C++ 字符串加字符
```
cout << ("s" + 'a');

```
这样一段代码，为什么输出是一个空行呢？
难道是 `string literal + character` 没有这个运算吗？
没有的话为什么输出是空行呢？
C++ 中， `string literal` 就是 `const char[n]` 吗？
表达式`"s"`的类型是`const char[]`，表达式 `'a'`的类型是`char`。
数组是不能进行内置加法运算的。所以`const char []`会被转换成`const char *`，这里的运算就变成了"指针+整型"(`char`是一种整型)。输出空行的运行结果实际上是数组越界引起的。
String literal
Narrow multibyte string literal. The type of an unprefixed string literal is const char[]
Additive operators
addition. For the built-in operator, lhs and rhs must be one of the following: Both have arithmetic or unscoped enumeration type. In this case, the usual arithmetic conversions are performed on both operands and determine the type of the result. One is a pointer to complete object type, the other has integral or unscoped enumeration type. In this case, the result type has the type of the pointer
Array-to-pointer decay
There is an implicit conversion from lvalues and rvalues of array type to rvalues of pointer type: it constructs a pointer to the first element of an array. This conversion is used whenever arrays appear in context where arrays are not expected, but pointers are.