Why Does This C Code Print 'ba' Instead of 'ab'?

Unveiling the Mystery of Left-to-Right Printing in C

The behavior of the following C code is puzzling:

myQueue.enqueue('a');
myQueue.enqueue('b');
cout << myQueue.dequeue() << myQueue.dequeue();

Surprisingly, it prints "ba" to the console instead of the expected "ab". Why does this occur?

To unravel this enigma, we need to understand the nature of the overloaded << operator. In C , the << operator takes an arbitrary number of arguments and returns a reference to an ostream object. In this case, cout is the ostream object.

The evaluation order of the arguments to << is not guaranteed by the C standard. This means that the compiler is free to evaluate the arguments in any order, including potentially evaluating them out of order.

In our example, the code compiles into an expression of the form:

cout << ( (myQueue.dequeue()) << (myQueue.dequeue()) );

The compiler inserts parentheses around each dequeue call and around the result of the recursive call. The result of the second dequeue call is then inserted into the parentheses surrounding the first.

The key to understanding the behavior is to realize that the arguments to the << operator are evaluated in right-to-left order. Therefore, the following is equivalent to the code above:

( (myQueue.dequeue()) << (myQueue.dequeue()) ) << cout;

In this order, the second dequeue call is evaluated first, resulting in 'b'. The result of this call is then passed to the << operator, which returns a reference to cout. The first dequeue call is then evaluated, resulting in 'a', which is passed to the reference returned by the previous << operator.

This explains why the code prints "ba" instead of "ab". The << operator first prints 'b', and then prints 'a'.

To ensure consistent left-to-right printing, parentheses should be used to explicitly control the evaluation order, as follows:

cout << (myQueue.dequeue()) << ' ' << (myQueue.dequeue());

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