Explain the output when converting Float to integer? [duplicate] - c

Question

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Is floating point math broken?
(31 answers)
Closed 8 years ago.
I exactly don't know how why output is coming
float f=1.4, t;
int d,s;
d=(int)f;
printf("d=%d\n",d);
t=f-d;
printf("t=%f\n",t);
t=t*10;
printf("t=%f\n",t);
s=(int)t;
printf("s=%d\n",s);
the output is
d=1
t=0.400000
t=4.000000
s=3
and similarly when f=1.1
the output is
d=1
t=0.100000
t=1.000000
s=1
Is this related to the way the integer and float is stored in the memory or something else ?

Answer 1

You have initialized f=1.4, not when you do
d=(int)f;
You are converting float to integer, and when float is converted to integer, all the number after period "." are truncated. Now d has 1 so
t=f-d;
will be 1.4 - 1 = 0.4
t=t*10;
t=0.4*10=4 and as t is float so it outputs 4.0000
Float represent trailing zeros at the end
s=(int)t;
Here again you are converting float to integer, Now here is tricky part, All the values above are rounded off, here t has value 3.99999976 so when converted to integer it shows 3 as a result
This is all because when you initialize t=1.4, in actual it is initialized to 1.39999998

Answer 2

During the first assignment
float f=1.4;
there is an approximation because 1.4 is intended as double (not float). Something like 1.39999999 is assigned to f.
Try to use
float f=1.4f;
and it should work as you expect.

Answer 3

Let's take it step-by-step and see how floating point and int interact.
Assume a typical platform where
float is IEEE 754 single-precision binary floating-point format and
double is IEEE 754 double-precision binary floating-point format
float f=1.4, t;
// 1.4 isn't exactly representable in FP & takes on the closest `double` value of
// 1.399999999999999911182158029987476766109466552734375
// which when assigned to a float becomes
// 1.39999997615814208984375
int d,s;
d=(int)f;
// d gets the truncated value of 1 and prints 1, no surprise.
printf("d=%d\n",d);
t=f-d;
// t gets the value 0.39999997615814208984375
// A (double) version of t, with the same value is passed to printf()
// This is printed out, rounded to 6 (default) decimal places after the '.' as
// 0.400000
printf("t=%f\n",t);
t=t*10;
// t is multiplied by exactly 10 and gets the value
// 3.9999997615814208984375
// A (double) version of t, with the same value is passed to printf()
// which prints out, rounded to 6 decimal places after the '.' as
// 4.00000.
printf("t=%f\n",t);
s=(int)t;
// s gets the truncated value of 3.9999997615814208984375
// which is 3 and prints out 3. - A bit of a surprise.
printf("s=%d\n",s);