有一个函数的原型是void f(CString &str);
调用处的代码是f("abc");结果编译不过，函数原型改成 void f(const CString &str); 编译就可以通过。
查了一下<<C++ Primer>>，发现nonconst引用类型只可以引用相同的类型的对象，const引用类型才可以引用类型不同的对象。
附上<<C++ Primer>>的解释：
A const reference can be initialized to an object of a different type or to an rvalue , such as a literal constant:
int i = 42;
// legal for const references only
const int &r = 42;
const int &r2 = r + i;
The same initializations are not legal for nonconst references. Rather, they result in compile-time errors. The reason is subtle and warrants an explanation.
This behavior is easiest to understand when we look at what happens when we bind a reference to an object of a different type. If we write
double dval = 3.14;
const int &ri = dval;
the compiler transforms this code into something like this:
int temp = dval; // create temporary int from the double
const int &ri = temp; // bind ri to that temporary
If ri were not const, then we could assign a new value to ri. Doing so would not change dval but would instead change temp. To the programmer expecting that assignments to ri would change dval, it would appear that the change did not work. Allowing only const references to be bound to values requiring temporaries avoids the problem entirely because a const reference is read-only.
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A nonconst reference may be attached only to an object of the same type as the reference itself. |
A const reference may be bound to an object of a different but related type or to an rvalue.