How to fetch an Integer variable as String in GoLang?
Last Updated :
22 Jun, 2020
To fetch an Integer variable as String, Go provides strconv package which has methods that return a string representation of int variable. There is nothing like an integer variable is converted to a string variable, rather, you have to store integer value as a string in a string variable. Following are the functions used to convert an integer to string:
1. strconv.Itoa(): To convert an int to a decimal string.
// s == "60"
s := strconv.Itoa(60)
2. strconv.FormatInt(): To format an int64 in a given base.
var n int64 = 60
s := strconv.FormatInt(n, 10) // s == "60" (decimal)
s := strconv.FormatInt(n, 16) // s == "3C" (hexadecimal)
3. strconv.FormatUint(): To return the string representation of x in the given base, i.e., 2 <= base <= 36.
fmt.Println(strconv.FormatUint(60, 2)) // 111100 (binary)
fmt.Println(strconv.FormatUint(60, 10)) // 60 (decimal)
Example 1:
package main
import (
"fmt"
"strconv"
)
func main() {
var var_int int
var_int = 23
s1 := strconv.Itoa(var_int)
fmt.Printf("%T %v\n", s1, s1)
s2 := strconv.FormatInt(int64(var_int), 10)
fmt.Printf("%T %v\n", s2, s2)
s3 := strconv.FormatUint(uint64(var_int), 10)
fmt.Printf("%T %v\n", s3, s3)
fmt.Println("Concatenating all strings: ", s1+s2+s3)
}
|
Output:
string 23
string 23
string 23
Concatenating all strings: 232323
Example 2:
package main
import (
"fmt"
"strconv"
)
func main() {
var var_int int
var_int = 50
s1 := strconv.Itoa(var_int)
fmt.Printf("%T %v\n", s1, s1)
s2 := strconv.FormatInt(int64(var_int), 2)
fmt.Printf("%T %v\n", s2, s2)
s3 := strconv.FormatUint(uint64(var_int), 16)
fmt.Printf("%T %v\n", s3, s3)
fmt.Println("Concatenating all strings: ", s1+s2+s3)
}
|
Output:
string 50
string 110010
string 32
Concatenating all strings: 5011001032
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...