"i < 0.7" is true following an "float i = 0.7" initialization above?

Question

Possible Duplicate:
Floating point comparison

#include<stdio.h>
#include<conio.h>
int main()
{
float i=0.7;
clrscr();
if(i < 0.7)
     printf("If Block");
else
     printf("Else Block");
getch();
return 0;
}

I dont understand whay the output will be "If block".....please help why the if part is executed?

Spend an hour or so and read [What Every Computer Scientist Should Know About Floating-Point Arithmetic](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html). It will be a real eye-opener if you're fuzzy on floating point approximation. — WhozCraig, Nov 27 '12 at 18:22

score 13 · Accepted Answer · edited May 08 '17 at 11:23

13

Actually i is 0.69999999998 in it's floating representation.

When you assign i=0.7 in memory 0.7 cannot be represented in double precision as you would have thought.

So the comparison between float and double leads to type promotion and in that case i is less than 0.7 which is double.

edited May 08 '17 at 11:23

gsamaras

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answered Nov 27 '12 at 18:19

Omkant

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shouldn't 0.7 always round to the same value, causing the comparison to consistently fail? – John Dvorak Nov 27 '12 at 18:21
3

@JanDvorak: no, the comparison is between a `float` and a `double`, so the `float` is promoted and compared as `double`. – Yakov Galka Nov 27 '12 at 18:22
0.7 is by default double – Omkant Nov 27 '12 at 18:23
The key ingredient is the implicit double=>float cast that rounds down, then compares to the original double. – John Dvorak Nov 27 '12 at 18:23
1

@Jan Dvorak: If the original code used `0.7f` instead of just `0.7`, the comparison would fail, since if would indeed be represented identically. But `0.7` by itself is a `double`, which is represented differently than that `float i`. – AnT stands with Russia Nov 27 '12 at 18:27
`(double)((float)0.7) != 0.7` -- there's loss of precision in the conversions. – pmg May 08 '17 at 11:52

"i < 0.7" is true following an "float i = 0.7" initialization above?

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