syntax questions

[chipp]

what's this mean

INPUT input = {0};

and

test test1 = {}; //test is a name of a structure 

does it mean like:

//default constructor of test
test (): variable_of_test(0){}

[Null]

test test1 = {}; and test test1 = {0}; sets all members of struct to 0

struct test
{
    int a;
    int b;
    char str[20];
};

int main()
{
    test t;
    cout <<t.a<<' '<<t.b<<' '<<t.str; // Will print garbage -1990867164 -2 b◄Ļu╝[’uÓ→@

    cout <<endl;

    test t2={}; // or {0}
    cout <<t2.a<<' '<<t2.b<<' '<<t2.str; // Will print 0 0

    cout <<endl;
    
   // Initisalize each member of struct
    test t3={1,2,"Hello"};
    cout <<t3.a<<' '<<t3.b<<' '<<t3.str; // Will print 1 2 Hello

}

does it mean like:
//default constructor of test
test (): variable_of_test(0){}

I don't know but it can be used to initialize const members:

struct test
{
   const int a;
};

int main()
{
    test t; // error: structure with unitialized const members
    test t={10}; // ok
}

[andywestken]

I thought that a brace-enclosed initializer list could only be used to initialise a 'plain old data' (i.e. C) struct. So when I see code like

test test1 = {0}

I understand test to be a 'pod' struct, with no need for an actual constructor.

Andy

Of course, C++0x is changing this rule, but my gcc quoted C++98 at me when I tried to use a brace-enclosed initializer list with a class which had a constructor.. I was told to use normal constructor notation.

[chipp]

I don't know but it can be used to initialize const members

it's the same, i just tested it. and afterall, if you could assign const that way, that means you can define const by the default constructor too...

[andywestken]

Out of interest, with which compiler?