Is "auto" safe from the integer overflow in C++17?

Question

If I write the code like this:

auto n = 2048 * 2048 * 5;
char* buf = new char[n];

So, Is auto deduction type safe from the integer overflow in C++17?

`2048 * 2048 * 5` is deduced to `int` no matter the overflow. The deduction only cares about types, not values. — Zereges, Sep 02 '17 at 17:10
`auto` doesn't magically fix overflows or change variable types. It just tells the compiler "please figure out the type here" instead of you having to write it yourself. — Jesper Juhl, Sep 02 '17 at 17:17
`auto` is a red herring. The expression on the right has to be valid on its own first. — Kerrek SB, Sep 02 '17 at 17:19
@Mahendra Yes, but to my knowledge there arent any changes in that version of the standard that would affect this behavior. hence my question. — Borgleader, Sep 02 '17 at 17:41
See also https://stackoverflow.com/questions/42892227/c-literal-integer-type — Zan Lynx, Sep 02 '17 at 19:09

score 7 · Accepted Answer · answered Sep 02 '17 at 17:27

2048 and 5 in C++ have a type, and that type is int. Multiplying two int's has a type and that type is int. There are values for which the result cannot fit in an int, and auto cannot prevent that.

What auto can prevent is accidentally narrowing the result, e.g.:

short x = 4 * 8192;

Is "auto" safe from the integer overflow in C++17?

1 Answers1