是否使用auto声明变量并使用原始文字定义的行为进行初始化？

Question

If I initialise a variable declared using auto with a primitive literal, are the results defined eg

auto i = 6; // Is this always going to evaluate to a int?
            // Or could it evaluate to some similar type like short?

Answer 1

The type of integer literal 6 is int , as you can check here .

Then, from the cppreference.com :

For variables, auto specifies that the type of the variable that is being declared will be automatically deduced from its initializer.

So i will have type of int , because 6 has type of int .

And this behaviour is absolutely well-defined.

Answer 2

Yes, that's exactly what auto is for.

For example, auto i; wouldn't make sense. The compiler uses 6 to deduce the type. And 6 is an integer literal.

Answer 3

Yes, because every literal has a well defined type. The type of literal 6 is int . So, your auto will be translated to int .

If int is 16 or 32 bits long is implementation defined, but that's doesn't produce undefined behaviour, because every simple int through your program has a same length. Besides that, irrespective of the length of a int according to your architecture, 6 is an int , not short or long .