# How can we truncate float value (float32, float64) to a particular precision in Golang?

Given a floating-point value, we have to truncate it to a particular precision in Golang.
Submitted by IncludeHelp, on October 24, 2021

In the Go programming language, a float value (float32, float64) can be truncated by using the format verb %f with precision value (i.e., %.nfn is the number of precisions).

Consider the below examples,

Example 1:

```// Golang program to truncate float value
// to a particular precision

package main

import (
"fmt"
)

func main() {
var x float32 = 123.456789
var y float64 = 123456789.90123456

// Printing the values without
// truncating
fmt.Println("Without truncating...")
fmt.Printf("x: %f, y: %f\n", x, y)

// Printing the truncatedvalues
fmt.Println("Truncated values...")
fmt.Printf("x: %.2f, y: %.2f\n", x, y)
fmt.Printf("x: %.3f, y: %.3f\n", x, y)
fmt.Printf("x: %.5f, y: %.5f\n", x, y)
}
```

Output:

```Without truncating...
x: 123.456787, y: 123456789.901235
Truncated values...
x: 123.46, y: 123456789.90
x: 123.457, y: 123456789.901
x: 123.45679, y: 123456789.90123
```

Example 2:

```// Golang program to truncate float value
// to a particular precision

package main

import (
"fmt"
)

func main() {
x := 10 / 3.0
fmt.Printf("%.2f", x)
}
```

Output:

```3.33
```

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