Multiplying floats gives infinite values

Question

I tried a lot of methods:

float a = 1234.34;
float b = 1234.52;
float r = a * b;
cout << r << endl;
float c = 1234.34 * 1234.52;
cout << c << endl;

cout << 1234.34 * 1234.52 << endl;
cout << (float)(1234.34 * 1234.52) << endl;

And all of these seem to be giving an infinite values..

1.52382e+006
1.52382e+006
1.52382e+006
1.52382e+006

What am I doing wrong here?

None of those seem to be giving an infinite values [sic]. Perhaps you're simply not familiar with the [E notation](http://en.wikipedia.org/wiki/Scientific_notation#E_notation) — eerorika, May 19 '14 at 14:00
That's not an infinite value. It's the floating-point representation of the product you're multiplying. Your code is working fine. — Bucket, May 19 '14 at 14:00
Your definition of infinite seems quite small. You are multiplying two numbers that are about 1,000 times each other. The answer should be around 1,000,000. 1.52382e6 is the same as 1.52382*10^6, which is 1.5 million. Seems like the computer works. — mtrw, May 19 '14 at 14:01
http://stackoverflow.com/questions/6301547/turn-off-scientific-notation-on-float — Joachim Isaksson, May 19 '14 at 14:02

score 6 · Accepted Answer · edited May 23 '17 at 11:50

6

The number in the result is nowhere close to approaching infinity.

The result is simply C++ showing the value in scientific notation.

Please check How to avoid scientific notation for large numbers?.

edited May 23 '17 at 11:50

Community

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answered May 19 '14 at 14:01

jordan

959
7
17

score 3 · Answer 2 · 2014-05-19T14:31:12.870

3

The values are not infinite, they're only in scientific notation.

1.52382e+006 evaluates to 1.52382 * (10)^6.

You may want to use doubles, and set the precision for the ostream:

#include <limits>    // std::numeric_limits
#include <iostream>  // std::cout

int main() {
  double f = 3.141592;

  std::cout.precision(std::numeric_limits<double>::digits10);
  std::cout << std::fixed;
  std::cout << f;

  return 0;
}

edited May 19 '14 at 14:31

answered May 19 '14 at 14:00

So, how can I get the value of multiplication the floats in numbers (normal)? – user3402189 May 19 '14 at 14:01

Mohit Jain · Answer 3 · 2014-05-19T14:22:15.027

The answer is correct. 1.52382e+006 is same as 1.52382 * 10⁶

If you don't want the output in E format, use std::fixed

#include <iostream>
#include <iomanip>
using namespace std;

int main() {
    float a = 1234.34;
    float b = 1234.52;
    float r = a * b;
    cout << std::fixed << r << endl;
    float c = 1234.34 * 1234.52;
    cout << std::fixed << c << endl;

    cout << std::fixed << 1234.34 * 1234.52 << endl;
    cout << std::fixed << (float)(1234.34 * 1234.52) << endl;
    return 0;
}

This code outputs the following

1523817.375000
1523817.375000
1523817.416800
1523817.375000

score 1 · Answer 4 · answered May 19 '14 at 14:02

1

What you get is the scientific notation of your results, 1523817.4168, which in in scientific notation is 1.52382E+6, where the E stands for "10^", meaning 1.52382 * 10 ^ 6, you can look it up.

answered May 19 '14 at 14:02

private_meta

561
4
19

score 1 · Answer 5 · answered May 19 '14 at 14:02

1

That is just scientific notation. It does not signify infinity.

answered May 19 '14 at 14:02

Benjamin Trent

7,378
3
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Multiplying floats gives infinite values

5 Answers5