Why does interpolating a const string result in a compiler error?

Question

Why does string interpolation in c# does not work with const strings? For example:

private const string WEB_API_ROOT = "/private/WebApi/"; private const string WEB_API_PROJECT = $"{WEB_API_ROOT}project.json"; 

From my point of view, everything is known at compile time. Or is that a feature that will be added later?

Compiler message:

The expression being assigned to 'DynamicWebApiBuilder.WEB_API_PROJECT' must be constant.

Thanks a lot!

Answer 1

Interpolated strings are simply converted to calls to string.Format. So your above line actually reads

private const string WEB_API_PROJECT = string.Format("{0}project.json", WEB_API_ROOT); 

And this is not compile time constant as a method call is included.

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On the other hand, string concatenation (of simple, constant string literals) can be done by the compiler, so this will work:

private const string WEB_API_ROOT = "/private/WebApi/"; private const string WEB_API_PROJECT = WEB_API_ROOT + "project.json"; 

or switch from const to static readonly:

private static readonly string WEB_API_PROJECT = $"{WEB_API_ROOT}project.json"; 

so the string is initialized (and string.Format called) at the first access to any member of the declaring type.

Answer 2

An additional explanation why string interpolation expressions are not considered constants is that they are not constant, even if all their inputs are constants. Specifically, they vary based on the current culture. Try executing the following code:

CultureInfo.CurrentCulture = CultureInfo.InvariantCulture;  Console.WriteLine($"{3.14}");  CultureInfo.CurrentCulture = new CultureInfo("cs-CZ");  Console.WriteLine($"{3.14}"); 

Its output is:

3.14 3,14 

Note that the output is different, even though the string interpolation expression is the same in both cases. So, with const string pi = $"{3.14}", it wouldn't be clear what code should the compiler generate.

UPDATE: In C# 10/.Net 6, string interpolations that only contains const strings can be const. So the code in the question is not an error anymore.