Calculator Program

[alex067]

So I have a small project where we have to construct a calculator program

However, the problem I have is:

ask for the user to enter their equation all at once (i.e., as "4 * 5", with each piece separated by whitespace) instead of prompting for each piece separately

My question is, what variable do I store the whole equation under? A string?

Also, my professor is asking us to use cin.peek to predetermine what the binary operator is being used (ie +, -. *), how do I include this?

[TwilightSpectre]

You don't need to store the whole input at once, though if you did it would have to be as a string. As for the rest of it, just use the stream extraction operator (std::cin >> ...) to input the first value. I don't really know why std::cin.peek is required, but you could just test for the operator after taking in the first number and using std::cin.ignore to get rid of the useless whitespace.

[alex067]

my professor just emailed me the following code

    cout << "Please enter an expression (like `4 * 5` or `ln 5`): ";
    while (    cin.peek() == ' '  || cin.peek() == '\t'
        || cin.peek() == '\n' || cin.peek() == '\v'
        || cin.peek() == '\f' || cin.peek() == '\r' )
    cin.get();
    
if (cin.peek() >= 'a' && cin.peek() <= 'z') {
cin >> op >> a;
} else if (cin.peek() >= '0' && cin.peek() <= '9') {
cin >> a >> op >> b; 
}

Can someone please explain to me what is exactly going on in this code?

[TwilightSpectre]

int a, b;
std::string op;
std::cout << "Please enter an expression: ";
std::cin >> a;

// pointless
while (std::cin.peek() == ' ')
    std::cin.ignore();

std::cin >> op;
std::cin >> b;

// process operator and values... 

Of course, you would need to modify this to work for things like 'ln 5', but you get the basic idea. Maybe you can think of a way to make std::cin.peek necessary, too...

EDIT:
Maybe you can use std::cin.peek to test to see if the first character is a number, and process the equation differently otherwise.

[alex067]

yes but my professor wants me to input the following code

   cout << "Please enter an expression (like `4 * 5` or `ln 5`): ";
    while (    cin.peek() == ' '  || cin.peek() == '\t'
        || cin.peek() == '\n' || cin.peek() == '\v'
        || cin.peek() == '\f' || cin.peek() == '\r' )
    cin.get();
    
if (cin.peek() >= 'a' && cin.peek() <= 'z') {
cin >> op >> a;
} else if (cin.peek() >= '0' && cin.peek() <= '9') {
cin >> a >> op >> b; 
}

I am at a loss here, what does this following exactly do?

I understand that cin.peek peeks a certain set of characters to determine what they are, so how does this work when trying to figure out what kind of equation the user input?

[giblit]

The code is like this:

Ask the user to enter an equation

lines 2->5 pretty much ignore any whitespaces at the start of an input so if you have something like

   \r\n    12

it becomes

12

since you are using std::cin >> after all and it stops at the next whitespace.

line 7 checks if the next character is a letter if so read in the operator and 1 variable.

line 9 is otherwise if the next character is an integer then read in a variable, operator, and another variable.

You figure out the equal based on what the operator is. I would assume the operator is a string and not a character so you can input ln otherwise you'd probably only be able to do +,-,*,/

[JLBorges]

> yes but my professor wants me to input the following code

This is what your professor is attempting to do:

a. skip over leading white-space
b. peek at the next (first non-white-space) character
      if it is alphabetic  => read operation, number: eg. 'ln 5'
      else if it is a digit => read number, operation, second number: eg. '4 * 5'

This is one of the sane ways of doing it:

#include <iostream>
#include <string>
#include <cctype>

int main()
{
    double number, second_number ;
    std::string operation ;

    std::cout << "Please enter an expression (like `4 * 5` or `ln 5`): ";

    // skip over leading white-space
    char c ;
    std::cin >> c ; // read the first non-whitespace character
    std::cin.putback(c) ; // and put it back

    // look at the first non-whitespace character
    if( std::isdigit(c) ) // if the first char is a digit
    {
        // input is of the form 'number operator number';
        // read it into number, operation, second_number
        std::cin >> number >> operation >> second_number ;

        std::cout << "number==" << number << " operation=='" << operation
                   << "' second_number==" << second_number << '\n' ;
    }

    else if( std::isalpha(c) ) // if the input starts with an alphabet
    {
        // input is of the form 'operator number';
        // read it into operation, number
        std::cin >> operation >> number ;

        std::cout << "operation=='" << operation << "' number==" << number << '\n' ;
    }
    
    // else // error in input: first non-whitespace character is neither a digit nor a letter
}

http://coliru.stacked-crooked.com/a/cc9f207dfdaa96d9

[36k1hbRD]

use this

#include <iostream>

//    writing this line here would be better
//instead of inside main().
using namespace std;

int main()
{
    //    for beautiful coding
    //always add your declarations
    //to the beginning.
    char o;
    int x;
    int y;

    //    it will keep looping as long
    //as x isnt equal to -1.
    while(x != -1)
    {
        cout << "enter a number: \n";
        cin >> x;
        cout << "\n";
        cout << "\n";
        cout << "1 = multiplication  ";
        cout << "2 = division  ";
        cout << "3 = addition  ";
        cout << "4 = subtraction  ";
        cout << "-----------------------------------------------------------------------------------------\n";
        cout << "enter your choice: \n";
        cin >> o;
        cout << "enter your last number: \n";
        cin >> y;
        //it should be == instead of =
        //haha sorry i didnt realized it :D
        if ((o == 1))
            cout << x * y << endl;
        else if ((o == 2))
            cout << x / y << endl;
        else if ((o == 3))
            cout << x + y << endl;
        else if ((o == 4))
            cout << x - y << endl;
        else
            cout << "you entered an invalid operator\a";
    }

    return 0;
}

[giblit]

There's a few problems wrong with that jdogsis.
1) it does not meet his requirements
2)your operator is a character and not a string
3) all your if statements are wrong characters have single quotes around them(' ')
4) they can enter just an operator and 1 number example ln 5 or log 10
5) What if they want to do something like -1 * 100 (anyways you never explicitly told the user what the exit key was)