ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x50(%rbp)
  movl   $0x1,-0x4c(%rbp)
  movl   $0x0,-0x48(%rbp)
  movl   $0x1,-0x44(%rbp)
  movl   $0x0,-0x40(%rbp)
  movl   $0x1,-0x3c(%rbp)
  movl   $0x0,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  movl   $0x0,-0x30(%rbp)
  lea    -0x50(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  lea    -0x44(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x3,-0x64(%rbp)
  movl   $0x3,-0x2c(%rbp)
  movl   $0x3,-0x28(%rbp)
  movl   $0x3,-0x24(%rbp)
  movq   $0x0,-0x60(%rbp)
  movl   $0x0,-0x68(%rbp)
  lea    -0x68(%rbp),%rdi
  lea    -0x60(%rbp),%rcx
  lea    -0x2c(%rbp),%rdx
  mov    -0x64(%rbp),%esi
  lea    -0x20(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <flipAndInvertImage@plt>
  mov    %rax,-0x58(%rbp)
  cmpq   $0x0,-0x58(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int a1[] = {1, 1, 0};
  int a2[] = {1, 0, 1};
  int a3[] = {0, 0, 0};
  int *A[] = {a1, a2, a3};
  int ARowSize = 3;
  int AColSizes[] = {3, 3, 3};
  int *columnSizes = NULL;
  int returnSize = 0;
  int **retval = flipAndInvertImage((int **)A, ARowSize, AColSizes, &columnSizes, &returnSize);
  assert(retval != NULL);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  movl   $0x0,0x18(%rsp)
  movl   $0x1,0x1c(%rsp)
  movl   $0x0,0x20(%rsp)
  movl   $0x1,0x24(%rsp)
  movl   $0x0,0x28(%rsp)
  movl   $0x0,0x2c(%rsp)
  movl   $0x0,0x30(%rsp)
  lea    0x10(%rsp),%rax
  mov    %rax,0x40(%rsp)
  lea    0x1c(%rsp),%rax
  mov    %rax,0x48(%rsp)
  lea    0x28(%rsp),%rax
  mov    %rax,0x50(%rsp)
  movl   $0x3,0x34(%rsp)
  movl   $0x3,0x38(%rsp)
  movl   $0x3,0x3c(%rsp)
  movq   $0x0,0x8(%rsp)
  movl   $0x0,0x4(%rsp)
  lea    0x8(%rsp),%rcx
  lea    0x34(%rsp),%rdx
  lea    0x40(%rsp),%rdi
  lea    0x4(%rsp),%r8
  mov    $0x3,%esi
  call   <flipAndInvertImage@plt>
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x68,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int a1[] = {1, 1, 0};
  int a2[] = {1, 0, 1};
  int a3[] = {0, 0, 0};
  int *A[] = {a1, a2, a3};
  int ARowSize = 3;
  int AColSizes[] = {3, 3, 3};
  int *columnSizes = NULL;
  int returnSize = 0;
  int **retval = flipAndInvertImage((int **)A, ARowSize, AColSizes, &columnSizes, &returnSize);
  assert(retval != NULL);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x68,%rsp
  mov    $0x3,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x100000001,%rax
  lea    0x40(%rsp),%rdi
  lea    0x8(%rsp),%rcx
  movl   $0x0,0x18(%rsp)
  lea    0x34(%rsp),%rdx
  lea    0x4(%rsp),%r8
  movq   $0x1,0x1c(%rsp)
  movl   $0x1,0x24(%rsp)
  movq   $0x0,0x28(%rsp)
  movl   $0x0,0x30(%rsp)
  movl   $0x3,0x3c(%rsp)
  mov    %rax,0x10(%rsp)
  lea    0x10(%rsp),%rax
  mov    %rax,0x40(%rsp)
  lea    0x1c(%rsp),%rax
  mov    %rax,0x48(%rsp)
  lea    0x28(%rsp),%rax
  mov    %rax,0x50(%rsp)
  movabs $0x300000003,%rax
  mov    %rax,0x34(%rsp)
  movq   $0x0,0x8(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <flipAndInvertImage@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x68,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int a1[] = {1, 1, 0};
  int a2[] = {1, 0, 1};
  int a3[] = {0, 0, 0};
  int *A[] = {a1, a2, a3};
  int ARowSize = 3;
  int AColSizes[] = {3, 3, 3};
  int *columnSizes = NULL;
  int returnSize = 0;
  int **retval = flipAndInvertImage((int **)A, ARowSize, AColSizes, &columnSizes, &returnSize);
  assert(retval != NULL);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  mov    $0x3,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x50(%rsp),%rdi
  lea    0x8(%rsp),%rcx
  mov    D12(%rip),%rax
  lea    0x40(%rsp),%rdx
  lea    0x4(%rsp),%r8
  movl   $0x0,0x18(%rsp)
  mov    %rax,0x10(%rsp)
  mov    D0(%rip),%rax
  movl   $0x1,0x28(%rsp)
  mov    %rax,0x20(%rsp)
  lea    0x10(%rsp),%rax
  movq   %rax,%xmm0
  lea    0x20(%rsp),%rax
  movq   $0x0,0x30(%rsp)
  movq   %rax,%xmm1
  lea    0x30(%rsp),%rax
  movl   $0x0,0x38(%rsp)
  mov    %rax,0x60(%rsp)
  mov    D10(%rip),%rax
  punpcklqdq %xmm1,%xmm0
  movl   $0x3,0x48(%rsp)
  mov    %rax,0x40(%rsp)
  movq   $0x0,0x8(%rsp)
  movl   $0x0,0x4(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <flipAndInvertImage@plt>
  xor    %edi,%edi
  test   %rax,%rax
  setne  %dil
  call   <assert@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x78,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D12
type_guessed: byte[8]
name: retrieve_data
arguments: tag: D0
type_guessed: byte[8]
name: retrieve_data
arguments: tag: D10
type_guessed: byte[8]

tool: {"D12": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D0": [0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D10": [0x03,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char **argv) {
  int a1[] = {1, 1, 0};
  int a2[] = {1, 0, 1};
  int a3[] = {0, 0, 0};
  int *A[] = {a1, a2, a3};
  int ARowSize = 3;
  int AColSizes[] = {3, 3, 3};
  int *columnSizes = NULL;
  int returnSize = 0;
  int **retval = flipAndInvertImage((int **)A, ARowSize, AColSizes, &columnSizes, &returnSize);
  assert(retval != NULL);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<adb_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <halt_printf@plt>
  mov    $0x0,%eax
  call   <show_adb_log@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
data_type: string

tool: D4: Adb Error


assistant: ```c
void adb_error(void) {
  halt_printf("Adb Error\n");
  show_adb_log();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<adb_error>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <halt_printf@plt>
  mov    $0x0,%eax
  call   <show_adb_log@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
datatype: string

tool: D8: Adb Error


assistant: ```c
void adb_error(void) {
  halt_printf("Adb Error\n");
  show_adb_log();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<adb_error>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <halt_printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <show_adb_log@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type_guessed: string

tool: D4: Adb Error


assistant: ```c
void adb_error(void) {
  halt_printf("Adb Error\n");
  show_adb_log();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<adb_error>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <halt_printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <show_adb_log@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
data_type: string

tool: D6: Adb Error


assistant: ```c
void adb_error(void) {
  halt_printf("Adb Error\n");
  show_adb_log();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_quoted_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L69
L56:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L73
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x74,%eax
  je     L114
  cmp    $0x74,%eax
  jg     L96
  cmp    $0x62,%eax
  je     L68
  cmp    $0x6e,%eax
  jne    L96
  mov    -0x10(%rbp),%rax
  movb   $0xa,(%rax)
  jmp    L95
L114:
  mov    -0x10(%rbp),%rax
  movb   $0x9,(%rax)
  jmp    L95
L68:
  mov    -0x10(%rbp),%rax
  movb   $0x8,(%rax)
  jmp    L95
L96:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  jmp    L13
L95:
  jmp    L13
L73:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
L13:
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L69:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L111
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L56
L111:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void parse_quoted_string(char *str) {
  char *to, *from;
  for (to = from = str; *from && *from != '"'; to++, from++) {
    if (*from == '\\') {
      from++;
      switch (*from) {
      case 'n':
        *to = '\n';
        break;
      case 't':
        *to = '\t';
        break;
      case 'b':
        *to = '\b';
        break;
      default:
        *to = *from;
      }
      continue;
    }
    *to = *from;
  }
  *to = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<parse_quoted_string>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L12
  cmp    $0x22,%al
  je     L12
  mov    %rdi,%rcx
  mov    $0x8,%esi
  jmp    L65
L9:
  cmp    $0x62,%al
  cmove  %esi,%eax
L5:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  lea    0x1(%rdx),%rcx
  movzbl 0x1(%rdx),%eax
  test   %al,%al
  je     L12
  cmp    $0x22,%al
  je     L12
L65:
  mov    %rcx,%rdx
  cmp    $0x5c,%al
  jne    L5
  lea    0x1(%rcx),%rdx
  movzbl 0x1(%rcx),%eax
  cmp    $0x6e,%al
  je     L22
  cmp    $0x74,%al
  jne    L9
  mov    $0x9,%eax
  jmp    L5
L22:
  mov    $0xa,%eax
  jmp    L5
L12:
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void parse_quoted_string(char *str) {
  char *to, *from;
  for (to = from = str; *from && *from != '"'; to++, from++) {
    if (*from == '\\') {
      from++;
      switch (*from) {
      case 'n':
        *to = '\n';
        break;
      case 't':
        *to = '\t';
        break;
      case 'b':
        *to = '\b';
        break;
      default:
        *to = *from;
      }
      continue;
    }
    *to = *from;
  }
  *to = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<parse_quoted_string>:
  endbr64
  movzbl (%rdi),%eax
  mov    %rdi,%rcx
  cmp    $0x22,%al
  je     L25
  test   %al,%al
  je     L25
  mov    $0x8,%esi
  jmp    L30
  nopl   0x0(%rax)
L28:
  mov    %rdi,%rdx
L72:
  mov    %al,(%rcx)
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rcx
  lea    0x1(%rdx),%rdi
  test   %al,%al
  je     L25
  cmp    $0x22,%al
  je     L25
L30:
  lea    0x1(%rdi),%rdx
  cmp    $0x5c,%al
  jne    L28
  movzbl 0x1(%rdi),%eax
  cmp    $0x6e,%al
  je     L103
  cmp    $0x74,%al
  je     L107
  cmp    $0x62,%al
  cmove  %esi,%eax
  jmp    L72
  nopl   0x0(%rax)
L25:
  movb   $0x0,(%rcx)
  ret
  nopl   0x0(%rax)
L107:
  mov    $0x9,%eax
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L103:
  mov    $0xa,%eax
  jmp    L72
```

assistant: ```c
void parse_quoted_string(char *str) {
  char *to, *from;
  for (to = from = str; *from && *from != '"'; to++, from++) {
    if (*from == '\\') {
      from++;
      switch (*from) {
      case 'n':
        *to = '\n';
        break;
      case 't':
        *to = '\t';
        break;
      case 'b':
        *to = '\b';
        break;
      default:
        *to = *from;
      }
      continue;
    }
    *to = *from;
  }
  *to = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<parse_quoted_string>:
  endbr64
  movzbl (%rdi),%eax
  mov    %rdi,%rcx
  cmp    $0x22,%al
  je     L24
  test   %al,%al
  je     L24
  mov    $0x8,%esi
  jmp    L85
  nopl   0x0(%rax)
L37:
  mov    %rdi,%rdx
L2:
  mov    %al,(%rcx)
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rcx
  lea    0x1(%rdx),%rdi
  test   %al,%al
  je     L24
  cmp    $0x22,%al
  je     L24
L85:
  lea    0x1(%rdi),%rdx
  cmp    $0x5c,%al
  jne    L37
  movzbl 0x1(%rdi),%eax
  cmp    $0x6e,%al
  je     L31
  cmp    $0x74,%al
  je     L82
  cmp    $0x62,%al
  cmove  %esi,%eax
  jmp    L2
  nopl   0x0(%rax)
L24:
  movb   $0x0,(%rcx)
  ret
  nopl   0x0(%rax)
L82:
  mov    $0x9,%eax
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L31:
  mov    $0xa,%eax
  jmp    L2
```

assistant: ```c
void parse_quoted_string(char *str) {
  char *to, *from;
  for (to = from = str; *from && *from != '"'; to++, from++) {
    if (*from == '\\') {
      from++;
      switch (*from) {
      case 'n':
        *to = '\n';
        break;
      case 't':
        *to = '\t';
        break;
      case 'b':
        *to = '\b';
        break;
      default:
        *to = *from;
      }
      continue;
    }
    *to = *from;
  }
  *to = '\0';
}
```
user: What is the c source code of the assembly code below?

```asm
<to2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L19
  mov    $0x0,%eax
  jmp    L113
L19:
  movl   $0x1,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L79
L58:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L79:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L58
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L132
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L113
L132:
  movl   $0x0,-0x18(%rbp)
  jmp    L111
L5:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x18(%rbp)
L111:
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L5
  mov    -0x8(%rbp),%rax
  movb   $0x31,(%rax)
  movl   $0x1,-0x18(%rbp)
  jmp    L54
L129:
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x24(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2sdl -0x10(%rbp),%xmm2
  movsd  D18(%rip),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L89
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L63
L76:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L63:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L76
  mov    -0xc(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x31,(%rax)
  addl   $0x1,-0x18(%rbp)
L54:
  cmpl   $0x1,-0x24(%rbp)
  jg     L129
  jmp    L124
L89:
  nop
L124:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
L113:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
data_type: string
name: extract_symbolic_data
arguments: label: D18
data_type: f64

tool: D12: malloc failed


tool: D18: 0.5

assistant: ```c
int to2(int n) {
  int i, j;
  int d;
  int siz;
  if (n == 0)
    return 0;
  for (j = 1, d = 0; d <= n; j++)
    d = pow(2, j);
  siz = j - 1;
  char *tmp;
  if (NULL == (tmp = malloc(siz * sizeof(char)))) {
    printf("malloc failed\n");
    return -1;
  }
  for (i = 0; i < siz; i++)
    tmp[i] = '0';
  tmp[0] = '1';
  for (i = 1; n > 1; i++) {
    n -= d * 0.5;
    if (n == 0)
      break;
    for (j = 0, d = 0; d <= n; j++)
      d = pow(2, j);
    tmp[siz - j + 1] = '1';
  }
  return atoi(tmp);
  free(tmp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<to2>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L6
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r14d
  js     L76
  mov    $0x1,%ebp
L94:
  mov    %ebp,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%ebx
  mov    %ebp,%r12d
  add    $0x1,%ebp
  cmp    %eax,%r14d
  jge    L94
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L102
  test   %r12d,%r12d
  jle    L60
  mov    %r12d,%edx
  add    %rax,%rdx
L127:
  movb   $0x30,(%rax)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L127
L60:
  movb   $0x31,0x0(%r13)
  cmp    $0x1,%r14d
  jle    L49
  pxor   %xmm0,%xmm0
  cvtsi2sd %r14d,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  mulsd  D18(%rip),%xmm1
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%ebp
  test   %ebp,%ebp
  jne    L159
  jmp    L49
L102:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L27
L14:
  mov    $0x0,%eax
  mov    $0x0,%ebx
L106:
  mov    %r12d,%edx
  sub    %ebx,%edx
  movslq %edx,%rdx
  movb   $0x31,0x1(%r13,%rdx,1)
  cmp    $0x1,%ebp
  jle    L49
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  D18(%rip),%xmm1
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%ebp
  test   %ebp,%ebp
  je     L49
L159:
  test   %ebp,%ebp
  js     L14
  mov    $0x0,%ebx
L42:
  mov    %ebx,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  add    $0x1,%ebx
  cmp    %ebp,%eax
  jle    L42
  jmp    L106
L76:
  mov    $0x0,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L102
  movb   $0x31,(%rax)
L49:
  mov    %r13,%rdi
  call   <atoi@plt>
L27:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  ret
```

tool_calls: name: read_data
arguments: tag: D18
type_infered: f64
name: read_data
arguments: tag: D7
type_infered: string

tool: D18: 0.5

tool: D7: malloc failed


assistant: ```c
int to2(int n) {
  int i, j;
  int d;
  int siz;
  if (n == 0)
    return 0;
  for (j = 1, d = 0; d <= n; j++)
    d = pow(2, j);
  siz = j - 1;
  char *tmp;
  if (NULL == (tmp = malloc(siz * sizeof(char)))) {
    printf("malloc failed\n");
    return -1;
  }
  for (i = 0; i < siz; i++)
    tmp[i] = '0';
  tmp[0] = '1';
  for (i = 1; n > 1; i++) {
    n -= d * 0.5;
    if (n == 0)
      break;
    for (j = 0, d = 0; d <= n; j++)
      d = pow(2, j);
    tmp[siz - j + 1] = '1';
  }
  return atoi(tmp);
  free(tmp);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<to2>:
  endbr64
  test   %edi,%edi
  je     L118
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  js     L66
  mov    %edi,%r14d
  mov    $0x1,%ebp
  nopw   0x0(%rax,%rax,1)
L91:
  mov    %ebp,%esi
  mov    $0x2,%edi
  mov    %ebp,%r12d
  call   <pow@plt>
  lea    0x1(%rbp),%ebp
  mov    %eax,%ebx
  cmp    %eax,%r14d
  jge    L91
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L60
  movslq %r12d,%rdx
  mov    $0x30,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  movb   $0x31,0x0(%r13)
  cmp    $0x1,%r14d
  jle    L65
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r14d,%xmm0
  cvtsi2sd %ebx,%xmm1
  nopl   (%rax)
L92:
  mulsd  D11(%rip),%xmm1
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%ebp
  test   %ebp,%ebp
  je     L65
  test   %ebp,%ebp
  js     L38
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L55:
  mov    %ebx,%esi
  mov    $0x2,%edi
  add    $0x1,%ebx
  call   <pow@plt>
  cmp    %ebp,%eax
  jle    L55
  mov    %r12d,%edx
  sub    %ebx,%edx
  add    $0x1,%edx
  movslq %edx,%rdx
  movb   $0x31,0x0(%r13,%rdx,1)
  cmp    $0x1,%ebp
  jle    L65
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebp,%xmm0
  cvtsi2sd %eax,%xmm1
  jmp    L92
  nopl   0x0(%rax,%rax,1)
L38:
  lea    0x1(%r12),%eax
  cltq
  movb   $0x31,0x0(%r13,%rax,1)
L65:
  pop    %rbx
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10d0 <atoi@plt>
L118:
  xor    %eax,%eax
  ret
L66:
  xor    %edi,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L60
  movb   $0x31,(%rax)
  jmp    L65
L60:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D11
type: f64
name: unpack_data
arguments: data_label: D7
type: string

tool: D11: 0.5

tool: D7: malloc failed


assistant: ```c
int to2(int n) {
  int i, j;
  int d;
  int siz;
  if (n == 0)
    return 0;
  for (j = 1, d = 0; d <= n; j++)
    d = pow(2, j);
  siz = j - 1;
  char *tmp;
  if (NULL == (tmp = malloc(siz * sizeof(char)))) {
    printf("malloc failed\n");
    return -1;
  }
  for (i = 0; i < siz; i++)
    tmp[i] = '0';
  tmp[0] = '1';
  for (i = 1; n > 1; i++) {
    n -= d * 0.5;
    if (n == 0)
      break;
    for (j = 0, d = 0; d <= n; j++)
      d = pow(2, j);
    tmp[siz - j + 1] = '1';
  }
  return atoi(tmp);
  free(tmp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<to2>:
  endbr64
  test   %edi,%edi
  je     L28
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  js     L83
  mov    %edi,%r14d
  mov    $0x1,%ebp
  nopw   0x0(%rax,%rax,1)
L71:
  mov    %ebp,%esi
  mov    $0x2,%edi
  mov    %ebp,%r12d
  call   <pow@plt>
  lea    0x1(%rbp),%ebp
  mov    %eax,%ebx
  cmp    %eax,%r14d
  jge    L71
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L24
  movslq %r12d,%rdx
  mov    $0x30,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  movb   $0x31,0x0(%r13)
  cmp    $0x1,%r14d
  jle    L41
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r14d,%xmm0
  cvtsi2sd %ebx,%xmm1
  nopl   (%rax)
L46:
  mulsd  D7(%rip),%xmm1
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%ebp
  test   %ebp,%ebp
  je     L41
  test   %ebp,%ebp
  js     L101
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L100:
  mov    %ebx,%esi
  mov    $0x2,%edi
  add    $0x1,%ebx
  call   <pow@plt>
  cmp    %ebp,%eax
  jle    L100
  mov    %r12d,%edx
  sub    %ebx,%edx
  add    $0x1,%edx
  movslq %edx,%rdx
  movb   $0x31,0x0(%r13,%rdx,1)
  cmp    $0x1,%ebp
  jle    L41
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebp,%xmm0
  cvtsi2sd %eax,%xmm1
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L101:
  lea    0x1(%r12),%eax
  cltq
  movb   $0x31,0x0(%r13,%rax,1)
L41:
  pop    %rbx
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10d0 <atoi@plt>
L28:
  xor    %eax,%eax
  ret
L83:
  xor    %edi,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L24
  movb   $0x31,(%rax)
  jmp    L41
L24:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: f64
name: read_data
arguments: tag: D16
type_infered: string

tool: D7: 0.5

tool: D16: malloc failed


assistant: ```c
int to2(int n) {
  int i, j;
  int d;
  int siz;
  if (n == 0)
    return 0;
  for (j = 1, d = 0; d <= n; j++)
    d = pow(2, j);
  siz = j - 1;
  char *tmp;
  if (NULL == (tmp = malloc(siz * sizeof(char)))) {
    printf("malloc failed\n");
    return -1;
  }
  for (i = 0; i < siz; i++)
    tmp[i] = '0';
  tmp[0] = '1';
  for (i = 1; n > 1; i++) {
    n -= d * 0.5;
    if (n == 0)
      break;
    for (j = 0, d = 0; d <= n; j++)
      d = pow(2, j);
    tmp[siz - j + 1] = '1';
  }
  return atoi(tmp);
  free(tmp);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<unlock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%eax
  call   <gettid@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L0
  mov    $0x0,%eax
  jmp    L10
L0:
  mov    -0x18(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <__sync_fetch_and_add@plt>
  mov    -0x18(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <futex_wake@plt>
  mov    $0x1,%eax
L10:
  leave
  ret
```

assistant: ```c
int unlock(unsigned int *lock) {
  unsigned int self = gettid();
  if (*lock != self)
    return 0;
  *lock = 0;
  __sync_fetch_and_add(lock, 0);
  futex_wake((int *)lock, 1);
  return 1;
}
```