C struct compound initialization without setting everything to zero-CodePudding

Is there any syntax similar to doing something like this

typedef struct foo
{
   int X;
   int Y;
   int Z;
} foo;

int main()
{
   // Initialize everything
   foo Variable = (foo){
    .X = 10;
    .Y = 15;
    .Z = 20;
   };

  // Assign the X and Y variables, but keep Z the same
  Variable = (foo){
   .X = 15;
   .Y = 20;
  // .Z stays the same
  };
}

Would be nice to have this type of QOL syntax instead of having to type

int main()
{
   foo Variable = {bla bla};

   // Keep having to type Variable. to access struct members
   Variable.X = 10;
   Variable.Y = 15;
}

This gets really annoying with structs in structs i.e.

Struct1->Struct2.Struct3.Element.X = 10;
Struct1->Struct2.Struct3.Element.Y = 15;

CodePudding user response：

No, C does not support this style of initialization or assignment.

If you want to access only a part of a structure, you need to express this explicitely.

EDIT:

You can get away with:

    Variable = (foo){
       .X = 15;
       .Y = 20;
       .Z = Variable.Z;
    };

At least an optimizing compiler will just generate the operations for the changing elements. But it is more source than single assignments.

CodePudding user response：

You can use the preprocessor to save your fingers (or copy/paste)...

    struct {
        int f1;
        int f2;
        struct {
            int b1;
            int b2;
            int b3;
        } bar;
    } foo = {
        1, 2, { 42, 43, 44 },
    };

    printf( "%d  %d  %d\n", foo.bar.b1, foo.bar.b2, foo.bar.b3 );
#define S foo.bar
    S.b1 = 7;
    S.b2 = 8;
    printf( "%d  %d  %d\n", S.b1, S.b2, S.b3 );
#undef S