The Go Programming Language Specification

Integer literals

An integer literal is a sequence of digits representing an integer constant.

Floating-point literals

A floating-point literal is a decimal representation of a floating-point constant. It has an integer part, a decimal point, a fractional part, and an exponent part. The integer and fractional part comprise decimal digits; the exponent part is an e or E followed by an optionally signed decimal exponent. One of the integer part or the fractional part may be elided; one of the decimal point or the exponent may be elided.

Arithmetic operators

For two integer values x and y, the integer quotient q = x / y and remainder r = x % y satisfy the following relationships:

x = q*y + r  and  |r| < |y|

with x / y truncated towards zero.

You wrote, using integer literals and arithmetic (x / y truncates towards zero):

package main

import (
    "fmt"
    "strconv"
)

func main() {
    var num float64
    num = 5 / 3 // float64(int(5)/int(3))
    fmt.Printf("%v\n", num)
    numString := strconv.FormatFloat(num, 'f', -1, 64)
    fmt.Println(numString)
}

Playground: https://play.golang.org/p/PBqSbpHvuSL

Output:

1
1

You should write, using floating-point literals and arithmetic:

package main

import (
    "fmt"
    "strconv"
)

func main() {
    var num float64
    num = 5.0 / 3.0 // float64(float64(5.0) / float64 (3.0))
    fmt.Printf("%v\n", num)
    numString := strconv.FormatFloat(num, 'f', -1, 64)
    fmt.Println(numString)
}

Playground: https://play.golang.org/p/Hp1nac358HK

Output:

1.6666666666666667
1.6666666666666667

package main
import "fmt"
func main() {
  var x float64 = 5.7
  var y int = int(x)
  fmt.Println(y)  // outputs "5"
}

Using math.Float32bits and math.Float64bits, you can see how Go represents the different decimal values as a IEEE 754 binary value:

Playground: https://play.golang.org/p/ZqzdCZLfvC

Result:

float32(0.1): 00111101110011001100110011001101
float32(0.2): 00111110010011001100110011001101
float32(0.3): 00111110100110011001100110011010
float64(0.1): 0011111110111001100110011001100110011001100110011001100110011010
float64(0.2): 0011111111001001100110011001100110011001100110011001100110011010
float64(0.3): 0011111111010011001100110011001100110011001100110011001100110011

If you convert these binary representation to decimal values and do your loop, you can see that for float32, the initial value of a will be:

0.20000000298023224
+ 0.10000000149011612
- 0.30000001192092896
= -7.4505806e-9

a negative value that can never never sum up to 1.

So, why does C behave different?

If you look at the binary pattern (and know slightly about how to represent binary values), you can see that Go rounds the last bit while I assume C just crops it instead.

So, in a sense, while neither Go nor C can represent 0.1 exactly in a float, Go uses the value closest to 0.1:

Go:   00111101110011001100110011001101 => 0.10000000149011612
C(?): 00111101110011001100110011001100 => 0.09999999403953552

Edit:

I posted a question about how C handles float constants, and from the answer it seems that any implementation of the C standard is allowed to do either. The implementation you tried it with just did it differently than Go.