Why can I assign an existing reference to a literal value in C++?

Question

Consider the following:

int ival = 1.01;
int &rval = 1.01; // error: non-const reference to a const value.
int &rval = ival;

rval = 1.01;

The first assignment of &rval to a literal value fails as expected. If I comment out that line the code compiles and runs. I understand why the initialization fails, but I'm confused why the assignment to rval works in the last line. I didn't think it was allowed to assign a reference to a literal value.

EDIT: Thanks for the quick answers. I'm tempted to delete this to hide my shame, but I think I'll leave it here so everyone else can point and laugh.

In my own defense, I'm working through the exercises in a book (C++ Primer) and this problem is about reference initialization. Still, it's pretty embarrassing to have so completely overlooked the point of a reference in the first place. :)

Answer 1

ival isn't a literal value, 1.01 is the literal value. It's been copied to ival which is a variable, which most definitely can have it's references assigned to another variable.