Question

FracNum DecNum-num int count, IntPart float FracPart; float temp temp-FracNum *16; FracPartstemp- (int) temp; //getting Fractional part from temp IntPart (int) temp; count-e ise while (FracPart >e && count<=6) //while fractional part e and count Less than or equal to 6 then //Now taking fractional part of the number //muLtiplying Fractional part with 16 //getting integer part from temp if (IntPart<10) FHex[i++] 48 IntPart; if integer part <10 then assigning equivalent number to it //if integer part 10 then assinging equivalent character[A- - - F] to it else FHex[i++] 55 IntPart; CEC+1 //incrementing count temp-FracPart *16; FracPartstemp- (int ) temp; //getting Fractional part from temp IntPart (int) temp count count+1 //again multiplying fractional part with 16 //getting integer part from temp

The code is to convert the fractional part of a decimal number to its hexadecimal form.

Can you explain the code, for example ,why the fractional part need to multiply 16 instead of devise 16？

Answer 1

There are two parts in a decimal number system. The integer part and fractional part.

When converting a decimal number to Hexadecimal number.

Example: 10.16

10 is the integer part, 0.16 is a fractional part. Both the parts are converted differently.

For integer part :

The answer is (A)₁₆, which is obtained by dividing the number by 16.

For fractional part :

The answer is approximately (0.28F5C...)₁₆ which is obtained by multiplying the number by 16.

So the answer as the whole is (10.16)₁₀ = (A.28F5C...)₁₆

The same steps are followed in the program, as it uses the fractional part therefore it is multiplied by 16.