Expecting Same Output ?

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1

I was looking at two C programs, These codes are giving different outputs but i was thinking of getting the same output, I have already asked same question other c developer forum, But I am not satisfied from that answers, can you please explain why they are giving such outputs.

enter code here Code 1

#include<stdio.h>
int main()
{
float x = 0.1;
if (x == 0.1)
printf(“IF”);
else if (x == 0.1f)
printf(“ELSE IF”);
else
printf(“ELSE”);
}

The output of the above program is ELSE IF

Code 2

include<stdio.h>
int main()
{
float x = 0.5;
if (x == 0.5)
printf(“IF”);
else if (x == 0.5f)
printf(“ELSE IF”);
else
printf(“ELSE”);
}

while this one gives IF as an output.

2

#include<stdio.h>
int main() { float x = 0.1; if (x == 0.1) printf(“IF”); else if (x == 0.1f) printf(“ELSE IF”); else printf(“ELSE”); }

The output of above program is “ELSE IF” which means the expression “x == 0.1″ returns false and expression “x == 0.1f” returns true.

Let consider the of following program to understand the reason behind the above output.

#include<stdio.h>
int main()
{
float x = 0.1;
printf("%d %d %d", sizeof(x), sizeof(0.1), sizeof(0.1f));
return 0;
}
The output of above program is “4 8 4” on a typical C compiler. It actually prints size of float, size of double and size of float.

The values used in an expression are considered as double (double precision floating point format http://en.wikipedia.org/wiki/Double_precision_floating-point_format ) unless a ‘f’ is specified at the end. So the expression “x==0.1″ has a double on right side and float which are stored in a single precision floating point format ( http://en.wikipedia.org/wiki/Single_precision_floating-point_format ) on left side. In such situations float is promoted to double (see this http://publib.boulder.ibm.com/infocenter/comphelp/v101v121/index.jsp?topic=/com.ibm.xlcpp101.aix.doc/language_ref/cplr066.html ). The double precision format uses uses more bits for precision than single precision format.
Note that the promotion of float to double can only cause mismatch when a value (like 0.1) uses more precision bits than the bits of single precision. For example, the following C program prints “IF”.

#include<stdio.h>
int main()
{
float x = 0.5;
if (x == 0.5)
printf(“IF”);
else if (x == 0.5f)
printf(“ELSE IF”);
else
printf(“ELSE”);
}
Output:

IF
You can refer Floating Point Representation – Basics ( http://www.geeksforgeeks.org/floating-point-representation-basics/ ) for representation of floating point numbers.

3

The simple answer is that 0.1 is not exactly represent-able within the precision of float and has to be truncated to the precision of float when assigned. For the comparison, the float is converted to double and compared with the literal 0.1 which was truncated to the precision of double.

If you change to double you will get True, as both will be truncated to the same value.

If you change to a number represent-able within the mantissa bits of the float (0.5 or 0.25 or 0.75 etc) you will also get True.