C++ - member variable aggregate initialization

Question

In the following code:

class Class
{
private:
LUID luid;

public:
Class()
{
luid = { 0, 0}; // A. Does not compile
LUID test = {0, 0}; // B. Compiles
test = {1,1}; // C. Does not compile
}

Why are A and C not right, but B is fine?

The error I get for A and C is:

error C2059: syntax error : '{'

error C2143: syntax error : missing ';' before '{'

error C2143: syntax error : missing ';' before '}'

I think it has to do with the C++ version, although I am not sure which version this is using besides it not being very new.

Well then time to find out what compiler you are using (name, version number, etc.)? — UnholySheep
– UnholySheep, Commented Oct 11, 2017 at 18:06
How is LUID defined? Do you have C++11 or higher turned on? — NathanOliver
– NathanOliver, Commented Oct 11, 2017 at 18:07
@UnholySheep The only common C++ compiler that outputs errors in that format is MS VC++. — Daniel H
– Daniel H, Commented Oct 11, 2017 at 18:45

Stephan Lechner · Accepted Answer · 2017-10-11 18:28:19Z

3

Statement LUID test = {0, 0} is an initialization of a local variable using an initialization list; this is valid, as it is used in the course of a variable definition. test = {0, 0}, in contrast, is an assignment, as test is defined elsewhere. Assigning initializer lists is supported only in particular cases (e.g. when assigning to a scalar or to a particular sort of class type (cf., for example, braced-init-list assignment).

Other cases, like, for example, arrays, cannot be assigned but just initialized:

typedef int LUID[2];

int main(){

    LUID t = { 10, 20 }; // compiles
    // t = { 10, 20};     // does not compile, since an array is not assignable

     return 0;
}

edited Oct 11, 2017 at 18:28

answered Oct 11, 2017 at 18:11

Stephan Lechner

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1 Comment

Daniel H Over a year ago

That doesn’t explain why it was giving a syntax error. It’s valid syntax even if it isn’t valid to do it for the LUID class.

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