CODEWITHSUNDEEP
c++c++11auto
void.pointer
void.pointer

Reputation: 26365

Usage of auto for global constants

Suppose I have something like this:

namespace {
  const unsigned MY_UINT = 100u;
  const float MY_FLOAT = 0.f;
  const char* MY_STRING = "Hello World";
}

Do I get expected behavior by using auto for these? I presume this is an improvement, but I'm not sure about this in practice.

namespace {
  auto MY_UINT = 100u;
  auto MY_FLOAT = 0.f;
  auto MY_STRING = "Hello World";
}

Are the two code examples semantically the same? Will these be const automatically? If not, should I specify auto const?

Upvotes: 1

Views: 133

Answers (3)

Dakorn
Dakorn

Reputation: 903

No, they are not the same.

When you type:

auto var = 123u;

It is the same as:

unsigned var = 123u;

To achive

const unsigned var = 123u;

You should write

const auto var = 123u;

Do not expect from auto too much. Some kind of magic guesses.

Upvotes: 1

Piotr Siupa
Piotr Siupa

Reputation: 4838

Yes, you must manually specify it const.

No, it is not a good practice. You generally shouldn't overuse auto. That keyword can be helpful if name of type is very long but typedef is often better solution. Auto obscures code. Moreover, this facilitates errors. Sometimes the type that deduce the compiler may not be the same as what you mean. You can use auto if you want but better don't do it without a reason.

Upvotes: 1

David G
David G

Reputation: 96810

auto's deduction rules are equivalent to by-value template argument deduction. Creating an object by value entails stripping references and top-level cv-qualifiers from the initializer. Your two examples are not equivalent. In particular, the primitive 100u is of type unsigned int, so that's what it is deduced as. Likewise, 0.f is of type float.

Adding const only makes sense if the variable itself will not be modified. If you want to make constants in your program, using constexpr might be better.

Upvotes: 3

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