ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_include_flags>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %esi,%r13d
  lea    D25(%rip),%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    0x20(%rdx),%r12
  call   <strcpy@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <strcat@plt>
  lea    D22(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  cmp    $0x5,%r13d
  jg     L28
L19:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  lea    D14(%rip),%r15
  mov    %rbp,%rdi
  mov    %r15,%rsi
  call   <strcat@plt>
  lea    -0x1(%r13),%eax
  mov    %eax,0xc(%rsp)
  cmp    $0x6,%r13d
  je     L14
  lea    0x28(%r14),%rbx
  mov    $0x5,%r12d
  nopw   0x0(%rax,%rax,1)
L11:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  add    $0x1,%r12d
  add    $0x8,%rbx
  call   <strcat@plt>
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  cmp    0xc(%rsp),%r12d
  jl     L11
  mov    %r12d,0xc(%rsp)
  cmp    %r12d,%r13d
  jle    L19
L14:
  movslq 0xc(%rsp),%rbx
  nopl   0x0(%rax)
L48:
  mov    (%r14,%rbx,8),%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <strcat@plt>
  cmp    %ebx,%r13d
  jg     L48
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D25
type: string
name: parse_data
arguments: data_tag: D22
type: string
name: parse_data
arguments: data_tag: D14
type: string

tool: D25: -I"

tool: D22: "

tool: D14:  

assistant: ```c
void get_include_flags(char *dst, int argc, char *argv[]) {
  const char *cutest_path = argv[4];
  int i;
  strcpy(dst, "-I\"");
  strcat(dst, cutest_path);
  strcat(dst, "\"");
  if (argc > 5) {
    strcat(dst, " ");
  }
  for (i = 5; i < argc; i++) {
    strcat(dst, argv[i]);
    if (i < argc - 1) {
      strcat(dst, " ");
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<g>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <rand@plt>
  and    $0x1,%eax
  test   %eax,%eax
  jne    L8
  mov    $0x0,%eax
  call   <g>
  jmp    L4
L8:
  mov    $0x0,%eax
  call   <lock@plt>
L4:
  nop
  pop    %rbp
  ret
```

assistant: ```c
int g() {
  if (rand() % 2 == 0) {
    g();
  } else {
    lock();
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<g>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <rand@plt>
  test   $0x1,%al
  jne    L10
  mov    $0x0,%eax
  call   <g>
L5:
  add    $0x8,%rsp
  ret
L10:
  mov    $0x0,%eax
  call   <lock@plt>
  jmp    L5
```

assistant: ```c
int g() {
  if (rand() % 2 == 0) {
    g();
  } else {
    lock();
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<g>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <rand@plt>
  test   $0x1,%al
  je     L3
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <lock@plt>
```

assistant: ```c
int g() {
  if (rand() % 2 == 0) {
    g();
  } else {
    lock();
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<g>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  call   <rand@plt>
  test   $0x1,%al
  je     L5
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <lock@plt>
```

assistant: ```c
int g() {
  if (rand() % 2 == 0) {
    g();
  } else {
    lock();
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L19
L35:
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L35
  subl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L38
L18:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L38:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L18
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j;
  char a;
  for (i = 0; s[i]; i++)
    ;
  i--;
  j = 0;
  while (j < i) {
    a = s[i];
    s[i] = s[j];
    s[j] = a;
    i--;
    j++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reverse>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L10
  mov    $0x0,%edx
L16:
  mov    %edx,%eax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L16
  test   %eax,%eax
  jle    L10
  cltq
  mov    $0x0,%edx
L2:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rdi,%rdx,1),%esi
  mov    %sil,(%rdi,%rax,1)
  mov    %cl,(%rdi,%rdx,1)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L2
L10:
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j;
  char a;
  for (i = 0; s[i]; i++)
    ;
  i--;
  j = 0;
  while (j < i) {
    a = s[i];
    s[i] = s[j];
    s[j] = a;
    i--;
    j++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L44
  xor    %edx,%edx
  nopl   (%rax)
L27:
  movslq %edx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L27
  test   %eax,%eax
  je     L44
  xor    %edx,%edx
  jmp    L35
  nopl   (%rax)
L0:
  movzbl (%rdi,%rdx,1),%ecx
L35:
  movzbl (%rdi,%rax,1),%esi
  mov    %cl,(%rdi,%rax,1)
  sub    $0x1,%rax
  mov    %sil,(%rdi,%rdx,1)
  add    $0x1,%rdx
  cmp    %eax,%edx
  jl     L0
L44:
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j;
  char a;
  for (i = 0; s[i]; i++)
    ;
  i--;
  j = 0;
  while (j < i) {
    a = s[i];
    s[i] = s[j];
    s[j] = a;
    i--;
    j++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<reverse>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L18
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L11:
  movslq %edx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L11
  test   %eax,%eax
  je     L18
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L35:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rdi,%rdx,1),%esi
  mov    %sil,(%rdi,%rax,1)
  sub    $0x1,%rax
  mov    %cl,(%rdi,%rdx,1)
  add    $0x1,%rdx
  cmp    %eax,%edx
  jl     L35
L18:
  ret
```

assistant: ```c
void reverse(char s[]) {
  int i, j;
  char a;
  for (i = 0; s[i]; i++)
    ;
  i--;
  j = 0;
  while (j < i) {
    a = s[i];
    s[i] = s[j];
    s[j] = a;
    i--;
    j++;
  }
}
```
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
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %ebx,%edx
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %ebx,%edx
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D12
datatype: string
name: disassemble_data
arguments: tag: D3
datatype: string

tool: D12: You roll [%d] [%d]


tool: D3: PC rolls [%d] [%d]


assistant: ```c
int main() {
  printf("You roll [%d] [%d]\n", roll(), roll());
  printf("PC rolls [%d] [%d]\n", roll(), roll());
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %eax,%esi
  mov    %ebx,%edx
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <roll@plt>
  mov    %eax,%esi
  mov    %ebx,%edx
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type: string
name: unpack_data
arguments: label: D9
type: string

tool: D8: You roll [%d] [%d]


tool: D9: PC rolls [%d] [%d]


assistant: ```c
int main() {
  printf("You roll [%d] [%d]\n", roll(), roll());
  printf("PC rolls [%d] [%d]\n", roll(), roll());
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <roll@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <roll@plt>
  mov    %r12d,%edx
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <roll@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <roll@plt>
  mov    %r12d,%edx
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
datatype: string
name: read_data
arguments: data_tag: D7
datatype: string

tool: D6: You roll [%d] [%d]


tool: D7: PC rolls [%d] [%d]


assistant: ```c
int main() {
  printf("You roll [%d] [%d]\n", roll(), roll());
  printf("PC rolls [%d] [%d]\n", roll(), roll());
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <roll@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <roll@plt>
  mov    %r12d,%edx
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <roll@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <roll@plt>
  mov    %r12d,%edx
  lea    D19(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_guessed: string
name: disassemble_data
arguments: data_label: D19
type_guessed: string

tool: D3: You roll [%d] [%d]


tool: D19: PC rolls [%d] [%d]


assistant: ```c
int main() {
  printf("You roll [%d] [%d]\n", roll(), roll());
  printf("PC rolls [%d] [%d]\n", roll(), roll());
  return 0;
}
```
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    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x420(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x400(%rbp)
  movq   $0x0,-0x410(%rbp)
  movq   $0x0,-0x408(%rbp)
  movq   $0x0,-0x418(%rbp)
  movq   $0x0,-0x3f8(%rbp)
  jmp    L20
L24:
  mov    -0x408(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rcx
  lea    -0x3f0(%rbp),%rdx
  mov    -0x418(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  add    %rcx,%rax
  sub    $0x30,%rax
  mov    -0x420(%rbp),%rcx
  cqto
  idiv   %rcx
  mov    %rdx,-0x408(%rbp)
  mov    -0x410(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rcx
  lea    -0x3f0(%rbp),%rdx
  mov    -0x418(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  add    %rcx,%rax
  sub    $0x30,%rax
  mov    -0x420(%rbp),%rsi
  cqto
  idiv   %rsi
  mov    %rax,-0x410(%rbp)
  mov    -0x410(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x408(%rbp),%rax
  mov    %rax,-0x410(%rbp)
  addq   $0x1,-0x418(%rbp)
L20:
  mov    -0x418(%rbp),%rax
  cmp    -0x400(%rbp),%rax
  jl     L24
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
data_type: string
name: disassemble_data
arguments: data_label: D5
data_type: string

tool: D6: %s

tool: D5: %lld

assistant: ```c
int main() {
  long long int len, i, j, num, div, rem, l;
  char a[1000], total[1000];
  scanf("%s", a);
  scanf("%lld", &num);
  len = strlen(a);
  div = 0;
  rem = 0;
  i = 0;
  l = 0;
  while (i < len) {
    rem = (rem * 10 + a[i] - '0') % num;
    div = (div * 10 + a[i] - '0') / num;
    printf("%lld", div);
    div = rem;
    i++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x408,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  lea    0x8(%rsp),%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jle    L8
  lea    (%rax,%rbx,1),%r12
  mov    $0x0,%ebp
  lea    D9(%rip),%r13
L16:
  lea    0x0(%rbp,%rbp,4),%rdx
  movsbq (%rbx),%rax
  lea    -0x30(%rax,%rdx,2),%rax
  cqto
  idivq  0x8(%rsp)
  mov    %rdx,%rbp
  mov    %rax,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L16
L8:
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x408,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
data_type: string
name: extract_symbolic_data
arguments: label: D9
data_type: string

tool: D5: %s

tool: D9: %lld

assistant: ```c
int main() {
  long long int len, i, j, num, div, rem, l;
  char a[1000], total[1000];
  scanf("%s", a);
  scanf("%lld", &num);
  len = strlen(a);
  div = 0;
  rem = 0;
  i = 0;
  l = 0;
  while (i < len) {
    rem = (rem * 10 + a[i] - '0') % num;
    div = (div * 10 + a[i] - '0') / num;
    printf("%lld", div);
    div = rem;
    i++;
  }
  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   %r13
  lea    D7(%rip),%rdi
  push   %r12
  lea    D17(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x408,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  je     L11
  lea    (%rbx,%rax,1),%r13
  xor    %ebp,%ebp
  nopl   0x0(%rax)
L14:
  movsbq (%rbx),%rax
  lea    0x0(%rbp,%rbp,4),%rdx
  mov    %r12,%rdi
  add    $0x1,%rbx
  lea    -0x30(%rax,%rdx,2),%rax
  cqto
  idivq  0x8(%rsp)
  mov    %rax,%rsi
  mov    %rdx,%rbp
  call   <printf@plt>
  cmp    %rbx,%r13
  jne    L14
L11:
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x408,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
type_guessed: string
name: retrieve_data
arguments: label: D17
type_guessed: string

tool: D7: %s

tool: D17: %lld

assistant: ```c
int main() {
  long long int len, i, j, num, div, rem, l;
  char a[1000], total[1000];
  scanf("%s", a);
  scanf("%lld", &num);
  len = strlen(a);
  div = 0;
  rem = 0;
  i = 0;
  l = 0;
  while (i < len) {
    rem = (rem * 10 + a[i] - '0') % num;
    div = (div * 10 + a[i] - '0') / num;
    printf("%lld", div);
    div = rem;
    i++;
  }
  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
<main>:
  endbr64
  push   %r13
  lea    D5(%rip),%rdi
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x408,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3f8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  je     L2
  lea    (%rbx,%rax,1),%r13
  xor    %ebp,%ebp
  nopl   0x0(%rax)
L18:
  movsbq (%rbx),%rax
  lea    0x0(%rbp,%rbp,4),%rdx
  mov    %r12,%rdi
  add    $0x1,%rbx
  lea    -0x30(%rax,%rdx,2),%rax
  cqto
  idivq  0x8(%rsp)
  mov    %rax,%rsi
  mov    %rdx,%rbp
  call   <printf@plt>
  cmp    %rbx,%r13
  jne    L18
L2:
  mov    0x3f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x408,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_infered: string
name: extract_symbolic_data
arguments: label: D10
type_infered: string

tool: D5: %s

tool: D10: %lld

assistant: ```c
int main() {
  long long int len, i, j, num, div, rem, l;
  char a[1000], total[1000];
  scanf("%s", a);
  scanf("%lld", &num);
  len = strlen(a);
  div = 0;
  rem = 0;
  i = 0;
  l = 0;
  while (i < len) {
    rem = (rem * 10 + a[i] - '0') % num;
    div = (div * 10 + a[i] - '0') / num;
    printf("%lld", div);
    div = rem;
    i++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rsi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  imul   %ecx,%eax
  cmp    %eax,%edx
  jne    L5
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L38
L5:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  imul   %ecx,%eax
  cmp    %eax,%edx
  jle    L29
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L38
L29:
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L38:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D13
datatype: string
name: disassemble_data
arguments: tag: D26
datatype: string
name: disassemble_data
arguments: tag: D36
datatype: string
name: disassemble_data
arguments: tag: D34
datatype: string

tool: D13: %d%d%d%d

tool: D26: 0


tool: D36: -1


tool: D34: 1


assistant: ```c
int main() {
  int C1, P1, C2, P2;
  scanf("%d%d%d%d", &C1, &P1, &C2, &P2);
  if (C1 * P1 == C2 * P2)
    printf("0\n");
  else if (C1 * P1 > C2 * P2)
    printf("-1\n");
  else
    printf("1\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  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
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  imul   0xc(%rsp),%edx
  mov    0x10(%rsp),%eax
  imul   0x14(%rsp),%eax
  cmp    %eax,%edx
  je     L26
  jle    L5
  lea    D22(%rip),%rdi
  call   <printf@plt>
L47:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L26:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  jmp    L47
L5:
  lea    D27(%rip),%rdi
  call   <printf@plt>
  jmp    L47
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D10
type_guessed: string
name: parse_data
arguments: label: D22
type_guessed: string
name: parse_data
arguments: label: D19
type_guessed: string
name: parse_data
arguments: label: D27
type_guessed: string

tool: D10: %d%d%d%d

tool: D22: -1


tool: D19: 0


tool: D27: 1


assistant: ```c
int main() {
  int C1, P1, C2, P2;
  scanf("%d%d%d%d", &C1, &P1, &C2, &P2);
  if (C1 * P1 == C2 * P2)
    printf("0\n");
  else if (C1 * P1 > C2 * P2)
    printf("-1\n");
  else
    printf("1\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D4(%rip),%rdi
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    0x10(%rsp),%eax
  imul   0xc(%rsp),%edx
  imul   0x14(%rsp),%eax
  cmp    %eax,%edx
  je     L28
  jg     L40
  lea    D18(%rip),%rdi
  call   <printf@plt>
L44:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L40:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L44
L28:
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L44
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type: string
name: unpack_data
arguments: data_label: D18
type: string
name: unpack_data
arguments: data_label: D17
type: string
name: unpack_data
arguments: data_label: D20
type: string

tool: D4: %d%d%d%d

tool: D18: 1


tool: D17: -1


tool: D20: 0


assistant: ```c
int main() {
  int C1, P1, C2, P2;
  scanf("%d%d%d%d", &C1, &P1, &C2, &P2);
  if (C1 * P1 == C2 * P2)
    printf("0\n");
  else if (C1 * P1 > C2 * P2)
    printf("-1\n");
  else
    printf("1\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D15(%rip),%rdi
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    0x10(%rsp),%eax
  imul   0xc(%rsp),%edx
  imul   0x14(%rsp),%eax
  cmp    %eax,%edx
  je     L24
  jg     L44
  lea    D30(%rip),%rdi
  call   <printf@plt>
L11:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L44:
  lea    D34(%rip),%rdi
  call   <printf@plt>
  jmp    L11
L24:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L11
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
data_type: string
name: read_data
arguments: tag: D30
data_type: string
name: read_data
arguments: tag: D34
data_type: string
name: read_data
arguments: tag: D4
data_type: string

tool: D15: %d%d%d%d

tool: D30: 1


tool: D34: -1


tool: D4: 0


assistant: ```c
int main() {
  int C1, P1, C2, P2;
  scanf("%d%d%d%d", &C1, &P1, &C2, &P2);
  if (C1 * P1 == C2 * P2)
    printf("0\n");
  else if (C1 * P1 > C2 * P2)
    printf("-1\n");
  else
    printf("1\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dcopy_codegen_4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x14,-0x8(%rbp)
  movl   $0x14,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L14
L13:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L32
L28:
  mov    -0x4(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x10(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x10(%rbp)
L32:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L28
  addl   $0x1,-0xc(%rbp)
L14:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L13
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dcopy_codegen_4(double *A, double *C) {
  const int n = 4 + 0 * 40 + 4 * 4;
  const int lda = 4 + 0 * 40 + 4 * 4;
  int ii, jj;
  for (jj = 0; jj < n; jj++) {
    for (ii = jj; ii < n; ii++) {
      C[ii + lda * jj] = A[ii + lda * jj];
    }
  }
}
```