Is there any difference between struct Data d = {0} and struct Data d = {}

Question

I have two types of structure variable initialization in my code.

Example

#include<iostream>
#include<string>
using namespace std;
struct Data{
   int arr[5];
   float x;

};
int main(){
   struct Data d = {0};
   struct Data d1 = {};
   cout<<d.arr[0]<<d.x;
   cout<<d1.arr[0]<<d1.x<<endl;
   return 0;
}

I am running the code ad getting 0 0 0 0 as my output. Please help me, is there any difference between both initialization.

Answer 1

According to the rule of aggregate initialization, the effect is the same here, i.e. all the members of the struct will be value-initialized (zero-initialized here for non-class types).

If the number of initializer clauses is less than the number of members and bases (since C++17) or initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default initializers, if provided in the class definition, and otherwise (since C++14) by empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates). If a member of a reference type is one of these remaining members, the program is ill-formed.

More precisely,

struct Data d = {0}; // initialize the 1st member of Data to 0, value-initialize(zero-initialize) the remaining members
struct Data d1 = {}; // value-initialize(zero-initialize) all the members of Data

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Note that the whole story is based on that Data is an aggregate type and its members are non-class types, otherwise the behavior would change according to the rule of list initialization.