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int main(){

    long double fraDecimal,fraBinary,bFractional = 0.0,dFractional,fraFactor=0.1;
    long int dIntegral,bIntegral=0;
    long int intFactor=1,remainder,temp,i;

    printf("Enter any fractional decimal number: ");
    scanf("%Lf",&fraDecimal);

    dIntegral = fraDecimal;
    dFractional =  fraDecimal - dIntegral;

    while(dIntegral!=0){
         remainder=dIntegral%2;
         bIntegral=bIntegral+remainder*intFactor;
         dIntegral=dIntegral/2;
         intFactor=intFactor*10;
    }

   for(i=1;i<=12;i++){

       dFractional = dFractional * 2;
       temp =  dFractional;

       bFractional = bFractional + fraFactor* temp;
       if(temp ==1)
             dFractional = dFractional - temp;

       fraFactor=fraFactor/10;
   }

   fraBinary =  bIntegral +  bFractional;
   printf("Equivalent binary value: %Lf",fraBinary);

   return 0;
}
kiran Biradar
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electra3.14
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  • refer this one [here](https://stackoverflow.com/questions/52660296/why-does-double-in-c-print-lesser-decimal-digits-that-c) – asio_guy Oct 05 '18 at 07:41
  • You need mention the width of decimal points you want to print as `%.[width]f `example: `%.4f`-> this will print the 4 decimal points. – danglingpointer Oct 05 '18 at 07:47

1 Answers1

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Instead of this:

printf("Equivalent binary value: %Lf",fraBinary);

Try this:

printf("Equivalent binary value: %.12Lf", fraBinary);

Also note that it's not always displaying the last number correctly. At least on my machine, where long double is 8 bytes. It seems like this is pushing long double's precision to the limit. And sure enough, this states that

The 53-bit significand precision gives from 15 to 17 significant decimal digits precision

Blaze
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  • For what input is the program not displaying the last number correctly, what is the output you observe, and what is the output you would expect to be correct? – Eric Postpischil Oct 05 '18 at 11:05
  • @EricPostpischil tried it with the input `44.44`, got `101100.011100001007`, expected `101100.011100001010`. I suppose 128 bit floats would be required to solve this issue, barring a more complicated workaround. – Blaze Oct 05 '18 at 11:12
  • Thanks, now I see the problem is they are accumulating “binary” digits in a “decimal” numbering inside a binary floating-point format, and they push it too far. It would have been simpler just to accumulate characters in a string. – Eric Postpischil Oct 05 '18 at 11:37
  • Forgot to mention, I have already tried %.12Lf but when I do that it gives me 12 decimal place but it doesn't calculate anything. for example if i give input 4.404 the output becomes 0.000000000000 for any input. – electra3.14 Oct 05 '18 at 23:55