int i{0}; //(1)
int i = 0; //(2)

Do I have correct understanding that in first case (1) the variable i will be created already with 0 as it's value, and in (2) it will be created without a value and then assigned a value 0, so (1) will always faster then (2)?

But seems like most(all?) modern compilers will still optimize (2) to be same as (1) under the hood?

CodePudding user response：

initializing variables with Brace Initialization performs additional checks at compile time (no effect on runtime). . Such as if you enter a float literal inside the curly braces it will throw an error. Initialization with the equal sign will be optimized by the compiler. Prefer brace initialization whenever possible. I hope that answers your question