ylfeng/ReF-Decompile-dataset huggingface dataset:
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    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L10
  lea    D9(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L10:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x6(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  mov    %ax,0x4(%rsp)
  call   <RandomFunc@plt>
  movzwl 0x6(%rsp),%esi
  cmp    $0x7fd9,%si
  je     L2
L24:
  xor    %eax,%eax
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L2:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  movzwl 0x6(%rsp),%esi
  jmp    L24
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
data_type: string
name: unpack_data
arguments: label: D13
data_type: string
name: unpack_data
arguments: label: D16
data_type: string

tool: D9: Call this program with %i arguments


tool: D13: %u


tool: D16: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned short input[1];
  unsigned short output[1];
  int randomFuns_i5;
  unsigned short randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned short)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 32729) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L27
  lea    D25(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L27:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x6(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  mov    %ax,0x4(%rsp)
  call   <RandomFunc@plt>
  movzwl 0x6(%rsp),%esi
  cmp    $0x7fd9,%si
  je     L2
L25:
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L2:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  movzwl 0x6(%rsp),%esi
  jmp    L25
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D25
type: string
name: parse_data
arguments: tag: D4
type: string
name: parse_data
arguments: tag: D14
type: string

tool: D25: Call this program with %i arguments


tool: D4: %u


tool: D14: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned short input[1];
  unsigned short output[1];
  int randomFuns_i5;
  unsigned short randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned short)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 32729) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<clock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D17(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
clock_t clock(void) {
  errno = ENOSYS;
  return (clock_t)-1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<clock>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
clock_t clock(void) {
  errno = ENOSYS;
  return (clock_t)-1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<clock>:
  endbr64
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
clock_t clock(void) {
  errno = ENOSYS;
  return (clock_t)-1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<clock>:
  endbr64
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D17(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
clock_t clock(void) {
  errno = ENOSYS;
  return (clock_t)-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
<router_solicitation>:
  endbr64
  push   %rbx
  sub    $0x50,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movl   $0x0,0x38(%rsp)
  movl   $0xff,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x2,0x3c(%rsp)
  movb   $0x60,(%rsi)
  movb   $0x0,0x1(%rsi)
  movb   $0x0,0x2(%rsi)
  movb   $0x0,0x3(%rsi)
  movb   $0x3a,0x6(%rsi)
  movb   $0xff,0x7(%rsi)
L9:
  mov    (%rdi,%rax,4),%ecx
  mov    %cl,0x8(%rbx,%rax,1)
  mov    (%rsp,%rax,4),%ecx
  mov    %cl,0x18(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    $0x10,%rax
  jne    L9
  movb   $0x85,0x28(%rbx)
  movb   $0x0,0x29(%rbx)
  movb   $0x0,0x2c(%rbx)
  movb   $0x0,0x2d(%rbx)
  movb   $0x0,0x2e(%rbx)
  movb   $0x0,0x2f(%rbx)
  movb   $0x1,0x30(%rbx)
  movb   $0x1,0x31(%rbx)
  mov    $0x2,%eax
L18:
  mov    (%rdx,%rax,4),%ecx
  mov    %cl,0x30(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    $0x8,%rax
  jne    L18
  movb   $0x0,0x4(%rbx)
  movb   $0x10,0x5(%rbx)
  mov    $0x38,%esi
  mov    %rbx,%rdi
  call   <checksum_icmpv6@plt>
  movzbl %ah,%edx
  mov    %dl,0x2a(%rbx)
  mov    %al,0x2b(%rbx)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    %rbx,%rax
  add    $0x50,%rsp
  pop    %rbx
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *router_solicitation(uint8_t *src_address, char *buffer, uint8_t *mac_address) {
  int offset = 0;
  uint8_t dst_address[16] = {0xff, 0x02, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0x02};
  uint16_t payload_length;
  uint16_t checksum = 0;
  int ix;
  buffer[0] = 0x60;
  buffer[1] = 0;
  buffer[2] = 0;
  buffer[3] = 0;
  offset += 4;
  offset += 2;
  buffer[offset] = 0x3a;
  offset += 1;
  buffer[offset] = 255;
  offset += 1;
  for (ix = 0; ix < 16; ix++) {
    buffer[offset + ix] = src_address[ix];
    buffer[offset + 16 + ix] = dst_address[ix];
  }
  offset += 32;
  buffer[offset] = 0x85;
  offset += 1;
  buffer[offset] = 0;
  offset += 1;
  offset += 2;
  buffer[offset] = 0;
  buffer[offset + 1] = 0;
  buffer[offset + 2] = 0;
  buffer[offset + 3] = 0;
  offset += 4;
  buffer[offset] = 1;
  offset += 1;
  buffer[offset] = 1;
  offset += 1;
  for (ix = 2; ix < 8; ix++) {
    buffer[offset + (ix - 2)] = mac_address[ix];
  }
  offset += 6;
  payload_length = (offset - 40);
  buffer[4] = (payload_length >> 8) & 255;
  buffer[5] = (payload_length >> 0) & 255;
  checksum = checksum_icmpv6(buffer, (offset));
  buffer[42] = (checksum >> 8) & 255;
  buffer[43] = (checksum >> 0) & 255;
  return buffer;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<router_solicitation>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  mov    $0xffffff3a,%r8d
  mov    %rsi,%r12
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x60,(%rsi)
  mov    %r8w,0x6(%rsi)
  mov    $0xff,%esi
  movl   $0x0,0x38(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x2,0x3c(%rsp)
  movups %xmm0,0x8(%rsp)
  movups %xmm0,0x18(%rsp)
  movups %xmm0,0x28(%rsp)
  jmp    L37
L34:
  mov    (%rsp,%rax,4),%esi
L37:
  mov    (%rdi,%rax,4),%ecx
  mov    %sil,0x18(%r12,%rax,1)
  mov    %cl,0x8(%r12,%rax,1)
  add    $0x1,%rax
  cmp    $0x10,%rax
  jne    L34
  mov    $0x85,%ecx
  mov    $0x101,%esi
  mov    $0x2,%eax
  movl   $0x0,0x2c(%r12)
  mov    %cx,0x28(%r12)
  mov    %si,0x30(%r12)
  xchg   %ax,%ax
L19:
  mov    (%rdx,%rax,4),%ecx
  mov    %cl,0x30(%r12,%rax,1)
  add    $0x1,%rax
  cmp    $0x8,%rax
  jne    L19
  mov    $0x1000,%eax
  mov    $0x38,%esi
  mov    %r12,%rdi
  mov    %ax,0x4(%r12)
  call   <checksum_icmpv6@plt>
  rol    $0x8,%ax
  mov    %ax,0x2a(%r12)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x50,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *router_solicitation(uint8_t *src_address, char *buffer, uint8_t *mac_address) {
  int offset = 0;
  uint8_t dst_address[16] = {0xff, 0x02, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0x02};
  uint16_t payload_length;
  uint16_t checksum = 0;
  int ix;
  buffer[0] = 0x60;
  buffer[1] = 0;
  buffer[2] = 0;
  buffer[3] = 0;
  offset += 4;
  offset += 2;
  buffer[offset] = 0x3a;
  offset += 1;
  buffer[offset] = 255;
  offset += 1;
  for (ix = 0; ix < 16; ix++) {
    buffer[offset + ix] = src_address[ix];
    buffer[offset + 16 + ix] = dst_address[ix];
  }
  offset += 32;
  buffer[offset] = 0x85;
  offset += 1;
  buffer[offset] = 0;
  offset += 1;
  offset += 2;
  buffer[offset] = 0;
  buffer[offset + 1] = 0;
  buffer[offset + 2] = 0;
  buffer[offset + 3] = 0;
  offset += 4;
  buffer[offset] = 1;
  offset += 1;
  buffer[offset] = 1;
  offset += 1;
  for (ix = 2; ix < 8; ix++) {
    buffer[offset + (ix - 2)] = mac_address[ix];
  }
  offset += 6;
  payload_length = (offset - 40);
  buffer[4] = (payload_length >> 8) & 255;
  buffer[5] = (payload_length >> 0) & 255;
  checksum = checksum_icmpv6(buffer, (offset));
  buffer[42] = (checksum >> 8) & 255;
  buffer[43] = (checksum >> 0) & 255;
  return buffer;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ace_tu_fs_does_file_contain_line_at>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movl   $0x0,-0x40(%rbp)
  movq   $0x0,-0x28(%rbp)
  movl   $0x0,-0x3c(%rbp)
  mov    -0x50(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x20(%rbp)
  jmp    L87
L46:
  addl   $0x1,-0x40(%rbp)
  mov    -0x30(%rbp),%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  sub    $0x1,%rax
  add    %rbx,%rax
  movb   $0x0,(%rax)
  mov    -0x40(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  je     L72
  cmpl   $0x0,-0x44(%rbp)
  jg     L16
L72:
  mov    $0x1,%eax
  jmp    L89
L16:
  mov    $0x0,%eax
L89:
  mov    %eax,-0x34(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  je     L87
  mov    -0x30(%rbp),%rax
  mov    -0x58(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L87
  movl   $0x1,-0x3c(%rbp)
  jmp    L19
L87:
  mov    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rcx
  lea    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <getline@plt>
  mov    %eax,-0x38(%rbp)
  cmpl   $0xffffffff,-0x38(%rbp)
  jne    L46
L19:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x30(%rbp),%rax
  test   %rax,%rax
  je     L50
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L50:
  mov    -0x3c(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type: string

tool: D0: r

assistant: ```c
bool ace_tu_fs_does_file_contain_line_at(int expected_line_number, char *filepath, char *str_without_linebreak) {
  FILE *file;
  char *line = NULL;
  int line_number = 0;
  ssize_t line_length;
  size_t buffer_length = 0;
  bool does_line_number_match;
  bool result = false;
  file = fopen(filepath, "r");
  while ((line_length = getline(&line, &buffer_length, file)) != -1) {
    line_number++;
    line[strlen(line) - 1] = '\0';
    does_line_number_match = (line_number == expected_line_number) || (expected_line_number <= 0);
    if (does_line_number_match && (strcmp(line, str_without_linebreak) == 0)) {
      result = true;
      break;
    }
  }
  fclose(file);
  if (line)
    free(line);
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ace_tu_fs_does_file_contain_line_at>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%r12d
  mov    %rsi,%rdi
  mov    %rdx,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  lea    D5(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%r13
  mov    $0x0,%ebx
  lea    0x10(%rsp),%r14
  jmp    L1
L57:
  mov    %r15,%rsi
  mov    0x8(%rsp),%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L39
L1:
  lea    0x8(%rsp),%rdi
  mov    %r13,%rdx
  mov    %r14,%rsi
  call   <getline@plt>
  cmp    $0xffffffff,%eax
  je     L65
  add    $0x1,%ebx
  mov    0x8(%rsp),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbp,%rax,1)
  cmp    %r12d,%ebx
  je     L57
  test   %r12d,%r12d
  jg     L1
  jmp    L57
L65:
  mov    $0x0,%ebx
L73:
  mov    %r13,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rdi
  test   %rdi,%rdi
  je     L85
  call   <free@plt>
L85:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    %ebx,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  mov    $0x1,%ebx
  jmp    L73
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string

tool: D5: r

assistant: ```c
bool ace_tu_fs_does_file_contain_line_at(int expected_line_number, char *filepath, char *str_without_linebreak) {
  FILE *file;
  char *line = NULL;
  int line_number = 0;
  ssize_t line_length;
  size_t buffer_length = 0;
  bool does_line_number_match;
  bool result = false;
  file = fopen(filepath, "r");
  while ((line_length = getline(&line, &buffer_length, file)) != -1) {
    line_number++;
    line[strlen(line) - 1] = '\0';
    does_line_number_match = (line_number == expected_line_number) || (expected_line_number <= 0);
    if (does_line_number_match && (strcmp(line, str_without_linebreak) == 0)) {
      result = true;
      break;
    }
  }
  fclose(file);
  if (line)
    free(line);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ace_tu_fs_does_file_contain_line_at>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    %rsi,%rdi
  lea    D5(%rip),%rsi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  mov    %rdx,0x8(%rsp)
  lea    0x20(%rsp),%r14
  lea    0x18(%rsp),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  call   <fopen@plt>
  mov    %rax,%r12
  nopl   0x0(%rax)
L60:
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <getline@plt>
  cmp    $0xffffffff,%eax
  je     L50
  mov    0x18(%rsp),%r15
  add    $0x1,%ebx
  mov    %r15,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r15,%rax,1)
  cmp    %ebp,%ebx
  je     L61
  test   %ebp,%ebp
  jg     L60
L61:
  mov    0x8(%rsp),%rsi
  mov    0x18(%rsp),%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L60
  mov    $0x1,%r13d
  jmp    L3
  nop
L50:
  xor    %r13d,%r13d
L3:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x18(%rsp),%rdi
  test   %rdi,%rdi
  je     L24
  call   <free@plt>
L24:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L68
  add    $0x38,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L68:
  call   <__stack_chk_fail@plt>
```

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

tool: D5: r

assistant: ```c
bool ace_tu_fs_does_file_contain_line_at(int expected_line_number, char *filepath, char *str_without_linebreak) {
  FILE *file;
  char *line = NULL;
  int line_number = 0;
  ssize_t line_length;
  size_t buffer_length = 0;
  bool does_line_number_match;
  bool result = false;
  file = fopen(filepath, "r");
  while ((line_length = getline(&line, &buffer_length, file)) != -1) {
    line_number++;
    line[strlen(line) - 1] = '\0';
    does_line_number_match = (line_number == expected_line_number) || (expected_line_number <= 0);
    if (does_line_number_match && (strcmp(line, str_without_linebreak) == 0)) {
      result = true;
      break;
    }
  }
  fclose(file);
  if (line)
    free(line);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ace_tu_fs_does_file_contain_line_at>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    %rsi,%rdi
  lea    D8(%rip),%rsi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  mov    %rdx,0x8(%rsp)
  lea    0x20(%rsp),%r14
  lea    0x18(%rsp),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  call   <fopen@plt>
  mov    %rax,%r12
  nopl   0x0(%rax)
L23:
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <getline@plt>
  cmp    $0xffffffff,%eax
  je     L48
  mov    0x18(%rsp),%r15
  add    $0x1,%ebx
  mov    %r15,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r15,%rax,1)
  cmp    %ebp,%ebx
  je     L15
  test   %ebp,%ebp
  jg     L23
L15:
  mov    0x8(%rsp),%rsi
  mov    0x18(%rsp),%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L23
  mov    $0x1,%r13d
  jmp    L16
  nop
L48:
  xor    %r13d,%r13d
L16:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x18(%rsp),%rdi
  test   %rdi,%rdi
  je     L42
  call   <free@plt>
L42:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x38,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D8
type: string

tool: D8: r

assistant: ```c
bool ace_tu_fs_does_file_contain_line_at(int expected_line_number, char *filepath, char *str_without_linebreak) {
  FILE *file;
  char *line = NULL;
  int line_number = 0;
  ssize_t line_length;
  size_t buffer_length = 0;
  bool does_line_number_match;
  bool result = false;
  file = fopen(filepath, "r");
  while ((line_length = getline(&line, &buffer_length, file)) != -1) {
    line_number++;
    line[strlen(line) - 1] = '\0';
    does_line_number_match = (line_number == expected_line_number) || (expected_line_number <= 0);
    if (does_line_number_match && (strcmp(line, str_without_linebreak) == 0)) {
      result = true;
      break;
    }
  }
  fclose(file);
  if (line)
    free(line);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<eratosthenes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpl   $0x1,-0x24(%rbp)
  jg     L69
  mov    $0x0,%eax
  jmp    L63
L69:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x24(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x1,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x1,(%rax)
  movl   $0x2,-0x14(%rbp)
  jmp    L1
L68:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L36
L59:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L40
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x1,(%rax)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
L40:
  mov    -0x14(%rbp),%eax
  add    %eax,-0x10(%rbp)
L36:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L59
L39:
  addl   $0x1,-0x14(%rbp)
L1:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L68
  mov    -0x8(%rbp),%rax
L63:
  leave
  ret
```

assistant: ```c
char *eratosthenes(int n, int *c) {
  char *sieve;
  int i, j, m;
  if (n < 2)
    return NULL;
  *c = n - 1;
  m = (int)sqrt((double)n);
  sieve = calloc(n + 1, sizeof(char));
  sieve[0] = 1;
  sieve[1] = 1;
  for (i = 2; i <= m; i++)
    if (!sieve[i])
      for (j = i * i; j <= n; j += i)
        if (!sieve[j]) {
          sieve[j] = 1;
          --(*c);
        }
  return sieve;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<eratosthenes>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%edi
  jle    L34
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%rbp
  lea    -0x1(%rdi),%eax
  mov    %eax,(%rsi)
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
  call   <sqrt@plt>
  mov    %rax,%r12
  lea    0x1(%rbx),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  movb   $0x1,(%rax)
  movb   $0x1,0x1(%rax)
  cmp    $0x1,%r12d
  jle    L2
  lea    -0x2(%r12),%esi
  add    $0x3,%rsi
  mov    $0x2,%ecx
  jmp    L51
L37:
  add    %rcx,%rdx
  cmp    %edx,%ebx
  jl     L17
L35:
  cmpb   $0x0,(%rax,%rdx,1)
  jne    L37
  movb   $0x1,(%rax,%rdx,1)
  subl   $0x1,0x0(%rbp)
  jmp    L37
L17:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  je     L2
L51:
  cmpb   $0x0,(%rax,%rcx,1)
  jne    L17
  mov    %ecx,%edx
  imul   %ecx,%edx
  cmp    %edx,%ebx
  jl     L17
  movslq %edx,%rdx
  jmp    L35
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L34:
  ret
```

assistant: ```c
char *eratosthenes(int n, int *c) {
  char *sieve;
  int i, j, m;
  if (n < 2)
    return NULL;
  *c = n - 1;
  m = (int)sqrt((double)n);
  sieve = calloc(n + 1, sizeof(char));
  sieve[0] = 1;
  sieve[1] = 1;
  for (i = 2; i <= m; i++)
    if (!sieve[i])
      for (j = i * i; j <= n; j += i)
        if (!sieve[j]) {
          sieve[j] = 1;
          --(*c);
        }
  return sieve;
}
```