Are uninitialized struct members always set to zero?

Question

Consider a C struct:

struct T {     int x;     int y; }; 

When this is partially initialized as in

struct T t = {42}; 

is t.y guaranteed to be 0 or is this an implementation decision of the compiler?

Answer 1

It's guaranteed to be 0 if it's partially initialized, just like array initializers. If it's uninitialized, it'll be unknown.

struct T t; // t.x, t.y will NOT be initialized to 0 (not guaranteed to)  struct T t = {42}; // t.y will be initialized to 0. 

Similarly:

int x[10]; // Won't be initialized.  int x[10] = {1}; // initialized to {1,0,0,...} 

Sample:

// a.c struct T { int x, y }; extern void f(void*); void partialInitialization() {   struct T t = {42};   f(&t); } void noInitialization() {   struct T t;   f(&t); }  // Compile with: gcc -O2 -S a.c  // a.s:  ; ... partialInitialzation: ; ... ; movl $0, -4(%ebp)     ;;;; initializes t.y to 0. ; movl $42, -8(%ebp) ; ... noInitialization: ; ... ; Nothing related to initialization. It just allocates memory on stack. 

Answer 2

item 8.5.1.7 of standard draft:

-7- If there are fewer initializers in the list than there are members in the aggregate, then each member not explicitly initialized shall be default-initialized (dcl.init). [Example:

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

initializes ss.a with 1, ss.b with "asdf", and ss.c with the value of an expression of the form int(), that is, 0. ]