Difference between = and {} syntaxes for initializing a variable in C++

Question

I have read quite a few C++ codes, and I have come across two methods of initialising a variable.

Method 1:

int score = 0;

Method 2:

int score {};

I know that int score {}; will initialise the score to 0, and so will int score = 0;

What is the difference between these two? I have read initialization: parenthesis vs. equals sign but that does not answer my question. I wish to know what is the difference between equal sign and curly brackets, not parenthesis. Which one should be used in which case?

songyuanyao · Accepted Answer · 2021-06-02 00:41:05Z

7

int score = 0; performs copy initialization, as the effect, score is initialized to the specified value 0.

Otherwise (if neither T nor the type of other are class types), standard conversions are used, if necessary, to convert the value of other to the cv-unqualified version of T.

int score {}; performs value initialization with braced initializer, which was supported since C++11, as the effect,

otherwise, the object is zero-initialized.

score is of built-in type int, it's zero-initialized at last, i.e. initialized to 0.

If T is a scalar type, the object's initial value is the integral constant zero explicitly converted to T.

edited Jun 2, 2021 at 0:41

answered Dec 19, 2019 at 7:44

songyuanyao

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6 Comments

Aditya Prakash Over a year ago

Is there an advantage in using one over the other?

Victor Over a year ago

An entire story should be written about brace initializers.

songyuanyao Over a year ago

@AdityaPrakash Since they have the same effect, I think it's just a coding style issue. For the 1st one, it could be used to initialize to the specified value; for the 2nd one, it could be used to initialize to the default value.

super Over a year ago

That depends on what you are initializing. In the example as written there is no difference.

M.M Over a year ago

Empty braces is value-initialization, which for the case of a variable of type int, resolves to zero-initialization

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b4l8 · Accepted Answer · 2019-12-19 08:13:16Z

0

You may have some interest in ISO/IEC 14882 8.5.1. It will tell you that a brace-or-equal-initializer can be assignment-expression or braced-init-list. In Method2, you are using an default initializer on a scalar type, witch should be set to zero.

answered Dec 19, 2019 at 8:13

b4l8

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