Syntax "Foo(int v) : val_(v) {}"?

Question

I am going through some code and I am having dificulty in comprehending the syntax:

  struct Foo {

  int val_;

  Foo(int v) : val_(v) {}  //--->THIS LINE

  };

The colon : seems like a member access operator.

Can I get a clear explanation the above line?

score 3 · Accepted Answer · answered May 02 '11 at 18:22

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Its initializing val_ to v.

See: Constructor Initialization Lists

answered May 02 '11 at 18:22

Daniel A. White

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Foo(**int v**) or foo() does it makes any differnce.Is parameter **int v** required for this initialization.if not why is that parameter there – munish May 02 '11 at 18:40
@munish Whatever you are passing into the constructor as the parameter v is the value to which `val_` is initialized, so yes, you do need the parameter for the constructor. – jonsca May 02 '11 at 18:46
1

Though you can use initializer lists with no constructor parameters. For example, `Foo() : val_(5) {}` would cause every default-constructed `Foo` to have 5 as the initial value of member `val_`. – aschepler May 03 '11 at 18:26

score 1 · Answer 2 · edited May 23 '17 at 12:03

1

Structures in C++ can have constructors just like classes do. This is initializing the public variable val_ to v just as if you had val_ = v; in the constructor body. See this thread for their benefits in terms of efficiency.

edited May 23 '17 at 12:03

Community

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answered May 02 '11 at 18:25

jonsca

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Syntax "Foo(int v) : val_(v) {}"?

2 Answers2