c++ Type declaration before function invocation?

Question

I'm looking through the source code for the v8 javascript engine (here's the github link), and on line 33 of parsing.cc, there's this statement within a function:

Parser parser(info);

I'm only used to seeing a type declaration like this before an assignment, such as:

Parser myparser = Parser(...);

So what does the first example do? Why not just invoke parser(info) without a type declaration?

Thank you in advance.

Doesn't that just invoke the `Parser` constructor? My C++ is rusty, but this just looks like a variable initialization. — Carcigenicate, Oct 03 '17 at 16:15
related/dupe: https://stackoverflow.com/questions/4470553/initialization-parenthesis-vs-equals-sign — NathanOliver, Oct 03 '17 at 16:22
I'd argue that it's not a duplicate. My question calls for people to help identify a syntactic form, whereas the referenced question asks for comparison between two forms. — James M. Lay, Oct 03 '17 at 16:26

score 4 · Accepted Answer · answered Oct 03 '17 at 16:16

The line

Parser parser(info);

constructs a Parser object by calling the constructor that takes info as the argument.

This method of constructing an object is called direct initialization.

You may also use the form

Parser parser = Parser(info);

to construct the object. This method of constructing an object is called copy initialization.

9Breaker · Answer 2 · 2017-10-03T16:43:31.553

0

You are constructing an object of type Parser on the stack. (or as the comments say depending on the block scope this could be a global variable, not able to discern from the info available but more than likely it is the stack :) Object creation on the stack/heap?

edited Oct 03 '17 at 16:43

answered Oct 03 '17 at 16:17

9Breaker

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How do you know it is on the stack? – juanchopanza Oct 03 '17 at 16:20
@juanchopanza If it isn't using "new" and going to the heap, then it would be the stack. – 9Breaker Oct 03 '17 at 16:27
No, it could also be static storage. – juanchopanza Oct 03 '17 at 16:28
Wouldn't that need the static qualifier for that? – 9Breaker Oct 03 '17 at 16:32
Nope. Who said anything about block scope? – StoryTeller - Unslander Monica Oct 03 '17 at 16:34
Good points, I was assuming... Could be global, so if it is global, then it would be stored elsewhere. I will edit that. But if it was a static and out of the block scope of a function, then it would still need the qualifier static on it. – 9Breaker Oct 03 '17 at 16:42

c++ Type declaration before function invocation?

2 Answers2