.2f\n", Avg);
  return 0;
}
```