ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  movss  D23(%rip),%xmm0
  lea    D17(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  movss  D23(%rip),%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movss  D39(%rip),%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
  movss  D24(%rip),%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0xa,%edi
  pop    %rbp
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D23
datatype: f32
name: unpack_data
arguments: data_label: D17
datatype: string
name: unpack_data
arguments: data_label: D39
datatype: f32
name: unpack_data
arguments: data_label: D24
datatype: f32

tool: D23: 5

tool: D17: value=%format	

tool: D39: 3

tool: D24: 8

assistant: ```c
void main() {
  float x = 5, x1 = 3, x2 = 8;
  printf("value=%format\t", (x));
  printf("value=%format\t", (x));
  putchar('\n');
  ;
  printf("value=%format\t", (x1));
  printf("value=%format\t", (x2));
  putchar('\n');
  ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<check_zero_sized_VLA>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rcx
  cmpl   $0x0,-0x24(%rbp)
  je     L57
  mov    %rcx,%rsp
  jmp    L40
L57:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x18(%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,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffff000,%rsi
  mov    %rsp,%rdx
  sub    %rsi,%rdx
L4:
  cmp    %rdx,%rsp
  je     L30
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L4
L30:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L38
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L38:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x10(%rbp)
  mov    %rcx,%rsp
L40:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void check_zero_sized_VLA(int x) {
  if (x)
    return;
  int vla[x];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<check_zero_sized_VLA>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rsp,%rax
  test   %edi,%edi
  jne    L16
  mov    %rax,%rsp
L6:
  leave
  ret
L16:
  mov    %rax,%rsp
  jmp    L6
```

assistant: ```c
void check_zero_sized_VLA(int x) {
  if (x)
    return;
  int vla[x];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<check_zero_sized_VLA>:
  endbr64
  ret
```

assistant: ```c
void check_zero_sized_VLA(int x) {
  if (x)
    return;
  int vla[x];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<check_zero_sized_VLA>:
  endbr64
  ret
```

assistant: ```c
void check_zero_sized_VLA(int x) {
  if (x)
    return;
  int vla[x];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<help_menu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <get_da_func@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D5
datatype: string

tool: D5: deglupta: Password mutator so serious it may break your hard drive!
	-h:                 help menu
	-i <file>:          input file (required)
	-o <file>:          output file to append to (default: stdout)
	-r <pass>:          resume output after "pass"
	-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)
	-m <num>:           aprox max output size (default: 4294967296 ie: 4G)
	                    Will be off by at most the length of the current password
	                    -m -1 results in: 18446744073709551615 byte file size limit



assistant: ```c
void help_menu() {
  printf("deglupta: Password mutator so serious it may break your hard drive!\n"
         "\t-h:                 help menu\n"
         "\t-i <file>:          input file (required)\n"
         "\t-o <file>:          output file to append to (default: stdout)\n"
         "\t-r <pass>:          resume output after \"pass\"\n"
         "\t-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)\n"
         "\t-m <num>:           aprox max output size (default: 4294967296 ie: 4G)\n"
         "\t                    Will be off by at most the length of the current password\n"
         "\t                    -m -1 results in: 18446744073709551615 byte file size limit\n"
         "\n");
  get_da_func(NULL);
  exit(1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<help_menu>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <get_da_func@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: string

tool: D5: deglupta: Password mutator so serious it may break your hard drive!
	-h:                 help menu
	-i <file>:          input file (required)
	-o <file>:          output file to append to (default: stdout)
	-r <pass>:          resume output after "pass"
	-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)
	-m <num>:           aprox max output size (default: 4294967296 ie: 4G)
	                    Will be off by at most the length of the current password
	                    -m -1 results in: 18446744073709551615 byte file size limit



assistant: ```c
void help_menu() {
  printf("deglupta: Password mutator so serious it may break your hard drive!\n"
         "\t-h:                 help menu\n"
         "\t-i <file>:          input file (required)\n"
         "\t-o <file>:          output file to append to (default: stdout)\n"
         "\t-r <pass>:          resume output after \"pass\"\n"
         "\t-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)\n"
         "\t-m <num>:           aprox max output size (default: 4294967296 ie: 4G)\n"
         "\t                    Will be off by at most the length of the current password\n"
         "\t                    -m -1 results in: 18446744073709551615 byte file size limit\n"
         "\n");
  get_da_func(NULL);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<help_menu>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <get_da_func@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type_infered: string

tool: D0: deglupta: Password mutator so serious it may break your hard drive!
	-h:                 help menu
	-i <file>:          input file (required)
	-o <file>:          output file to append to (default: stdout)
	-r <pass>:          resume output after "pass"
	-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)
	-m <num>:           aprox max output size (default: 4294967296 ie: 4G)
	                    Will be off by at most the length of the current password
	                    -m -1 results in: 18446744073709551615 byte file size limit



assistant: ```c
void help_menu() {
  printf("deglupta: Password mutator so serious it may break your hard drive!\n"
         "\t-h:                 help menu\n"
         "\t-i <file>:          input file (required)\n"
         "\t-o <file>:          output file to append to (default: stdout)\n"
         "\t-r <pass>:          resume output after \"pass\"\n"
         "\t-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)\n"
         "\t-m <num>:           aprox max output size (default: 4294967296 ie: 4G)\n"
         "\t                    Will be off by at most the length of the current password\n"
         "\t                    -m -1 results in: 18446744073709551615 byte file size limit\n"
         "\n");
  get_da_func(NULL);
  exit(1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<help_menu>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <get_da_func@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
type_infered: string

tool: D9: deglupta: Password mutator so serious it may break your hard drive!
	-h:                 help menu
	-i <file>:          input file (required)
	-o <file>:          output file to append to (default: stdout)
	-r <pass>:          resume output after "pass"
	-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)
	-m <num>:           aprox max output size (default: 4294967296 ie: 4G)
	                    Will be off by at most the length of the current password
	                    -m -1 results in: 18446744073709551615 byte file size limit



assistant: ```c
void help_menu() {
  printf("deglupta: Password mutator so serious it may break your hard drive!\n"
         "\t-h:                 help menu\n"
         "\t-i <file>:          input file (required)\n"
         "\t-o <file>:          output file to append to (default: stdout)\n"
         "\t-r <pass>:          resume output after \"pass\"\n"
         "\t-f  <str>:          list of functions to Travers, (default: 3,1,2,4,6,6,5)\n"
         "\t-m <num>:           aprox max output size (default: 4294967296 ie: 4G)\n"
         "\t                    Will be off by at most the length of the current password\n"
         "\t                    -m -1 results in: 18446744073709551615 byte file size limit\n"
         "\n");
  get_da_func(NULL);
  exit(1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<calculate_checksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    $0x0,%r12d
  movl   $0x0,-0x24(%rbp)
  mov    -0x3c(%rbp),%eax
  and    $0x7,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x3c(%rbp),%eax
  sar    $0x3,%eax
  movslq %eax,%rbx
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L33
L10:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x24(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_m_psadbw@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <_mm_extract_pi16@plt>
  add    %rax,%r12
  sub    $0x1,%rbx
  addq   $0x4,-0x18(%rbp)
L33:
  test   %rbx,%rbx
  jg     L10
  mov    -0x3c(%rbp),%eax
  sub    -0x20(%rbp),%eax
  movslq %eax,%rbx
  jmp    L24
L8:
  mov    %rbx,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  add    %rax,%r12
  add    $0x1,%rbx
L24:
  mov    -0x3c(%rbp),%eax
  cltq
  cmp    %rax,%rbx
  jl     L8
  mov    %r12d,%eax
  movzbl %al,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int calculate_checksum(const char *buf, int buf_len) {
  register long int sum = 0;
  register long int ii;
  int extra;
  __m64 zero = (__m64)0LL;
  __m64 dest;
  __m64 *pointer;
  extra = buf_len & 0x07;
  ii = buf_len >> 3;
  pointer = (__m64 *)buf;
  while (ii > 0) {
    dest = _m_psadbw(*pointer, zero);
    sum += _mm_extract_pi16(dest, 0);
    ii--;
    pointer++;
  }
  for (ii = buf_len - extra; ii < buf_len; ii++)
    sum += buf[ii];
  return (int)(sum & 0xFF);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<calculate_checksum>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %esi,%r13d
  sar    $0x3,%r13d
  movslq %r13d,%r13
  test   %r13,%r13
  jle    L19
  mov    %rdi,%r14
  mov    $0x0,%ebx
L21:
  mov    $0x0,%esi
  mov    (%r14),%edi
  call   <_m_psadbw@plt>
  mov    %eax,%edi
  mov    $0x0,%esi
  call   <_mm_extract_pi16@plt>
  add    %rax,%rbx
  add    $0x4,%r14
  sub    $0x1,%r13
  jne    L21
L23:
  mov    %r12d,%eax
  and    $0xfffffff8,%eax
  cmp    %r12d,%eax
  jge    L3
  cltq
  add    %rbp,%rax
  movslq %r12d,%r12
  add    %r12,%rbp
L18:
  movsbq (%rax),%rdx
  add    %rdx,%rbx
  add    $0x1,%rax
  cmp    %rbp,%rax
  jne    L18
L3:
  movzbl %bl,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  mov    $0x0,%ebx
  jmp    L23
```

assistant: ```c
int calculate_checksum(const char *buf, int buf_len) {
  register long int sum = 0;
  register long int ii;
  int extra;
  __m64 zero = (__m64)0LL;
  __m64 dest;
  __m64 *pointer;
  extra = buf_len & 0x07;
  ii = buf_len >> 3;
  pointer = (__m64 *)buf;
  while (ii > 0) {
    dest = _m_psadbw(*pointer, zero);
    sum += _mm_extract_pi16(dest, 0);
    ii--;
    pointer++;
  }
  for (ii = buf_len - extra; ii < buf_len; ii++)
    sum += buf[ii];
  return (int)(sum & 0xFF);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calculate_checksum>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  movslq %esi,%r12
  mov    %r12d,%eax
  push   %rbp
  mov    %rdi,%rbp
  sar    $0x3,%eax
  push   %rbx
  test   %eax,%eax
  jle    L29
  movslq %eax,%r13
  mov    %rdi,%r14
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L21:
  mov    (%r14),%edi
  xor    %esi,%esi
  add    $0x4,%r14
  call   <_m_psadbw@plt>
  xor    %esi,%esi
  mov    %eax,%edi
  call   <_mm_extract_pi16@plt>
  add    %rax,%rbx
  sub    $0x1,%r13
  jne    L21
L1:
  mov    %r12d,%eax
  and    $0xfffffff8,%eax
  cmp    %r12d,%eax
  jge    L12
  cltq
  add    %rbp,%rax
  add    %r12,%rbp
  nopl   0x0(%rax,%rax,1)
L48:
  movsbq (%rax),%rdx
  add    $0x1,%rax
  add    %rdx,%rbx
  cmp    %rax,%rbp
  jne    L48
L12:
  movzbl %bl,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L29:
  xor    %ebx,%ebx
  jmp    L1
```

assistant: ```c
int calculate_checksum(const char *buf, int buf_len) {
  register long int sum = 0;
  register long int ii;
  int extra;
  __m64 zero = (__m64)0LL;
  __m64 dest;
  __m64 *pointer;
  extra = buf_len & 0x07;
  ii = buf_len >> 3;
  pointer = (__m64 *)buf;
  while (ii > 0) {
    dest = _m_psadbw(*pointer, zero);
    sum += _mm_extract_pi16(dest, 0);
    ii--;
    pointer++;
  }
  for (ii = buf_len - extra; ii < buf_len; ii++)
    sum += buf[ii];
  return (int)(sum & 0xFF);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dir_exits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  je     L0
  mov    $0x0,%eax
  jmp    L24
L0:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  je     L32
  mov    $0x1,%eax
  jmp    L24
L32:
  mov    $0x0,%eax
L24:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
bool dir_exits(const char *path) {
  struct stat st;
  if (stat(path, &st) != 0) {
    return false;
  }
  if (S_ISDIR(st.st_mode)) {
    return true;
  } else {
    return false;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dir_exits>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L27
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x18,%rsp
  ret
L27:
  mov    0x4(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L7
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool dir_exits(const char *path) {
  struct stat st;
  if (stat(path, &st) != 0) {
    return false;
  }
  if (S_ISDIR(st.st_mode)) {
    return true;
  } else {
    return false;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<dir_exits>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L17
  mov    0x4(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L17:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool dir_exits(const char *path) {
  struct stat st;
  if (stat(path, &st) != 0) {
    return false;
  }
  if (S_ISDIR(st.st_mode)) {
    return true;
  } else {
    return false;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dir_exits>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L17
  mov    0x4(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L17:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x18,%rsp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool dir_exits(const char *path) {
  struct stat st;
  if (stat(path, &st) != 0) {
    return false;
  }
  if (S_ISDIR(st.st_mode)) {
    return true;
  } else {
    return false;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<null_pointer_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type: string

tool: D3: %i

assistant: ```c
void null_pointer_read(void) {
  int *i = 0;
  printf("%i", *i);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<null_pointer_read>:
  endbr64
  sub    $0x8,%rsp
  mov    0x0,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_guessed: string

tool: D2: %i

assistant: ```c
void null_pointer_read(void) {
  int *i = 0;
  printf("%i", *i);
}
```
user: What is the c source code of the assembly code below?

```asm
<null_pointer_read>:
  endbr64
  mov    0x0,%eax
  ud2
```

assistant: ```c
void null_pointer_read(void) {
  int *i = 0;
  printf("%i", *i);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<null_pointer_read>:
  endbr64
  mov    0x0,%eax
  ud2
```

assistant: ```c
void null_pointer_read(void) {
  int *i = 0;
  printf("%i", *i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    -0x1b000(%rsp),%r11
L158:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L158
  sub    $0x780,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1b778(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x1b76c,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0xc,%ebx
  jmp    L93
L32:
  add    $0x1,%rbx
  cmp    $0x6ddc,%rbx
  je     L126
L93:
  mov    %ebx,%edi
  call   <sum_of_proper_divisors@plt>
  cmp    %ebx,%eax
  jle    L32
  cmp    $0x6ddb,%eax
  jg     L32
  movl   $0x1,(%rsp,%rbx,4)
  jmp    L32
L126:
  mov    $0x6dda,%eax
  mov    %rsp,%rdx
L124:
  mov    %eax,%r9d
  cmpl   $0x1,(%rdx,%rax,4)
  je     L61
  sub    $0x1,%rax
  cmp    $0xffffffffffffffff,%rax
  jne    L124
  mov    $0x6ddb,%r9d
L61:
  mov    $0x1,%ecx
  sub    %r9d,%ecx
  mov    $0x1,%edi
  mov    $0x0,%r8d
  jmp    L83
L15:
  add    $0x4,%rdx
  sub    $0x1,%eax
  cmp    %ecx,%eax
  je     L11
L70:
  cmpl   $0x1,(%rdx)
  jne    L15
  test   %eax,%eax
  je     L11
  js     L15
  movslq %eax,%rsi
  cmpl   $0x1,(%rsp,%rsi,4)
  jne    L15
  jmp    L80
L11:
  add    %edi,%r8d
L80:
  add    $0x1,%edi
  add    $0x1,%ecx
  cmp    $0x6ddb,%edi
  je     L48
L83:
  mov    %rsp,%rdx
  mov    %edi,%eax
  test   %r9d,%r9d
  jg     L70
  jmp    L11
L48:
  mov    %r8d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x1b778(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L132
  mov    $0x0,%eax
  add    $0x1b780,%rsp
  pop    %rbx
  ret
L132:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D3
data_type: string

tool: D3: %i

assistant: ```c
int main(int argc, char *argv[]) {
  int abundants[28123];
  memset(abundants, 0, sizeof(int) * 28123);
  for (int i = 12; i <= 28123; i++) {
    int sum = sum_of_proper_divisors(i);
    if (i < sum && sum <= 28123) {
      abundants[i] = 1;
    }
  }
  int last_abundant = 28123;
  for (int i = 28123 - 1; i >= 0; i--) {
    if (abundants[i] == 1) {
      last_abundant = i;
      break;
    }
  }
  int sum = 0;
  for (int i = 1; i < 28123; i++) {
    for (int e = 0; e < last_abundant; e++) {
      if (abundants[e] == 1) {
        int compliment = i - e;
        if (compliment == 0)
          goto found;
        if (compliment < 0)
          continue;
        if (abundants[compliment] == 1) {
          goto not_found;
        }
      }
    }
  found:
    sum += i;
  not_found:;
  }
  printf("%i", sum);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x1b000(%rsp),%r11
L5:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L5
  sub    $0x788,%rsp
  mov    $0x1b76c,%edx
  xor    %esi,%esi
  mov    $0xc,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x1b778(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
L173:
  mov    %ebp,%edi
  call   <sum_of_proper_divisors@plt>
  cmp    %ebp,%eax
  jle    L148
  cmp    $0x6ddb,%eax
  jg     L148
  movl   $0x1,(%rbx,%rbp,4)
L148:
  add    $0x1,%rbp
  cmp    $0x6ddc,%rbp
  jne    L173
  mov    $0x6dda,%eax
  jmp    L147
L12:
  sub    $0x1,%rax
  jb     L24
L147:
  cmpl   $0x1,(%rbx,%rax,4)
  mov    %eax,%r10d
  jne    L12
L92:
  mov    $0x1,%esi
  lea    0x4(%rsp),%r8
  mov    $0x1,%edi
  xor    %r9d,%r9d
  sub    %r10d,%esi
  nopl   (%rax)
L58:
  mov    %rbx,%rdx
  mov    %edi,%eax
  mov    %r8,%rcx
  test   %r10d,%r10d
  jne    L22
  jmp    L79
  nopw   0x0(%rax,%rax,1)
L93:
  sub    $0x1,%eax
  add    $0x4,%rdx
  sub    $0x4,%rcx
  cmp    %esi,%eax
  je     L79
L22:
  cmpl   $0x1,(%rdx)
  jne    L93
  test   %eax,%eax
  je     L79
  js     L93
  cmpl   $0x1,(%rcx)
  jne    L93
  nop
L75:
  add    $0x1,%edi
  add    $0x1,%esi
  add    $0x4,%r8
  cmp    $0x6ddb,%edi
  jne    L58
  mov    %r9d,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x1b778(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L164
  add    $0x1b788,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L79:
  add    %edi,%r9d
  jmp    L75
L24:
  mov    $0x6ddb,%r10d
  jmp    L92
L164:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
type_guessed: string

tool: D9: %i

assistant: ```c
int main(int argc, char *argv[]) {
  int abundants[28123];
  memset(abundants, 0, sizeof(int) * 28123);
  for (int i = 12; i <= 28123; i++) {
    int sum = sum_of_proper_divisors(i);
    if (i < sum && sum <= 28123) {
      abundants[i] = 1;
    }
  }
  int last_abundant = 28123;
  for (int i = 28123 - 1; i >= 0; i--) {
    if (abundants[i] == 1) {
      last_abundant = i;
      break;
    }
  }
  int sum = 0;
  for (int i = 1; i < 28123; i++) {
    for (int e = 0; e < last_abundant; e++) {
      if (abundants[e] == 1) {
        int compliment = i - e;
        if (compliment == 0)
          goto found;
        if (compliment < 0)
          continue;
        if (abundants[compliment] == 1) {
          goto not_found;
        }
      }
    }
  found:
    sum += i;
  not_found:;
  }
  printf("%i", sum);
}
```