C++: Variable initialization using x{0}

Question

I saw a initialization syntax which is new for me. I searched on google and here but I couldn't find something useful.

int a = 0;
int a = {0};
int a{0}; // <- this is new for me

Why do I need third style while others exist? What is the difference between each exactly?

Thanks.

The third one solves what Scott Meyers called "C++'s most vexing parse." Under certain circumstances, `int a(var)` could be misinterpreted as a function declaration rather than constructing a variable. http://en.wikipedia.org/wiki/Most_vexing_parse — Joe Z, Dec 05 '13 at 06:00

score 4 · Accepted Answer · edited May 23 '17 at 11:49

4

You may be interested by C++11 initializer lists. They might not explain the third example, but they are useful, especially for real class objects.

Your code int a{0}; is called uniform initialization in C++11. See also most vexing parse wikipage (as commented by Joe Z).

Take time to at least read the C++11 wikipage. The new features of C++11 makes it almost a different language than C++03.

edited May 23 '17 at 11:49

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answered Dec 05 '13 at 05:59

Basile Starynkevitch

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score 2 · Answer 2 · answered Dec 05 '13 at 06:00

This form of initialization is referred to as list initialization in C++11.

When used with variables of built-in type, list initialization is different in one way: you can't list initialize variables of built-in type if the initializer might lead to the loss of information.

double pi = 3.1415926;
int a(pi); //fine
int a{pi}; //compile error

C++: Variable initialization using x{0}

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