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

```asm
<ft_ulitoa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x1,%esi
  call   <ft_n_count@plt>
  mov    %eax,%r12d
  lea    0x1(%rax),%edi
  call   <ft_strnew@plt>
  mov    %rax,%rbx
  test   %r12d,%r12d
  jle    L9
  movslq %r12d,%rax
  lea    -0x1(%rbx,%rax,1),%rsi
  lea    -0x2(%rbx,%rax,1),%rdi
  lea    -0x1(%r12),%eax
  sub    %rax,%rdi
  mov    %rbp,%rcx
  movabs $0xcccccccccccccccd,%r8
L6:
  mov    %rcx,%rax
  mul    %r8
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  add    $0x30,%ecx
  mov    %cl,(%rsi)
  mov    %rdx,%rcx
  sub    $0x1,%rsi
  cmp    %rdi,%rsi
  jne    L6
L9:
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <ft_n_count@plt>
  lea    0x1(%rax),%edi
  call   <ft_strnew@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_ulitoa(unsigned long int n) {
  int len;
  unsigned long int nbr;
  char *str2;
  int i;
  char *str;
  len = 1;
  nbr = n;
  i = ft_n_count(n, len);
  str2 = ft_strnew(i + 1);
  while (i > 0) {
    str2[i - 1] = ((char)((nbr % 10) + 48));
    i--;
    nbr /= 10;
  }
  str = ft_strnew(ft_n_count(n, len) + 1);
  ft_strcpy(str, str2);
  free(str2);
  return (str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_ulitoa>:
  endbr64
  push   %r12
  mov    $0x1,%esi
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <ft_n_count@plt>
  lea    0x1(%rax),%edi
  mov    %eax,%ebx
  call   <ft_strnew@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L16
  movabs $0xcccccccccccccccd,%r8
  movslq %ebx,%rax
  mov    %r12,%rcx
  lea    -0x2(%rbp,%rax,1),%rdi
  lea    -0x1(%rbp,%rax,1),%rsi
  lea    -0x1(%rbx),%eax
  sub    %rax,%rdi
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    %rcx,%rax
  sub    $0x1,%rsi
  mul    %r8
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  add    $0x30,%ecx
  mov    %cl,0x1(%rsi)
  mov    %rdx,%rcx
  cmp    %rdi,%rsi
  jne    L15
L16:
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <ft_n_count@plt>
  lea    0x1(%rax),%edi
  call   <ft_strnew@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <ft_strcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_ulitoa(unsigned long int n) {
  int len;
  unsigned long int nbr;
  char *str2;
  int i;
  char *str;
  len = 1;
  nbr = n;
  i = ft_n_count(n, len);
  str2 = ft_strnew(i + 1);
  while (i > 0) {
    str2[i - 1] = ((char)((nbr % 10) + 48));
    i--;
    nbr /= 10;
  }
  str = ft_strnew(ft_n_count(n, len) + 1);
  ft_strcpy(str, str2);
  free(str2);
  return (str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_ulitoa>:
  endbr64
  push   %r12
  mov    $0x1,%esi
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <ft_n_count@plt>
  lea    0x1(%rax),%edi
  mov    %eax,%ebx
  call   <ft_strnew@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L12
  movabs $0xcccccccccccccccd,%r8
  movslq %ebx,%rax
  mov    %r12,%rcx
  lea    -0x2(%rbp,%rax,1),%rdi
  lea    -0x1(%rbp,%rax,1),%rsi
  lea    -0x1(%rbx),%eax
  sub    %rax,%rdi
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %rcx,%rax
  sub    $0x1,%rsi
  mul    %r8
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  add    $0x30,%ecx
  mov    %cl,0x1(%rsi)
  mov    %rdx,%rcx
  cmp    %rsi,%rdi
  jne    L14
L12:
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <ft_n_count@plt>
  lea    0x1(%rax),%edi
  call   <ft_strnew@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <ft_strcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_ulitoa(unsigned long int n) {
  int len;
  unsigned long int nbr;
  char *str2;
  int i;
  char *str;
  len = 1;
  nbr = n;
  i = ft_n_count(n, len);
  str2 = ft_strnew(i + 1);
  while (i > 0) {
    str2[i - 1] = ((char)((nbr % 10) + 48));
    i--;
    nbr /= 10;
  }
  str = ft_strnew(ft_n_count(n, len) + 1);
  ft_strcpy(str, str2);
  free(str2);
  return (str);
}
```
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
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf70,%rsp
  mov    %edi,-0x1f64(%rbp)
  mov    %rsi,-0x1f70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1f70(%rbp),%rdx
  mov    -0x1f64(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <parse_cmd@plt>
  mov    %eax,-0x1f58(%rbp)
  cmpl   $0xffffffff,-0x1f58(%rbp)
  jne    L12
  movl   $0xa,-0x1f58(%rbp)
L12:
  lea    -0x1f50(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getlines@plt>
  mov    %eax,-0x1f54(%rbp)
  mov    -0x1f58(%rbp),%edx
  mov    -0x1f54(%rbp),%ecx
  lea    -0x1f50(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <output_tail_lines@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int tail_num, lines_num;
  char *lines[1000];
  tail_num = parse_cmd(argc, argv);
  if (tail_num == -1) {
    tail_num = 10;
  }
  lines_num = getlines(lines, 1000);
  output_tail_lines(lines, lines_num, tail_num);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1f48(%rsp)
  xor    %eax,%eax
  call   <parse_cmd@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  mov    $0xa,%eax
  cmove  %eax,%ebx
  mov    %rsp,%rbp
  mov    $0x3e8,%esi
  mov    %rbp,%rdi
  call   <getlines@plt>
  mov    %eax,%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <output_tail_lines@plt>
  mov    0x1f48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x1f58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int tail_num, lines_num;
  char *lines[1000];
  tail_num = parse_cmd(argc, argv);
  if (tail_num == -1) {
    tail_num = 10;
  }
  lines_num = getlines(lines, 1000);
  output_tail_lines(lines, lines_num, tail_num);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1f48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <parse_cmd@plt>
  mov    $0x3e8,%esi
  mov    %rbp,%rdi
  cmp    $0xffffffff,%eax
  mov    %eax,%r12d
  mov    $0xa,%eax
  cmove  %eax,%r12d
  call   <getlines@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  mov    %r12d,%edx
  call   <output_tail_lines@plt>
  mov    0x1f48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x1f58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int tail_num, lines_num;
  char *lines[1000];
  tail_num = parse_cmd(argc, argv);
  if (tail_num == -1) {
    tail_num = 10;
  }
  lines_num = getlines(lines, 1000);
  output_tail_lines(lines, lines_num, tail_num);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1f48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <parse_cmd@plt>
  mov    $0x3e8,%esi
  mov    %rbp,%rdi
  cmp    $0xffffffff,%eax
  mov    %eax,%r12d
  mov    $0xa,%eax
  cmove  %eax,%r12d
  call   <getlines@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  mov    %r12d,%edx
  call   <output_tail_lines@plt>
  mov    0x1f48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x1f58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int tail_num, lines_num;
  char *lines[1000];
  tail_num = parse_cmd(argc, argv);
  if (tail_num == -1) {
    tail_num = 10;
  }
  lines_num = getlines(lines, 1000);
  output_tail_lines(lines, lines_num, tail_num);
  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
<count1sposnos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L70
  cmpq   $0x1,-0x18(%rbp)
  je     L70
  cmpq   $0x2,-0x18(%rbp)
  je     L70
  cmpq   $0x3,-0x18(%rbp)
  jne    L1
L70:
  movl   $0x0,-0x1c(%rbp)
L1:
  cmpq   $0x0,-0x18(%rbp)
  jne    L69
  mov    $0x0,%eax
  jmp    L128
L69:
  cmpq   $0x1,-0x18(%rbp)
  je     L56
  cmpq   $0x2,-0x18(%rbp)
  jne    L118
L56:
  mov    $0x1,%eax
  jmp    L128
L118:
  cmpq   $0x3,-0x18(%rbp)
  jne    L43
  mov    $0x2,%eax
  jmp    L128
L43:
  mov    -0x18(%rbp),%rax
  cqto
  shr    $0x3f,%rdx
  add    %rdx,%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  cmp    $0x1,%rax
  jne    L11
  mov    -0x1c(%rbp),%eax
  xor    $0x2,%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%rax
  cqto
  idiv   %rsi
  mov    %rax,%rbx
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rcx
  shr    $0x3f,%rcx
  add    %rcx,%rdx
  sar    %rdx
  mov    %eax,%esi
  mov    %rdx,%rdi
  call   <count1sposnos>
  add    %rax,%rax
  add    %rax,%rbx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,%rdi
  call   <bitcount@plt>
  add    %rbx,%rax
  jmp    L128
L11:
  mov    -0x1c(%rbp),%eax
  xor    $0x2,%eax
  movslq %eax,%rdi
  mov    -0x18(%rbp),%rax
  cqto
  idiv   %rdi
  mov    %rax,%rbx
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rcx
  shr    $0x3f,%rcx
  add    %rcx,%rdx
  sar    %rdx
  mov    %eax,%esi
  mov    %rdx,%rdi
  call   <count1sposnos>
  add    %rax,%rax
  add    %rbx,%rax
L128:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long count1sposnos(long n, int pwr2) {
  if (n == 0 || n == 1 || n == 2 || n == 3)
    pwr2 = 0;
  if (n == 0)
    return 0;
  else if (n == 1 || n == 2)
    return 1;
  else if (n == 3)
    return 2;
  else if (n % 2 == 1) {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++) + bitcount(n / 2));
  } else {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++));
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<count1sposnos>:
  endbr64
  mov    %rdi,%rcx
  cmp    $0x3,%rdi
  ja     L30
  mov    %rdi,%rax
  test   %rdi,%rdi
  jne    L17
L21:
  ret
L17:
  mov    $0x0,%esi
L30:
  lea    -0x1(%rcx),%rdx
  mov    $0x1,%eax
  cmp    $0x1,%rdx
  jbe    L21
  cmp    $0x3,%rcx
  je     L2
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rcx,%rdx
  shr    $0x3f,%rdx
  lea    (%rcx,%rdx,1),%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  cmp    $0x1,%rax
  je     L45
  mov    %esi,%edi
  xor    $0x2,%edi
  movslq %edi,%rdi
  mov    %rcx,%rax
  cqto
  idiv   %rdi
  mov    %rax,%rbx
  mov    %rcx,%rdi
  shr    $0x3f,%rdi
  add    %rcx,%rdi
  sar    %rdi
  call   <count1sposnos>
  lea    (%rbx,%rax,2),%rax
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L45:
  mov    %esi,%edi
  xor    $0x2,%edi
  movslq %edi,%rdi
  mov    %rcx,%rax
  cqto
  idiv   %rdi
  mov    %rax,%rbx
  mov    %rcx,%rbp
  shr    $0x3f,%rbp
  add    %rcx,%rbp
  sar    %rbp
  mov    %rbp,%rdi
  call   <count1sposnos>
  lea    (%rbx,%rax,2),%rbx
  mov    %rbp,%rdi
  call   <bitcount@plt>
  add    %rbx,%rax
  jmp    L9
L2:
  mov    $0x2,%eax
  ret
```

assistant: ```c
long count1sposnos(long n, int pwr2) {
  if (n == 0 || n == 1 || n == 2 || n == 3)
    pwr2 = 0;
  if (n == 0)
    return 0;
  else if (n == 1 || n == 2)
    return 1;
  else if (n == 3)
    return 2;
  else if (n % 2 == 1) {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++) + bitcount(n / 2));
  } else {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++));
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<count1sposnos>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  jmp    L37
  nopl   0x0(%rax)
L36:
  xor    %esi,%esi
L46:
  lea    -0x1(%rdi),%rax
  cmp    $0x1,%rax
  jbe    L6
  cmp    $0x3,%rdi
  je     L16
  mov    %esi,%ecx
  mov    %rdi,%rax
  xor    $0x2,%ecx
  cqto
  movslq %ecx,%rcx
  idiv   %rcx
  mov    %rax,%rbx
  mov    %rdi,%rax
  shr    $0x3f,%rax
  lea    (%rdi,%rax,1),%rdx
  and    $0x1,%edx
  sub    %rax,%rdx
  cmp    $0x1,%rdx
  je     L49
  mov    %rdi,%rax
  imul   %r12,%rbx
  add    %r12,%r12
  shr    $0x3f,%rax
  add    %rdi,%rax
  sar    %rax
  add    %rbx,%r13
  mov    %rax,%rdi
L37:
  cmp    $0x3,%rdi
  ja     L46
  test   %rdi,%rdi
  jne    L36
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L6:
  add    $0x8,%rsp
  add    %r12,%r13
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L16:
  add    $0x8,%rsp
  lea    0x0(%r13,%r12,2),%r13
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L49:
  sar    %rdi
  mov    %rdi,%rbp
  call   <count1sposnos>
  mov    %rbp,%rdi
  lea    (%rbx,%rax,2),%rbx
  call   <bitcount@plt>
  add    $0x8,%rsp
  add    %rax,%rbx
  imul   %r12,%rbx
  add    %rbx,%r13
  pop    %rbx
  pop    %rbp
  mov    %r13,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
long count1sposnos(long n, int pwr2) {
  if (n == 0 || n == 1 || n == 2 || n == 3)
    pwr2 = 0;
  if (n == 0)
    return 0;
  else if (n == 1 || n == 2)
    return 1;
  else if (n == 3)
    return 2;
  else if (n % 2 == 1) {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++) + bitcount(n / 2));
  } else {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++));
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<count1sposnos>:
  endbr64
  push   %r15
  mov    %esi,%r8d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  jmp    L107
L120:
  xor    %r8d,%r8d
L113:
  lea    -0x1(%rdi),%rax
  cmp    $0x1,%rax
  jbe    L142
  cmp    $0x3,%rdi
  je     L176
  mov    %rdi,%rax
  mov    %r8d,%esi
  shr    $0x3f,%rax
  xor    $0x2,%esi
  lea    (%rdi,%rax,1),%rbx
  movslq %esi,%rsi
  and    $0x1,%ebx
  sub    %rax,%rbx
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rax,%r12
  cmp    $0x1,%rbx
  je     L23
  mov    %rdi,%rax
  imul   %r13,%r12
  add    %r13,%r13
  shr    $0x3f,%rax
  add    %rdi,%rax
  sar    %rax
  add    %r12,%r14
  mov    %rax,%rdi
L107:
  cmp    $0x3,%rdi
  ja     L113
  test   %rdi,%rdi
  jne    L120
L3:
  add    $0x18,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L142:
  add    %r13,%r14
  jmp    L3
  nopl   (%rax)
L176:
  lea    (%r14,%r13,2),%r14
  jmp    L3
  cs nopw 0x0(%rax,%rax,1)
L23:
  sar    %rdi
  xor    %ecx,%ecx
  mov    %rdi,%rbp
  nopl   0x0(%rax,%rax,1)
L59:
  cmp    $0x3,%rdi
  ja     L33
  test   %rdi,%rdi
  je     L159
  cmp    $0x3,%rdi
  jne    L123
L116:
  lea    (%rcx,%rbx,2),%rcx
L159:
  mov    %rbp,%rdi
  lea    (%r12,%rcx,2),%rbx
  call   <bitcount@plt>
  add    %rax,%rbx
  imul   %r13,%rbx
  add    %rbx,%r14
  jmp    L3
  nopl   (%rax)
L33:
  test   %rdi,%rdi
  je     L159
  cmp    $0x3,%rdi
  je     L116
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rdi,%rdx
  shr    $0x3f,%rdx
  mov    %rax,%r15
  lea    (%rdi,%rdx,1),%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  cmp    $0x1,%rax
  je     L136
  mov    %rdi,%rax
  imul   %rbx,%r15
  add    %rbx,%rbx
  shr    $0x3f,%rax
  add    %rdi,%rax
  sar    %rax
  add    %r15,%rcx
  mov    %rax,%rdi
  jmp    L59
L123:
  add    %rbx,%rcx
  jmp    L159
L136:
  sar    %rdi
  mov    %r8d,%esi
  mov    %rcx,0x8(%rsp)
  mov    %rdi,(%rsp)
  call   <count1sposnos>
  mov    (%rsp),%rdi
  lea    (%r15,%rax,2),%r15
  call   <bitcount@plt>
  mov    0x8(%rsp),%rcx
  add    %rax,%r15
  imul   %rbx,%r15
  add    %r15,%rcx
  jmp    L159
```

assistant: ```c
long count1sposnos(long n, int pwr2) {
  if (n == 0 || n == 1 || n == 2 || n == 3)
    pwr2 = 0;
  if (n == 0)
    return 0;
  else if (n == 1 || n == 2)
    return 1;
  else if (n == 3)
    return 2;
  else if (n % 2 == 1) {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++) + bitcount(n / 2));
  } else {
    return (n / (2 ^ pwr2) + 2 * count1sposnos(n / 2, pwr2++));
  }
}
```
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
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D24
data_type: string
name: read_data
arguments: tag: D20
data_type: string
name: read_data
arguments: tag: D16
data_type: string

tool: D24: %d

tool: D20: 
%d

tool: D16: 
%d


assistant: ```c
int main() {
  int a, square, cube;
  scanf("%d", &a);
  square = a * a;
  cube = a * a * a;
  printf("\n%d", square);
  printf("\n%d\n", cube);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%esi
  imul   %eax,%esi
  imul   %esi,%eax
  mov    %eax,%ebx
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
datatype: string
name: retrieve_data
arguments: tag: D26
datatype: string
name: retrieve_data
arguments: tag: D8
datatype: string

tool: D4: %d

tool: D26: 
%d

tool: D8: 
%d


assistant: ```c
int main() {
  int a, square, cube;
  scanf("%d", &a);
  square = a * a;
  cube = a * a * a;
  printf("\n%d", square);
  printf("\n%d\n", cube);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D22(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  imul   %eax,%esi
  imul   %esi,%eax
  mov    %eax,%ebp
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D22
data_type: string
name: unpack_data
arguments: data_label: D13
data_type: string
name: unpack_data
arguments: data_label: D28
data_type: string

tool: D22: %d

tool: D13: 
%d

tool: D28: 
%d


assistant: ```c
int main() {
  int a, square, cube;
  scanf("%d", &a);
  square = a * a;
  cube = a * a * a;
  printf("\n%d", square);
  printf("\n%d\n", cube);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  imul   %eax,%esi
  imul   %esi,%eax
  mov    %eax,%ebp
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D18
datatype: string
name: retrieve_data
arguments: tag: D2
datatype: string
name: retrieve_data
arguments: tag: D21
datatype: string

tool: D18: %d

tool: D2: 
%d

tool: D21: 
%d


assistant: ```c
int main() {
  int a, square, cube;
  scanf("%d", &a);
  square = a * a;
  cube = a * a * a;
  printf("\n%d", square);
  printf("\n%d\n", cube);
  return 0;
}
```