How do I print out only the decimal places for a number in C?

So here is the question:

Create a program that will read in a floating point number and output the decimal part of the number. e.g. given 3.14159, it will output .14159

I do not know the %f for only showing the decimal part of the number the user enters. What is it to only show the decimal places of a number?

Please explain your answer.

Code:

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int main()
    {
    float n1;

    printf("Please enter a number with 4 decimal places:\n");
    scanf("%f", &n1);
    printf("The decimal places you entered are %f .\n", n1);

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floor is in math.h

Code:

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printf("The decimal places you entered are %f .\n", n1-floor(n1));

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%f may need to be %.4f

Tim S.

Quote:

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Originally Posted by stahta01

floor is in math.h

Code:

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printf("The decimal places you entered are %f .\n", n1-floor(n1));

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%f may need to be %.4f

Tim S.

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I've tried that and it still prints out the whole number in front of the decimal places. Is there anyway to - the portion of the whole number so I am just left with the decimals?

convert to int and subtract the int value from the float

Quote:

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Originally Posted by Elkvis

convert to int and subtract the int value from the float

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How would I do that?

Quote:

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Originally Posted by Elkvis

convert to int and subtract the int value from the float

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Quote:

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Originally Posted by benrogers

How would I do that?

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Quote:

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Originally Posted by stahta01

floor is in math.h

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I am obligated to type something here so the post goes through.

Thanks a bunch.

Code:

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


int main()
{
    float n1;
   
    printf("Please enter a number with at least 4 decimal places:\n");
    scanf("%f", &n1);
    printf("The decimal places you entered are %f .\n", n1-floor(n1));


    return 0;
}

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works perfectly!