C++代码：
char *ss = "2cc5";来看看stackoverflow里面的一个回答：
“
Why? Well, C and C++ differ in the type of the string literal. In C the type is array of char and in C++ it is constant array of char. In any case, you are not allowed to change the characters of the string literal, so the const in C++ is not really a restriction
but more of a type safety thing. A conversion from const char* to char* is generally not possible without an explicit cast for safety reasons. But for backwards compatibility with C the language C++ still allows assigning a string literal to a char* and gives
you a warning about this conversion being deprecated.
”
简单来说，是因为C++不同于C，C++里面的string是const的char数组，上面的C++代码，右边是一个string，把它赋值给了左边的char数组，所以编译器会报这个警告，注意如果试图对ss数组进行某一位修改：
ss[0] = 'H';我试了下用这句代码修改ss数组某一位，编译可以通过，但是运行会崩溃，因为ss[0]指向const string的第一个元素，而任意试图对const string进行修改的操作都是错误的。
但是，如果重新赋值一个const string给ss，确是可以的：
ss = "Hello";因为这样就相当于新建了一个"Hello"的const string，然后把它的地址复制给ss，也就相当于ss这个指针指向的内存发生了变化而已，并非试图修改原来"2cc5"的const string.
总之，既然都这样提醒了，就别这样干就行了，把char*定义为const的就行了：
const char *ss = "2cc5";