ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  movl   $0xa,0xc(%rsp)
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  lea    D9(%rip),%r14
  lea    D25(%rip),%r13
  lea    D1(%rip),%r12
  lea    D43(%rip),%r15
  jmp    L43
L13:
  add    $0x1,%ebx
L71:
  subl   $0x1,0xc(%rsp)
  je     L45
L43:
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  jmp    L39
L42:
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
L39:
  mov    0x14(%rsp),%eax
  cmp    $0x1,%eax
  je     L13
  cmp    $0x2,%eax
  jne    L42
  add    $0x1,%ebp
  jmp    L71
L45:
  mov    %ebx,%esi
  lea    D44(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x8,%ebx
  ja     L32
L55:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  jmp    L55
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_infered: string
name: extract_symbolic_data
arguments: data_label: D25
type_infered: string
name: extract_symbolic_data
arguments: data_label: D1
type_infered: string
name: extract_symbolic_data
arguments: data_label: D43
type_infered: string
name: extract_symbolic_data
arguments: data_label: D44
type_infered: string
name: extract_symbolic_data
arguments: data_label: D24
type_infered: string
name: extract_symbolic_data
arguments: data_label: D23
type_infered: string

tool: D9: Enter result ( 1=pass,2=fail ): 

tool: D25: %s

tool: D1: %d

tool: D43: You did not enter n accepted input!!!Please try again!!! 

tool: D44: Passed %u


tool: D24: Failed %u


tool: D23: Bonus to instructor!

assistant: ```c
int main(void) {
  unsigned int passes = 0;
  unsigned int failures = 0;
  unsigned int student = 1;
  int result = 0;
  int flag = -1;
  while (student <= 10) {
    printf("%s", "Enter result ( 1=pass,2=fail ): ");
    scanf("%d", &result);
    while (flag != 1) {
      if (result == 1) {
        passes = passes + 1;
        flag = 1;
      } else if (result == 2) {
        failures = failures + 1;
        flag = 1;
      } else {
        flag = -1;
        printf("%s\n", "You did not enter n accepted input!!!Please try again!!! ");
        printf("%s", "Enter result ( 1=pass,2=fail ): ");
        scanf("%d", &result);
      }
    }
    student = student + 1;
    flag = -1;
  }
  printf("Passed %u\n", passes);
  printf("Failed %u\n", failures);
  if (passes > 8) {
    puts("Bonus to instructor!");
  }
  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   %r15
  xor    %r15d,%r15d
  push   %r14
  lea    D8(%rip),%r14
  push   %r13
  lea    D49(%rip),%r13
  push   %r12
  lea    D41(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D25(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  lea    0x14(%rsp),%rbp
  movl   $0xa,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  xchg   %ax,%ax
L79:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    $0x1,%eax
  je     L28
  cmp    $0x2,%eax
  jne    L1
  addl   $0x1,0x8(%rsp)
L18:
  subl   $0x1,0xc(%rsp)
  jne    L79
  mov    %r15d,%esi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <printf@plt>
  cmp    $0x8,%r15d
  ja     L17
L35:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L78
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L1:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L79
  nopl   (%rax)
L28:
  add    $0x1,%r15d
  jmp    L18
L17:
  lea    D20(%rip),%rdi
  call   <puts@plt>
  jmp    L35
L78:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D8
data_type: string
name: read_data
arguments: data_tag: D49
data_type: string
name: read_data
arguments: data_tag: D41
data_type: string
name: read_data
arguments: data_tag: D25
data_type: string
name: read_data
arguments: data_tag: D10
data_type: string
name: read_data
arguments: data_tag: D30
data_type: string
name: read_data
arguments: data_tag: D20
data_type: string

tool: D8: You did not enter n accepted input!!!Please try again!!! 

tool: D49: Enter result ( 1=pass,2=fail ): 

tool: D41: %s

tool: D25: %d

tool: D10: Passed %u


tool: D30: Failed %u


tool: D20: Bonus to instructor!

assistant: ```c
int main(void) {
  unsigned int passes = 0;
  unsigned int failures = 0;
  unsigned int student = 1;
  int result = 0;
  int flag = -1;
  while (student <= 10) {
    printf("%s", "Enter result ( 1=pass,2=fail ): ");
    scanf("%d", &result);
    while (flag != 1) {
      if (result == 1) {
        passes = passes + 1;
        flag = 1;
      } else if (result == 2) {
        failures = failures + 1;
        flag = 1;
      } else {
        flag = -1;
        printf("%s\n", "You did not enter n accepted input!!!Please try again!!! ");
        printf("%s", "Enter result ( 1=pass,2=fail ): ");
        scanf("%d", &result);
      }
    }
    student = student + 1;
    flag = -1;
  }
  printf("Passed %u\n", passes);
  printf("Failed %u\n", failures);
  if (passes > 8) {
    puts("Bonus to instructor!");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  lea    D63(%rip),%r14
  push   %r13
  lea    D54(%rip),%r13
  push   %r12
  lea    D65(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D59(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  lea    0x14(%rsp),%rbp
  movl   $0xa,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  xchg   %ax,%ax
L61:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    $0x1,%eax
  je     L59
  cmp    $0x2,%eax
  jne    L12
  addl   $0x1,0x8(%rsp)
L68:
  subl   $0x1,0xc(%rsp)
  jne    L61
  mov    %r15d,%esi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  xor    %eax,%eax
  lea    D46(%rip),%rdi
  call   <printf@plt>
  cmp    $0x8,%r15d
  ja     L77
L30:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L12:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L61
  nopl   (%rax)
L59:
  add    $0x1,%r15d
  jmp    L68
L77:
  lea    D67(%rip),%rdi
  call   <puts@plt>
  jmp    L30
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D63
type_infered: string
name: parse_data
arguments: label: D54
type_infered: string
name: parse_data
arguments: label: D65
type_infered: string
name: parse_data
arguments: label: D59
type_infered: string
name: parse_data
arguments: label: D12
type_infered: string
name: parse_data
arguments: label: D46
type_infered: string
name: parse_data
arguments: label: D67
type_infered: string

tool: D63: You did not enter n accepted input!!!Please try again!!! 

tool: D54: Enter result ( 1=pass,2=fail ): 

tool: D65: %s

tool: D59: %d

tool: D12: Passed %u


tool: D46: Failed %u


tool: D67: Bonus to instructor!

assistant: ```c
int main(void) {
  unsigned int passes = 0;
  unsigned int failures = 0;
  unsigned int student = 1;
  int result = 0;
  int flag = -1;
  while (student <= 10) {
    printf("%s", "Enter result ( 1=pass,2=fail ): ");
    scanf("%d", &result);
    while (flag != 1) {
      if (result == 1) {
        passes = passes + 1;
        flag = 1;
      } else if (result == 2) {
        failures = failures + 1;
        flag = 1;
      } else {
        flag = -1;
        printf("%s\n", "You did not enter n accepted input!!!Please try again!!! ");
        printf("%s", "Enter result ( 1=pass,2=fail ): ");
        scanf("%d", &result);
      }
    }
    student = student + 1;
    flag = -1;
  }
  printf("Passed %u\n", passes);
  printf("Failed %u\n", failures);
  if (passes > 8) {
    puts("Bonus to instructor!");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <push@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <push@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  mov    D19(%rip),%rdx
  mov    (%rdx),%edx
  cmp    %edx,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <push@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  cmp    $0x3,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  mov    D19(%rip),%rdx
  mov    (%rdx),%edx
  cmp    %edx,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void test() {
  struct node *stack = NULL;
  push(&stack, 1);
  push(&stack, 2);
  assert(2 == pop(&stack));
  assert(1 == pop(&stack));
  assert(INT_MIN == pop(&stack));
  push(&stack, 3);
  assert(3 == pop(&stack));
  assert(INT_MIN == pop(&stack));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rbx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    $0x2,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  cmp    $0x2,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  cmp    $0x1,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  mov    D1(%rip),%rbp
  cmp    %eax,0x0(%rbp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  cmp    $0x3,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  cmp    %eax,0x0(%rbp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test() {
  struct node *stack = NULL;
  push(&stack, 1);
  push(&stack, 2);
  assert(2 == pop(&stack));
  assert(1 == pop(&stack));
  assert(INT_MIN == pop(&stack));
  push(&stack, 3);
  assert(3 == pop(&stack));
  assert(INT_MIN == pop(&stack));
}
```
user: What is the c source code of the assembly code below:

```asm
<test>:
  endbr64
  push   %rbp
  mov    $0x1,%esi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rdi
  call   <push@plt>
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    $0x2,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    D5(%rip),%rbx
  xor    %edi,%edi
  cmp    %eax,(%rbx)
  sete   %dil
  call   <assert@plt>
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    $0x3,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    %eax,(%rbx)
  sete   %dil
  call   <assert@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test() {
  struct node *stack = NULL;
  push(&stack, 1);
  push(&stack, 2);
  assert(2 == pop(&stack));
  assert(1 == pop(&stack));
  assert(INT_MIN == pop(&stack));
  push(&stack, 3);
  assert(3 == pop(&stack));
  assert(INT_MIN == pop(&stack));
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test>:
  endbr64
  push   %rbp
  mov    $0x1,%esi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rdi
  call   <push@plt>
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    $0x2,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    D0(%rip),%rbx
  xor    %edi,%edi
  cmp    %eax,(%rbx)
  sete   %dil
  call   <assert@plt>
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    $0x3,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  xor    %edi,%edi
  cmp    %eax,(%rbx)
  sete   %dil
  call   <assert@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test() {
  struct node *stack = NULL;
  push(&stack, 1);
  push(&stack, 2);
  assert(2 == pop(&stack));
  assert(1 == pop(&stack));
  assert(INT_MIN == pop(&stack));
  push(&stack, 3);
  assert(3 == pop(&stack));
  assert(INT_MIN == pop(&stack));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fn_qtg_1_4294967295_cl_9_10_13_13_32_32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L19:
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L13:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <fn_cl_9_10_13_13_32_32@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L19
  cmpl   $0x0,-0x8(%rbp)
  jle    L31
  mov    -0xc(%rbp),%eax
  jmp    L29
L31:
  mov    $0xffffffff,%eax
L29:
  leave
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_9_10_13_13_32_32(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_9_10_13_13_32_32(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fn_qtg_1_4294967295_cl_9_10_13_13_32_32>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  jmp    L1
L7:
  add    %eax,%ebx
  add    $0x1,%ebp
L1:
  movslq %ebx,%rdi
  add    %r12,%rdi
  call   <fn_cl_9_10_13_13_32_32@plt>
  cmp    $0xffffffff,%eax
  jne    L7
  test   %ebp,%ebp
  cmovle %eax,%ebx
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_9_10_13_13_32_32(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_9_10_13_13_32_32(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<fn_qtg_1_4294967295_cl_9_10_13_13_32_32>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L16:
  add    %eax,%r12d
  add    $0x1,%ebx
L15:
  movslq %r12d,%rdi
  add    %rbp,%rdi
  call   <fn_cl_9_10_13_13_32_32@plt>
  cmp    $0xffffffff,%eax
  jne    L16
  test   %ebx,%ebx
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_9_10_13_13_32_32(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_9_10_13_13_32_32(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fn_qtg_1_4294967295_cl_9_10_13_13_32_32>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L19:
  add    %eax,%r12d
  add    $0x1,%ebx
L5:
  movslq %r12d,%rdi
  add    %rbp,%rdi
  call   <fn_cl_9_10_13_13_32_32@plt>
  cmp    $0xffffffff,%eax
  jne    L19
  test   %ebx,%ebx
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int fn_qtg_1_4294967295_cl_9_10_13_13_32_32(const char *text) {
  int pos = 0, p = 0, count = 0;
  while ((p = fn_cl_9_10_13_13_32_32(text + pos)) != -1) {
    pos += p;
    count++;
  }
  if (count >= 1)
    return pos;
  return -1;
}
```
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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <count_set_bits@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
type: string
name: disassemble_data
arguments: label: D7
type: string

tool: D5: %d

tool: D7: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", count_set_bits(n));
  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
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <count_set_bits@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
type: string
name: unpack_data
arguments: label: D1
type: string

tool: D10: %d

tool: D1: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", count_set_bits(n));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D1(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <count_set_bits@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: string
name: parse_data
arguments: data_label: D13
type_guessed: string

tool: D1: %d

tool: D13: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", count_set_bits(n));
  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
  sub    $0x18,%rsp
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <count_set_bits@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D13
type_infered: string
name: read_data
arguments: label: D1
type_infered: string

tool: D13: %d

tool: D1: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", count_set_bits(n));
  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
  mov    %rsp,%rbp
  mov    $0x20,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x9,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xa,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xc,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xd,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xb,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x5c,%edi
  call   <isspace@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  assert(isspace(' '));
  assert(isspace('\t'));
  assert(isspace('\n'));
  assert(isspace('\f'));
  assert(isspace('\r'));
  assert(isspace('\v'));
  assert(!isspace('\\'));
  return 0;
}
```
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    $0x20,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x9,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xa,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xc,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xd,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xb,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x5c,%edi
  call   <isspace@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(isspace(' '));
  assert(isspace('\t'));
  assert(isspace('\n'));
  assert(isspace('\f'));
  assert(isspace('\r'));
  assert(isspace('\v'));
  assert(!isspace('\\'));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x20,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x9,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xa,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xc,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xd,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xb,%edi
  call   <isspace@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x5c,%edi
  call   <isspace@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(isspace(' '));
  assert(isspace('\t'));
  assert(isspace('\n'));
  assert(isspace('\f'));
  assert(isspace('\r'));
  assert(isspace('\v'));
  assert(!isspace('\\'));
  return 0;
}
```