A simple square root calculator

Hello all my dear friends,

This is a very simple calculator I made to find the square root of an entered number.

It works fine and errors free

The calculator keeps running until the user enter 0 to quit the program

I hope the code is free of mistakes

Code:

---


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

int main()
{
    float x;

  do {
    printf("Square Root Calculator\n\n");
    printf("\nNote:\nThe square root of 1 is 1, and 0 is 0\n\n");

    system("pause");
    system("cls");

    printf("\nEnter a number to find the sqauare root, 1 to go back to main or 0 to exit:\n");

    scanf("%f", &x);

    if(x != 0 && x > 1){

    x = sqrt(x);

    printf("\n\n\nThe sqaure root of the entered number is %.1f\n\n\n", x);
    }

    else if(x == 1){

        system ("cls");
        continue;

    }
    else if(x == 0){
        exit(0);
    }

    system ("pause");
    system ("cls");

  } while (x != 0);

}

---

Please try it and give me your feedback


Thank you :)

What happens if I input 0.1?

Quote:

---

Originally Posted by claudiu

What happens if I input 0.1?

---

You are right, it brings the user to the main screen.

so the new code after correcting (Bold):

Code:

---


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

int main()
{
    float x;

  do {
    printf("Square Root Calculator\n\n");
    printf("\nNote:\nThe square root of 1 is 1, and 0 is 0\n\n");

    system("pause");
    system("cls");

    printf("\nEnter a number to find the sqauare root, 1 to go back to main or 0 to exit:\n");

    scanf("%f", &x);

    if(x != 0 && x != 1){

    x = sqrt(x);

    printf("\n\n\nThe sqaure root of the entered number is %.1f\n\n\n", x);
    }

    else if(x == 1){

        system ("cls");
        continue;

    }
    else if(x == 0){
        exit(0);
    }

    system ("pause");
    system ("cls");

  } while (x != 0);

}

---

Thank you for your note, claudio :)

Why do you explicitly look for x==1? What's the problem with including that in the general conditions? sqrt(1) is 1.

Also, ideally your stopping condition should be something like x = -1 or x <0. 0 also has a valid sq root. It's 0.

Quote:

---

Originally Posted by claudiu

Why do you explicitly look for x==1? What's the problem with including that in the general conditions? sqrt(1) is 1..

Also, ideally your stopping condition should be something like x = -1 or x <0. 0 also has a valid sq root. It's 0.

---

I already gave the user a note telling them that the square root of 0 and 1, because the program will use 0 and 1 to do functions like 1 to go back and 0 to exit.


I will try to make some changes as in your reply above and post the code again

Now, it's too late here in malaysia (3:21 AM) and I have classes tomorrow,

so thank you claudio, and good night

here i my code find squareroot without math.h:)

Code:

---

#include<stdio.h>

int main(){

        int i;
        double number,x;

        printf("\nenter number:");
        scanf("%lf",&number);

        x=20;

        for(i=0;i<1000;i++){
                x=0.5*(x+(number/x));

        }
        printf("\nresult:%f\n",x);
        return 0;

}

---

O_o... Where did you find that code?!

Quote:

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Originally Posted by Sipher

O_o... Where did you find that code?!

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i wrote with using this sequence

http://img138.imageshack.us/img138/1696/90511389.jpg

if you want to find exactly result, you have to find limit, but 1000 loop is also give very very close answer:)