CODEWITHSUNDEEP
c++initializationconditional-operator
evan hu
evan hu

Reputation: 23

Can I use conditional operator to initialize C style string literal?

I'm trying to use the ternary operator (?:) to initialize a character array to either one string or another

char answ[] = ans > 0 ? "Anton" : "Danik";

Where ans is just an integer obtained earlier and I keep getting the error:

initialization with '{...}' expected for aggregate object

Is it that you simply can't initialize arrays through a ternary operator?

I also tried this:

char answ[] = { ans>0 ? "Anton" : "Danik" };

What gave the error:

value of type "const char *" cannot be used to initialize an entity of type "char"

Upvotes: 1

Views: 744

Answers (1)

Jorenar
Jorenar

Reputation: 2884

Is it that you simply can't initialize arrays through a ternary operator?

Indeed, you can't.

In your example, you don't want to have just array. You want to have string literal - a specific array. Unfortunately, compiler doesn't treat answ as one, because of you try to use conditional operator for initialization. Compiler treats it directly as array of chars.

But there is different method of declaring C strings - using const and pointer.

const char* answ = ans > 0 ? "Anton" : "Danik";

The downside of this approach is, well, const - you cannot modify this string.

That's why, if you are using C++, you should use its strings - std::string :

std::string answ = ans > 0 ? "Anton" : "Danik";

In C, you could do:

char answ[6];
ans > 0 ? strcpy(answ, "Anton") : strcpy(answ, "Danik");

But at this point, ternary operator is just less verbose than normal if-else, so don't do that.

Upvotes: 3

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