C++ ternary operators, Any difference?

Question

I was reading about the difference of the following:

1)

std::string test = level > 10 ? "Master" : "Beginner";

2)

std::string test;
if (level > 10)
    test = "Master";
else
    test = "Beginner";

And in the second case I was told that in the second option a temporary value is being create which I don't understand.

I learnt that when we declare a variable without initialisation it takes garbage, so the compiler just holds a memory location and doesn't "work" to change its value.

Does this answer your question? [Ternary operator ?: vs if...else](https://stackoverflow.com/questions/3565368/ternary-operator-vs-if-else) — Tony Tannous, May 25 '20 at 11:09
there is no "garbage" in your example, `std::string` has a constructor that gets called. Strictly speaking also in general the value of something unitialized is not "garbage", you are just not allowed to use it and if you do you get something that can look like garbage — 463035818_is_not_an_ai, May 25 '20 at 11:13

Nelfeal · Answer 1 · 2020-05-25T11:20:23.223

Syntactically, the first case is somewhat equivalent to this:

constexpr const char* f(int level) {
    if (level > 10)
        return "Master";
    else
        return "Beginner";
}

std::string test = f(level);

Here, test is immediately initialized with either "Master" or "Beginner". In the second case, test is first default-initialized, then either value is assigned to it.

However, this is not necessarily true after optimizations: the compiler can very well generate the exact same code in both cases.

when we declare a variable without initialisation it takes garbage

That is only true for fundamental and POD types, not classes like std::string which have a default-constructor. Look up default-initialization for details.

*"However, this is not necessarily true after optimizations: the compiler can very well generate the exact same code in both cases"* - In the general case of a user defined copy c'tor and assignment operator, a compiler cannot prove the two invocations mean exactly the same thing, and so likely won't generate the same code. — StoryTeller - Unslander Monica, May 25 '20 at 13:29

songyuanyao · Answer 2 · 2020-05-25T11:28:42.630

5

There're no temporary objects constructed in both cases.

In the 1st case, test is copy-initialized from the const char* returned by the ternary operator directly, via the constructor of std::string taking const char*.

In the 2nd case, test is default-initialized firstly, then assgined from the const char* in the if or else branch via the assignment operator of std::string. (BTW: After default-initialization test won't take garbage, it'll be initialized by the default constructor of std::string as empty string.)

edited May 25 '20 at 11:28

answered May 25 '20 at 11:15

songyuanyao

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But when I access it I get strange values – May 25 '20 at 18:04
@smith_brown I tried it [here](https://wandbox.org/permlink/6pu9AsqhoSiBY0fe) and seems working fine. What do you mean *getting strange values*? Could you make a [Minimal, Reproducible Example](https://stackoverflow.com/help/minimal-reproducible-example) for it? – songyuanyao May 26 '20 at 00:52

score 0 · Answer 3 · answered May 25 '20 at 19:34

There is no reason not to use the first form, but the compiler would have grievous performance issues if it didn't produce the same code for either. Having said that, the first form is clear and easy to understand, and might give the optimizer less work to do, so you might as well use it. It clearly states the intent to initialize the variable to one of two given values, and the code that expresses intent most clearly often produces the best generated code (as well as being easiest to understand.)

C++ ternary operators, Any difference?

3 Answers3