Golang strconv.ParseFloat fails to parse the string correctly when using 32 for bitSize parameter

Why the converted float64 variable has 1.590000033378601 as value?
1.59 seems small enough to be fitted into 32 bits:

package main

import (
    "fmt"
	"strconv"
)

func main() {
	str := `1.59`
    float, err := strconv.ParseFloat(str, 32)
	if err != nil {
    	fmt.Println("err: ", err)
	}
    fmt.Printf("%T -> %+v\n", float, float)
}

Go playground link

It is precision issue of 32bit variable. This is not an issue of Go. See following URLs.

IEEE 754 - Wikipedia

Single-precision floating-point format - Wikipedia

You will understand what happen with 1.59 in this C code.

#include <stdio.h>

int
main(int argc, char* argv[]) {
  float f = 1.59;
  if (f == 1.59) {
	puts("float: equal!");
  } else {
	puts("float: not equal!");
  }

  double d = 1.59;
  if (d == 1.59) {
	puts("double: equal!");
  } else {
	puts("double: not equal!");
  }
  return 0;
}

http://ideone.com/WobYbU

1.590000033378601 is 1.59 in 32bit variables. So you can use the value with cast to float32. i.e. float32(1.590000033378601)

package main

import (
	"fmt"
	"strconv"
)

func main() {
	str := `1.59`
	f64, err := strconv.ParseFloat(str, 32)
	if err != nil {
		fmt.Println("err: ", err)
	}
	f32 := float32(f64)
	fmt.Printf("%T -> %+v\n", f32, f32)
}

Go Playground

UPDATE

On Most of computer, floating point value is stored as following elements.

• Sign (zero or one)
• Significand (Coefficient)
• Exponent

For example, representations 0.75 is +1.1 x 2^-1

+ is Sign, .1 is Significand, -1 is Exponent. This is stored like below on 32bit memory space.

Sign                                 Significand
+-+               +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
  +-+-+-+-+-+-+-+-+
      Exponent

For example, 0.75 is stored like below

0-01111110-10000000000000000000000

     +- Significand
     |
+ 1[.1] x 2 ^ -1
|         |    |
+- Sign   |    +- Exponent
          |
          +------ Base

> + 1.1 x 2^-1 = 1 x 2^0 + 1 x 2^-1 x 2^-1 = 0.75

Since floating point value is represented as above, it is different between the mathematical value and the computer's value. This is precision issue.

The decimal number 1.59 looks short, but that's only because we are using the <b>decimal number system</b>.

When the computer tries to represent the same number in <b>binary scientific notation</b> (a * 2^b) it requires a lot more fractional digits: 1.1001011100001010001111010111000010100011110101110001 * 2^0.

If you round this value to 32 bits, store it in a variable of type float32 and then print as decimal, you get 1.59, but storing the same rounded value in a float64 variable and printing as decimal, you get 1.590000033378601.

Function strconv.ParseFloat returns float64, but since you have specified 32 bits of precision, you are supposed to cast the value to float32 prior to using.