ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ini_acf_iireq_src_24m_5m>:
  endbr64
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movl   $0x2ab,(%rsp)
  movl   $0x77,0x4(%rsp)
  movl   $0xc6,0x8(%rsp)
  movl   $0x1e,0xc(%rsp)
  movl   $0x295,0x10(%rsp)
  movl   $0xa5,0x14(%rsp)
  movl   $0x2fa,0x18(%rsp)
  movl   $0x6d,0x1c(%rsp)
  movl   $0x27f,0x20(%rsp)
  movl   $0xd2,0x24(%rsp)
  movl   $0x297,0x28(%rsp)
  movl   $0x5e,0x2c(%rsp)
  movl   $0x272,0x30(%rsp)
  movl   $0xed,0x34(%rsp)
  movl   $0x27b,0x38(%rsp)
  movl   $0x4b,0x3c(%rsp)
  movl   $0x26a,0x40(%rsp)
  movl   $0xfb,0x44(%rsp)
  movl   $0x272,0x48(%rsp)
  movl   $0x4d,0x4c(%rsp)
  movl   $0x3f3fff,0x50(%rsp)
  movl   $0x3db7cf,0x54(%rsp)
  movl   $0x3c2f9e,0x58(%rsp)
  movl   $0x3aa76c,0x5c(%rsp)
  movl   $0x39173b,0x60(%rsp)
  movl   $0x377f08,0x64(%rsp)
  movl   $0x35ded5,0x68(%rsp)
  movl   $0x343ea1,0x6c(%rsp)
  movl   $0x328e6c,0x70(%rsp)
  movl   $0x30de36,0x74(%rsp)
  movl   $0x2f15ff,0x78(%rsp)
  movl   $0x2d45c6,0x7c(%rsp)
  movl   $0x2b658a,0x80(%rsp)
  movl   $0x29754d,0x84(%rsp)
  movl   $0x27650d,0x88(%rsp)
  movl   $0x253cca,0x8c(%rsp)
  movl   $0x22f483,0x90(%rsp)
  movl   $0x208438,0x94(%rsp)
  movl   $0x1de3e7,0x98(%rsp)
  movl   $0x1b0b90,0x9c(%rsp)
  movl   $0x17eb30,0xa0(%rsp)
  movl   $0x1472c7,0xa4(%rsp)
  movl   $0x107a51,0xa8(%rsp)
  movl   $0xbd9c8,0xac(%rsp)
  movl   $0x7c927,0xb0(%rsp)
  movl   $0x4a8c9,0xb4(%rsp)
  movl   $0x1205f,0xb8(%rsp)
  movl   $0x3cf7e4,0xbc(%rsp)
  movl   $0x37e752,0xc0(%rsp)
  movl   $0x31669a,0xc4(%rsp)
  movl   $0x31662c,0xc8(%rsp)
  movl   $0x31662c,0xcc(%rsp)
  movl   $0x62c,0xd0(%rsp)
  lea    0x50(%rsp),%rsi
  mov    %rsp,%rdi
  call   <program_acf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0xe8,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ini_acf_iireq_src_24m_5m(void) {
  int acf1[] = {0x2ab, 0x077, 0x0c6, 0x01e, 0x295, 0x0a5, 0x2fa, 0x06d, 0x27f, 0x0d2, 0x297, 0x05e, 0x272, 0x0ed, 0x27b, 0x04b, 0x26a, 0x0fb, 0x272, 0x04d};
  int acf2[] = {0x3f3fff, 0x3db7cf, 0x3c2f9e, 0x3aa76c, 0x39173b, 0x377f08, 0x35ded5, 0x343ea1, 0x328e6c, 0x30de36, 0x2f15ff, 0x2d45c6, 0x2b658a, 0x29754d, 0x27650d, 0x253cca, 0x22f483, 0x208438, 0x1de3e7, 0x1b0b90, 0x17eb30, 0x1472c7, 0x107a51, 0x0bd9c8, 0x07c927, 0x04a8c9, 0x01205f, 0x3cf7e4, 0x37e752, 0x31669a, 0x31662c, 0x31662c, 0x00062c};
  program_acf(acf1, acf2);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ini_acf_iireq_src_24m_5m>:
  endbr64
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  movabs $0x77000002ab,%rax
  lea    0x50(%rsp),%rsi
  mov    %rsp,%rdi
  movl   $0x62c,0xd0(%rsp)
  mov    %rax,(%rsp)
  movabs $0x1e000000c6,%rax
  mov    %rax,0x8(%rsp)
  movabs $0xa500000295,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x6d000002fa,%rax
  mov    %rax,0x18(%rsp)
  movabs $0xd20000027f,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x5e00000297,%rax
  mov    %rax,0x28(%rsp)
  movabs $0xed00000272,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x4b0000027b,%rax
  mov    %rax,0x38(%rsp)
  movabs $0xfb0000026a,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x4d00000272,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x3db7cf003f3fff,%rax
  mov    %rax,0x50(%rsp)
  movabs $0x3aa76c003c2f9e,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x377f080039173b,%rax
  mov    %rax,0x60(%rsp)
  movabs $0x343ea10035ded5,%rax
  mov    %rax,0x68(%rsp)
  movabs $0x30de3600328e6c,%rax
  mov    %rax,0x70(%rsp)
  movabs $0x2d45c6002f15ff,%rax
  mov    %rax,0x78(%rsp)
  movabs $0x29754d002b658a,%rax
  mov    %rax,0x80(%rsp)
  movabs $0x253cca0027650d,%rax
  mov    %rax,0x88(%rsp)
  movabs $0x2084380022f483,%rax
  mov    %rax,0x90(%rsp)
  movabs $0x1b0b90001de3e7,%rax
  mov    %rax,0x98(%rsp)
  movabs $0x1472c70017eb30,%rax
  mov    %rax,0xa0(%rsp)
  movabs $0xbd9c800107a51,%rax
  mov    %rax,0xa8(%rsp)
  movabs $0x4a8c90007c927,%rax
  mov    %rax,0xb0(%rsp)
  movabs $0x3cf7e40001205f,%rax
  mov    %rax,0xb8(%rsp)
  movabs $0x31669a0037e752,%rax
  mov    %rax,0xc0(%rsp)
  movabs $0x31662c0031662c,%rax
  mov    %rax,0xc8(%rsp)
  call   <program_acf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0xe8,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ini_acf_iireq_src_24m_5m(void) {
  int acf1[] = {0x2ab, 0x077, 0x0c6, 0x01e, 0x295, 0x0a5, 0x2fa, 0x06d, 0x27f, 0x0d2, 0x297, 0x05e, 0x272, 0x0ed, 0x27b, 0x04b, 0x26a, 0x0fb, 0x272, 0x04d};
  int acf2[] = {0x3f3fff, 0x3db7cf, 0x3c2f9e, 0x3aa76c, 0x39173b, 0x377f08, 0x35ded5, 0x343ea1, 0x328e6c, 0x30de36, 0x2f15ff, 0x2d45c6, 0x2b658a, 0x29754d, 0x27650d, 0x253cca, 0x22f483, 0x208438, 0x1de3e7, 0x1b0b90, 0x17eb30, 0x1472c7, 0x107a51, 0x0bd9c8, 0x07c927, 0x04a8c9, 0x01205f, 0x3cf7e4, 0x37e752, 0x31669a, 0x31662c, 0x31662c, 0x00062c};
  program_acf(acf1, acf2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cause_segfault>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movq   $0x12345678,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cause_segfault() {
  int *p = (int *)0x12345678;
  *p = 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cause_segfault>:
  endbr64
  movl   $0x0,0x12345678
  ret
```

assistant: ```c
void cause_segfault() {
  int *p = (int *)0x12345678;
  *p = 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cause_segfault>:
  endbr64
  movl   $0x0,0x12345678
  ret
```

assistant: ```c
void cause_segfault() {
  int *p = (int *)0x12345678;
  *p = 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<cause_segfault>:
  endbr64
  movl   $0x0,0x12345678
  ret
```

assistant: ```c
void cause_segfault() {
  int *p = (int *)0x12345678;
  *p = 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  je     L29
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L15
L29:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <readInput@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L8
  mov    $0x1,%eax
  jmp    L15
L8:
  mov    $0x0,%eax
L15:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
type_infered: string

tool: D8: USAGE: ./a.out <INPUT.txt>


assistant: ```c
int main(int argc, char **argv) {
  if (argc != 2) {
    printf("USAGE: ./a.out <INPUT.txt>\n");
    return 0;
  }
  int eCheck = readInput(argv[1]);
  if (eCheck == 1)
    return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L6
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L4:
  add    $0x8,%rsp
  ret
L6:
  mov    0x8(%rsi),%rdi
  call   <readInput@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L4
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D6: USAGE: ./a.out <INPUT.txt>


assistant: ```c
int main(int argc, char **argv) {
  if (argc != 2) {
    printf("USAGE: ./a.out <INPUT.txt>\n");
    return 0;
  }
  int eCheck = readInput(argv[1]);
  if (eCheck == 1)
    return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L19
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
L17:
  add    $0x8,%rsp
  ret
L19:
  mov    0x8(%rsi),%rdi
  call   <readInput@plt>
  sub    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L17
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: string

tool: D0: USAGE: ./a.out <INPUT.txt>


assistant: ```c
int main(int argc, char **argv) {
  if (argc != 2) {
    printf("USAGE: ./a.out <INPUT.txt>\n");
    return 0;
  }
  int eCheck = readInput(argv[1]);
  if (eCheck == 1)
    return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L14
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
L19:
  add    $0x8,%rsp
  ret
L14:
  mov    0x8(%rsi),%rdi
  call   <readInput@plt>
  sub    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  jmp    L19
```

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

tool: D9: USAGE: ./a.out <INPUT.txt>


assistant: ```c
int main(int argc, char **argv) {
  if (argc != 2) {
    printf("USAGE: ./a.out <INPUT.txt>\n");
    return 0;
  }
  int eCheck = readInput(argv[1]);
  if (eCheck == 1)
    return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0xa,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x64,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x2710,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D84(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D90(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D85(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1b,%edx
  lea    D84(%rip),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <printPredefinedMacros@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D70
datatype: string
name: retrieve_data
arguments: tag: D23
datatype: string
name: retrieve_data
arguments: tag: D99
datatype: string
name: retrieve_data
arguments: tag: D57
datatype: string
name: retrieve_data
arguments: tag: D84
datatype: string
name: retrieve_data
arguments: tag: D90
datatype: string
name: retrieve_data
arguments: tag: D42
datatype: string
name: retrieve_data
arguments: tag: D85
datatype: string
name: retrieve_data
arguments: tag: D54
datatype: string

tool: D70: n =10 , M =  %d


tool: D23: SQUARE(10) = %d


tool: D99: "Ztiany" and "Beautifully Grail" Together !

tool: D57: token100 = %d


tool: D84: Ztiany

tool: D90: MY_NAME = %s


tool: D42: -------------------------------------------------------

tool: D85: Hello PR

tool: D54: name=%s,age=%d


assistant: ```c
int main() {
  int n = 10;
  int m = (n * n + 3 * n);
  printf("n =10 , M =  %d\n", m);
  int a = ((10) * (10));
  printf("SQUARE(10) = %d\n", a);
  printf("\"Ztiany\""
         " and "
         "\"Beautifully Grail\""
         " Together !\n");
  int token100 = 10000;
  printf("token"
         "100"
         " = %d\n",
         token100);
  char *name = "Ztiany";
  printf("MY_NAME = %s\n", name);
  printf("-------------------------------------------------------\n");
  printf("Hello PR\n");
  printf("name=%s,age=%d\n", "Ztiany", 27);
  printPredefinedMacros();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x82,%esi
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x64,%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <puts@plt>
  mov    $0x2710,%esi
  lea    D64(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D49(%rip),%rbx
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  lea    D37(%rip),%rdi
  call   <puts@plt>
  mov    $0x1b,%edx
  mov    %rbx,%rsi
  lea    D72(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <printPredefinedMacros@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D35
type_guessed: string
name: parse_data
arguments: data_tag: D11
type_guessed: string
name: parse_data
arguments: data_tag: D60
type_guessed: string
name: parse_data
arguments: data_tag: D64
type_guessed: string
name: parse_data
arguments: data_tag: D49
type_guessed: string
name: parse_data
arguments: data_tag: D9
type_guessed: string
name: parse_data
arguments: data_tag: D1
type_guessed: string
name: parse_data
arguments: data_tag: D37
type_guessed: string
name: parse_data
arguments: data_tag: D72
type_guessed: string

tool: D35: n =10 , M =  %d


tool: D11: SQUARE(10) = %d


tool: D60: "Ztiany" and "Beautifully Grail" Together !

tool: D64: token100 = %d


tool: D49: Ztiany

tool: D9: MY_NAME = %s


tool: D1: -------------------------------------------------------

tool: D37: Hello PR

tool: D72: name=%s,age=%d


assistant: ```c
int main() {
  int n = 10;
  int m = (n * n + 3 * n);
  printf("n =10 , M =  %d\n", m);
  int a = ((10) * (10));
  printf("SQUARE(10) = %d\n", a);
  printf("\"Ztiany\""
         " and "
         "\"Beautifully Grail\""
         " Together !\n");
  int token100 = 10000;
  printf("token"
         "100"
         " = %d\n",
         token100);
  char *name = "Ztiany";
  printf("MY_NAME = %s\n", name);
  printf("-------------------------------------------------------\n");
  printf("Hello PR\n");
  printf("name=%s,age=%d\n", "Ztiany", 27);
  printPredefinedMacros();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x82,%esi
  lea    D71(%rip),%rdi
  xor    %eax,%eax
  lea    D76(%rip),%rbp
  call   <printf@plt>
  mov    $0x64,%esi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D55(%rip),%rdi
  call   <puts@plt>
  mov    $0x2710,%esi
  lea    D53(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D58(%rip),%rdi
  call   <puts@plt>
  lea    D80(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    $0x1b,%edx
  xor    %eax,%eax
  lea    D42(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <printPredefinedMacros@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D71
datatype: string
name: unpack_data
arguments: data_tag: D76
datatype: string
name: unpack_data
arguments: data_tag: D25
datatype: string
name: unpack_data
arguments: data_tag: D55
datatype: string
name: unpack_data
arguments: data_tag: D53
datatype: string
name: unpack_data
arguments: data_tag: D37
datatype: string
name: unpack_data
arguments: data_tag: D58
datatype: string
name: unpack_data
arguments: data_tag: D80
datatype: string
name: unpack_data
arguments: data_tag: D42
datatype: string

tool: D71: n =10 , M =  %d


tool: D76: Ztiany

tool: D25: SQUARE(10) = %d


tool: D55: "Ztiany" and "Beautifully Grail" Together !

tool: D53: token100 = %d


tool: D37: MY_NAME = %s


tool: D58: -------------------------------------------------------

tool: D80: Hello PR

tool: D42: name=%s,age=%d


assistant: ```c
int main() {
  int n = 10;
  int m = (n * n + 3 * n);
  printf("n =10 , M =  %d\n", m);
  int a = ((10) * (10));
  printf("SQUARE(10) = %d\n", a);
  printf("\"Ztiany\""
         " and "
         "\"Beautifully Grail\""
         " Together !\n");
  int token100 = 10000;
  printf("token"
         "100"
         " = %d\n",
         token100);
  char *name = "Ztiany";
  printf("MY_NAME = %s\n", name);
  printf("-------------------------------------------------------\n");
  printf("Hello PR\n");
  printf("name=%s,age=%d\n", "Ztiany", 27);
  printPredefinedMacros();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x82,%esi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  lea    D61(%rip),%rbp
  call   <printf@plt>
  mov    $0x64,%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D74(%rip),%rdi
  call   <puts@plt>
  mov    $0x2710,%esi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D69(%rip),%rdi
  call   <puts@plt>
  lea    D57(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    $0x1b,%edx
  xor    %eax,%eax
  lea    D39(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <printPredefinedMacros@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
datatype: string
name: retrieve_data
arguments: data_label: D61
datatype: string
name: retrieve_data
arguments: data_label: D9
datatype: string
name: retrieve_data
arguments: data_label: D74
datatype: string
name: retrieve_data
arguments: data_label: D30
datatype: string
name: retrieve_data
arguments: data_label: D29
datatype: string
name: retrieve_data
arguments: data_label: D69
datatype: string
name: retrieve_data
arguments: data_label: D57
datatype: string
name: retrieve_data
arguments: data_label: D39
datatype: string

tool: D8: n =10 , M =  %d


tool: D61: Ztiany

tool: D9: SQUARE(10) = %d


tool: D74: "Ztiany" and "Beautifully Grail" Together !

tool: D30: token100 = %d


tool: D29: MY_NAME = %s


tool: D69: -------------------------------------------------------

tool: D57: Hello PR

tool: D39: name=%s,age=%d


assistant: ```c
int main() {
  int n = 10;
  int m = (n * n + 3 * n);
  printf("n =10 , M =  %d\n", m);
  int a = ((10) * (10));
  printf("SQUARE(10) = %d\n", a);
  printf("\"Ztiany\""
         " and "
         "\"Beautifully Grail\""
         " Together !\n");
  int token100 = 10000;
  printf("token"
         "100"
         " = %d\n",
         token100);
  char *name = "Ztiany";
  printf("MY_NAME = %s\n", name);
  printf("-------------------------------------------------------\n");
  printf("Hello PR\n");
  printf("name=%s,age=%d\n", "Ztiany", 27);
  printPredefinedMacros();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<vector_normalize_3d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <vector_magnitude_3d@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  divsd  -0x8(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void vector_normalize_3d(double v[3]) {
  double mag = vector_magnitude_3d(v);
  v[0] /= mag;
  v[1] /= mag;
  v[2] /= mag;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<vector_normalize_3d>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <vector_magnitude_3d@plt>
  movsd  (%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rbx)
  movsd  0x8(%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rbx)
  movsd  0x10(%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void vector_normalize_3d(double v[3]) {
  double mag = vector_magnitude_3d(v);
  v[0] /= mag;
  v[1] /= mag;
  v[2] /= mag;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vector_normalize_3d>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <vector_magnitude_3d@plt>
  movsd  (%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rbx)
  movsd  0x8(%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rbx)
  movsd  0x10(%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void vector_normalize_3d(double v[3]) {
  double mag = vector_magnitude_3d(v);
  v[0] /= mag;
  v[1] /= mag;
  v[2] /= mag;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vector_normalize_3d>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <vector_magnitude_3d@plt>
  movupd (%rbx),%xmm1
  movapd %xmm0,%xmm2
  unpcklpd %xmm2,%xmm2
  divpd  %xmm2,%xmm1
  movups %xmm1,(%rbx)
  movsd  0x10(%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void vector_normalize_3d(double v[3]) {
  double mag = vector_magnitude_3d(v);
  v[0] /= mag;
  v[1] /= mag;
  v[2] /= mag;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<norm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <uni@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <uni@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  sub    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  movsd  %xmm0,-0x18(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  add    %eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <cos@plt>
  mulsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double norm() {
  double u;
  double v;
  u = uni();
  v = uni();
  return (sqrt(-2 * log(u)) * cos(2 * M_PI * v));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<norm>:
  endbr64
  push   %r14
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <uni@plt>
  movq   %xmm0,%rbx
  mov    $0x0,%eax
  call   <uni@plt>
  movq   %xmm0,%r14
  movq   %rbx,%xmm0
  call   <log@plt>
  neg    %eax
  lea    (%rax,%rax,1),%edi
  call   <sqrt@plt>
  movq   %xmm0,%rbx
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  add    %eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movq   %r14,%xmm1
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  movq   %rbx,%xmm2
  mulsd  %xmm2,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %r14
  ret
```

assistant: ```c
double norm() {
  double u;
  double v;
  u = uni();
  v = uni();
  return (sqrt(-2 * log(u)) * cos(2 * M_PI * v));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<norm>:
  endbr64
  sub    $0x18,%rsp
  xor    %eax,%eax
  call   <uni@plt>
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  call   <uni@plt>
  movsd  (%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <log@plt>
  neg    %eax
  lea    (%rax,%rax,1),%edi
  call   <sqrt@plt>
  mov    D6(%rip),%rax
  movsd  %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  mov    (%rax),%eax
  add    %eax,%eax
  cvtsi2sd %eax,%xmm0
  mulsd  0x8(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  mulsd  (%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
double norm() {
  double u;
  double v;
  u = uni();
  v = uni();
  return (sqrt(-2 * log(u)) * cos(2 * M_PI * v));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<norm>:
  endbr64
  sub    $0x18,%rsp
  xor    %eax,%eax
  call   <uni@plt>
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  call   <uni@plt>
  movsd  (%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <log@plt>
  neg    %eax
  lea    (%rax,%rax,1),%edi
  call   <sqrt@plt>
  mov    D0(%rip),%rax
  movsd  %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  mov    (%rax),%eax
  add    %eax,%eax
  cvtsi2sd %eax,%xmm0
  mulsd  0x8(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <cos@plt>
  mulsd  (%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
double norm() {
  double u;
  double v;
  u = uni();
  v = uni();
  return (sqrt(-2 * log(u)) * cos(2 * M_PI * v));
}
```