CS253: Software Development with C++

Fall 2021

DRY

Show Lecture.DRY as a slide show.

CS253 DRY

☔ ☔ ☔
Keep it DRY

Definition

D R Y
Don’t Repeat Yourself

Definition

W E T
Write Everything Twice

Repetition is bad

Example

Here’s code to see if a number is present in a vector. Apparently, the programmer didn’t know about set or the find() algorithm.

bool present(const vector<int> &vec, int target) {
    for (auto v : vec)
        if (v == target)
            return true;
    return false;
}

int main() {
    vector a = {11,22,33};
    cout << boolalpha << present(a, 22) << '\n'
         << present(a, 44) << "\n\n";
}
true
false

Example

We also need to find if a value is absent. That’s easy—copy & paste, change == to !=.

bool present(const vector<int> &vec, int target) {
    for (auto v : vec)
        if (v == target)
            return true;
    return false;
}

bool absent(const vector<int> &vec, int target) {
    for (auto v : vec)
        if (v != target)    // Changed == to !=
            return true;
    return false;
}

int main() {
    vector a = {11,22,33};
    cout << boolalpha << present(a, 22) << '\n'
         << present(a, 44) << '\n'
         << absent(a, 33) << "\n\n";
}
true
false
true

absent() is incorrect; 33 is not absent.

Try, try, again

Let’s try this:

bool present(const vector<int> &vec, int target) {
    for (auto v : vec)
        if (v == target)
            return true;
    return false;
}

bool absent(const vector<int> &vec, int target) {
    for (auto v : vec)
        if (v == target)
            return false;   // changed true to false
    return true;            // changed false to true
}

int main() {
    vector a = {11,22,33};
    cout << boolalpha << present(a, 22) << '\n'
         << present(a, 44) << '\n'
         << absent(a, 33) << "\n\n";
}
true
false
false

Better, but WET.

Not good enough

Sure, the previous solution is correct, but it’s not maintainable. What if we have to modify present(), say, to treat negative values as equal to positive values? Will we remember to also modify absent()?

Better

bool present(const vector<int> &vec, int target) {
    for (auto v : vec)
        if (v == target)
            return true;
    return false;
}

bool absent(const vector<int> &vec, int target) {
    return !present(vec, target);
}

int main() {
    vector a = {11,22,33};
    cout << boolalpha << present(a, 22) << '\n'
         << present(a, 44) << '\n'
         << absent(a, 33) << "\n\n";
}
true
false
false

Using references

As we all know, a reference creates an alias for another thing. For example:

int a = 42;
int &r = a;
cout << "r=" << r << " a=" << a << '\n';
a = 99;
cout << "r=" << r << " a=" << a << '\n';
r = 56;
cout << "r=" << r << " a=" << a << '\n';
r=42 a=42
r=99 a=99
r=56 a=56

Loops

Let’s say that we want to change all the fives in this vector to negative fives:

vector values = {3,1,4,1,5,9,2,6,5,3,5};
for (size_t i=0; i<values.size(); i++)
    if (values[i] == 5)
        values[i] = -5;
for (size_t i=0; i<values.size(); i++)
    cout << values[i] << ' ';
3 1 4 1 -5 9 2 6 -5 3 -5 

That works, but it’s a lot of repetition—quite a few instances of values[i].

Better loops

Let’s use a more modern loop to write out the numbers:

vector values = {3,1,4,1,5,9,2,6,5,3,5};
for (size_t i=0; i<values.size(); i++)
    if (values[i] == 5)
        values[i] = -5;
for (int v : values)
    cout << v << ' ';
3 1 4 1 -5 9 2 6 -5 3 -5 

Hooray! One fewer values[i]!

Betterer loops

Let’s use that same technique to change 5 to −5:

vector values = {3,1,4,1,5,9,2,6,5,3,5};
for (int v : values)
    if (v == 5)
        v = -5;
for (int v : values)
    cout << v << ' ';
3 1 4 1 5 9 2 6 5 3 5 

That didn’t work. Why?

Best loops

REFERENCES!

vector values = {3,1,4,1,5,9,2,6,5,3,5};
for (int &v : values)
    if (v == 5)
        v = -5;
for (int v : values)
    cout << v << ' ';
3 1 4 1 -5 9 2 6 -5 3 -5 

I need &v in the first loop, because I’m changing the number. I’m not changing anything in the second loop, so a reference isn’t required.

C🙈ns🙈rsh🙈p

Ducks

Consider this file:

% cat ~cs253/pub/ducks
Huey (red)
Dewey (blue)
Louie (green)

I want code that reads the lines of that file into a vector<string>, and converts the lower-case vowels to asterisks, so it looks like censorship.

Attempt #1

const string home = getpwnam("cs253")->pw_dir;
ifstream in(home+"/pub/ducks");
vector<string> ducks;
for (string line; getline(in, line); ducks.push_back(line));
for (size_t i=0; i<ducks.size(); i++)
    for (size_t j=0; j<ducks[i].size(); j++)
        if (ducks[i][j]=='a' || ducks[i][j]=='e' ||
            ducks[i][j]=='i' || ducks[i][j]=='o' ||
            ducks[i][j]=='u')
            ducks[i][j] = '*';
for (size_t i=0; i<ducks.size(); i++)
    cout << ducks[i] << '\n';
H**y (r*d)
D*w*y (bl**)
L**** (gr**n)

That was horrible. So soon, we’ve forgotten all that we’ve learned.

Attempt #2

const string home = getpwnam("cs253")->pw_dir;
ifstream in(home+"/pub/ducks");
vector<string> ducks;
for (string line; getline(in, line); ducks.push_back(line));
for (string &s : ducks)
    for (size_t j=0; j<s.size(); j++)
        if (s[j]=='a' || s[j]=='e' ||
            s[j]=='i' || s[j]=='o' ||
            s[j]=='u')
            s[j] = '*';
for (string s : ducks)
    cout << s << '\n';
H**y (r*d)
D*w*y (bl**)
L**** (gr**n)

Better, but still awful. What do we see a lot of?

Attempt #3

const string home = getpwnam("cs253")->pw_dir;
ifstream in(home+"/pub/ducks");
vector<string> ducks;
for (string line; getline(in, line); ducks.push_back(line));
for (string &s : ducks)
    for (char &c : s)
        if (c=='a' || c=='e' || c=='i' || c=='o' || c=='u')
            c = '*';
for (string s : ducks)
    cout << s << '\n';
H**y (r*d)
D*w*y (bl**)
L**** (gr**n)

Finally, it’s DRY! Still, we’re copying strings in the last loop.

Attempt #4

const string home = getpwnam("cs253")->pw_dir;
ifstream in(home+"/pub/ducks");
vector<string> ducks;
for (string line; getline(in, line); ducks.push_back(line));
for (string &s : ducks)
    for (char &c : s)
        if (c=='a' || c=='e' || c=='i' || c=='o' || c=='u')
            c = '*';
for (const string &s : ducks)
    cout << s << '\n';
H**y (r*d)
D*w*y (bl**)
L**** (gr**n)

DRY and efficient.

Attempt #5

const string home = getpwnam("cs253")->pw_dir;
ifstream in(home+"/pub/ducks");
vector<string> ducks;
for (string line; getline(in, line); ducks.push_back(line));
for (string &s : ducks)
    for (size_t p=0; (p = s.find_first_of("aeiou",p)) != s.npos; p++)
        s[p] = '*';
for (const string &s : ducks)
    cout << s << '\n';
H**y (r*d)
D*w*y (bl**)
L**** (gr**n)

A different approach.

Summary