ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  push   %rbx
  sub    $0xa8,%rsp
  movdqa D56(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movb   $0x0,0x28(%rsp)
  movabs $0x7372656d6d617267,%rax
  movaps %xmm0,(%rsp)
  movdqa D79(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  call   <printf@plt>
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movsbl (%rsp),%esi
  test   %sil,%sil
  je     L23
  lea    0x1(%rsp),%rbx
  lea    D66(%rip),%rbp
  nopl   0x0(%rax)
L8:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  movsbl -0x1(%rbx),%esi
  test   %sil,%sil
  jne    L8
L23:
  mov    $0xa,%edi
  lea    0x30(%rsp),%rbp
  call   <putchar@plt>
  lea    D51(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
type: string
name: retrieve_data
arguments: data_label: D56
type: byte[16]
name: retrieve_data
arguments: data_label: D79
type: byte[16]
name: retrieve_data
arguments: data_label: D4
type: string
name: retrieve_data
arguments: data_label: D66
type: string
name: retrieve_data
arguments: data_label: D51
type: string
name: retrieve_data
arguments: data_label: D14
type: string
name: retrieve_data
arguments: data_label: D34
type: string

tool: D3: s1 is : %s


tool: {"D56": [0x4e,0x6f,0x77,0x20,0x69,0x73,0x20,0x74,0x68,0x65,0x20,0x74,0x69,0x6d,0x65,0x20]}

tool: {"D79": [0x66,0x6f,0x72,0x20,0x61,0x6c,0x6c,0x20,0x67,0x6f,0x6f,0x64,0x20,0x70,0x72,0x6f]}

tool: D4: Printing one char at a time: 

tool: D66: %c 

tool: D51: Enter a string: 

tool: D14: %100[^
]

tool: D34: s2 is: %s


assistant: ```c
int main() {
  char s1[] = "Now is the time for all good programmers";
  int i;
  printf("s1 is : %s\n", s1);
  printf("Printing one char at a time: ");
  for (i = 0; s1[i] != '\0'; i++) {
    printf("%c ", s1[i]);
  }
  printf("\n");
  char s2[100];
  printf("Enter a string: ");
  scanf("%100[^\n]", s2);
  printf("s2 is: %s\n", s2);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D10(%rip),%rdi
  push   %rbx
  sub    $0xa8,%rsp
  movdqa D74(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movb   $0x0,0x28(%rsp)
  movabs $0x7372656d6d617267,%rax
  movaps %xmm0,(%rsp)
  movdqa D62(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  call   <printf@plt>
  lea    D64(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movsbl (%rsp),%esi
  test   %sil,%sil
  je     L27
  lea    0x1(%rsp),%rbx
  lea    D60(%rip),%rbp
  nopl   0x0(%rax)
L18:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  movsbl -0x1(%rbx),%esi
  test   %sil,%sil
  jne    L18
L27:
  mov    $0xa,%edi
  lea    0x30(%rsp),%rbp
  call   <putchar@plt>
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D76(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0xa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D10
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D74
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D62
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D64
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D60
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D32
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D76
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string

tool: D10: s1 is : %s


tool: {"D74": [0x4e,0x6f,0x77,0x20,0x69,0x73,0x20,0x74,0x68,0x65,0x20,0x74,0x69,0x6d,0x65,0x20]}

tool: {"D62": [0x66,0x6f,0x72,0x20,0x61,0x6c,0x6c,0x20,0x67,0x6f,0x6f,0x64,0x20,0x70,0x72,0x6f]}

tool: D64: Printing one char at a time: 

tool: D60: %c 

tool: D32: Enter a string: 

tool: D76: %100[^
]

tool: D19: s2 is: %s


assistant: ```c
int main() {
  char s1[] = "Now is the time for all good programmers";
  int i;
  printf("s1 is : %s\n", s1);
  printf("Printing one char at a time: ");
  for (i = 0; s1[i] != '\0'; i++) {
    printf("%c ", s1[i]);
  }
  printf("\n");
  char s2[100];
  printf("Enter a string: ");
  scanf("%100[^\n]", s2);
  printf("s2 is: %s\n", s2);
  return 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
<redhttp_strndup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L6
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L6
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L6:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *redhttp_strndup(const char *str1, size_t str1_len) {
  char *str2 = NULL;
  if (str1) {
    str2 = malloc(str1_len + 1);
    if (str2) {
      strncpy(str2, str1, str1_len);
      str2[str1_len] = '\0';
    }
  }
  return str2;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<redhttp_strndup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L8
  mov    %rdi,%rbp
  mov    %rsi,%r12
  lea    0x1(%rsi),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L8
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%rbx,%r12,1)
L8:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *redhttp_strndup(const char *str1, size_t str1_len) {
  char *str2 = NULL;
  if (str1) {
    str2 = malloc(str1_len + 1);
    if (str2) {
      strncpy(str2, str1, str1_len);
      str2[str1_len] = '\0';
    }
  }
  return str2;
}
```
user: What is the c source code of the assembly code below?

```asm
<redhttp_strndup>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L17
  mov    %rdi,%rbp
  lea    0x1(%rsi),%rdi
  mov    %rsi,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L17
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L17:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *redhttp_strndup(const char *str1, size_t str1_len) {
  char *str2 = NULL;
  if (str1) {
    str2 = malloc(str1_len + 1);
    if (str2) {
      strncpy(str2, str1, str1_len);
      str2[str1_len] = '\0';
    }
  }
  return str2;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<redhttp_strndup>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L14
  mov    %rdi,%rbp
  lea    0x1(%rsi),%rdi
  mov    %rsi,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L14
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L14:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *redhttp_strndup(const char *str1, size_t str1_len) {
  char *str2 = NULL;
  if (str1) {
    str2 = malloc(str1_len + 1);
    if (str2) {
      strncpy(str2, str1, str1_len);
      str2[str1_len] = '\0';
    }
  }
  return str2;
}
```
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    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x6c(%rbp),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x68(%rbp),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x64(%rbp),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x6c(%rbp),%edx
  mov    -0x68(%rbp),%eax
  add    %eax,%edx
  mov    -0x64(%rbp),%eax
  cmp    %eax,%edx
  jg     L48
  mov    -0x68(%rbp),%edx
  mov    -0x64(%rbp),%eax
  add    %eax,%edx
  mov    -0x6c(%rbp),%eax
  cmp    %eax,%edx
  jg     L48
  mov    -0x6c(%rbp),%edx
  mov    -0x64(%rbp),%eax
  add    %eax,%edx
  mov    -0x68(%rbp),%eax
  cmp    %eax,%edx
  jle    L1
L48:
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L5
L1:
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L5:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D35
type_guessed: string
name: disassemble_data
arguments: data_tag: D44
type_guessed: string
name: disassemble_data
arguments: data_tag: D50
type_guessed: string
name: disassemble_data
arguments: data_tag: D39
type_guessed: string
name: disassemble_data
arguments: data_tag: D33
type_guessed: string
name: disassemble_data
arguments: data_tag: D77
type_guessed: string

tool: D35: nhap a: 

tool: D44: %d

tool: D50: 
 nhap b: 

tool: D39: 
 nhap c: 

tool: D33: 
 a b c khong the la canh cua 1 tam giac

tool: D77: 
 a b c la canh cua 1 tam giac

assistant: ```c
int main(int argc, char *argv[]) {
  char str[80];
  int a, b, c;
  printf("nhap a: ");
  scanf("%d", &a);
  printf("\n nhap b: ");
  scanf("%d", &b);
  printf("\n nhap c: ");
  scanf("%d", &c);
  if (a + b > c || b + c > a || a + c > b) {
    printf("\n a b c khong the la canh cua 1 tam giac");
  } else {
    printf("\n a b c la canh cua 1 tam giac");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D56(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    0x10(%rsp),%edx
  mov    0x14(%rsp),%ecx
  lea    (%rax,%rdx,1),%esi
  cmp    %ecx,%esi
  jg     L21
  lea    (%rdx,%rcx,1),%esi
  cmp    %esi,%eax
  jl     L21
  add    %ecx,%eax
  cmp    %eax,%edx
  jge    L36
L21:
  lea    D55(%rip),%rdi
  call   <printf@plt>
L34:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L36:
  lea    D23(%rip),%rdi
  call   <printf@plt>
  jmp    L34
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D56
type: string
name: unpack_data
arguments: label: D7
type: string
name: unpack_data
arguments: label: D18
type: string
name: unpack_data
arguments: label: D36
type: string
name: unpack_data
arguments: label: D55
type: string
name: unpack_data
arguments: label: D23
type: string

tool: D56: nhap a: 

tool: D7: %d

tool: D18: 
 nhap b: 

tool: D36: 
 nhap c: 

tool: D55: 
 a b c khong the la canh cua 1 tam giac

tool: D23: 
 a b c la canh cua 1 tam giac

assistant: ```c
int main(int argc, char *argv[]) {
  char str[80];
  int a, b, c;
  printf("nhap a: ");
  scanf("%d", &a);
  printf("\n nhap b: ");
  scanf("%d", &b);
  printf("\n nhap c: ");
  scanf("%d", &c);
  if (a + b > c || b + c > a || a + c > b) {
    printf("\n a b c khong the la canh cua 1 tam giac");
  } else {
    printf("\n a b c la canh cua 1 tam giac");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D12(%rip),%rdi
  lea    D28(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D51(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    0x10(%rsp),%edx
  mov    0x14(%rsp),%ecx
  lea    (%rax,%rdx,1),%esi
  cmp    %ecx,%esi
  jg     L40
  lea    (%rdx,%rcx,1),%esi
  cmp    %esi,%eax
  jge    L18
L40:
  lea    D27(%rip),%rdi
  call   <printf@plt>
L20:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L18:
  add    %ecx,%eax
  cmp    %eax,%edx
  jl     L40
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D12
type: string
name: retrieve_data
arguments: data_tag: D28
type: string
name: retrieve_data
arguments: data_tag: D51
type: string
name: retrieve_data
arguments: data_tag: D39
type: string
name: retrieve_data
arguments: data_tag: D27
type: string
name: retrieve_data
arguments: data_tag: D18
type: string

tool: D12: nhap a: 

tool: D28: %d

tool: D51: 
 nhap b: 

tool: D39: 
 nhap c: 

tool: D27: 
 a b c khong the la canh cua 1 tam giac

tool: D18: 
 a b c la canh cua 1 tam giac

assistant: ```c
int main(int argc, char *argv[]) {
  char str[80];
  int a, b, c;
  printf("nhap a: ");
  scanf("%d", &a);
  printf("\n nhap b: ");
  scanf("%d", &b);
  printf("\n nhap c: ");
  scanf("%d", &c);
  if (a + b > c || b + c > a || a + c > b) {
    printf("\n a b c khong the la canh cua 1 tam giac");
  } else {
    printf("\n a b c la canh cua 1 tam giac");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rdi
  lea    D59(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    0x10(%rsp),%edx
  mov    0x14(%rsp),%ecx
  lea    (%rax,%rdx,1),%esi
  cmp    %ecx,%esi
  jg     L24
  lea    (%rdx,%rcx,1),%esi
  cmp    %esi,%eax
  jge    L0
L24:
  lea    D13(%rip),%rdi
  call   <printf@plt>
L23:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  add    %ecx,%eax
  cmp    %eax,%edx
  jl     L24
  lea    D21(%rip),%rdi
  call   <printf@plt>
  jmp    L23
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D15
type_infered: string
name: retrieve_data
arguments: data_tag: D59
type_infered: string
name: retrieve_data
arguments: data_tag: D33
type_infered: string
name: retrieve_data
arguments: data_tag: D47
type_infered: string
name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D21
type_infered: string

tool: D15: nhap a: 

tool: D59: %d

tool: D33: 
 nhap b: 

tool: D47: 
 nhap c: 

tool: D13: 
 a b c khong the la canh cua 1 tam giac

tool: D21: 
 a b c la canh cua 1 tam giac

assistant: ```c
int main(int argc, char *argv[]) {
  char str[80];
  int a, b, c;
  printf("nhap a: ");
  scanf("%d", &a);
  printf("\n nhap b: ");
  scanf("%d", &b);
  printf("\n nhap c: ");
  scanf("%d", &c);
  if (a + b > c || b + c > a || a + c > b) {
    printf("\n a b c khong the la canh cua 1 tam giac");
  } else {
    printf("\n a b c la canh cua 1 tam giac");
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sumWptr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L3:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sumWptr(int a[], int len) {
  int sum = 0;
  for (int i = 0; i < len; i++) {
    sum += *(a + i);
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sumWptr>:
  endbr64
  test   %esi,%esi
  jle    L15
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%edx
L12:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L12
L26:
  mov    %edx,%eax
  ret
L15:
  mov    $0x0,%edx
  jmp    L26
```

assistant: ```c
int sumWptr(int a[], int len) {
  int sum = 0;
  for (int i = 0; i < len; i++) {
    sum += *(a + i);
  }
  return sum;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sumWptr>:
  endbr64
  test   %esi,%esi
  jle    L0
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L13:
  add    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L13
  ret
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sumWptr(int a[], int len) {
  int sum = 0;
  for (int i = 0; i < len; i++) {
    sum += *(a + i);
  }
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<sumWptr>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L38
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L56
  mov    %esi,%ecx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L2:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L2
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%sil
  je     L51
L67:
  movslq %ecx,%rdi
  lea    0x0(,%rdi,4),%r8
  add    (%rdx,%rdi,4),%eax
  lea    0x1(%rcx),%edi
  cmp    %edi,%esi
  jle    L63
  add    $0x2,%ecx
  add    0x4(%rdx,%r8,1),%eax
  cmp    %ecx,%esi
  jle    L63
  add    0x8(%rdx,%r8,1),%eax
  ret
  nopl   0x0(%rax)
L38:
  xor    %eax,%eax
L63:
  ret
  nopl   0x0(%rax,%rax,1)
L51:
  ret
L56:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L67
```

assistant: ```c
int sumWptr(int a[], int len) {
  int sum = 0;
  for (int i = 0; i < len; i++) {
    sum += *(a + i);
  }
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cpuid_edx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rsi
  mov    -0x24(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <vmi_cpuid@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
unsigned int cpuid_edx(unsigned int op) {
  unsigned int eax, ebx, ecx, edx;
  vmi_cpuid(op, &eax, &ebx, &ecx, &edx);
  return edx;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cpuid_edx>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <vmi_cpuid@plt>
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int cpuid_edx(unsigned int op) {
  unsigned int eax, ebx, ecx, edx;
  vmi_cpuid(op, &eax, &ebx, &ecx, &edx);
  return edx;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cpuid_edx>:
  endbr64
  sub    $0x28,%rsp
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <vmi_cpuid@plt>
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int cpuid_edx(unsigned int op) {
  unsigned int eax, ebx, ecx, edx;
  vmi_cpuid(op, &eax, &ebx, &ecx, &edx);
  return edx;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpuid_edx>:
  endbr64
  sub    $0x28,%rsp
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <vmi_cpuid@plt>
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int cpuid_edx(unsigned int op) {
  unsigned int eax, ebx, ecx, edx;
  vmi_cpuid(op, &eax, &ebx, &ecx, &edx);
  return edx;
}
```
user: Translate the below assembly code into C programming language:

```asm
<chiProbability>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <chiSquared@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <chiSquaredDistribution@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string

tool: D1: chisq: %4.2f dof: %d


assistant: ```c
double chiProbability(double **observed, int categories, int dataTypes) {
  int dof = (categories - 1) * (dataTypes - 1);
  double chi;
  chi = chiSquared(observed, categories, dataTypes);
  printf("chisq: %4.2f dof: %d\n", chi, dof);
  return chiSquaredDistribution(chi, dof);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<chiProbability>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    -0x1(%rsi),%ebp
  lea    -0x1(%rdx),%eax
  imul   %eax,%ebp
  call   <chiSquared@plt>
  movq   %xmm0,%rbx
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  movq   %rbx,%xmm0
  call   <chiSquaredDistribution@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
datatype: string

tool: D2: chisq: %4.2f dof: %d


assistant: ```c
double chiProbability(double **observed, int categories, int dataTypes) {
  int dof = (categories - 1) * (dataTypes - 1);
  double chi;
  chi = chiSquared(observed, categories, dataTypes);
  printf("chisq: %4.2f dof: %d\n", chi, dof);
  return chiSquaredDistribution(chi, dof);
}
```