CODEWITHSUNDEEP
c
newya
newya

Reputation: 77

Addition of two numbers

I need help to solve a really simple problem. I wrote code that adds two numbers, but only in type float. So when I write 2+2, it gives me 4.0. What do I need to do in order to get only 4? But in the same time if I write 5.2+5.3 to get 10.5?

It's homework, and should not include an if statement.

I tried all variable types, but it just gives me unrealistic numbers. I would really appreciate if someone helped.

Code

#include <stdio.h>

/*Addition of two numbers*/

 int main()
{
    float a;
    float b;
    float x;

   printf("Enter the first number:\n");
   scanf("%f", &a);

   printf("Enter the second number:\n");
   scanf("%f", &b);

   x = a + b;

   /*Printing decimal number*/

   printf("Result: %.1f + %.1f = %.1f", a, b, x);

   return 0;

}

Upvotes: 5

Views: 272

Answers (2)

Adrian Mole
Adrian Mole

Reputation: 51825

If you want to be 'smart' (and I use that term loosely, rather than advisedly - and neither am I trying to detract anything from the great answer posted by @Sebi), and format your output according to the (maximum) precision of the input(s), then this will work for 'fractional' parts that aren't zero:

#include <stdio.h>

int main()
{
    float a, b, x, test;
    int aPrec = 0, bPrec = 0, xPrec = 0, iTest;

    printf("Enter the first number:\n");
    scanf("%f", &a);
    test = (float)fabs(a); // Just in case it's -ve!
    iTest = (int)(test + 0.5);
    while (test != (float)(iTest)) {
        ++aPrec;
        test *= 10.0;
        iTest = (int)(test + 0.5);
    }

    printf("Enter the second number:\n");
    scanf("%f", &b);
    test = (float)fabs(b); // Just in case it's -ve!
    iTest = (int)(test + 0.5);
    while (test != (float)(iTest)) {
        ++bPrec;
        test *= 10.0;
        iTest = (int)(test + 0.5);
    }

    x = a + b;
    xPrec = aPrec > bPrec ? aPrec : bPrec; // Or, you could use max()!!
    // The "*" precision specifiers get their values from arguments immediately …
    /// … preceding the relevant 'float' value!
    printf("Result: %.*f + %.*f = %.*f", aPrec, a, bPrec, b, xPrec, x);
    return 0;
}

What the code is doing, is to keep multiplying each input by 10 until it sees that this 'test' value is then equal to its integral (truncated) value; this is the "precision" value for each input, a and b. Now, being smart (again, I use the term loosely), it thinks that the output precision should be the greater of the two inputs.

If you don't know how the "%.*f format specifier works, I can add more details, of course.

Upvotes: 1

Sebi
Sebi

Reputation: 588

I think what you want to do is to use the g format specifier for the printf instruction. This is used for printing the shortest representation. You can read more about format specifiers here: http://www.cplusplus.com/reference/cstdio/printf/

This code prints how you described: 5 + 5 = 10 and 5.1 + 5.2 = 10.3

#include <stdio.h>

int  main()
{
  float a;
  float b;
  float x;

  printf("Enter the first number:\n");
  scanf("%f", &a);

  printf("Enter the second number:\n");
  scanf("%f", &b);

  x = a + b;

  /*Printing decimal number*/

  printf("Result: %.1f + %.1f = %g\n", a, b, x);

  return 0;

}

Upvotes: 10

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