c++ - IDA Ctrl+F5生成的伪代码中字符串是几进制的？

Question

比如这个字节，它的值是几进制的？该如何转为正常的字符串？

char *sub_1386F()
{
  if ( byte_146C8 != 104 )
  {
    byte_146C8 = 104;
    byte_146C9 = 116;
    byte_146CA = 116;
    byte_146CB = 112;
    byte_146CC = 58;
  }
  return &byte_146C8;
}

麻烦懂的大哥解释下这个伪代码。谢谢

Answer 1

Actually, it is a decimal ASCII code. In C++, if you want to convert the ASCII code value into characters, you can use "" (backslash) + ASCII code.
For example:

#include<iostream>
using namespace std;
int main(){
   char a;//定义一个char类型变量
   a=104;
   cout<<a<<endl;
   cout<<"4"<<endl; //你在运行时会发现两个都是一样的
}