c++ - What does {0} mean when initializing an object? -

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when {0} used initialize object, mean? can't find references {0} anywhere, , because of curly braces google searches not helpful.

example code:

shellexecuteinfo sexi = {0}; // do? sexi.cbsize = sizeof(shellexecuteinfo); sexi.hwnd = null; sexi.fmask = see_mask_nocloseprocess; sexi.lpfile = lpfile.c_str(); sexi.lpparameters = args; sexi.nshow = nshow;  if(shellexecuteex(&sexi)) {     dword wait = waitforsingleobject(sexi.hprocess, infinite);     if(wait == wait_object_0)         getexitcodeprocess(sexi.hprocess, &returncode); }

without it, above code crash on runtime.

what's happening here called aggregate initialization. here (abbreviated) definition of aggregate section 8.5.1 of iso spec:

an aggregate array or class no user-declared constructors, no private or protected non-static data members, no base classes, , no virtual functions.

now, using {0} initialize aggregate trick 0 entire thing. because when using aggregate initialization you don't have specify members , spec requires unspecified members default initialized, means set 0 simple types.

here relevant quote spec:

if there fewer initializers in list there members in aggregate, each member not explicitly initialized shall default-initialized. example:

struct s { int a; char* b; int c; }; s ss = { 1, "asdf" };

initializes ss.a 1, ss.b "asdf", , ss.c value of expression of form int(), is, 0.

you can find complete spec on topic here


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