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

```asm
<atben_setup>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <rf_setup@plt>
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <mlockall@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <ben_setup@plt>
```

assistant: ```c
void atben_setup(struct atrf_dsc *dsc, int size, int trim) {
  rf_setup(dsc, size, trim);
  mlockall(MCL_CURRENT);
  ben_setup(dsc);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<atben_setup>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <rf_setup@plt>
  mov    D0(%rip),%rax
  mov    (%rax),%edi
  call   <mlockall@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <ben_setup@plt>
```

assistant: ```c
void atben_setup(struct atrf_dsc *dsc, int size, int trim) {
  rf_setup(dsc, size, trim);
  mlockall(MCL_CURRENT);
  ben_setup(dsc);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ft_strmapi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  cltq
  shl    $0x3,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L21
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  jmp    L23
L32:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%rdx
  mov    %edx,%ecx
  mov    -0x18(%rbp),%rsi
  mov    -0x20(%rbp),%rdx
  lea    (%rsi,%rdx,1),%rbx
  mov    -0x30(%rbp),%rdx
  mov    %eax,%esi
  mov    %ecx,%edi
  call   *%rdx
  mov    %al,(%rbx)
  addq   $0x1,-0x20(%rbp)
L23:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L32
  mov    -0x18(%rbp),%rax
  jmp    L3
L21:
  mov    $0x0,%eax
L3:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *ft_strmapi(const char *s, char (*f)(unsigned int, char)) {
  size_t i;
  char *str_new;
  i = 0;
  str_new = malloc(sizeof(str_new) * ft_strlen((char *)s) + 1);
  if (str_new) {
    ft_strcpy(str_new, s);
    while (s[i]) {
      str_new[i] = (*f)(i, s[i]);
      i++;
    }
    return (str_new);
  } else
    return (NULL);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_strmapi>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  call   <ft_strlen@plt>
  lea    0x1(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L21
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  movzbl (%r12),%esi
  test   %sil,%sil
  je     L21
  mov    $0x0,%ebx
L10:
  movsbl %sil,%esi
  mov    %ebx,%edi
  call   *%r13
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
  movzbl (%r12,%rbx,1),%esi
  test   %sil,%sil
  jne    L10
L21:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *ft_strmapi(const char *s, char (*f)(unsigned int, char)) {
  size_t i;
  char *str_new;
  i = 0;
  str_new = malloc(sizeof(str_new) * ft_strlen((char *)s) + 1);
  if (str_new) {
    ft_strcpy(str_new, s);
    while (s[i]) {
      str_new[i] = (*f)(i, s[i]);
      i++;
    }
    return (str_new);
  } else
    return (NULL);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ft_strmapi>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  lea    0x1(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L7
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  movsbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L7
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %ebx,%edi
  call   *%r13
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L8
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *ft_strmapi(const char *s, char (*f)(unsigned int, char)) {
  size_t i;
  char *str_new;
  i = 0;
  str_new = malloc(sizeof(str_new) * ft_strlen((char *)s) + 1);
  if (str_new) {
    ft_strcpy(str_new, s);
    while (s[i]) {
      str_new[i] = (*f)(i, s[i]);
      i++;
    }
    return (str_new);
  } else
    return (NULL);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_strmapi>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  lea    0x1(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L23
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  movsbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L23
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %ebx,%edi
  call   *%r13
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L20
L23:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *ft_strmapi(const char *s, char (*f)(unsigned int, char)) {
  size_t i;
  char *str_new;
  i = 0;
  str_new = malloc(sizeof(str_new) * ft_strlen((char *)s) + 1);
  if (str_new) {
    ft_strcpy(str_new, s);
    while (s[i]) {
      str_new[i] = (*f)(i, s[i]);
      i++;
    }
    return (str_new);
  } else
    return (NULL);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x8(%rbp)
  jmp    L12
L24:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L5
L13:
  mov    -0x4(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  subl   $0x1,-0xc(%rbp)
L5:
  cmpl   $0x1,-0xc(%rbp)
  jne    L13
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L12:
  cmpl   $0x5,-0x8(%rbp)
  jle    L24
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D13
type_infered: string
name: parse_data
arguments: label: D2
type_infered: string

tool: D13: The number	!factorial

tool: D2: %d		%d


assistant: ```c
int main(void) {
  int number = 0;
  int counter = 0;
  printf("The number\t!factorial\n");
  for (counter = 1; counter <= 5; ++counter) {
    number = counter;
    int factorial = 1;
    while (number != 1) {
      factorial = factorial * number;
      --number;
    }
    printf("%d\t\t%d\n", counter, factorial);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    $0x2,%ebp
  mov    $0x1,%ebx
  lea    D10(%rip),%r12
  jmp    L37
L34:
  mov    $0x1,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L11:
  add    $0x1,%ebx
  add    $0x1,%ebp
L37:
  mov    %ebx,%esi
  mov    %ebx,%eax
  mov    $0x1,%edx
  cmp    $0x1,%ebx
  je     L34
L2:
  imul   %eax,%edx
  sub    $0x1,%eax
  cmp    $0x1,%eax
  jne    L2
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x5,%ebp
  jle    L11
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D17
type: string
name: retrieve_data
arguments: label: D10
type: string

tool: D17: The number	!factorial

tool: D10: %d		%d


assistant: ```c
int main(void) {
  int number = 0;
  int counter = 0;
  printf("The number\t!factorial\n");
  for (counter = 1; counter <= 5; ++counter) {
    number = counter;
    int factorial = 1;
    while (number != 1) {
      factorial = factorial * number;
      --number;
    }
    printf("%d\t\t%d\n", counter, factorial);
  }
  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
  push   %rbp
  lea    D5(%rip),%rdi
  lea    D0(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  mov    $0x1,%esi
  mov    $0x1,%edx
  lea    0x1(%rsi),%ebx
  mov    %esi,%eax
  cmp    $0x1,%esi
  je     L46
  nopl   0x0(%rax)
L40:
  imul   %eax,%edx
  sub    $0x1,%eax
  cmp    $0x1,%eax
  jne    L40
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x6,%ebx
  je     L45
  mov    %ebx,%esi
L39:
  lea    0x1(%rsi),%ebx
  mov    %esi,%eax
  mov    $0x1,%edx
  cmp    $0x1,%esi
  jne    L40
L46:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  jmp    L39
  nopl   0x0(%rax)
L45:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
type: string
name: retrieve_data
arguments: tag: D0
type: string

tool: D5: The number	!factorial

tool: D0: %d		%d


assistant: ```c
int main(void) {
  int number = 0;
  int counter = 0;
  printf("The number\t!factorial\n");
  for (counter = 1; counter <= 5; ++counter) {
    number = counter;
    int factorial = 1;
    while (number != 1) {
      factorial = factorial * number;
      --number;
    }
    printf("%d\t\t%d\n", counter, factorial);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D9(%rip),%rdi
  lea    D7(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <puts@plt>
  mov    $0x1,%esi
  lea    0x1(%rsi),%ebp
  cmp    $0x1,%esi
  je     L3
L21:
  cmp    $0x1,%ebx
  je     L62
  mov    %ebx,%edx
  lea    -0x2(%rsi),%eax
  imul   %esi,%edx
  cmp    $0x3,%esi
  je     L8
  imul   %eax,%edx
  lea    -0x3(%rsi),%eax
  cmp    $0x4,%esi
  je     L8
  imul   %eax,%edx
L6:
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0x6,%ebp
  je     L45
L49:
  mov    %ebp,%esi
  add    $0x1,%ebx
  lea    0x1(%rsi),%ebp
  cmp    $0x1,%esi
  jne    L21
L3:
  mov    $0x1,%edx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L49
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L49
  nopl   0x0(%rax)
L45:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L62:
  mov    %esi,%edx
  jmp    L6
```

tool_calls: name: unpack_data
arguments: data_tag: D9
data_type: string
name: unpack_data
arguments: data_tag: D7
data_type: string

tool: D9: The number	!factorial

tool: D7: %d		%d


assistant: ```c
int main(void) {
  int number = 0;
  int counter = 0;
  printf("The number\t!factorial\n");
  for (counter = 1; counter <= 5; ++counter) {
    number = counter;
    int factorial = 1;
    while (number != 1) {
      factorial = factorial * number;
      --number;
    }
    printf("%d\t\t%d\n", counter, factorial);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L7:
  cmp    %rdx,%rsp
  je     L13
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L7
L13:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L97
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L97:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x1,-0x30(%rbp)
  jmp    L117
L28:
  mov    -0x20(%rbp),%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  movl   $0x0,(%rax,%rdx,4)
  addl   $0x1,-0x30(%rbp)
L117:
  mov    -0x30(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jle    L28
  movl   $0x0,-0x30(%rbp)
  jmp    L88
L47:
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x20(%rbp),%rdx
  movslq %eax,%rcx
  mov    (%rdx,%rcx,4),%edx
  lea    0x1(%rdx),%ecx
  mov    -0x20(%rbp),%rdx
  cltq
  mov    %ecx,(%rdx,%rax,4)
  addl   $0x1,-0x30(%rbp)
L88:
  mov    -0x30(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jl     L47
  movl   $0x1,-0x30(%rbp)
  jmp    L25
L72:
  mov    -0x20(%rbp),%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  test   %eax,%eax
  je     L54
  movl   $0x0,-0x2c(%rbp)
  jmp    L125
L6:
  mov    -0x30(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x2c(%rbp)
L125:
  mov    -0x20(%rbp),%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L6
L54:
  addl   $0x1,-0x30(%rbp)
L25:
  mov    -0x30(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jle    L72
  mov    %rbx,%rsp
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L121
  call   <__stack_chk_fail@plt>
L121:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type: string

tool: D2: %d  

assistant: ```c
void count_sort(int ar[], int l, int n) {
  int Arr[l + 1], i, j;
  for (i = 1; i <= l; i++)
    Arr[i] = 0;
  for (i = 0; i < n; i++) {
    Arr[ar[i]]++;
  }
  for (i = 1; i <= l; i++) {
    if (Arr[i] != 0) {
      for (j = 0; j < Arr[i]; j++) {
        printf("%d  ", i);
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<count_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,-0x44(%rbp)
  mov    %edx,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0x1(%rsi),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L24:
  cmp    %rdx,%rsp
  je     L77
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L24
L77:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L28
  orq    $0x0,-0x8(%rsp,%rax,1)
L28:
  mov    %rsp,%r14
  mov    -0x44(%rbp),%esi
  test   %esi,%esi
  jle    L93
  lea    0x4(%r14),%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%r14,%rdx,4),%rdx
L104:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L104
  test   %ecx,%ecx
  jle    L70
L57:
  mov    $0x0,%eax
L19:
  movslq (%rdi,%rax,4),%rdx
  addl   $0x1,(%r14,%rdx,4)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L19
  cmpl   $0x0,-0x44(%rbp)
  jle    L138
L70:
  mov    $0x1,%r12d
  jmp    L84
L122:
  add    $0x1,%r12
  cmp    %r12d,-0x44(%rbp)
  jl     L138
L84:
  mov    %r12d,%r15d
  mov    (%r14,%r12,4),%r13d
  test   %r13d,%r13d
  jle    L122
  mov    $0x0,%ebx
L16:
  mov    %r15d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r13d
  jne    L16
  jmp    L122
L93:
  test   %ecx,%ecx
  jg     L57
L138:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: string

tool: D3: %d  

assistant: ```c
void count_sort(int ar[], int l, int n) {
  int Arr[l + 1], i, j;
  for (i = 1; i <= l; i++)
    Arr[i] = 0;
  for (i = 0; i < n; i++) {
    Arr[ar[i]]++;
  }
  for (i = 1; i <= l; i++) {
    if (Arr[i] != 0) {
      for (j = 0; j < Arr[i]; j++) {
        printf("%d  ", i);
      }
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<count_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %edx,%ebx
  sub    $0x28,%rsp
  mov    %esi,-0x4c(%rbp)
  mov    %rsp,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0x1(%rsi),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L32
L114:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L114
L32:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L18
L132:
  mov    -0x4c(%rbp),%eax
  mov    %rsp,%rcx
  mov    %rcx,-0x48(%rbp)
  test   %eax,%eax
  jle    L29
  movslq %eax,%rdx
  xor    %esi,%esi
  lea    0x4(%rcx),%rdi
  shl    $0x2,%rdx
  call   <memset@plt>
  test   %ebx,%ebx
  jle    L115
L86:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L133:
  movslq (%r12,%rax,4),%rdx
  mov    -0x48(%rbp),%rcx
  add    $0x1,%rax
  addl   $0x1,(%rcx,%rdx,4)
  cmp    %eax,%ebx
  jg     L133
  mov    -0x4c(%rbp),%eax
  test   %eax,%eax
  jle    L84
L115:
  mov    $0x1,%r15d
  lea    D2(%rip),%r14
L54:
  mov    -0x48(%rbp),%rax
  mov    %r15d,%r13d
  mov    (%rax,%r15,4),%r12d
  test   %r12d,%r12d
  jle    L127
  xor    %ebx,%ebx
L2:
  mov    %r13d,%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L2
L127:
  add    $0x1,%r15
  cmp    %r15d,-0x4c(%rbp)
  jge    L54
L84:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L18:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L132
L29:
  test   %ebx,%ebx
  jg     L86
  jmp    L84
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D2
datatype: string

tool: D2: %d  

assistant: ```c
void count_sort(int ar[], int l, int n) {
  int Arr[l + 1], i, j;
  for (i = 1; i <= l; i++)
    Arr[i] = 0;
  for (i = 0; i < n; i++) {
    Arr[ar[i]]++;
  }
  for (i = 1; i <= l; i++) {
    if (Arr[i] != 0) {
      for (j = 0; j < Arr[i]; j++) {
        printf("%d  ", i);
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %edx,%ebx
  sub    $0x28,%rsp
  mov    %esi,-0x4c(%rbp)
  mov    %rsp,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0x1(%rsi),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L21
L99:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L99
L21:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L39
L129:
  mov    -0x4c(%rbp),%eax
  mov    %rsp,%rcx
  mov    %rcx,-0x48(%rbp)
  test   %eax,%eax
  jle    L97
  movslq %eax,%rdx
  xor    %esi,%esi
  lea    0x4(%rcx),%rdi
  shl    $0x2,%rdx
  call   <memset@plt>
  test   %ebx,%ebx
  jle    L111
L61:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L112:
  movslq (%r12,%rax,4),%rdx
  mov    -0x48(%rbp),%rcx
  add    $0x1,%rax
  addl   $0x1,(%rcx,%rdx,4)
  cmp    %eax,%ebx
  jg     L112
  mov    -0x4c(%rbp),%eax
  test   %eax,%eax
  jle    L137
L111:
  mov    $0x1,%r15d
  lea    D8(%rip),%r14
L8:
  mov    -0x48(%rbp),%rax
  mov    %r15d,%r13d
  mov    (%rax,%r15,4),%r12d
  test   %r12d,%r12d
  jle    L59
  xor    %ebx,%ebx
L138:
  mov    %r13d,%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L138
L59:
  add    $0x1,%r15
  cmp    %r15d,-0x4c(%rbp)
  jge    L8
L137:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L39:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L129
L97:
  test   %ebx,%ebx
  jg     L61
  jmp    L137
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
data_type: string

tool: D8: %d  

assistant: ```c
void count_sort(int ar[], int l, int n) {
  int Arr[l + 1], i, j;
  for (i = 1; i <= l; i++)
    Arr[i] = 0;
  for (i = 0; i < n; i++) {
    Arr[ar[i]]++;
  }
  for (i = 1; i <= l; i++) {
    if (Arr[i] != 0) {
      for (j = 0; j < Arr[i]; j++) {
        printf("%d  ", i);
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strdup_vprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movq   $0x50,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
L43:
  mov    -0x2c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <va_copy@plt>
  mov    -0x20(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rsi
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  cmpl   $0x0,-0x1c(%rbp)
  js     L64
  mov    -0x18(%rbp),%rax
  cmp    %eax,-0x1c(%rbp)
  jge    L64
  mov    -0x10(%rbp),%rax
  jmp    L44
L64:
  cmpl   $0x0,-0x1c(%rbp)
  js     L74
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  jmp    L82
L74:
  shlq   -0x18(%rbp)
L82:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  cmpq   $0x0,-0x10(%rbp)
  je     L10
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x10(%rbp)
L10:
  mov    $0x0,%eax
  jmp    L44
L6:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L43
L44:
  leave
  ret
```

assistant: ```c
char *strdup_vprintf(const char *fmt, va_list ap) {
  size_t len = 80;
  char *buf;
  buf = malloc(len);
  for (;;) {
    int n;
    char *nbuf;
    va_list ap2;
    va_copy(ap2, ap);
    n = vsnprintf(buf, len, fmt, ap2);
    va_end(ap2);
    if (n >= 0 && n < (int)len)
      return buf;
    if (n >= 0)
      len = n + 1;
    else
      len *= 2;
    nbuf = realloc(buf, len);
    if (!nbuf) {
      {
        if (buf) {
          free(buf);
          buf = NULL;
        }
      };
      return NULL;
    }
    buf = nbuf;
  }
}
```