append1.cppimplements this experiment.
Timer.docimplements the timer required by all experiments (there is no
For this experiment, the test is to append just one value to the end of an existing container:
theList.push_back(1); theVector.push_back(1);One of these statements is the
Perform the teststatement in the test algorithms.
However, as we saw in the previous experiment, creating a new container comes at a cost. Consequently, in this experiment, we actually have two costs: creating a new container and appending the new element. We really only want to bother with the second operation, so we'll have to ignore the first.
Compile and run the program on the sample range of n.
Enter the results of this experiment into the spreadsheet that you started for Experiment #1. Since we want to ignore the time that it takes to create a new container (computed in Experiment #1), subtract each result for this experiment from the result you got in Experiment #1 for each value of n. These differences will be the time it takes to do the append. Graph these differences.
Using the categories in Experiment #1 and your new graph, answer these questions:
Question #14.2.1: How would you categorize the time it takes to append one value to the end of a
listof size n? Justify your answer.
Question #14.2.2: How would you categorize the time it takes to append one value to the end of a
vectorof size n? Justify your answer.
Let's explore what's going on behind the scenes.
The program begins with a
list of size
list<int> aList(n);For simplicity, suppose that
nis 3, in which case we have a
The program then calls
push_back() to append a 1 to
aList.push_back(1);To append a 1 to the
push_back()operation performs these three steps:
listby first going through the previous-node pointer in the head node and updating the pointers in the header node, the last node, and the new node. Finally it updates the size of the
list, and we end up with this:
The time for these three steps is independent of the number of
values in the list; it doesn't matter how large the
to begin with; adding one more always takes the same amount of time.
The program also defines a
vector of size n:
vector is declared this way, the size and capacity are
the same. So if we again consider the case where n is 3, we
aVector as follows:
The program then calls
push_back() to append a single value
aVector.push_back(1);Since the size and capacity of
aVectorare the same, the
push_back()operation follows these steps:
All of these steps take constant time except for the step that copies
the values from the old array to the new array; this copying takes
time proportional to n---for each of the
copy the value into the new array.