C++11 type deduction vs const char *

Question

In GotW 94, Herb Sutter draws a distinction between the "classic C++" declaration

const char* s = "Hello";

and the "modern" style

auto s = "Hello";

He tells us that there's a "subtle difference in the type of s, where the auto style is more correct". [Edited to add: comments suggest that this might not be a fair representation of what Sutter actually meant; see discussion below.]

But... what's the difference? I was under the impression that a const char * is the correct way to refer to a string literal. Further, when I asked my debugger (lldb), it seems to think the types are actually the same:

* thread #1: tid = 0x1756c2, 0x0000000100000f8f test`main + 31 at test.cc:4, queue = 'com.apple.main-thread', stop reason = breakpoint 1.1
    frame #0: 0x0000000100000f8f test`main + 31 at test.cc:4
   1    int main(void) {
   2        const char* s = "Hello";
   3        auto t = "Hello";
-> 4        return 0;
   5    }
(lldb) fr v
(const char *) s = 0x0000000100000f91 "Hello"
(const char *) t = 0x0000000100000f91 "Hello"

Where's the subtle difference Sutter refers to?

There's no difference. He might be thinking that `auto` would deduce an array instead, but it won't. — T.C., Dec 18 '14 at 15:55
Sutter actually states "for example, can you spot a subtle difference in the type of s, where the auto style is more correct?" Your paraphrase gives a completely different meaning to the statement. — , Dec 18 '14 at 15:56
@kiss-o-matic: No, array-to-pointer conversion is applied during type deduction. — Mike Seymour, Dec 18 '14 at 15:57
Note also that `s` and `t` have been coalesced (they have the same address). — Paul R, Dec 18 '14 at 16:01
On nth reading, @remyabel may be correct, but I'm struggling to grasp exactly what Sutter's point is, if it's not what I interpreted it as! — user2862505, Dec 18 '14 at 16:03
@PaulRubel: No, the type of an ordinary string literal is always "array of `const char`", from which `const char*` will be deduced by `auto`. — Mike Seymour, Dec 18 '14 at 16:14
@remyabel he heavily implied that there's a difference with that statement. And if there really isn't a difference, I'm not sure what point he was trying to make. — Mark Ransom, Dec 18 '14 at 16:14
I checked if Herb's statement could be a simple typo, but it looks like it was just a misunderstanding of his about the effect of `auto`. — Cheers and hth. - Alf, Dec 18 '14 at 18:00

score 9 · Answer 1 · answered Dec 31 '14 at 00:21

I posted this to Herb Sutter:

What do you think is the “subtle difference in the type of s, where the auto style is more correct”? See C++11 type deduction vs const char *

And here's his reply:

I think I had in mind that the auto version deduced an array, which it doesn’t. If I had something else in mind, I’ve forgotten it now, so I’ve removed that phrase and added your moniker to the Acknowledgments. Thanks for the feedback!

score 8 · Accepted Answer · answered Dec 18 '14 at 16:06

8

You're not entirely clear on what Herb stated (context is important), but anyway the types are the same.

answered Dec 18 '14 at 16:06

Cheers and hth. - Alf

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4

He didn't say any more (in that article) than the quote given in the question. I'm pretty sure you're right, and there's no difference, subtle or otherwise. – Mike Seymour Dec 18 '14 at 16:10
1

Thanks to this, and to the comment by @remyabel above. I'm sure you're correct that I didn't grasp exactly what his point was, but even after a several re-readings I'm not sure what he actually _is_ saying. That the auto version is more maintainable if there is a subtle change to the types, perhaps. – user2862505 Dec 18 '14 at 16:14

C++11 type deduction vs const char *

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