ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<_one_mult>:
  endbr64
  mov    %rdi,%r9
  mov    %edx,%r8d
  mov    %rcx,%rdi
  test   %edx,%edx
  je     L5
  cmp    $0x1,%edx
  je     L19
  movslq %esi,%rax
  sub    $0x1,%rax
  add    %rax,%r9
  add    %rax,%rdi
  test   %esi,%esi
  jle    L32
  lea    -0x1(%rsi),%r10d
  xor    %ecx,%ecx
  xor    %eax,%eax
  not    %r10
  nopw   0x0(%rax,%rax,1)
L22:
  movsbl (%r9,%rcx,1),%edx
  imul   %r8d,%edx
  add    %eax,%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  mov    %dl,(%rdi,%rcx,1)
  sub    $0x1,%rcx
  cmp    %rcx,%r10
  jne    L22
  add    %r10,%rdi
  test   %eax,%eax
  je     L32
  mov    %al,(%rdi)
L32:
  ret
  nopl   (%rax)
L5:
  mov    %esi,%edx
  xor    %esi,%esi
  jmp    1060 <memset@plt>
  nopl   0x0(%rax)
L19:
  mov    %esi,%edx
  mov    %r9,%rsi
  jmp    1070 <memcpy@plt>
```

assistant: ```c
void _one_mult(char *num, int size, int digit, char *result) {
  int carry, value;
  char *nptr;
  char *rptr;
  if (digit == 0)
    memset(result, 0, size);
  else {
    if (digit == 1)
      memcpy(result, num, size);
    else {
      nptr = num + size - 1;
      rptr = result + size - 1;
      carry = 0;
      while (size-- > 0) {
        value = *nptr-- * digit + carry;
        *rptr-- = value % 10;
        carry = value / 10;
      }
      if (carry != 0)
        *rptr = carry;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_one_mult>:
  endbr64
  mov    %rdi,%r9
  mov    %edx,%r8d
  mov    %rcx,%rdi
  test   %edx,%edx
  je     L41
  cmp    $0x1,%edx
  je     L33
  movslq %esi,%rax
  sub    $0x1,%rax
  add    %rax,%r9
  add    %rax,%rdi
  test   %esi,%esi
  jle    L10
  lea    -0x1(%rsi),%r10d
  xor    %ecx,%ecx
  xor    %eax,%eax
  not    %r10
  nopw   0x0(%rax,%rax,1)
L19:
  movsbl (%r9,%rcx,1),%edx
  imul   %r8d,%edx
  add    %eax,%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  mov    %dl,(%rdi,%rcx,1)
  sub    $0x1,%rcx
  cmp    %rcx,%r10
  jne    L19
  add    %r10,%rdi
  test   %eax,%eax
  je     L10
  mov    %al,(%rdi)
L10:
  ret
  nopl   (%rax)
L41:
  mov    %esi,%edx
  xor    %esi,%esi
  jmp    1060 <memset@plt>
  nopl   0x0(%rax)
L33:
  mov    %esi,%edx
  mov    %r9,%rsi
  jmp    1070 <memcpy@plt>
```

assistant: ```c
void _one_mult(char *num, int size, int digit, char *result) {
  int carry, value;
  char *nptr;
  char *rptr;
  if (digit == 0)
    memset(result, 0, size);
  else {
    if (digit == 1)
      memcpy(result, num, size);
    else {
      nptr = num + size - 1;
      rptr = result + size - 1;
      carry = 0;
      while (size-- > 0) {
        value = *nptr-- * digit + carry;
        *rptr-- = value % 10;
        carry = value / 10;
      }
      if (carry != 0)
        *rptr = carry;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<lineck>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3837363534333231,%rax
  mov    %rax,-0x12(%rbp)
  movw   $0x39,-0xa(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L15
L22:
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  lea    -0x12(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L11
  mov    $0x0,%eax
  jmp    L27
L11:
  mov    -0x20(%rbp),%rax
  movb   $0x30,(%rax)
L15:
  cmpl   $0x8,-0x24(%rbp)
  jle    L22
  mov    $0x1,%eax
L27:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L34
  call   <__stack_chk_fail@plt>
L34:
  leave
  ret
```

assistant: ```c
int lineck(const char *line) {
  char *chr, test[10] = "123456789";
  int c = 0;
  while (c < 9) {
    if (!(chr = strchr(test, line[c++])))
      return 0;
    *chr = '0';
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lineck>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x3837363534333231,%rax
  mov    %rax,0xe(%rsp)
  movw   $0x39,0x16(%rsp)
  mov    %rdi,%rbx
  lea    0x9(%rdi),%rbp
  lea    0xe(%rsp),%r12
L31:
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L23
  movb   $0x30,(%rax)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L31
  mov    $0x1,%eax
L6:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L38
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  mov    $0x0,%eax
  jmp    L6
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int lineck(const char *line) {
  char *chr, test[10] = "123456789";
  int c = 0;
  while (c < 9) {
    if (!(chr = strchr(test, line[c++])))
      return 0;
    *chr = '0';
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lineck>:
  endbr64
  push   %r12
  lea    0x9(%rdi),%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x3837363534333231,%rax
  lea    0xe(%rsp),%rbp
  mov    %rax,0xe(%rsp)
  mov    $0x39,%eax
  mov    %ax,0x16(%rsp)
  nop
L39:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L8
  add    $0x1,%rbx
  movb   $0x30,(%rax)
  cmp    %r12,%rbx
  jne    L39
  mov    $0x1,%eax
L19:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  jmp    L19
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int lineck(const char *line) {
  char *chr, test[10] = "123456789";
  int c = 0;
  while (c < 9) {
    if (!(chr = strchr(test, line[c++])))
      return 0;
    *chr = '0';
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lineck>:
  endbr64
  push   %r12
  lea    0x9(%rdi),%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x3837363534333231,%rax
  lea    0xe(%rsp),%rbp
  mov    %rax,0xe(%rsp)
  mov    $0x39,%eax
  mov    %ax,0x16(%rsp)
  nop
L32:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L3
  add    $0x1,%rbx
  movb   $0x30,(%rax)
  cmp    %r12,%rbx
  jne    L32
  mov    $0x1,%eax
L27:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L35
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  jmp    L27
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int lineck(const char *line) {
  char *chr, test[10] = "123456789";
  int c = 0;
  while (c < 9) {
    if (!(chr = strchr(test, line[c++])))
      return 0;
    *chr = '0';
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sc_strcat_ex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <sc_strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <sc_strlen@plt>
  mov    %eax,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L48
  cmpq   $0x0,-0x20(%rbp)
  je     L48
  movl   $0x0,-0xc(%rbp)
  jmp    L7
L58:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L7:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L58
  jmp    L49
L38:
  mov    -0xc(%rbp),%eax
  sub    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L49:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L38
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L23
L48:
  nop
L23:
  leave
  ret
```

assistant: ```c
void sc_strcat_ex(char *A, char *B, char *C) {
  int walker = 0;
  int length_of_A = sc_strlen(A);
  int length_of_B = sc_strlen(B);
  if (!A || !B)
    return;
  for (walker = 0; walker < length_of_A; walker++)
    C[walker] = A[walker];
  for (; walker < length_of_A + length_of_B; walker++)
    C[walker] = B[walker - length_of_A];
  C[walker] = '\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
<sc_strcat_ex>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  call   <sc_strlen@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <sc_strlen@plt>
  test   %r13,%r13
  je     L70
  test   %rbp,%rbp
  je     L70
  test   %r12d,%r12d
  jle    L62
  mov    %r12d,%esi
  mov    $0x0,%edx
L45:
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L45
  mov    %r12d,%edx
L56:
  add    %r12d,%eax
  cmp    %edx,%eax
  jle    L55
  movslq %edx,%rdx
  movslq %r12d,%r12
  sub    %r12,%rbp
L60:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L60
L25:
  cltq
  movb   $0x0,(%rbx,%rax,1)
L70:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L62:
  mov    $0x0,%edx
  jmp    L56
L55:
  mov    %edx,%eax
  jmp    L25
```

assistant: ```c
void sc_strcat_ex(char *A, char *B, char *C) {
  int walker = 0;
  int length_of_A = sc_strlen(A);
  int length_of_B = sc_strlen(B);
  if (!A || !B)
    return;
  for (walker = 0; walker < length_of_A; walker++)
    C[walker] = A[walker];
  for (; walker < length_of_A + length_of_B; walker++)
    C[walker] = B[walker - length_of_A];
  C[walker] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sc_strcat_ex>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <sc_strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <sc_strlen@plt>
  test   %r12,%r12
  je     L65
  test   %rbp,%rbp
  je     L65
  test   %r13d,%r13d
  jle    L45
  movslq %r13d,%rsi
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L30:
  movzbl (%r12,%rdx,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L30
  add    %rbx,%rdx
  mov    %r13d,%ecx
L15:
  add    %r13d,%eax
  cmp    %eax,%ecx
  jge    L68
  movslq %r13d,%rsi
  movslq %ecx,%rdx
  sub    %rsi,%rbp
  nopw   0x0(%rax,%rax,1)
L39:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L39
  cltq
  lea    (%rbx,%rax,1),%rdx
L68:
  movb   $0x0,(%rdx)
L65:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L45:
  mov    %rbx,%rdx
  xor    %ecx,%ecx
  jmp    L15
```

assistant: ```c
void sc_strcat_ex(char *A, char *B, char *C) {
  int walker = 0;
  int length_of_A = sc_strlen(A);
  int length_of_B = sc_strlen(B);
  if (!A || !B)
    return;
  for (walker = 0; walker < length_of_A; walker++)
    C[walker] = A[walker];
  for (; walker < length_of_A + length_of_B; walker++)
    C[walker] = B[walker - length_of_A];
  C[walker] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<y_m_d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <int2unico@plt>
  mov    -0x18(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <int2unico@plt>
  mov    -0x18(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <int2unico@plt>
  mov    -0x18(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_infered: string
name: retrieve_data
arguments: data_tag: D8
type_infered: string
name: retrieve_data
arguments: data_tag: D5
type_infered: string

tool: D2: 5E74

tool: D8: 6708

tool: D5: 65E5

assistant: ```c
void y_m_d(int y, int m, int d, char ch[]) {
  int2unico(y, ch);
  strcat(ch, "5E74");
  int2unico(m, ch);
  strcat(ch, "6708");
  int2unico(d, ch);
  strcat(ch, "65E5");
}
```
user: What is the c source code of the assembly code below:

```asm
<y_m_d>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r12d
  mov    %edx,%ebp
  mov    %rcx,%rbx
  mov    %rcx,%rsi
  call   <int2unico@plt>
  lea    D7(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <int2unico@plt>
  lea    D24(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <int2unico@plt>
  lea    D28(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type_guessed: string
name: parse_data
arguments: label: D24
type_guessed: string
name: parse_data
arguments: label: D28
type_guessed: string

tool: D7: 5E74

tool: D24: 6708

tool: D28: 65E5

assistant: ```c
void y_m_d(int y, int m, int d, char ch[]) {
  int2unico(y, ch);
  strcat(ch, "5E74");
  int2unico(m, ch);
  strcat(ch, "6708");
  int2unico(d, ch);
  strcat(ch, "65E5");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<y_m_d>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  mov    %rcx,%rsi
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rcx,%rbp
  call   <int2unico@plt>
  mov    %rbp,%rdi
  lea    D3(%rip),%rsi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <int2unico@plt>
  mov    %rbp,%rdi
  lea    D8(%rip),%rsi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <int2unico@plt>
  mov    %rbp,%rdi
  lea    D5(%rip),%rsi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <strcat@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D8
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string

tool: D3: 5E74

tool: D8: 6708

tool: D5: 65E5

assistant: ```c
void y_m_d(int y, int m, int d, char ch[]) {
  int2unico(y, ch);
  strcat(ch, "5E74");
  int2unico(m, ch);
  strcat(ch, "6708");
  int2unico(d, ch);
  strcat(ch, "65E5");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<y_m_d>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  mov    %rcx,%rsi
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rcx,%rbp
  call   <int2unico@plt>
  mov    %rbp,%rdi
  lea    D28(%rip),%rsi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <int2unico@plt>
  mov    %rbp,%rdi
  lea    D2(%rip),%rsi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <int2unico@plt>
  mov    %rbp,%rdi
  lea    D5(%rip),%rsi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <strcat@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D28
type_infered: string
name: extract_symbolic_data
arguments: data_label: D2
type_infered: string
name: extract_symbolic_data
arguments: data_label: D5
type_infered: string

tool: D28: 5E74

tool: D2: 6708

tool: D5: 65E5

assistant: ```c
void y_m_d(int y, int m, int d, char ch[]) {
  int2unico(y, ch);
  strcat(ch, "5E74");
  int2unico(m, ch);
  strcat(ch, "6708");
  int2unico(d, ch);
  strcat(ch, "65E5");
}
```
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
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L19
L49:
  cmpl   $0x1,-0x10(%rbp)
  jg     L42
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L38
L42:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L38:
  addl   $0x1,-0x10(%rbp)
L19:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L49
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D16
datatype: string
name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D4
datatype: string
name: disassemble_data
arguments: data_tag: D22
datatype: string

tool: D16: Bir sayi giriniz = 

tool: D24: %d

tool: D4: %d %d 

tool: D22: %d 

assistant: ```c
int main() {
  int sayi;
  int m = 0;
  int e = 1;
  int k = 1;
  printf("Bir sayi giriniz = ");
  scanf("%d", &sayi);
  int i;
  for (i = 1; i < sayi; i++) {
    if (i < 2) {
      printf("%d %d ", e, k);
    } else {
      m = e + k;
      printf("%d ", m);
      k = e;
      e = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L58
  mov    $0x1,%ebx
  mov    $0x1,%r13d
  mov    $0x1,%ebp
  lea    D1(%rip),%r14
  lea    D9(%rip),%r15
  jmp    L25
L15:
  lea    0x0(%rbp,%r13,1),%r12d
  mov    %r12d,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%r13d
  mov    %r12d,%ebp
L10:
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jle    L58
L25:
  cmp    $0x1,%ebx
  jg     L15
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L10
L58:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D39
datatype: string
name: retrieve_data
arguments: data_tag: D25
datatype: string
name: retrieve_data
arguments: data_tag: D1
datatype: string
name: retrieve_data
arguments: data_tag: D9
datatype: string

tool: D39: Bir sayi giriniz = 

tool: D25: %d

tool: D1: %d 

tool: D9: %d %d 

assistant: ```c
int main() {
  int sayi;
  int m = 0;
  int e = 1;
  int k = 1;
  printf("Bir sayi giriniz = ");
  scanf("%d", &sayi);
  int i;
  for (i = 1; i < sayi; i++) {
    if (i < 2) {
      printf("%d %d ", e, k);
    } else {
      m = e + k;
      printf("%d ", m);
      k = e;
      e = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D28(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L56
  mov    $0x1,%ebx
  mov    $0x1,%r13d
  mov    $0x1,%ebp
  lea    D7(%rip),%r14
  lea    D22(%rip),%r15
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L49:
  lea    0x0(%rbp,%r13,1),%r12d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %ebp,%r13d
  mov    %r12d,%esi
  mov    %r12d,%ebp
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jle    L56
L6:
  cmp    $0x1,%ebx
  jne    L49
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jg     L6
L56:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D28
type_guessed: string
name: parse_data
arguments: tag: D33
type_guessed: string
name: parse_data
arguments: tag: D7
type_guessed: string
name: parse_data
arguments: tag: D22
type_guessed: string

tool: D28: Bir sayi giriniz = 

tool: D33: %d

tool: D7: %d 

tool: D22: %d %d 

assistant: ```c
int main() {
  int sayi;
  int m = 0;
  int e = 1;
  int k = 1;
  printf("Bir sayi giriniz = ");
  scanf("%d", &sayi);
  int i;
  for (i = 1; i < sayi; i++) {
    if (i < 2) {
      printf("%d %d ", e, k);
    } else {
      m = e + k;
      printf("%d ", m);
      k = e;
      e = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  lea    D9(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L21
  mov    $0x1,%ebx
  mov    $0x1,%r13d
  mov    $0x1,%ebp
  lea    D18(%rip),%r15
  lea    D2(%rip),%r14
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L20:
  mov    %r12d,%ebp
L5:
  cmp    $0x1,%ebx
  jne    L25
  xor    %eax,%eax
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r15,%rdi
  call   <printf@plt>
  cmpl   $0x2,0x4(%rsp)
  jle    L21
  mov    $0x2,%ebx
L25:
  lea    0x0(%rbp,%r13,1),%r12d
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  mov    %r12d,%esi
  mov    %ebp,%r13d
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L20
L21:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
datatype: string
name: read_data
arguments: data_tag: D39
datatype: string
name: read_data
arguments: data_tag: D18
datatype: string
name: read_data
arguments: data_tag: D2
datatype: string

tool: D9: Bir sayi giriniz = 

tool: D39: %d

tool: D18: %d %d 

tool: D2: %d 

assistant: ```c
int main() {
  int sayi;
  int m = 0;
  int e = 1;
  int k = 1;
  printf("Bir sayi giriniz = ");
  scanf("%d", &sayi);
  int i;
  for (i = 1; i < sayi; i++) {
    if (i < 2) {
      printf("%d %d ", e, k);
    } else {
      m = e + k;
      printf("%d ", m);
      k = e;
      e = m;
    }
  }
  printf("\n");
  return 0;
}
```