Digital calculator :D

Nothing. Just posting this for fun. it takes around 30 seconds to make.

I accomplished alot of things. I'm learning Direct3d right now.

Well, about 5 months ago. I was trying to make a small console digital calculator. I didn't even finished it and it was around 1000+ line of codes.

Now, today, i just said, why not try the digital calculator again with my C++ skills.

Haha, i just realized how easy it was.

Here's the code.

Code:

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#include<iostream>
#include<string>
using namespace std;
int main()
{
float value1;
string variable = "";
float value2;
while(1)
{
cout<<">>";
cin >> value1;
cin >> variable;
cin >> value2;
cin.get();
if(variable == "+")
{
cout <<value1 + value2;
}
else if(variable == "-")
{
cout <<value1 - value2;
}
else if(variable == "/")
{
cout <<value1 / value2;
}
else if(variable == "*")
{
cout <<value1 * value2;
}
cin.get();
system("cls");
}
cin.get();
return 0;
}


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Oh well, back to Directx. Cya.

By the way, if your wondering how to use it.
Example:


5 * 5
5 + 5

Dont forget the spaces.

5(space) * (space) 5

Just a comment - I think the "variable" should be more appropriately called the "operator" :).

You somehow also messed up indentation of the code you posted (but I do assume you properly indented the code when you wrote it).
Simple, but effective. You can't ask for more, can you? ;)

I think it would take no more lines in straight C.

It probably would not. It's a too simple application.

Quote:

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I think it would take no more lines in straight C.

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And the point is?

Quote:

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Originally Posted by nonoob

I think it would take no more lines in straight C.

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I think the 1000+ lines vs. <100 lines was not C vs. C++, but 0 months experience vs. 5 months experience.

Example:

Code:

---

#include <stdio.h>
#include <math.h>

float mul(float a, float b) { return a*b; }
float add(float a, float b) { return a+b; }
float sub(float a, float b) { return a-b; }
float div(float a, float b) { return a*b; }
float safe(float a, float b, float (*func)(float,float)) { return (func)?func(a,b):NAN; }

int main(void)
{
  float (*operators)(float,float)[] = {mul,add,0,sub,0,div}, a, b;
  char op;

  printf(">>");
  while(sscanf("&#37;f %c %f", &a, &op, &b) == 3)
  {
    if(op >= '*' && op <= '/')
      printf("%f\n", safe(a, b, operators[op-'*']));
    else
      fputs("Invalid entry! (Example: 5 * 5)", stderr);
    printf(">>");
  }

  return 0;
}

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Happy, nonoob?