The reader may have noticed that a carbonyl group, by virtue of its resonance structures, acts as a good electron withdrawing group when attached to an alkene. However, when the carbonyl group is next to the alkene they are conjugated, and there are effectively three different reaction possibilities;
1. 1,2- addition to the alkene.
2. 1,2- addition to the carbonyl.
3. 1,4-conjugate addition to the entire system.
It is the latter possibility that we are now going to focus upon. An example of this is shown below;

It is particularly important to note that because of keto-enol tautomerism, the product will often end up looking exactly like the product of a 1,2- addition to the alkene. Indeed, as it is often impossible to tell which pathway has been taken, it is generally safer to assume that 1,4- conjugate addition has occurred rather than 1,2- addition, because the anion is stabilised throughout.

The more interesting question arises when competition between the C=C and the C=O is considered - that is competition between 1,4- conjugate addition and 1,2- addition on the C=O. 

1. If the reaction is reversible, then thermodynamics will usually prevail and cause the conjugate addition to predominate. The reasoning behind this is that a C=O is more stable than a C=C; the 1,4- addition leaves the C=O and the 1,2- leaves the C=C, hence 1,4- is preferred.
2. Steric hindrance at one of the sites will promote attack at the other.
3. If the carbonyl has a leaving group attached the situation becomes even more complicated because nucleophilic substitution can also occur, so there is competition particularly between the nucleophilic addition to the alkene, and the aforementioned nucleophilic substitution on the carbonyl. One rule of thumb that can help is that 'soft' nucleophiles (as described here) will tend to attack the 'soft' alkene end, and 'hard' nucleophiles will tend to attack the 'hard' carbonyl system. This is by no means a hard and fast rule though.