CODEWITHSUNDEEP
c++csyntax
Jared Updike
Jared Updike

Reputation: 7277

C (or C++?) Syntax: STRUCTTYPE varname = {0};

Normally one would declare/allocate a struct on the stack with:

STRUCTTYPE varname;

What does this syntax mean in C (or is this C++ only, or perhaps specific to VC++)?

STRUCTTYPE varname = {0};

where STRUCTTYPE is the name of a stuct type, like RECT or something. This code compiles and it seems to just zero out all the bytes of the struct but I'd like to know for sure if anyone has a reference. Also, is there a name for this construct?

Upvotes: 4

Views: 1166

Answers (2)

R.. GitHub STOP HELPING ICE
R.. GitHub STOP HELPING ICE

Reputation: 215211

In C, the initializer {0} is valid for initializing any variable of any type to all-zero-values. In C99, this also allows you to assign "zero" to any modifiable lvalue of any type using compound literal syntax:

type foo;
/* ... */
foo = (type){0};

Unfortunately some compilers will give you warnings for having the "wrong number" of braces around an initializer if the type is a basic type (e.g. int x = {0};) or a nested structure/array type (e.g. struct { int i[2]; } x = {0};). I would consider such warnings harmful and turn them off.

Upvotes: 4

Fred Nurk
Fred Nurk

Reputation: 14212

This is aggregate initialization and is both valid C and valid C++.

C++ additionally allows you to omit all initializers (e.g. the zero), but for both languages, objects without an initializer are value-initialized or zero-initialized:

// C++ code:
struct A {
  int n;
  std::string s;
  double f;
};

A a = {};  // This is the same as = {0, std::string(), 0.0}; with the
// caveat that the default constructor is used instead of a temporary
// then copy construction.

// C does not have constructors, of course, and zero-initializes for
// any unspecified initializer.  (Zero-initialization and value-
// initialization are identical in C++ for types with trivial ctors.)

Upvotes: 8

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