user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D30(%rip),%rsi
  lea    D47(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x8100000080,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x8300000082,%rax
  mov    %rax,0x38(%rsp)
  call   <fopen@plt>
  lea    D26(%rip),%rsi
  lea    D6(%rip),%rdi
  mov    %rax,%r12
  call   <fopen@plt>
  mov    %rax,0x10(%rsp)
  test   %r12,%r12
  je     L81
  test   %rax,%rax
  je     L81
L65:
  lea    0x28(%rsp),%rbp
  mov    %r12,%rcx
  mov    $0x4,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  mov    0x28(%rsp),%r14d
  test   %r14d,%r14d
  lea    0x7(%r14),%edi
  mov    %r14d,0x1c(%rsp)
  cmovns %r14d,%edi
  mov    %edi,%r13d
  and    $0xfffffff8,%edi
  call   <malloc@plt>
  sar    $0x3,%r13d
  mov    %rax,0x8(%rsp)
  cmp    $0x7,%r14d
  jle    L83
  mov    %rax,%rbx
  xor    %r15d,%r15d
  lea    0x30(%rsp),%r14
  cs nopw 0x0(%rax,%rax,1)
L58:
  mov    %r12,%rcx
  mov    $0x8,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  add    $0x1,%r15d
  call   <decrypt@plt>
  mov    %rbx,%rdi
  mov    $0x8,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  add    $0x8,%rbx
  cmp    %r15d,%r13d
  jg     L58
  lea    -0x1(%r13),%eax
  cmpl   $0x7,0x1c(%rsp)
  mov    $0x8,%edx
  mov    0x8(%rsp),%rcx
  lea    0x8(,%rax,8),%rax
  cmovle %rdx,%rax
  lea    (%rcx,%rax,1),%r14
L48:
  mov    0x10(%rsp),%rcx
  mov    0x8(%rsp),%rdi
  mov    %r13d,%edx
  mov    $0x8,%esi
  call   <fwrite@plt>
  testb  $0x7,0x1c(%rsp)
  jne    L80
L26:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x10(%rsp),%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L81:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L65
L80:
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  lea    0x30(%rsp),%rsi
  mov    %rbp,%rdi
  call   <decrypt@plt>
  mov    0x1c(%rsp),%eax
  mov    $0x8,%ecx
  mov    0x8(%rsp),%rdi
  mov    $0x8,%esi
  cltd
  idiv   %ecx
  mov    0x10(%rsp),%rcx
  call   <fwrite@plt>
  jmp    L26
L83:
  mov    0x8(%rsp),%r14
  jmp    L48
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D30
data_type: string
name: disassemble_data
arguments: label: D47
data_type: string
name: disassemble_data
arguments: label: D26
data_type: string
name: disassemble_data
arguments: label: D6
data_type: string
name: disassemble_data
arguments: label: D7
data_type: string

tool: D30: rb

tool: D47: imgno8.enc

tool: D26: wb

tool: D6: imagen2.png

tool: D7: Error: No se pudo abrir el fichero


assistant: ```c
int main() {
  unsigned int v[2], k[4] = {128, 129, 130, 131};
  int n_bloques, tam;
  FILE *f_encriptado, *f_desencriptado;
  unsigned int *ptr2, *ptr2_ref;
  f_encriptado = fopen("imgno8.enc", "rb");
  f_desencriptado = fopen("imagen2.png", "wb");
  if (f_encriptado == NULL || f_desencriptado == NULL) {
    printf("Error: No se pudo abrir el fichero\n");
  }
  fread(v, 1, sizeof(int), f_encriptado);
  tam = *((int *)v);
  n_bloques = tam / 8;
  ptr2 = malloc(sizeof(v) * n_bloques);
  ptr2_ref = ptr2;
  for (int i = 0; i < n_bloques; i++) {
    fread(v, 1, 8, f_encriptado);
    decrypt(v, k);
    memcpy(ptr2, v, 8);
    ptr2 += 2;
  }
  fwrite(ptr2_ref, 8, n_bloques, f_desencriptado);
  if (tam % 8 != 0) {
    memcpy(ptr2, v, 8);
    decrypt(v, k);
    fwrite(ptr2_ref, 8, tam % 8, f_desencriptado);
  }
  fclose(f_encriptado);
  fclose(f_desencriptado);
  free(ptr2_ref);
  return 0;
}
```