#define was changed to constexpr auto

Question

I defined some code in c++, ex:

#define array_width 3;

Visual Studio will suggest changing to:

constexpr auto array_width = 3;

what's the reason to change? and what is the benefit?

Thanks.

You mean `#define array_width 3`, but you will not get an error on that line, no. You'll get a sea of syntactic errors for each use. Now that is one reason to use the second. — LogicStuff, May 20 '19 at 09:54

score 2 · Accepted Answer · answered May 20 '19 at 09:51

The main reason for these suggestions is that the preprocessor does nothing but simple textual replacement (no type checking or similar things a compiler performs). There are many potential pitfalls when using the preprocessor - when you can avoid it, do so. `constexpr´ is one of the building blocks that allow for fewer macros these days.

To back this with an authority: From S. Meyers, Effective C++, Item 2 ("Prefer consts, enums, and inlines to #defines"):

Things to Remember

For simple constants, prefer const objects or enums to #defines

[...]

From S. Meyers, Effective Modern C++, Item 15 ("Use constexpr whenever possible"):

Things to Remember

constexpr objects are const and are initialized with values known during compilation.

[...]

constexpr objects and functions may be used in a wider range of contexts than non-constexpr objects and functions.

score 1 · Answer 2 · answered May 20 '19 at 09:46

Macros work by substituting text. With macro the following example code will be ill-formed:

struct foo
{
    int array_width{};
};

So in modern C++ one should prefer to avoid macros when there are alternatives available. Also it is a good idea to use UNIQUE_PREFIX_UPPER_CASE naming convention for macros to avoid possible clashes with normal code.

#define was changed to constexpr auto

2 Answers2