ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<mult_const_i_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L13:
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L20:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  imul   -0x1c(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L34:
  mov    0x10(%rbp),%eax
  sub    -0x30(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jle    L20
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x2c(%rbp),%eax
  sub    -0x20(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L13
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int **mult_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2 + 1; j++)
      res[i][j] = cn * vec[i][j];
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mult_const_i_2>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    %ecx,%r8d
  js     L28
  mov    %esi,%r10d
  mov    %rdx,%rsi
  mov    %r8d,%r8d
  lea    0x8(%rdx,%r8,8),%rbp
  mov    0x18(%rsp),%r11d
  sub    %r9d,%r11d
  lea    0x1(%r11),%eax
  lea    0x4(,%rax,4),%r8
  jmp    L3
L35:
  mov    (%rdi),%rdx
  mov    (%rsi),%rcx
  mov    %r10d,%r9d
  imul   (%rcx,%rax,1),%r9d
  mov    %r9d,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L35
L51:
  add    $0x8,%rsi
  add    $0x8,%rdi
  cmp    %rbp,%rsi
  je     L28
L3:
  mov    $0x0,%eax
  cmp    $0xffffffff,%r11d
  jge    L35
  jmp    L51
L28:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int **mult_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2 + 1; j++)
      res[i][j] = cn * vec[i][j];
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mult_const_i_2>:
  endbr64
  mov    %rdi,%r11
  sub    %ecx,%r8d
  js     L15
  push   %rbp
  movslq %r8d,%r8
  mov    %esi,%edi
  push   %rbx
  mov    0x18(%rsp),%r10d
  lea    0x8(,%r8,8),%rbp
  mov    %rdx,%rbx
  sub    %r9d,%r10d
  xor    %r9d,%r9d
  lea    0x1(%r10),%r8d
  nopl   (%rax)
L4:
  cmp    $0xffffffff,%r10d
  jl     L10
  mov    (%rbx,%r9,1),%rsi
  mov    (%r11,%r9,1),%rcx
  xor    %eax,%eax
L23:
  mov    (%rsi,%rax,4),%edx
  imul   %edi,%edx
  mov    %edx,(%rcx,%rax,4)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdx,%r8
  jne    L23
L10:
  add    $0x8,%r9
  cmp    %rbp,%r9
  jne    L4
  mov    %r11,%rax
  pop    %rbx
  pop    %rbp
  ret
L15:
  mov    %rdi,%rax
  ret
```

assistant: ```c
int **mult_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2 + 1; j++)
      res[i][j] = cn * vec[i][j];
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mult_const_i_2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    %ecx,%r8d
  js     L61
  mov    0x38(%rsp),%r11d
  mov    %rdx,%rbp
  sub    %r9d,%r11d
  mov    %r11d,%edx
  add    $0x1,%edx
  mov    %edx,-0x14(%rsp)
  js     L61
  movslq %r8d,%r8
  lea    0x2(%r11),%ebx
  mov    %esi,%r10d
  xor    %edi,%edi
  lea    0x8(,%r8,8),%rax
  mov    %ebx,-0x10(%rsp)
  movd   %r10d,%xmm4
  mov    %edx,%r9d
  mov    %rax,-0x20(%rsp)
  mov    %ebx,%eax
  and    $0xfffffffc,%ebx
  pshufd $0x0,%xmm4,%xmm2
  shr    $0x2,%eax
  lea    0x2(%rbx),%r14d
  movdqa %xmm2,%xmm3
  movslq %ebx,%r13
  lea    -0x1(%rax),%esi
  lea    0x1(%rbx),%eax
  movslq %r14d,%r14
  shl    $0x2,%r13
  mov    %eax,-0xc(%rsp)
  add    $0x1,%rsi
  movslq %eax,%r15
  shl    $0x2,%r14
  shl    $0x4,%rsi
  shl    $0x2,%r15
  psrlq  $0x20,%xmm3
  nopl   0x0(%rax,%rax,1)
L59:
  mov    0x0(%rbp,%rdi,1),%rax
  mov    (%r12,%rdi,1),%rdx
  lea    0x4(%rax),%r8
  mov    %rdx,%rcx
  sub    %r8,%rcx
  cmp    $0x8,%rcx
  jbe    L12
  xor    %ecx,%ecx
  cmpl   $0x2,-0x14(%rsp)
  jbe    L12
  nopw   0x0(%rax,%rax,1)
L45:
  movdqu (%rax,%rcx,1),%xmm0
  movdqu (%rax,%rcx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movups %xmm1,(%rdx,%rcx,1)
  add    $0x10,%rcx
  cmp    %rsi,%rcx
  jne    L45
  cmp    %ebx,-0x10(%rsp)
  je     L8
  mov    (%rax,%r13,1),%ecx
  imul   %r10d,%ecx
  mov    %ecx,(%rdx,%r13,1)
  cmp    %ebx,%r11d
  jl     L8
  mov    (%rax,%r15,1),%ecx
  imul   %r10d,%ecx
  mov    %ecx,(%rdx,%r15,1)
  cmp    -0xc(%rsp),%r11d
  jl     L8
  mov    (%rax,%r14,1),%ecx
  imul   %r10d,%ecx
  mov    %ecx,(%rdx,%r14,1)
L8:
  add    $0x8,%rdi
  cmp    -0x20(%rsp),%rdi
  jne    L59
L61:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L12:
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L101:
  mov    (%rax,%rcx,4),%r8d
  imul   %r10d,%r8d
  mov    %r8d,(%rdx,%rcx,4)
  mov    %rcx,%r8
  add    $0x1,%rcx
  cmp    %r9,%r8
  jne    L101
  jmp    L8
```

assistant: ```c
int **mult_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2 + 1; j++)
      res[i][j] = cn * vec[i][j];
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hash_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movq   $0x1505,-0x10(%rbp)
  jmp    L10
L3:
  mov    -0x10(%rbp),%rax
  shl    $0x5,%rax
  add    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbq %al,%rax
  xor    %rax,-0x10(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L3
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long hash_str(const char *key, size_t key_length) {
  char *end = (char *)key + key_length;
  unsigned long h = 5381;
  while (key < end) {
    h += h << 5;
    h ^= (unsigned long)*key++;
  }
  return h;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<hash_str>:
  endbr64
  add    %rdi,%rsi
  cmp    %rsi,%rdi
  jae    L19
  mov    $0x1505,%eax
L18:
  mov    %rax,%rdx
  shl    $0x5,%rdx
  add    %rdx,%rax
  add    $0x1,%rdi
  movsbq -0x1(%rdi),%rdx
  xor    %rdx,%rax
  cmp    %rdi,%rsi
  jne    L18
  ret
L19:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned long hash_str(const char *key, size_t key_length) {
  char *end = (char *)key + key_length;
  unsigned long h = 5381;
  while (key < end) {
    h += h << 5;
    h ^= (unsigned long)*key++;
  }
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash_str>:
  endbr64
  add    %rdi,%rsi
  mov    $0x1505,%eax
  cmp    %rsi,%rdi
  jae    L15
  nopl   0x0(%rax)
L0:
  mov    %rax,%rdx
  add    $0x1,%rdi
  shl    $0x5,%rdx
  add    %rdx,%rax
  movsbq -0x1(%rdi),%rdx
  xor    %rdx,%rax
  cmp    %rdi,%rsi
  jne    L0
  ret
  nopl   0x0(%rax)
L15:
  ret
```

assistant: ```c
unsigned long hash_str(const char *key, size_t key_length) {
  char *end = (char *)key + key_length;
  unsigned long h = 5381;
  while (key < end) {
    h += h << 5;
    h ^= (unsigned long)*key++;
  }
  return h;
}
```
user: What is the c source code of the assembly code below:

```asm
<hash_str>:
  endbr64
  add    %rdi,%rsi
  mov    $0x1505,%eax
  cmp    %rsi,%rdi
  jae    L6
  nopl   0x0(%rax)
L0:
  mov    %rax,%rdx
  add    $0x1,%rdi
  shl    $0x5,%rdx
  add    %rdx,%rax
  movsbq -0x1(%rdi),%rdx
  xor    %rdx,%rax
  cmp    %rdi,%rsi
  jne    L0
  ret
  nopl   0x0(%rax)
L6:
  ret
```

assistant: ```c
unsigned long hash_str(const char *key, size_t key_length) {
  char *end = (char *)key + key_length;
  unsigned long h = 5381;
  while (key < end) {
    h += h << 5;
    h ^= (unsigned long)*key++;
  }
  return h;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xffffffff,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L42
  mov    $0x0,%eax
  call   <abort@plt>
L42:
  mov    $0x0,%edi
  call   <foo@plt>
  test   %eax,%eax
  je     L23
  mov    $0x0,%eax
  call   <abort@plt>
L23:
  mov    $0x1,%edi
  call   <foo@plt>
  cmp    $0x1,%eax
  je     L36
  mov    $0x0,%eax
  call   <abort@plt>
L36:
  mov    $0x2,%edi
  call   <foo@plt>
  cmp    $0x2,%eax
  je     L14
  mov    $0x0,%eax
  call   <abort@plt>
L14:
  mov    $0x3,%edi
  call   <foo@plt>
  cmp    $0x3,%eax
  je     L46
  mov    $0x0,%eax
  call   <abort@plt>
L46:
  mov    $0x4,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L41
  mov    $0x0,%eax
  call   <abort@plt>
L41:
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L34
  mov    $0x0,%eax
  call   <abort@plt>
L34:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (foo(-1) != 4)
    abort();
  if (foo(0) != 0)
    abort();
  if (foo(1) != 1)
    abort();
  if (foo(2) != 2)
    abort();
  if (foo(3) != 3)
    abort();
  if (foo(4) != 4)
    abort();
  if (foo(5) != 4)
    abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  jne    L102
L96:
  mov    $0x0,%edi
  call   <foo@plt>
  test   %eax,%eax
  jne    L15
L22:
  mov    $0x1,%edi
  call   <foo@plt>
  cmp    $0x1,%eax
  jne    L29
L95:
  mov    $0x2,%edi
  call   <foo@plt>
  cmp    $0x2,%eax
  jne    L30
L48:
  mov    $0x3,%edi
  call   <foo@plt>
  cmp    $0x3,%eax
  jne    L115
L66:
  mov    $0x4,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  jne    L104
L14:
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  jne    L3
L57:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L102:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L96
L15:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L22
L29:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L95
L30:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L48
L115:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L66
L104:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L14
L3:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L57
```

assistant: ```c
int main() {
  if (foo(-1) != 4)
    abort();
  if (foo(0) != 0)
    abort();
  if (foo(1) != 1)
    abort();
  if (foo(2) != 2)
    abort();
  if (foo(3) != 3)
    abort();
  if (foo(4) != 4)
    abort();
  if (foo(5) != 4)
    abort();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L28
  xor    %eax,%eax
  call   <abort@plt>
L28:
  xor    %edi,%edi
  call   <foo@plt>
  test   %eax,%eax
  jne    L36
L51:
  mov    $0x1,%edi
  call   <foo@plt>
  cmp    $0x1,%eax
  je     L23
  xor    %eax,%eax
  call   <abort@plt>
L23:
  mov    $0x2,%edi
  call   <foo@plt>
  cmp    $0x2,%eax
  je     L64
  xor    %eax,%eax
  call   <abort@plt>
L64:
  mov    $0x3,%edi
  call   <foo@plt>
  cmp    $0x3,%eax
  je     L6
  xor    %eax,%eax
  call   <abort@plt>
L6:
  mov    $0x4,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L39
  xor    %eax,%eax
  call   <abort@plt>
L39:
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L14
  xor    %eax,%eax
  call   <abort@plt>
L14:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L36:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L51
```

assistant: ```c
int main() {
  if (foo(-1) != 4)
    abort();
  if (foo(0) != 0)
    abort();
  if (foo(1) != 1)
    abort();
  if (foo(2) != 2)
    abort();
  if (foo(3) != 3)
    abort();
  if (foo(4) != 4)
    abort();
  if (foo(5) != 4)
    abort();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L67
  xor    %eax,%eax
  call   <abort@plt>
L67:
  xor    %edi,%edi
  call   <foo@plt>
  test   %eax,%eax
  jne    L71
L58:
  mov    $0x1,%edi
  call   <foo@plt>
  cmp    $0x1,%eax
  je     L75
  xor    %eax,%eax
  call   <abort@plt>
L75:
  mov    $0x2,%edi
  call   <foo@plt>
  cmp    $0x2,%eax
  je     L46
  xor    %eax,%eax
  call   <abort@plt>
L46:
  mov    $0x3,%edi
  call   <foo@plt>
  cmp    $0x3,%eax
  je     L62
  xor    %eax,%eax
  call   <abort@plt>
L62:
  mov    $0x4,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L14
  xor    %eax,%eax
  call   <abort@plt>
L14:
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    $0x4,%eax
  je     L53
  xor    %eax,%eax
  call   <abort@plt>
L53:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L71:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L58
```

assistant: ```c
int main() {
  if (foo(-1) != 4)
    abort();
  if (foo(0) != 0)
    abort();
  if (foo(1) != 1)
    abort();
  if (foo(2) != 2)
    abort();
  if (foo(3) != 3)
    abort();
  if (foo(4) != 4)
    abort();
  if (foo(5) != 4)
    abort();
  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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  cmp    %eax,%edx
  jne    L8
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <valid@plt>
  jmp    L17
L8:
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L17:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D47
data_type: string
name: parse_data
arguments: data_label: D34
data_type: string
name: parse_data
arguments: data_label: D25
data_type: string

tool: D47: %d

tool: D34: %s

tool: D25: Not Valid

assistant: ```c
int main() {
  int m;
  char *pow;
  scanf("%d", &m);
  pow = (char *)malloc((m + 1) * sizeof(char));
  scanf("%s", pow);
  int n;
  char *pow2;
  scanf("%d", &n);
  pow2 = (char *)malloc((n + 1) * sizeof(char));
  scanf("%s", pow2);
  if (n == m) {
    valid(pow, pow2, n);
  } else {
    printf("Not Valid");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D20(%rip),%rbp
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  lea    D5(%rip),%r12
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    (%rsp),%edx
  je     L14
  lea    D13(%rip),%rdi
  call   <printf@plt>
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <valid@plt>
  jmp    L21
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D20
datatype: string
name: read_data
arguments: data_tag: D5
datatype: string
name: read_data
arguments: data_tag: D13
datatype: string

tool: D20: %d

tool: D5: %s

tool: D13: Not Valid

assistant: ```c
int main() {
  int m;
  char *pow;
  scanf("%d", &m);
  pow = (char *)malloc((m + 1) * sizeof(char));
  scanf("%s", pow);
  int n;
  char *pow2;
  scanf("%d", &n);
  pow2 = (char *)malloc((n + 1) * sizeof(char));
  scanf("%s", pow2);
  if (n == m) {
    valid(pow, pow2, n);
  } else {
    printf("Not Valid");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  lea    D1(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    (%rsp),%edx
  je     L15
  lea    D23(%rip),%rdi
  call   <printf@plt>
L25:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <valid@plt>
  jmp    L25
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
datatype: string
name: extract_symbolic_data
arguments: data_label: D1
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string

tool: D17: %s

tool: D1: %d

tool: D23: Not Valid

assistant: ```c
int main() {
  int m;
  char *pow;
  scanf("%d", &m);
  pow = (char *)malloc((m + 1) * sizeof(char));
  scanf("%s", pow);
  int n;
  char *pow2;
  scanf("%d", &n);
  pow2 = (char *)malloc((n + 1) * sizeof(char));
  scanf("%s", pow2);
  if (n == m) {
    valid(pow, pow2, n);
  } else {
    printf("Not Valid");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D22(%rip),%r13
  push   %r12
  lea    D12(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    (%rsp),%edx
  je     L21
  lea    D4(%rip),%rdi
  call   <printf@plt>
L24:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <valid@plt>
  jmp    L24
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D22
datatype: string
name: retrieve_data
arguments: data_tag: D12
datatype: string
name: retrieve_data
arguments: data_tag: D4
datatype: string

tool: D22: %s

tool: D12: %d

tool: D4: Not Valid

assistant: ```c
int main() {
  int m;
  char *pow;
  scanf("%d", &m);
  pow = (char *)malloc((m + 1) * sizeof(char));
  scanf("%s", pow);
  int n;
  char *pow2;
  scanf("%d", &n);
  pow2 = (char *)malloc((n + 1) * sizeof(char));
  scanf("%s", pow2);
  if (n == m) {
    valid(pow, pow2, n);
  } else {
    printf("Not Valid");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D61(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x2,-0xc(%rbp)
  movsd  D75(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x3,-0x10(%rbp)
  jmp    L5
L1:
  mov    -0xc(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0xc(%rbp),%xmm1
  movsd  D11(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L5:
  cmpl   $0x9,-0x10(%rbp)
  jle    L1
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D70
type_guessed: string
name: parse_data
arguments: data_tag: D49
type_guessed: string
name: parse_data
arguments: data_tag: D45
type_guessed: string
name: parse_data
arguments: data_tag: D2
type_guessed: string
name: parse_data
arguments: data_tag: D61
type_guessed: string
name: parse_data
arguments: data_tag: D75
type_guessed: f64
name: parse_data
arguments: data_tag: D11
type_guessed: f64
name: parse_data
arguments: data_tag: D3
type_guessed: string

tool: D70: n e

tool: D49: - -----------

tool: D45: 0 1

tool: D2: 1 2

tool: D61: 2 2.5

tool: D75: 2.5

tool: D11: 1

tool: D3: %d %.9lf


assistant: ```c
int main() {
  printf("n e\n");
  printf("- -----------\n");
  printf("0 1\n");
  printf("1 2\n");
  printf("2 2.5\n");
  int index, temp = 2;
  double e = 2.5;
  for (index = 3; index < 10; index++) {
    temp *= index;
    e += 1.0 / temp;
    printf("%d %.9lf\n", index, e);
  }
  return 0;
}
```