Home >Backend Development >C++ >Does `new unsigned int` Initialize Memory to Zero in C ?
In the following code snippet:
#include <iostream> int main(){ unsigned int* wsk2 = new unsigned int(5); std::cout << "wsk2: " << wsk2 << " " << *wsk2 << std::endl; delete wsk2; wsk2 = new unsigned int; std::cout << "wsk2: " << wsk2 << " " << *wsk2 << std::endl; return 0; }
The expected result is that the memory is not initialized to zero, but the output is:
wsk2: 0x928e008 5 wsk2: 0x928e008 0
It seems that the operator new is initializing the memory to zero, but it is actually not.
How it works:
There are two versions of the operator new:
wsk = new unsigned int; // default initialized (ie nothing happens) wsk = new unsigned int(); // zero initialized (ie set to 0)
The default initialization does not initialize the memory, while the zero initialization sets the memory to zero.
It also works for arrays:
wsa = new unsigned int[5]; // default initialized (ie nothing happens) wsa = new unsigned int[5](); // zero initialized (ie all elements set to 0)
To confirm that the memory is actually zeroed out, we can use placement new with a known piece of memory:
#include <new> #include <iostream> int main() { unsigned int wsa[5] = {1,2,3,4,5}; // Use placement new (to use a know piece of memory). // In the way described above. // unsigned int* wsp = new (wsa) unsigned int[5](); std::cout << wsa[0] << "\n"; // If these are zero then it worked as described. std::cout << wsa[1] << "\n"; // If they contain the numbers 1 - 5 then it failed. std::cout << wsa[2] << "\n"; std::cout << wsa[3] << "\n"; std::cout << wsa[4] << "\n"; }
The output of this code is:
0 0 0 0 0
Which confirms that the memory is indeed zeroed out by the zero-initialization version of the operator new.
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