CODEWITHSUNDEEP
c++cout
Cătălina Sîrbu
Cătălina Sîrbu

Reputation: 1283

Increment address in cout expression renders unexpected output

I don't understand what is happening here. There must be something related to "flush" termen but I would like an explanation. 

int arr[4] {3,8,1,6};

cout<<arr[0];
cout<<arr[1];
cout<<arr[2];
cout<<arr[3];

cout<<endl;

cout<<&arr[0]<<endl;
cout<<&arr[1]<<'\n';
cout<<&arr[2]<<endl;
cout<<&arr[3]<<endl;

cout<<&arr[0]<<endl;

int *j = &arr[0];
cout << *j << *(++j) << *(++j) << *(++j); // HERE IS THE PROBLEM

Why this last cout instruction tends to output the numbers reversed? It seems to me like backtracking but I'm not sure.

I was expecting 3816 as output, instead I got 6618.

Upvotes: 1

Views: 102

Answers (1)

anastaciu
anastaciu

Reputation: 23832

In the expression

cout << *j << *(++j) << *(++j) << *(++j);

The order of evaluation is not well defined, so the behaviour is undefined.

Examples of undefined behavior are memory accesses outside of array bounds, signed integer overflow, null pointer dereference, more than one modifications of the same scalar in an expression without any intermediate sequence point (until C++11) that are unsequenced (since C++11)

I ran it and it outputed 3816. Note that with warning flags this is detected by the compiler.

If you try:

cout << *j;
cout << *(++j);
cout << *(++j); 
cout << *(++j);

This well sequenced expression will render the expected output.

Upvotes: 5

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