Outputted as same as inputted

**main()** · 05-12-2010

Hi,

I kept it up too late last night trying to find an answer for my question which is:

Is it possible for a float number to be displayed as it was entered?

I mean if I enter 2.23, the output should be 2.23, and if I enter 2.123, the output should be 2.123, bothing entering using the same program, and no rounding-up rules, just showing the output the same way as the input.

Thank you

**whiteflags** · 05-12-2010

If you know how the input is formatted it is possible. You just print it out with printf in the same format you read it in.

**hk_mp5kpdw** · 05-12-2010

Enter/process/output it as a string.

**GL.Sam** · 05-12-2010

Actual value you read are not get changed in any way, if you up to the range limits of course. What may change is how it is displayed.

So, your 2.23 would look differently when used with different format strings.

**MK27** · 05-12-2010

Important to understand the limitations in floating point. The only numbers that can be represented exactly are inverse power of two fractions: 1/2 (0.5), 1/4(0.25), 1/8(0.125), etc. Not included: 0.1

Code:

#include <stdio.h>

int main() {
        float i;
        for (i=0.0f; i<20; i+=0.1f) {
                printf("%f\n",i);
        }
        return 0;
}

The output is not what you expect.

This is not to say you cannot get what you want, just you need to understand the problem first.

**main()** · 05-12-2010

Ok, this is for more explaining:

int main()
{
float num1, num2, add,

printf("Input two numbers to add");
scanf("%f%f", &num1, &num2);

add = num1 + num2;

printf("result of adding %f and %f is %f", num1, num2, add);

return 0;

}

when I run the above program, and input for example 2.23 for num1 and 2.43 for num2, the output is like this:

result of adding 2.230000 and 2.430000 is 4.66000

I want the output to be the same as the input , input 2.23 output 2.23 not 2.230000, as well as any float number like 2.123 to be output like 2.123 not 2.123000

HOW?

and Why is the rounding-up six decimal points?

**quzah** · 05-12-2010

It's not rounding, it has a default precision. You can specify a fixed precision, but if you want it to be the exact way it was entered, you're best off with strings.

Quzah.

**KBriggs** · 05-12-2010

Use %.2f if you want 2 decimal places. To different numbers of decimal places depending on the input, you'll have to figure out a way of determining how many significant decimal places your number has between the reading in and the outputting.

On the other hand, using strings makes it all quite simple.

**main()** · 05-12-2010

Originally Posted by quzah

It's not rounding, it has a default precision. You can specify a fixed precision, but if you want it to be the exact way it was entered, you're best off with strings.

Quzah.

Can you please explain more? example would be better if possible.

**quzah** · 05-12-2010

Explain which part, I said more than one thing there. There's a FAQ on reading input as a string.

Cprogramming.com FAQ > Get a line of text from the user/keyboard (C)
Cprogramming.com FAQ > How do I get a number from the user (C)

Quzah.

**main()** · 05-12-2010

Originally Posted by KBriggs

Use %.2f if you want 2 decimal places. To different numbers of decimal places depending on the input, you'll have to figure out a way of determining how many significant decimal places your number has between the reading in and the outputting.

On the other hand, using strings makes it all quite simple.

How to use string for reading in floats? example please

**MK27** · 05-12-2010

Originally Posted by KBriggs

Use %.2f if you want 2 decimal places.

This will format the output. However, it is not dynamic, meaning 3.666 will appear as 3.67.

As Quzah says, you will have to store the data as a string, and NOT a float, if you want that.

**quzah** · 05-12-2010

I suppose we could just use %g.

Of course, if you enter "1.230", it won't print "1.230", it will print "1.23". So again, strings.

Quzah.

**MK27** · 05-12-2010

Here:

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h> // needed for strtof

int main(int argc, const char *argv[]) {
	float n = 1.234;
	char str[32], *p;

	if (argc > 1) n = strtof(argv[1],NULL);

	sprintf(str, "%f", n);
	p = &str[strlen(str)-1];
	while (*p == '0') *(p--) = '\0';

	printf("%s\n", str);

	return 0;
}

If you are using gcc strtof() requires -std=c99. This contains a potentially uncontrollable overflow though since sprintf can't limit the input length with %f.

**main()** · 05-12-2010

Thank you all guys so much for your very helpful responses, I solved the problem using strings

This forum is the best ever

Thank you again

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