CODEWITHSUNDEEP
c++temporary
Bollinger
Bollinger

Reputation: 63

Temporary Object confusion

Have a look at this code snippet

struct S{ int i; int j;};

int main()
{
   assert(S().i ==  S().j) // is it guaranteed ?
}

Why?

Upvotes: 6

Views: 248

Answers (3)

Zac Howland
Zac Howland

Reputation: 15872

Technically, yes. They will be initialized to 0 (at least under a non-debug build for most compilers. Visual Studio's compiler will usually initialize uninitialized variables to a specific pattern in debug builds).

However, if you were sitting in a code review, don't be surprised if you get yelled at for not explicitly initializing your variables.

Upvotes: -1

j_kubik
j_kubik

Reputation: 6181

From C++ Standard ISO/IEC 14882:2003(E) point 3.6.2

Objects with static storage duration (3.7.1) shall be zero-initialized (8.5) before any other initialization takes place.

So this is valid as both variables are zero-initialized.

Upvotes: 0

Prasoon Saurav
Prasoon Saurav

Reputation: 92854

is it guaranteed ?

Yes it is guaranteed. The values of S().i and S().j would be 0. () implies value initialization. (that means i and j would be zero-initialized because S is a class without a user defined default constructor)

Upvotes: 10

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