ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D21(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L28
  mov    $0x0,%r12d
  je     L68
L12:
  mov    (%rsp),%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  cmp    0x4(%rsp),%ebx
  jg     L23
  mov    $0x1,%ebp
  lea    D4(%rip),%r14
  lea    D16(%rip),%r13
  jmp    L97
  cs nopw 0x0(%rax,%rax,1)
L40:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jl     L23
L97:
  movslq (%rsp),%rax
  mov    %ebx,%edx
  imul   %rax,%rbp
  mov    %rax,%rsi
  mov    %rbp,%rcx
  test   %r12w,%r12w
  je     L40
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L97
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  neg    %eax
  mov    $0x1,%r12d
  mov    %eax,0x4(%rsp)
  jmp    L12
L68:
  mov    $0x1,%esi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L23
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D21
datatype: string
name: read_data
arguments: tag: D19
datatype: string
name: read_data
arguments: tag: D32
datatype: string
name: read_data
arguments: tag: D4
datatype: string
name: read_data
arguments: tag: D16
datatype: string
name: read_data
arguments: tag: D26
datatype: string

tool: D21: Please, enter a base and an exponent in the format "b^e":  

tool: D19: %d ^ %d

tool: D32: 
%d^0:	1


tool: D4: %d^%d:	1/%ld


tool: D16: %d^%d:	%ld


tool: D26: Result: %ld


assistant: ```c
int main(void) {
  int base, exp, i;
  short int negative_exponent = 0;
  unsigned long int result = 1;
  printf("Please, enter a base and an exponent in the format \"b^e\":  ");
  scanf("%d ^ %d", &base, &exp);
  if (exp < 0) {
    negative_exponent = 1;
    exp = -exp;
  } else if (exp == 0) {
    result = 1;
    printf("Result: %ld\n", result);
    return 0;
  }
  printf("\n%d^0:\t1\n", base);
  while (i <= exp) {
    result *= base;
    if (negative_exponent == 0) {
      printf("%d^%d:\t%ld\n", base, i, result);
    } else {
      printf("%d^%d:\t1/%ld\n", base, i, result);
    }
    i++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D15(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D44(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L13
  je     L67
  mov    (%rsp),%esi
  lea    D69(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  mov    $0x1,%ebp
  lea    D22(%rip),%r12
  call   <printf@plt>
  cmp    0x4(%rsp),%ebx
  jg     L11
  nop
L30:
  movslq (%rsp),%rax
  mov    %ebx,%edx
  mov    %r12,%rdi
  add    $0x1,%ebx
  imul   %rax,%rbp
  mov    %rax,%rsi
  xor    %eax,%eax
  mov    %rbp,%rcx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L30
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  neg    %eax
  mov    (%rsp),%esi
  lea    D69(%rip),%rdi
  xor    %ebx,%ebx
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  mov    $0x1,%ebp
  lea    D36(%rip),%r12
  call   <printf@plt>
  cmp    0x4(%rsp),%ebx
  jg     L11
  nopl   0x0(%rax)
L6:
  movslq (%rsp),%rax
  mov    %ebx,%edx
  mov    %r12,%rdi
  add    $0x1,%ebx
  imul   %rax,%rbp
  mov    %rax,%rsi
  xor    %eax,%eax
  mov    %rbp,%rcx
  call   <printf@plt>
  cmp    0x4(%rsp),%ebx
  jle    L6
  jmp    L11
L67:
  mov    $0x1,%esi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L11
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
data_type: string
name: retrieve_data
arguments: tag: D44
data_type: string
name: retrieve_data
arguments: tag: D69
data_type: string
name: retrieve_data
arguments: tag: D22
data_type: string
name: retrieve_data
arguments: tag: D36
data_type: string
name: retrieve_data
arguments: tag: D30
data_type: string

tool: D15: Please, enter a base and an exponent in the format "b^e":  

tool: D44: %d ^ %d

tool: D69: 
%d^0:	1


tool: D22: %d^%d:	%ld


tool: D36: %d^%d:	1/%ld


tool: D30: Result: %ld


assistant: ```c
int main(void) {
  int base, exp, i;
  short int negative_exponent = 0;
  unsigned long int result = 1;
  printf("Please, enter a base and an exponent in the format \"b^e\":  ");
  scanf("%d ^ %d", &base, &exp);
  if (exp < 0) {
    negative_exponent = 1;
    exp = -exp;
  } else if (exp == 0) {
    result = 1;
    printf("Result: %ld\n", result);
    return 0;
  }
  printf("\n%d^0:\t1\n", base);
  while (i <= exp) {
    result *= base;
    if (negative_exponent == 0) {
      printf("%d^%d:\t%ld\n", base, i, result);
    } else {
      printf("%d^%d:\t1/%ld\n", base, i, result);
    }
    i++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<setRedBlueBlink>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x8e,%esi
  mov    %eax,%edi
  call   <sendConf@plt>
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    $0x11,%esi
  mov    %eax,%edi
  call   <sendBlink@plt>
  nop
  leave
  ret
```

assistant: ```c
void setRedBlueBlink(int sockfd, int timeOn, int period) {
  sendConf(sockfd, (1 << 7) | (1 << 3) | (3 << 1));
  sendBlink(sockfd, 0x11, timeOn, period);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<setRedBlueBlink>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edx,%r12d
  mov    $0x8e,%esi
  call   <sendConf@plt>
  mov    %r12d,%ecx
  mov    %ebp,%edx
  mov    $0x11,%esi
  mov    %ebx,%edi
  call   <sendBlink@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void setRedBlueBlink(int sockfd, int timeOn, int period) {
  sendConf(sockfd, (1 << 7) | (1 << 3) | (3 << 1));
  sendBlink(sockfd, 0x11, timeOn, period);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<setRedBlueBlink>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  mov    $0x8e,%esi
  push   %rbp
  mov    %edi,%ebp
  call   <sendConf@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%edi
  mov    $0x11,%esi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <sendBlink@plt>
```

assistant: ```c
void setRedBlueBlink(int sockfd, int timeOn, int period) {
  sendConf(sockfd, (1 << 7) | (1 << 3) | (3 << 1));
  sendBlink(sockfd, 0x11, timeOn, period);
}
```
user: What is the c source code of the assembly code below?

```asm
<setRedBlueBlink>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  mov    $0x8e,%esi
  push   %rbp
  mov    %edi,%ebp
  call   <sendConf@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%edi
  mov    $0x11,%esi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <sendBlink@plt>
```

assistant: ```c
void setRedBlueBlink(int sockfd, int timeOn, int period) {
  sendConf(sockfd, (1 << 7) | (1 << 3) | (3 << 1));
  sendBlink(sockfd, 0x11, timeOn, period);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TestIndirectJump>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xb,-0x4(%rbp)
  ja     L9
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D7(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D7(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L9
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L9
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L9
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L9
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
L9:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D60
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D26
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D41
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D31
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D67
type_guessed: string

tool: D60: Prime 1

tool: D26: Prime 2

tool: D41: Prime 3

tool: D31: Prime 4

tool: D67: Prime 5

assistant: ```c
void TestIndirectJump(int Arg) {
  switch (Arg) {
  case 2:
    puts("Prime 1");
    break;
  case 3:
    puts("Prime 2");
    break;
  case 5:
    puts("Prime 3");
    break;
  case 7:
    puts("Prime 4");
    break;
  case 11:
    puts("Prime 5");
    break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<TestIndirectJump>:
  endbr64
  cmp    $0xb,%edi
  ja     L9
  sub    $0x8,%rsp
  mov    %edi,%edi
  lea    D58(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D38(%rip),%rdi
  call   <puts@plt>
L26:
  add    $0x8,%rsp
  ret
  lea    D47(%rip),%rdi
  call   <puts@plt>
  jmp    L26
  lea    D57(%rip),%rdi
  call   <puts@plt>
  jmp    L26
  lea    D24(%rip),%rdi
  call   <puts@plt>
  jmp    L26
  lea    D32(%rip),%rdi
  call   <puts@plt>
  jmp    L26
L9:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D38
datatype: string
name: extract_symbolic_data
arguments: data_label: D47
datatype: string
name: extract_symbolic_data
arguments: data_label: D57
datatype: string
name: extract_symbolic_data
arguments: data_label: D24
datatype: string
name: extract_symbolic_data
arguments: data_label: D32
datatype: string

tool: D38: Prime 1

tool: D47: Prime 2

tool: D57: Prime 3

tool: D24: Prime 4

tool: D32: Prime 5

assistant: ```c
void TestIndirectJump(int Arg) {
  switch (Arg) {
  case 2:
    puts("Prime 1");
    break;
  case 3:
    puts("Prime 2");
    break;
  case 5:
    puts("Prime 3");
    break;
  case 7:
    puts("Prime 4");
    break;
  case 11:
    puts("Prime 5");
    break;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<TestIndirectJump>:
  endbr64
  cmp    $0xb,%edi
  ja     L2
  lea    D38(%rip),%rdx
  mov    %edi,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
L2:
  ret
  nopl   0x0(%rax)
  lea    D25(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D53(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D40(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D13(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D28(%rip),%rdi
  jmp    1050 <puts@plt>
```

tool_calls: name: read_data
arguments: label: D25
type: string
name: read_data
arguments: label: D53
type: string
name: read_data
arguments: label: D40
type: string
name: read_data
arguments: label: D13
type: string
name: read_data
arguments: label: D28
type: string

tool: D25: Prime 1

tool: D53: Prime 2

tool: D40: Prime 3

tool: D13: Prime 4

tool: D28: Prime 5

assistant: ```c
void TestIndirectJump(int Arg) {
  switch (Arg) {
  case 2:
    puts("Prime 1");
    break;
  case 3:
    puts("Prime 2");
    break;
  case 5:
    puts("Prime 3");
    break;
  case 7:
    puts("Prime 4");
    break;
  case 11:
    puts("Prime 5");
    break;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<TestIndirectJump>:
  endbr64
  cmp    $0xb,%edi
  ja     L4
  lea    D25(%rip),%rdx
  mov    %edi,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
L4:
  ret
  nopl   0x0(%rax)
  lea    D43(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D37(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D10(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D50(%rip),%rdi
  jmp    1050 <puts@plt>
  nopl   0x0(%rax)
  lea    D19(%rip),%rdi
  jmp    1050 <puts@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D43
type: string
name: retrieve_data
arguments: data_label: D37
type: string
name: retrieve_data
arguments: data_label: D10
type: string
name: retrieve_data
arguments: data_label: D50
type: string
name: retrieve_data
arguments: data_label: D19
type: string

tool: D43: Prime 1

tool: D37: Prime 2

tool: D10: Prime 3

tool: D50: Prime 4

tool: D19: Prime 5

assistant: ```c
void TestIndirectJump(int Arg) {
  switch (Arg) {
  case 2:
    puts("Prime 1");
    break;
  case 3:
    puts("Prime 2");
    break;
  case 5:
    puts("Prime 3");
    break;
  case 7:
    puts("Prime 4");
    break;
  case 11:
    puts("Prime 5");
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L49
L15:
  movl   $0x1,-0x10(%rbp)
  mov    -0x14(%rbp),%ecx
  mov    $0x1,%eax
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  jmp    L7
L33:
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L7:
  cmpl   $0x0,-0xc(%rbp)
  jne    L33
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L49:
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L15
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D16
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string

tool: D16: %d


tool: D11: %d

assistant: ```c
int main(int argc, const char *argv[]) {
  int a;
  while (scanf("%d", &a) == 1) {
    int digits = 1;
    int remainder = 1 % a;
    while (remainder != 0) {
      remainder = (1 + 10 * remainder) % a;
      digits++;
    }
    printf("%d\n", digits);
  }
  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
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  lea    D8(%rip),%rbp
  mov    $0x1,%ebx
  lea    D7(%rip),%r13
  jmp    L30
L52:
  mov    %edi,%esi
L20:
  mov    %r13,%rdi
  call   <printf@plt>
L30:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  jne    L45
  mov    0x4(%rsp),%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %edx,%esi
  test   %edx,%edx
  je     L52
  mov    %edx,%eax
L34:
  lea    (%rax,%rax,4),%eax
  lea    0x1(%rax,%rax,1),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  add    $0x1,%esi
  test   %edx,%edx
  jne    L34
  jmp    L20
L45:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D7
data_type: string

tool: D8: %d

tool: D7: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int a;
  while (scanf("%d", &a) == 1) {
    int digits = 1;
    int remainder = 1 % a;
    while (remainder != 0) {
      remainder = (1 + 10 * remainder) % a;
      digits++;
    }
    printf("%d\n", digits);
  }
  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   %r12
  lea    D16(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D10(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  nop
L11:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %eax,%esi
  cmp    $0x1,%eax
  jne    L18
  mov    %esi,%eax
  mov    0x4(%rsp),%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L21
  mov    %edx,%esi
  nopl   0x0(%rax)
L12:
  lea    (%rdx,%rdx,4),%eax
  add    $0x1,%esi
  lea    0x1(%rax,%rax,1),%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L12
L21:
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L18:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D16
type: string
name: parse_data
arguments: tag: D10
type: string

tool: D16: %d


tool: D10: %d

assistant: ```c
int main(int argc, const char *argv[]) {
  int a;
  while (scanf("%d", &a) == 1) {
    int digits = 1;
    int remainder = 1 % a;
    while (remainder != 0) {
      remainder = (1 + 10 * remainder) % a;
      digits++;
    }
    printf("%d\n", digits);
  }
  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   %r12
  lea    D14(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  nop
L48:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %eax,%esi
  cmp    $0x1,%eax
  jne    L2
  mov    %esi,%eax
  mov    0x4(%rsp),%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L5
  mov    %edx,%esi
  nopl   0x0(%rax)
L9:
  lea    (%rdx,%rdx,4),%eax
  add    $0x1,%esi
  lea    0x1(%rax,%rax,1),%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L9
L5:
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L48
  nopl   0x0(%rax,%rax,1)
L2:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D14
datatype: string
name: read_data
arguments: label: D1
datatype: string

tool: D14: %d


tool: D1: %d

assistant: ```c
int main(int argc, const char *argv[]) {
  int a;
  while (scanf("%d", &a) == 1) {
    int digits = 1;
    int remainder = 1 % a;
    while (remainder != 0) {
      remainder = (1 + 10 * remainder) % a;
      digits++;
    }
    printf("%d\n", digits);
  }
  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
  cmpl   $0x5,-0x8(%rbp)
  jne    L2
  movl   $0x7,-0x4(%rbp)
  nop
L2:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main(void) {
  int x;
  int y;
  if (x == 5)
    y = 7;
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main(void) {
  int x;
  int y;
  if (x == 5)
    y = 7;
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main(void) {
  int x;
  int y;
  if (x == 5)
    y = 7;
  return;
}
```
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
  ret
```

assistant: ```c
void main(void) {
  int x;
  int y;
  if (x == 5)
    y = 7;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<new_str_lpad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  jmp    L36
L31:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0x2c(%rbp),%edx
  mov    %dl,(%rax)
L36:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L31
L35:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    -0x20(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L42
  mov    -0x28(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  jmp    L9
L46:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0x2c(%rbp),%edx
  mov    %dl,(%rax)
L9:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L46
L42:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L35
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void new_str_lpad(char *dst, const char *src, size_t width, unsigned char pad) {
  int i;
  i = (int)width;
  while (i--)
    *dst++ = pad;
  do {
    *dst++ = *src;
    if (*src == '\n') {
      i = (int)width;
      while (i--)
        *dst++ = pad;
    }
  } while (*src++);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<new_str_lpad>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rdi
  mov    %ecx,%esi
  mov    %edx,%r10d
  lea    -0x1(%rdx),%r9d
  test   %edx,%edx
  je     L2
  lea    -0x1(%rdx),%ecx
  lea    0x1(%rax,%rcx,1),%r8
  mov    %rax,%rcx
L34:
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %r8,%rcx
  jne    L34
  mov    %r9d,%ecx
  lea    0x1(%rax,%rcx,1),%rax
L2:
  lea    -0x1(%rdx),%r8d
  add    $0x2,%r8
  mov    %r9d,%r9d
  add    $0x1,%r9
  jmp    L10
L6:
  test   %r10d,%r10d
  je     L86
  lea    (%rax,%r8,1),%rdx
  mov    %rcx,%rax
L30:
  add    $0x1,%rax
  mov    %sil,-0x1(%rax)
  cmp    %rdx,%rax
  jne    L30
  lea    (%rcx,%r9,1),%rax
L21:
  add    $0x1,%rdi
  cmpb   $0x0,-0x1(%rdi)
  je     L68
L10:
  lea    0x1(%rax),%rcx
  movzbl (%rdi),%edx
  mov    %dl,(%rax)
  cmp    $0xa,%dl
  je     L6
  mov    %rcx,%rax
  jmp    L21
L86:
  mov    %rcx,%rax
  jmp    L21
L68:
  ret
```

assistant: ```c
void new_str_lpad(char *dst, const char *src, size_t width, unsigned char pad) {
  int i;
  i = (int)width;
  while (i--)
    *dst++ = pad;
  do {
    *dst++ = *src;
    if (*src == '\n') {
      i = (int)width;
      while (i--)
        *dst++ = pad;
    }
  } while (*src++);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<new_str_lpad>:
  endbr64
  push   %r14
  movsbl %cl,%r14d
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  lea    -0x1(%rdx),%ebp
  push   %rbx
  mov    %rsi,%rbx
  test   %edx,%edx
  je     L54
  mov    %ebp,%r13d
  movsbl %cl,%esi
  add    $0x1,%r13
  mov    %r13,%rdx
  call   <memset@plt>
  mov    %rax,%rdi
  add    %r13,%rdi
  jmp    L54
  nopw   0x0(%rax,%rax,1)
L59:
  add    $0x1,%rbx
  cmpb   $0x0,-0x1(%rbx)
  je     L49
L54:
  movzbl (%rbx),%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  cmp    $0xa,%al
  jne    L59
  test   %r12d,%r12d
  je     L59
  mov    %ebp,%r13d
  mov    %r14d,%esi
  add    $0x1,%rbx
  add    $0x1,%r13
  mov    %r13,%rdx
  call   <memset@plt>
  mov    %rax,%rdi
  add    %r13,%rdi
  cmpb   $0x0,-0x1(%rbx)
  jne    L54
L49:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void new_str_lpad(char *dst, const char *src, size_t width, unsigned char pad) {
  int i;
  i = (int)width;
  while (i--)
    *dst++ = pad;
  do {
    *dst++ = *src;
    if (*src == '\n') {
      i = (int)width;
      while (i--)
        *dst++ = pad;
    }
  } while (*src++);
}
```