simple precision test getting inf output in C

Question

Trying to test precision making a sum from 1 to a million but my output keeps coming out as inf.

    #include <stdio.h>



int main()
{
float x=0.0 ;
double y =0.0;
int  i;
for(i=1; i<100000; i = i+1)
{
x=(float)  x+( 1.0f/(3.0f*(float)(i*i)));
y=(double) y+ ( 1.0/(3.0*(double)(i*i)));
}
printf("Natural Order \n");
printf("Single Precision: ");
printf("%f", x);
printf("\n");
printf("Double Precision: ");
printf("%lf", y);
printf("\n");
}

I have changed i range many times but still getting inf as my output.

what do you want to do with i^2 ? If you meant to square i, then use i*i — gtatr, Feb 02 '16 at 16:28
I have fixed that but when i print out x and y I still get "inf" — Kassandra Sandoval, Feb 02 '16 at 16:30
Yes. Im so sorry. Thank you. I realized right after I posted my edits. — Kassandra Sandoval, Feb 02 '16 at 16:35
If `i^2` is meant to be `i`-squared, use `i * i` instead... `^` is bitwise XOR. — Dmitri, Feb 02 '16 at 16:35
[Why is my power operator (^) not working?](http://stackoverflow.com/q/4843304/995714) — phuclv, Feb 02 '16 at 16:39

jofel · Answer 1 · 2016-02-02T16:36:07.733

3

i^2 is not i-squared, but a binary manipulation (xor). Use e.g. i*i instead. There is also the pow function for higher powers.

edited Feb 02 '16 at 16:36

answered Feb 02 '16 at 16:27

jofel

3,297
17
31

there is a `(float)` or `(double)` in front of the `1/3`, so it is not an integer division. – mch Feb 02 '16 at 16:32
@mch Ok, thanks. Operator precedence is a little bit complicated in c/c++. – jofel Feb 02 '16 at 16:35

LPs · Answer 2 · 2016-02-02T19:44:25.940

3

What do you want is probably the following:

int main()
{
    float x=0.0 ;
    double y =0.0;
    int  i;

    for(i=1; i<10; i = i+1)
    {
       x= x+( 1.0f / ( 3.0f * (float)(i)*(float)(i)   ));
       y= y+(  1.0 / ( 3.0  * (double)(i)*(double)(i) ));
    }

    printf("Single Precision: ");
    printf("%f", x);
    printf("\n");
    printf("Double Precision: ");
    printf("%f", y);
    printf("\n");

    return 0;
}

c is not like excel: ^ is not the way to pow a number/variable.

In c languace ^ is XOR bitwise operation

Your code does i XOR i that is 0 when i=2. Then the printf outputs inf because of the division by zero.

I did a lot of code optimization changing LD A, 0 with XOR A ...Z80 talking :)

Take note that printf %f format promote passed variable to double.

edited Feb 02 '16 at 19:44

answered Feb 02 '16 at 16:28

LPs

16,045
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even x that is originally a float? – Kassandra Sandoval Feb 02 '16 at 16:59
If you are talking about printf promotion: yes it is. – LPs Feb 02 '16 at 17:00
i was testing my code with your corrections and after I set the i range past 1000 (ex. i<10000), my output becomes "inf". Is it because i*i is too big? Wouldn't the equation be zero then? – Kassandra Sandoval Feb 02 '16 at 17:12
@KassandraSandoval Yes. because of `i*i` is too much for an `int` variable. I changed posted code. In that way you could have results with larger `i` values. – LPs Feb 02 '16 at 19:38

antiduh · Accepted Answer · 2016-02-02T17:43:27.650

3

The expression i^2 means i BITWISE-XOR 2, not repeated multiplication.

Some values of i^2 are zero - when i=2 - which means you sometimes divide by zero.

Replace (i^2) with (i*i).

edited Feb 02 '16 at 17:43

answered Feb 02 '16 at 16:29

antiduh

11,853
4
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1

Not for some value: only for i=2 – LPs Feb 02 '16 at 17:01
@LPs - You're confusing properties of the expression with the conditions necessary to invoke those properties. The expression "some values of `i^2` are zero" is absolutely true - the condition that satisfies that expression is `i=2`. Since I've stated both the property and its condition, your criticism is pointless. – antiduh Feb 02 '16 at 17:09
I wrote before your edit. You wrote: _Some values of i^2 are zero (odd values of i)_ Now the answer is ok. – LPs Feb 03 '16 at 07:10

score -1 · Answer 4 · edited May 23 '17 at 12:07

-1

printf("Double Precision: ");
printf("%lf", y);               //double has %lf format specifier.

Actually, according to the link float will automatically be promoted to double when needed, but depending on a compiler is a bad habit. Plus the code will not be clear to other readers.

edited May 23 '17 at 12:07

Community

1
1

answered Feb 02 '16 at 16:56

Mayank Verma

633
6
19

`%lf` is used for `scanf` only. `Printf` always promote `%f` passed variable to `double` and does not care to `l` part of format. – LPs Feb 02 '16 at 19:42
What I said was the code will not be clear to other readers. When did I say `printf` won't promote to `double`? Where am I wrong if I said the code won't be clear? – Mayank Verma Feb 03 '16 at 08:05
`float` is promoted to `double` always, not when needed. – LPs Feb 03 '16 at 08:17

simple precision test getting inf output in C

4 Answers4