C++ | For Loop Efficiency? - c++

Question

I've been learning C++ and obviously before hitting loops, the tuition manual I've been reading provided me with this scenario:
Expand the password checking program from earlier in this chapter and make it take multiple usernames, each with their own password, and ensure that the right username is used for the right password. Provide the ability to prompt users again if the first login attempt failed. Think about how easy (or hard) it is to do this for a lot of usernames and passwords.
Without any knowledge yet of loops and whatnot that'll repeat any incorrect values prompting the user to input the correct information until they do so, I've attempted to complete the task and ended up with this code here:
#include < iostream >
using namespace std;
int main()
{
int inputCodeOne, inputCodeTwo = 0;
bool correctPassOne=false,correctPassTwo;
cout << "Please enter your first Code: ";
cin >> inputCodeOne;
if(inputCodeOne==1111||inputCodeOne==2222||inputCodeOne==3333)
{
correctPassOne = true;
}
if (correctPassOne)
{
cout << "Please enter your second Code: ";
cin >> inputCodeTwo;
if (inputCodeOne == 1111 && inputCodeTwo == 100)
{
cout << "Password Correct! Welcome back David";
return 0;
}
else if (inputCodeOne == 2222 && inputCodeTwo == 200)
{
cout << "Password Correct! Welcome back Darren";
return 0;
}
else if (inputCodeOne == 3333 && inputCodeTwo == 300)
{
cout << "Password Correct! Welcome back Jake";
return 0;
}
correctPassTwo = false;
if(!correctPassTwo)
{
cout << "Please re-enter your second Code: ";
cin >> inputCodeTwo;
if (inputCodeOne == 1111 && inputCodeTwo == 100)
{
cout << "Password Correct! Welcome back David";
return 0;
}
else if (inputCodeOne == 2222 && inputCodeTwo == 200)
{
cout << "Password Correct! Welcome back Darren";
return 0;
}
else if (inputCodeOne == 3333 && inputCodeTwo == 300)
{
cout << "Password Correct! Welcome back Jake";
return 0;
}
cout << "ACCESS DENIED";
return 0;
}
}
else
{
cout << "Please re-enter your first Code: ";
cin >> inputCodeOne;
if (inputCodeOne == 1111 && inputCodeTwo == 100)
{
cout << "Password Correct! Welcome back David";
return 0;
}
else if (inputCodeOne == 2222 && inputCodeTwo == 200)
{
cout << "Password Correct! Welcome back Darren";
return 0;
}
else if (inputCodeOne == 3333 && inputCodeTwo == 300)
{
cout << "Password Correct! Welcome back Jake";
return 0;
}
else
{
cout << "Please enter your second Code: ";
cin >> inputCodeTwo;
if (inputCodeOne == 1111 && inputCodeTwo == 100)
{
cout << "Password Correct! Welcome back David";
return 0;
}
else if (inputCodeOne == 2222 && inputCodeTwo == 200)
{
cout << "Password Correct! Welcome back Darren";
return 0;
}
else if (inputCodeOne == 3333 && inputCodeTwo == 300)
{
cout << "Password Correct! Welcome back Jake";
return 0;
}
correctPassTwo = false;
if (!correctPassTwo)
{
cout << "Please re-enter your second Code: ";
cin >> inputCodeTwo;
if (inputCodeOne == 1111 && inputCodeTwo == 100)
{
cout << "Password Correct! Welcome back David";
return 0;
}
else if (inputCodeOne == 2222 && inputCodeTwo == 200)
{
cout << "Password Correct! Welcome back Darren";
return 0;
}
else if (inputCodeOne == 3333 && inputCodeTwo == 300)
{
cout << "Password Correct! Welcome back Jake";
return 0;
}
else
{
cout << "ACCESS DENIED";
return 0;
}
}
}
}
}
Apologies for how messy the code probably is, but I wanted to know if there were any ways to make this more space efficient whilst providing the same result.

Answer 1

Optimization -- First Pass
Create indentifiers or named constants
This allows you to avoid duplication and making typos. Also allows for compiler to make more optimizations.
const int PASSWORD1a = 1111;
const int PASSWORD2a = 2222;
const int PASSWORD3a = 3333;
const int PASSWORD1b = 100;
const int PASSWORD2b = 200;
const int PASSWORD3b = 300;
Group passwords together.
Keeping the passwords paired together will make the process more generic.
You could use the existing std::pair or create your own:
struct Password_Entry
{
int first;
int second;
};
Next, create a table of valid password pairs:
const Password_Entry valid_passwords[] =
{
{PASSWORD1a, PASSWORD1b},
{PASSWORD2a, PASSWORD2b},
{PASSWORD3a, PASSWORD3b},
};
const size_t quantity_valid_passwords =
sizeof(valid_passwords) / sizeof(valid_passwords[0]);
Search the table for valid passwords
int inputCode1;
int inputCode2;
bool passwords_are_valid = false;
std::cout << "Enter first password: ";
std::cin >> inputCode1;
for (unsigned int index = 0; index < quantity_valid_passwords; ++index)
{
if (inputCode1 == valid_passwords[i].first)
{
std::cout << "Enter second password: ";
std::cin >> inputCode2;
if (inputCode2 == valid_passwords[i].second)
{
passwords_are_valid = true;
}
}
}
Summary
The above code is table driven. The code to search the table is generic and depends on the data in the table. The quantity of entries can be changed without having to modify the remaining code.
Pairing the first password with the second, in a structure, allows for more optimal data storage and code space.
Using named constants allows for the value to only be specified once. If you need to change the value, you only make one change. You don't make the risk of skipping past one or more when making the changes.
Optimization -- Second Pass
Person name
Printing of the person's name can be optimized by adding another field or member to the structure:
struct Password_Entry
{
int first;
int second;
char * name;
};
The table now becomes:
const Password_Entry valid_passwords[] =
{
{PASSWORD1a, PASSWORD1b, "David"},
{PASSWORD2a, PASSWORD2b, "Darren"},
{PASSWORD3a, PASSWORD3b, "Jake"},
};
const size_t quantity_valid_passwords =
sizeof(valid_passwords) / sizeof(valid_passwords[0]);
The search / validation code is changed as:
std::cout << "Enter second password: ";
std::cin >> inputCode2;
if (inputCode2 == valid_passwords[i].second)
{
passwords_are_valid = true;
std::cout << "Password Correct! Welcome Back "
<< valid_passwords[i].name
<< "!\n";
}
Optimization -- Third Pass
Consolidation of duplicate text
There is duplicate text, which means that more space can be squeezed out:
char const * const text_enter[] = "Enter ";
char const * const text_password[] = "password";
The code can be changed as:
std::cout << text_enter << "first " << text_password << ": ";
//...
std::cout << text_enter << "second " << text_password << ": ";
//...
std::cout << "Correct " << text_password << "! Welcome Back "
<< valid_passwords[index].name << "\n";
Block Writing
You may be able to squeeze some code space by block writing instead of using formatted writing. All the output is text, so there is no need to format; the text can be directly output. You'll have to compare assembly language listings of before this optimization and after to measure the space difference.
This technique may also show some speed improvements.
Use the following:
std::cout.write(text_enter, sizeof(text_enter) - 1U); // -1 so the terminating nul is not output.
std::cout.write(text_password, sizeof(text_password) - 1U);
std::cout.write(": ", 2);
Likewise, replace the other std::cout << with std::cout.write like the code above.
Optimization -- Fourth Pass
Don't use std::cout.
The std::cout may carry extra baggage with it. You can save some code space by using an alternative.
Replace std::cout with fwrite(stdout, /*...*/).
The fwrite function contains minimal code to write to the given stream. No extra code to format or convert. Plain and simple, write the data to the stream.
Some compilers may be lazy and insert a "one-size-fits-all" library rather than only the code for fwrite.
You may be able to squeeze out more code space by accessing the low level drivers of your operating system directly. Your program is small enough that it doesn't need buffering and some other overhead from the C++ and OS streams. At this point, the space savings may be negligible to the development time spent achieving this. Depends on your platform. On memory constrained systems, this may be a worthwhile endeavor. On most desktop systems, this is not worth your development time, as they have more memory than the constrained systems.

Answer 2

As Jesper commented, yes there are many ways. I'd encourage you to learn about loops first and then revisit this.
Loops just allow you to do the same thing again, anywhere from no times at all (loop doesn't run) to forever (infinite loop).
Anytime you find yourself typing the same thing again or copy-pasting, there is a very high chance that using a loop (or moving code into a function, a class, etc.) would be a better option.
Also, your code doesn't appear to behave as intended. If you initially type the incorrect input, correctPassOne is false, and you going into the first else and prompted to "re-enter your first code:".
The first set of if else statements will never pass. It's checking if inputCodeTwo is correct but it has never been given by the user at this point.
It seems like you recognized this and put the else statement after that which then prompts for inputCodeTwo.
The problem is now that if inputCodeOne was incorrect the second time, inputCodeTwo doesn't matter. So you'll prompt for inputCodeTwo with no chance of success.
On lines 103, 104 you type
correctPassTwo = false;
if(!correctPassTwo) { }
If 104 is ever reached, 103 will have been reached as well, so the assignment of false and the conditional are unnecessary.
Again if inputCodeOne was incorrect twice, this code block is unnecessary. You're now prompting the user to give the second code twice, with no chance of success occurring. Because inputCodeOne is still wrong.
With loops you could do something like this:
Ask for inputCodeOne as many times as you'd like, until it's correct or until the limit is reached.
If the limit is reached, print "ACCESS DENIED" and end program.
If the password is correct, you can go onto the inputCodeTwo.
Same thing, ask for inputCodeTwo as many times as you'd like.
If the two codes match, print your password correct message.
Otherwise, after a certain number of attempts, you can terminate the program.
You can do that with 25% of the initial code.