Have a look at this code snippet
struct S{ int i; int j;};
int main()
{
assert(S().i == S().j) // is it guaranteed ?
}
Why?
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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)
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Prasoon Saurav
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In particular, adding `S() {}` (a do-nothing user-defined ctor) to `struct S` will cause `i` and `j` to be left uninitialised, meaning the `assert()` is likely to trigger. – j_random_hacker Dec 07 '10 at 17:16
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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.
j_kubik
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3I dont see any object with static storage allocation here. Expression `S()` creates a **temporary object** local to the scope of `assert()`. – Prasoon Saurav Dec 07 '10 at 16:50
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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.
Zac Howland
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