ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D10(%rip),%rdi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L48
  xor    %ebp,%ebp
  mov    $0x2,%r12d
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L30:
  mov    %eax,%r12d
L18:
  mov    %r12d,%edi
  call   <IsPrime@plt>
  cmp    $0x1,%rax
  lea    0x1(%r12),%eax
  sbb    $0xffffffff,%ebp
  cmp    %ebp,0x4(%rsp)
  jg     L30
L28:
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  mov    %r12d,%edx
  mov    %ebp,%esi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L48:
  mov    $0x1,%r12d
  xor    %ebp,%ebp
  jmp    L28
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
data_type: string
name: unpack_data
arguments: label: D38
data_type: string
name: unpack_data
arguments: label: D20
data_type: string
name: unpack_data
arguments: label: D13
data_type: string

tool: D10: Enter an ordinal number: 

tool: D38: %d

tool: D20: The %dth prime is %d.


tool: D13: PAUSE

assistant: ```c
int main(void) {
  int i, flag, prime_counter = 0, nth_prime;
  printf("Enter an ordinal number: ");
  scanf("%d", &nth_prime);
  i = 2;
  flag = 1;
  while (flag && prime_counter < nth_prime) {
    if (IsPrime(i))
      prime_counter++;
    i++;
  }
  printf("The %dth prime is %d.\n", prime_counter, --i);
  system("PAUSE");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D33(%rip),%rdi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L9
  xor    %ebp,%ebp
  mov    $0x2,%r12d
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %eax,%r12d
L1:
  mov    %r12d,%edi
  call   <IsPrime@plt>
  cmp    $0x1,%rax
  lea    0x1(%r12),%eax
  sbb    $0xffffffff,%ebp
  cmp    %ebp,0x4(%rsp)
  jg     L27
L35:
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  mov    %r12d,%edx
  mov    %ebp,%esi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L9:
  mov    $0x1,%r12d
  xor    %ebp,%ebp
  jmp    L35
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D33
data_type: string
name: disassemble_data
arguments: data_tag: D35
data_type: string
name: disassemble_data
arguments: data_tag: D9
data_type: string
name: disassemble_data
arguments: data_tag: D32
data_type: string

tool: D33: Enter an ordinal number: 

tool: D35: %d

tool: D9: The %dth prime is %d.


tool: D32: PAUSE

assistant: ```c
int main(void) {
  int i, flag, prime_counter = 0, nth_prime;
  printf("Enter an ordinal number: ");
  scanf("%d", &nth_prime);
  i = 2;
  flag = 1;
  while (flag && prime_counter < nth_prime) {
    if (IsPrime(i))
      prime_counter++;
    i++;
  }
  printf("The %dth prime is %d.\n", prime_counter, --i);
  system("PAUSE");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<make_md5_salt_para>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L3
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x10,%eax
  jg     L3
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  lea    D28(%rip),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    D22(%rip),%rax
  jmp    L0
L3:
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x0,%eax
L0:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D28
data_type: string
name: extract_symbolic_data
arguments: tag: D39
data_type: string

tool: D28: $1$%s$

tool: D39: Invalid Salt, please use up to 16 random alphanumeric characters


assistant: ```c
char *make_md5_salt_para(char *saltpara) {
  static char salt[21];
  if (saltpara && (strlen(saltpara) <= 16)) {
    sprintf(salt, "$1$%s$", saltpara);
    return salt;
  }
  printf("Invalid Salt, please use up to 16 random alphanumeric characters\n");
  exit(1);
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<make_md5_salt_para>:
  endbr64
  push   %rbx
  test   %rdi,%rdi
  je     L26
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmp    $0x10,%eax
  jle    L9
L26:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x0,%eax
L21:
  pop    %rbx
  ret
L9:
  mov    %rbx,%rdx
  lea    D6(%rip),%rsi
  lea    D10(%rip),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  jmp    L21
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D24
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D6
type_guessed: string

tool: D24: Invalid Salt, please use up to 16 random alphanumeric characters


tool: D6: $1$%s$

assistant: ```c
char *make_md5_salt_para(char *saltpara) {
  static char salt[21];
  if (saltpara && (strlen(saltpara) <= 16)) {
    sprintf(salt, "$1$%s$", saltpara);
    return salt;
  }
  printf("Invalid Salt, please use up to 16 random alphanumeric characters\n");
  exit(1);
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<make_md5_salt_para>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  je     L11
  mov    %rdi,%r12
  call   <strlen@plt>
  cmp    $0x10,%eax
  jle    L5
L11:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r12,%rdx
  lea    D3(%rip),%r12
  lea    D23(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D19
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string

tool: D19: Invalid Salt, please use up to 16 random alphanumeric characters


tool: D23: $1$%s$

assistant: ```c
char *make_md5_salt_para(char *saltpara) {
  static char salt[21];
  if (saltpara && (strlen(saltpara) <= 16)) {
    sprintf(salt, "$1$%s$", saltpara);
    return salt;
  }
  printf("Invalid Salt, please use up to 16 random alphanumeric characters\n");
  exit(1);
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<make_md5_salt_para>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  je     L4
  mov    %rdi,%r12
  call   <strlen@plt>
  cmp    $0x10,%eax
  jle    L13
L4:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r12,%rdx
  lea    D21(%rip),%r12
  lea    D8(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type: string
name: extract_symbolic_data
arguments: data_tag: D8
type: string

tool: D7: Invalid Salt, please use up to 16 random alphanumeric characters


tool: D8: $1$%s$

assistant: ```c
char *make_md5_salt_para(char *saltpara) {
  static char salt[21];
  if (saltpara && (strlen(saltpara) <= 16)) {
    sprintf(salt, "$1$%s$", saltpara);
    return salt;
  }
  printf("Invalid Salt, please use up to 16 random alphanumeric characters\n");
  exit(1);
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L19:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <fun@plt>
L13:
  addl   $0x1,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L19
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 1, j = 0;
  while (++i) {
    fun(i);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%ebx
L3:
  mov    %ebx,%edi
  call   <fun@plt>
  add    $0x1,%ebx
  jne    L3
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 1, j = 0;
  while (++i) {
    fun(i);
  }
  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    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L8:
  add    $0x1,%ebx
  mov    %ebx,%edi
  call   <fun@plt>
  jmp    L8
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 1, j = 0;
  while (++i) {
    fun(i);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L5:
  add    $0x1,%ebx
  mov    %ebx,%edi
  call   <fun@plt>
  jmp    L5
```

assistant: ```c
int main(int argc, const char *argv[]) {
  int i = 1, j = 0;
  while (++i) {
    fun(i);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<foo2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  movsd  %xmm0,-0x10(%rbp)
  movsd  %xmm1,-0x18(%rbp)
  movsd  %xmm2,-0x20(%rbp)
  movsd  %xmm3,-0x28(%rbp)
  movsd  %xmm4,-0x30(%rbp)
  movsd  %xmm5,-0x38(%rbp)
  movsd  %xmm6,-0x40(%rbp)
  movsd  %xmm7,-0x48(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm2
  movsd  -0x28(%rbp),%xmm3
  movsd  -0x30(%rbp),%xmm4
  movsd  -0x38(%rbp),%xmm5
  movsd  -0x40(%rbp),%xmm6
  movsd  -0x48(%rbp),%xmm7
  mov    -0x8(%rbp),%rax
  call   *%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long foo2(unsigned long long addr, double a0, double a1, double a2, double a3, double a4, double a5, double a6, double a7) {
  register double b0 asm("xmm0");
  register double b1 asm("xmm1");
  register double b2 asm("xmm2");
  register double b3 asm("xmm3");
  register double b4 asm("xmm4");
  register double b5 asm("xmm5");
  register double b6 asm("xmm6");
  register double b7 asm("xmm7");
  register unsigned long long result asm("rax");
  b0 = a0;
  b1 = a1;
  b2 = a2;
  b3 = a3;
  b4 = a4;
  b5 = a5;
  b6 = a6;
  b7 = a7;
  asm("call *%1" : "=r"(result) : "r"(addr), "x"(b0), "x"(b1), "x"(b2), "x"(b3), "x"(b4), "x"(b5), "x"(b6), "x"(b7));
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo2>:
  endbr64
  call   *%rdi
  ret
```

assistant: ```c
unsigned long long foo2(unsigned long long addr, double a0, double a1, double a2, double a3, double a4, double a5, double a6, double a7) {
  register double b0 asm("xmm0");
  register double b1 asm("xmm1");
  register double b2 asm("xmm2");
  register double b3 asm("xmm3");
  register double b4 asm("xmm4");
  register double b5 asm("xmm5");
  register double b6 asm("xmm6");
  register double b7 asm("xmm7");
  register unsigned long long result asm("rax");
  b0 = a0;
  b1 = a1;
  b2 = a2;
  b3 = a3;
  b4 = a4;
  b5 = a5;
  b6 = a6;
  b7 = a7;
  asm("call *%1" : "=r"(result) : "r"(addr), "x"(b0), "x"(b1), "x"(b2), "x"(b3), "x"(b4), "x"(b5), "x"(b6), "x"(b7));
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<foo2>:
  endbr64
  call   *%rdi
  ret
```

assistant: ```c
unsigned long long foo2(unsigned long long addr, double a0, double a1, double a2, double a3, double a4, double a5, double a6, double a7) {
  register double b0 asm("xmm0");
  register double b1 asm("xmm1");
  register double b2 asm("xmm2");
  register double b3 asm("xmm3");
  register double b4 asm("xmm4");
  register double b5 asm("xmm5");
  register double b6 asm("xmm6");
  register double b7 asm("xmm7");
  register unsigned long long result asm("rax");
  b0 = a0;
  b1 = a1;
  b2 = a2;
  b3 = a3;
  b4 = a4;
  b5 = a5;
  b6 = a6;
  b7 = a7;
  asm("call *%1" : "=r"(result) : "r"(addr), "x"(b0), "x"(b1), "x"(b2), "x"(b3), "x"(b4), "x"(b5), "x"(b6), "x"(b7));
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo2>:
  endbr64
  call   *%rdi
  ret
```

assistant: ```c
unsigned long long foo2(unsigned long long addr, double a0, double a1, double a2, double a3, double a4, double a5, double a6, double a7) {
  register double b0 asm("xmm0");
  register double b1 asm("xmm1");
  register double b2 asm("xmm2");
  register double b3 asm("xmm3");
  register double b4 asm("xmm4");
  register double b5 asm("xmm5");
  register double b6 asm("xmm6");
  register double b7 asm("xmm7");
  register unsigned long long result asm("rax");
  b0 = a0;
  b1 = a1;
  b2 = a2;
  b3 = a3;
  b4 = a4;
  b5 = a5;
  b6 = a6;
  b7 = a7;
  asm("call *%1" : "=r"(result) : "r"(addr), "x"(b0), "x"(b1), "x"(b2), "x"(b3), "x"(b4), "x"(b5), "x"(b6), "x"(b7));
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<poly1305aes_ppro_authenticate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8d,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <aes_ppro@plt>
  mov    -0x30(%rbp),%rax
  lea    0x10(%rax),%rsi
  mov    -0x44(%rbp),%edi
  mov    -0x40(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <poly1305_ppro@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void poly1305aes_ppro_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_ppro(aeskn, (kr + 0), n);
  poly1305_ppro(out, (kr + 16), aeskn, m, l);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<poly1305aes_ppro_authenticate>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rcx,%r12
  mov    %r8d,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_ppro@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_ppro@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_ppro_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_ppro(aeskn, (kr + 0), n);
  poly1305_ppro(out, (kr + 16), aeskn, m, l);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<poly1305aes_ppro_authenticate>:
  endbr64
  push   %r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_ppro@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_ppro@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_ppro_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_ppro(aeskn, (kr + 0), n);
  poly1305_ppro(out, (kr + 16), aeskn, m, l);
}
```
user: What is the c source code of the assembly code below?

```asm
<poly1305aes_ppro_authenticate>:
  endbr64
  push   %r14
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <aes_ppro@plt>
  lea    0x10(%rbx),%rsi
  mov    %r13d,%r8d
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rdi
  call   <poly1305_ppro@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void poly1305aes_ppro_authenticate(unsigned char out[16], const unsigned char kr[32], const unsigned char n[16], const unsigned char m[], unsigned int l) {
  unsigned char aeskn[16];
  aes_ppro(aeskn, (kr + 0), n);
  poly1305_ppro(out, (kr + 16), aeskn, m, l);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L56:
  mov    -0xc(%rbp),%eax
  shl    $0x2,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L49
L4:
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L49:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  test   %eax,%eax
  jne    L4
  cmpl   $0x1,-0x4(%rbp)
  jg     L7
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L21
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
  movl   $0x0,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  jmp    L7
L21:
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L7:
  cmpl   $0x4,-0x8(%rbp)
  jle    L56
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D36
type_infered: string
name: extract_symbolic_data
arguments: label: D30
type_infered: string
name: extract_symbolic_data
arguments: label: D22
type_infered: string
name: extract_symbolic_data
arguments: label: D18
type_infered: string

tool: D36:  

tool: D30: |

tool: D22: ____


tool: D18: 


assistant: ```c
int main() {
  int j = 0;
  int spaces = 0;
  int steps = 0;
  int step = 0;
  while (steps < 5) {
    j = spaces * 4;
    while (j--) {
      printf(" ");
    }
    if (step < 2) {
      printf("|");
      step++;
      if (!(step % 2)) {
        printf("____\n");
        spaces++;
        step = 0;
        steps++;
      } else {
        printf("\n");
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%r12d
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  lea    D33(%rip),%rbp
L13:
  lea    0x0(,%r13,4),%r15d
  jmp    L47
L87:
  cmp    $0x1,%r12d
  jle    L70
L47:
  cmp    $0x4,%r14d
  jg     L30
  test   %r13d,%r13d
  je     L87
  mov    %r15d,%ebx
L14:
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L14
  jmp    L87
L70:
  lea    D29(%rip),%rdi
  call   <printf@plt>
  add    $0x1,%r12d
  test   $0x1,%r12b
  jne    L9
  lea    D11(%rip),%rdi
  call   <printf@plt>
  add    $0x1,%r13d
  add    $0x1,%r14d
  mov    $0x0,%r12d
  jmp    L13
L9:
  lea    D28(%rip),%rdi
  call   <printf@plt>
  jmp    L13
L30:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D33
type: string
name: retrieve_data
arguments: data_label: D29
type: string
name: retrieve_data
arguments: data_label: D11
type: string
name: retrieve_data
arguments: data_label: D28
type: string

tool: D33:  

tool: D29: |

tool: D11: ____


tool: D28: 


assistant: ```c
int main() {
  int j = 0;
  int spaces = 0;
  int steps = 0;
  int step = 0;
  while (steps < 5) {
    j = spaces * 4;
    while (j--) {
      printf(" ");
    }
    if (step < 2) {
      printf("|");
      step++;
      if (!(step % 2)) {
        printf("____\n");
        spaces++;
        step = 0;
        steps++;
      } else {
        printf("\n");
      }
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D30(%rip),%r15
  push   %r14
  lea    D15(%rip),%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
L33:
  lea    0x0(,%r12,4),%ebx
  cmp    $0x5,%r12d
  je     L3
  test   %r12d,%r12d
  je     L1
L42:
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L42
L1:
  mov    %r14,%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  je     L7
  lea    D10(%rip),%rdi
  add    $0x1,%r12d
  xor    %r13d,%r13d
  call   <printf@plt>
  jmp    L33
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  mov    %r15,%rdi
  mov    $0x1,%r13d
  call   <printf@plt>
  jmp    L33
```

tool_calls: name: parse_data
arguments: data_label: D30
type_infered: string
name: parse_data
arguments: data_label: D15
type_infered: string
name: parse_data
arguments: data_label: D6
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string

tool: D30: 


tool: D15: |

tool: D6:  

tool: D10: ____


assistant: ```c
int main() {
  int j = 0;
  int spaces = 0;
  int steps = 0;
  int step = 0;
  while (steps < 5) {
    j = spaces * 4;
    while (j--) {
      printf(" ");
    }
    if (step < 2) {
      printf("|");
      step++;
      if (!(step % 2)) {
        printf("____\n");
        spaces++;
        step = 0;
        steps++;
      } else {
        printf("\n");
      }
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r15
  lea    D2(%rip),%r15
  push   %r14
  lea    D36(%rip),%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D17(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
L31:
  lea    0x0(,%r12,4),%ebx
  cmp    $0x5,%r12d
  je     L28
  test   %r12d,%r12d
  je     L52
L50:
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L50
L52:
  mov    %r14,%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  je     L20
  lea    D26(%rip),%rdi
  add    $0x1,%r12d
  xor    %r13d,%r13d
  call   <printf@plt>
  jmp    L31
L28:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  mov    %r15,%rdi
  mov    $0x1,%r13d
  call   <printf@plt>
  jmp    L31
```

tool_calls: name: unpack_data
arguments: data_tag: D2
data_type: string
name: unpack_data
arguments: data_tag: D36
data_type: string
name: unpack_data
arguments: data_tag: D17
data_type: string
name: unpack_data
arguments: data_tag: D26
data_type: string

tool: D2: 


tool: D36: |

tool: D17:  

tool: D26: ____


assistant: ```c
int main() {
  int j = 0;
  int spaces = 0;
  int steps = 0;
  int step = 0;
  while (steps < 5) {
    j = spaces * 4;
    while (j--) {
      printf(" ");
    }
    if (step < 2) {
      printf("|");
      step++;
      if (!(step % 2)) {
        printf("____\n");
        spaces++;
        step = 0;
        steps++;
      } else {
        printf("\n");
      }
    }
  }
  return 0;
}
```