ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x7b,-0x24(%rbp)
  movsd  D1(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movabs $0x6375654420656854,%rax
  mov    %rax,-0x13(%rbp)
  movw   $0x2165,-0xb(%rbp)
  movb   $0x0,-0x9(%rbp)
  movb   $0x74,-0x25(%rbp)
  movsbl -0x25(%rbp),%ecx
  lea    -0x13(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x24(%rbp),%eax
  movq   %rsi,%xmm0
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <minprintf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_infered: byte[8]
name: disassemble_data
arguments: tag: D17
type_infered: string

tool: {"D1": [0xee,0xeb,0xc0,0x39,0x23,0x8a,0x10,0xc0]}

tool: D17: int: %d, double: %f, string: %s, char: %c


assistant: ```c
main() {
  int i = 123;
  double d = -4.1349;
  char s[] = "The Deuce!";
  char c = 't';
  minprintf("int: %d, double: %f, string: %s, char: %c\n", i, d, s, c);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x6375654420656854,%rax
  mov    %rax,0xd(%rsp)
  movw   $0x2165,0x15(%rsp)
  movb   $0x0,0x17(%rsp)
  lea    0xd(%rsp),%rdx
  mov    $0x74,%ecx
  movsd  D19(%rip),%xmm0
  mov    $0x7b,%esi
  lea    D18(%rip),%rdi
  call   <minprintf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D19
type_guessed: byte[8]
name: read_data
arguments: label: D18
type_guessed: string

tool: {"D19": [0xee,0xeb,0xc0,0x39,0x23,0x8a,0x10,0xc0]}

tool: D18: int: %d, double: %f, string: %s, char: %c


assistant: ```c
main() {
  int i = 123;
  double d = -4.1349;
  char s[] = "The Deuce!";
  char c = 't';
  minprintf("int: %d, double: %f, string: %s, char: %c\n", i, d, s, c);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  movsd  D7(%rip),%xmm0
  mov    $0x74,%ecx
  mov    $0x7b,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x6375654420656854,%rax
  lea    0xd(%rsp),%rdx
  lea    D15(%rip),%rdi
  movb   $0x0,0x17(%rsp)
  mov    %rax,0xd(%rsp)
  mov    $0x2165,%eax
  mov    %ax,0x15(%rsp)
  call   <minprintf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: byte[8]
name: unpack_data
arguments: tag: D15
type_guessed: string

tool: {"D7": [0xee,0xeb,0xc0,0x39,0x23,0x8a,0x10,0xc0]}

tool: D15: int: %d, double: %f, string: %s, char: %c


assistant: ```c
main() {
  int i = 123;
  double d = -4.1349;
  char s[] = "The Deuce!";
  char c = 't';
  minprintf("int: %d, double: %f, string: %s, char: %c\n", i, d, s, c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  movsd  D8(%rip),%xmm0
  mov    $0x74,%ecx
  mov    $0x7b,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x6375654420656854,%rax
  lea    0xd(%rsp),%rdx
  lea    D14(%rip),%rdi
  movb   $0x0,0x17(%rsp)
  mov    %rax,0xd(%rsp)
  mov    $0x2165,%eax
  mov    %ax,0x15(%rsp)
  call   <minprintf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_guessed: byte[8]
name: extract_symbolic_data
arguments: tag: D14
type_guessed: string

tool: {"D8": [0xee,0xeb,0xc0,0x39,0x23,0x8a,0x10,0xc0]}

tool: D14: int: %d, double: %f, string: %s, char: %c


assistant: ```c
main() {
  int i = 123;
  double d = -4.1349;
  char s[] = "The Deuce!";
  char c = 't';
  minprintf("int: %d, double: %f, string: %s, char: %c\n", i, d, s, c);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dwt_interleave_rows>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jg     L47
  mov    -0x14(%rbp),%eax
  jmp    L5
L47:
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L11
  mov    -0x8(%rbp),%eax
  imul   -0xc(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L45
L11:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %eax,%ecx
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  imul   -0x1c(%rbp),%eax
  add    %ecx,%eax
  jmp    L5
L45:
  mov    -0x8(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0x4(%rbp),%eax
  cltd
  idiv   %edi
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  lea    (%rdx,%rax,1),%esi
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x4(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  imul   -0x1c(%rbp),%eax
  add    %esi,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int dwt_interleave_rows(const int rowid, const int height, const int stride) {
  if (height <= stride)
    return rowid;
  const int per_pass = ((height + stride - 1) / stride);
  const int long_passes = height % stride;
  if (long_passes == 0 || rowid < long_passes * per_pass)
    return (rowid / per_pass) + stride * (rowid % per_pass);
  const int rowid2 = rowid - long_passes * per_pass;
  return long_passes + (rowid2 / (per_pass - 1)) + stride * (rowid2 % (per_pass - 1));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dwt_interleave_rows>:
  endbr64
  mov    %edi,%ecx
  mov    %edx,%edi
  mov    %ecx,%eax
  cmp    %edx,%esi
  jle    L11
  lea    -0x1(%rsi,%rdx,1),%eax
  cltd
  idiv   %edi
  mov    %eax,%r8d
  mov    %eax,%r10d
  mov    %esi,%eax
  cltd
  idiv   %edi
  mov    %edx,%r9d
  test   %edx,%edx
  je     L22
  mov    %r8d,%edx
  imul   %r9d,%edx
  cmp    %ecx,%edx
  jg     L22
  mov    %ecx,%eax
  sub    %edx,%eax
  sub    $0x1,%r8d
  cltd
  idiv   %r8d
  add    %r9d,%eax
  imul   %edi,%edx
  add    %edx,%eax
L11:
  ret
L22:
  mov    %ecx,%eax
  cltd
  idiv   %r10d
  imul   %edi,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int dwt_interleave_rows(const int rowid, const int height, const int stride) {
  if (height <= stride)
    return rowid;
  const int per_pass = ((height + stride - 1) / stride);
  const int long_passes = height % stride;
  if (long_passes == 0 || rowid < long_passes * per_pass)
    return (rowid / per_pass) + stride * (rowid % per_pass);
  const int rowid2 = rowid - long_passes * per_pass;
  return long_passes + (rowid2 / (per_pass - 1)) + stride * (rowid2 % (per_pass - 1));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dwt_interleave_rows>:
  endbr64
  mov    %edx,%ecx
  mov    %edi,%r8d
  cmp    %edx,%esi
  jle    L16
  lea    -0x1(%rsi,%rdx,1),%eax
  cltd
  idiv   %ecx
  mov    %eax,%r9d
  mov    %esi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%r10d
  test   %edx,%edx
  je     L1
  mov    %r9d,%edx
  imul   %r10d,%edx
  cmp    %edi,%edx
  jg     L1
  mov    %edi,%eax
  sub    $0x1,%r9d
  sub    %edx,%eax
  cltd
  idiv   %r9d
  imul   %ecx,%edx
  add    %r10d,%eax
  lea    (%rax,%rdx,1),%r8d
L16:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %edi,%eax
  cltd
  idiv   %r9d
  imul   %ecx,%edx
  lea    (%rdx,%rax,1),%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int dwt_interleave_rows(const int rowid, const int height, const int stride) {
  if (height <= stride)
    return rowid;
  const int per_pass = ((height + stride - 1) / stride);
  const int long_passes = height % stride;
  if (long_passes == 0 || rowid < long_passes * per_pass)
    return (rowid / per_pass) + stride * (rowid % per_pass);
  const int rowid2 = rowid - long_passes * per_pass;
  return long_passes + (rowid2 / (per_pass - 1)) + stride * (rowid2 % (per_pass - 1));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dwt_interleave_rows>:
  endbr64
  mov    %edx,%ecx
  mov    %edi,%r8d
  cmp    %edx,%esi
  jle    L6
  lea    -0x1(%rsi,%rdx,1),%eax
  cltd
  idiv   %ecx
  mov    %eax,%r9d
  mov    %esi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%r10d
  test   %edx,%edx
  je     L3
  mov    %r9d,%edx
  imul   %r10d,%edx
  cmp    %edi,%edx
  jg     L3
  mov    %edi,%eax
  sub    $0x1,%r9d
  sub    %edx,%eax
  cltd
  idiv   %r9d
  imul   %ecx,%edx
  add    %r10d,%eax
  lea    (%rax,%rdx,1),%r8d
L6:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %edi,%eax
  cltd
  idiv   %r9d
  imul   %ecx,%edx
  lea    (%rdx,%rax,1),%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int dwt_interleave_rows(const int rowid, const int height, const int stride) {
  if (height <= stride)
    return rowid;
  const int per_pass = ((height + stride - 1) / stride);
  const int long_passes = height % stride;
  if (long_passes == 0 || rowid < long_passes * per_pass)
    return (rowid / per_pass) + stride * (rowid % per_pass);
  const int rowid2 = rowid - long_passes * per_pass;
  return long_passes + (rowid2 / (per_pass - 1)) + stride * (rowid2 % (per_pass - 1));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xrop3_59_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  not    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  or     -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_59_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~D;
  stk2 = T | stk3;
  stk1 = S & stk2;
  stk1 = ~stk1;
  return stk1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_59_0_0>:
  endbr64
  movzbl %dil,%edi
  mov    %esi,%eax
  or     $0xffffff00,%eax
  mov    %edi,%edx
  shl    $0x8,%edx
  or     %edi,%edx
  and    %edx,%eax
  not    %eax
  ret
```

assistant: ```c
unsigned xrop3_59_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~D;
  stk2 = T | stk3;
  stk1 = S & stk2;
  stk1 = ~stk1;
  return stk1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xrop3_59_0_0>:
  endbr64
  movzbl %dil,%edi
  mov    %esi,%eax
  mov    %edi,%edx
  or     $0xffffff00,%eax
  shl    $0x8,%edx
  or     %edi,%edx
  and    %edx,%eax
  not    %eax
  ret
```

assistant: ```c
unsigned xrop3_59_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~D;
  stk2 = T | stk3;
  stk1 = S & stk2;
  stk1 = ~stk1;
  return stk1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xrop3_59_0_0>:
  endbr64
  movzbl %dil,%edi
  mov    %esi,%eax
  mov    %edi,%edx
  or     $0xffffff00,%eax
  shl    $0x8,%edx
  or     %edi,%edx
  and    %edx,%eax
  not    %eax
  ret
```

assistant: ```c
unsigned xrop3_59_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~D;
  stk2 = T | stk3;
  stk1 = S & stk2;
  stk1 = ~stk1;
  return stk1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<iRobot_Motors>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x8a,-0xa(%rbp)
  movzwl -0x14(%rbp),%eax
  mov    %al,-0x9(%rbp)
  lea    -0xa(%rbp),%rax
  mov    %rax,%rdi
  call   <SerialWriteString@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void iRobot_Motors(unsigned short motors) {
  char data[] = {138, motors};
  SerialWriteString(data);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<iRobot_Motors>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x8a,0x6(%rsp)
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  call   <SerialWriteString@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void iRobot_Motors(unsigned short motors) {
  char data[] = {138, motors};
  SerialWriteString(data);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<iRobot_Motors>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  movb   $0x8a,0x6(%rsp)
  call   <SerialWriteString@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void iRobot_Motors(unsigned short motors) {
  char data[] = {138, motors};
  SerialWriteString(data);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<iRobot_Motors>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  movb   $0x8a,0x6(%rsp)
  call   <SerialWriteString@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void iRobot_Motors(unsigned short motors) {
  char data[] = {138, motors};
  SerialWriteString(data);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<yyerror>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

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

tool: D0: %s


assistant: ```c
void yyerror(char *msg) {
  printf("%s\n", msg);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<yyerror>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D8: %s


assistant: ```c
void yyerror(char *msg) {
  printf("%s\n", msg);
  exit(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<yyerror>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: read_data
arguments: tag: D0
type_infered: string

tool: D0: %s


assistant: ```c
void yyerror(char *msg) {
  printf("%s\n", msg);
  exit(1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<yyerror>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_infered: string

tool: D2: %s


assistant: ```c
void yyerror(char *msg) {
  printf("%s\n", msg);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  je     L0
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L87
L0:
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L100
L87:
  movl   $0x2,-0x8(%rbp)
  cmpl   $0x1,-0x18(%rbp)
  jne    L52
  negl   -0x14(%rbp)
  movl   $0xfffffffe,-0x8(%rbp)
L52:
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x1,-0x18(%rbp)
  jne    L61
  jmp    L116
L69:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L135
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L100
L135:
  subl   $0x1,-0x8(%rbp)
L116:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L69
L61:
  cmpl   $0x1,-0x18(%rbp)
  je     L43
  jmp    L5
L122:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L110
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L100
L110:
  addl   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L122
L43:
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L100:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D22
type_infered: string
name: read_data
arguments: data_label: D11
type_infered: string

tool: D22: 
1

tool: D11: 
0

assistant: ```c
void is_prime(int x, int flag) {
  int y, c;
  if (x == 1 || x == -1) {
    printf("\n1");
    return 0;
  }
  y = 2;
  if (flag == 1) {
    x = -x;
    y = -2;
  }
  c = 0;
  if (flag == 1) {
    while (y > x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y--;
    }
  }
  if (flag != 1) {
    while (y < x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y++;
    }
  }
  printf("\n1");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_prime>:
  endbr64
  sub    $0x8,%rsp
  lea    0x1(%rdi),%eax
  test   $0xfffffffd,%eax
  je     L52
  cmp    $0x1,%esi
  je     L82
  cmp    $0x2,%edi
  jle    L48
  test   $0x1,%dil
  je     L69
  mov    $0x2,%ecx
L54:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  je     L48
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L54
L69:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L52:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L82:
  mov    %edi,%esi
  neg    %esi
  cmp    $0x2,%edi
  jle    L48
  test   $0x1,%sil
  je     L63
  mov    $0xfffffffe,%ecx
L105:
  sub    $0x1,%ecx
  cmp    %ecx,%esi
  jge    L48
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L105
L63:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L48:
  lea    D24(%rip),%rdi
  call   <printf@plt>
L15:
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_guessed: string
name: disassemble_data
arguments: tag: D24
type_guessed: string

tool: D4: 
0

tool: D24: 
1

assistant: ```c
void is_prime(int x, int flag) {
  int y, c;
  if (x == 1 || x == -1) {
    printf("\n1");
    return 0;
  }
  y = 2;
  if (flag == 1) {
    x = -x;
    y = -2;
  }
  c = 0;
  if (flag == 1) {
    while (y > x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y--;
    }
  }
  if (flag != 1) {
    while (y < x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y++;
    }
  }
  printf("\n1");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<is_prime>:
  endbr64
  lea    0x1(%rdi),%eax
  and    $0xfffffffd,%eax
  je     L4
  cmp    $0x1,%esi
  je     L5
  cmp    $0x2,%edi
  jle    L4
  test   $0x1,%dil
  je     L118
  mov    $0x2,%ecx
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L100:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L118
L18:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L100
L4:
  lea    D9(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L118:
  lea    D10(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L5:
  mov    %edi,%esi
  neg    %esi
  cmp    $0x2,%edi
  jle    L4
  test   $0x1,%sil
  je     L118
  mov    $0xfffffffe,%ecx
  jmp    L115
  xchg   %ax,%ax
L40:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L118
L115:
  sub    $0x1,%ecx
  cmp    %ecx,%esi
  jl     L40
  lea    D9(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_guessed: string
name: parse_data
arguments: data_tag: D10
type_guessed: string

tool: D9: 
1

tool: D10: 
0

assistant: ```c
void is_prime(int x, int flag) {
  int y, c;
  if (x == 1 || x == -1) {
    printf("\n1");
    return 0;
  }
  y = 2;
  if (flag == 1) {
    x = -x;
    y = -2;
  }
  c = 0;
  if (flag == 1) {
    while (y > x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y--;
    }
  }
  if (flag != 1) {
    while (y < x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y++;
    }
  }
  printf("\n1");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_prime>:
  endbr64
  lea    0x1(%rdi),%eax
  and    $0xfffffffd,%eax
  je     L96
  cmp    $0x1,%esi
  je     L71
  cmp    $0x2,%edi
  jle    L96
  test   $0x1,%dil
  je     L109
  mov    $0x2,%ecx
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L109
L33:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L12
L96:
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L109:
  lea    D14(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L71:
  mov    %edi,%esi
  neg    %esi
  cmp    $0x2,%edi
  jle    L96
  test   $0x1,%sil
  je     L109
  mov    $0xfffffffe,%ecx
  jmp    L119
  xchg   %ax,%ax
L3:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L109
L119:
  sub    $0x1,%ecx
  cmp    %ecx,%esi
  jl     L3
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_infered: string
name: retrieve_data
arguments: data_label: D14
type_infered: string

tool: D2: 
1

tool: D14: 
0

assistant: ```c
void is_prime(int x, int flag) {
  int y, c;
  if (x == 1 || x == -1) {
    printf("\n1");
    return 0;
  }
  y = 2;
  if (flag == 1) {
    x = -x;
    y = -2;
  }
  c = 0;
  if (flag == 1) {
    while (y > x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y--;
    }
  }
  if (flag != 1) {
    while (y < x) {
      c = x % y;
      if (c == 0) {
        printf("\n0");
        return 0;
      }
      y++;
    }
  }
  printf("\n1");
}
```