How to convert Float to String in Java?

ss
In last post, we have learned how to convert string to float in java. In this post, we are going to learn how to convert float to string in Java.

Consider the situation, if we want to display float value is any text area or if we want to do any manipulation using float value, conversion of float to string is need to be known.

Conversion of float to string in java can be done in  following ways:


It is a static method of float class. It provides a string representation of float value.
If float value is NaN, then result in String is NaN. Else, it will provides string representation of float. For more detailed about this method.

public static String toString(float f)

Parameter
 
f - float value

Returns

a string representation of float value.


It is a static method of String class. It returns string representation of float argument.

public static String valueOf(float f)

Parameters

f - a float argument

Returns

string representation of float argument

Below code will convert float to string in java using above two mwthods:




Related Posts

Click here for all data type conversions in Java

 Please let us know your views in comment section.

Subscribe us to receive updates instantly.

Happy Coding.!!!

The intention behind this blog is to help all my fellow programmers. Also to provide solutions for various coding platform problems with solution and explanation. I will add many tutorials in future. I am not a professional blogger but whenever time permits I will post. I hope you’ll enjoy the content and find it useful! Check out my Writer blog for a more valuable quotes. For any queries or suggestions, please feel free to reach out to me. Happy coding!

Comments

Post a Comment

Popular posts from this blog

First Unique Character in a String

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase letters. class Solution {     public int firstUniqChar(String s) {         int length = s.length();         String str;         for(int i=0;i<length;i++){             if(i==length-1){                 str = s.substring(0, i);             }             else{                 str = s.substring(0,i)+s.substring(i+1);             }             if(!str.contains(s.charAt(i)+""))                 return i;         }         return -1;     } } Try it on Leetcode Here, we are going to use substring method. 1) If character is at end, then take substring from 0 till before last character of string. 2) Else make a temporary string, by removing exactly that particular character using index. [e.g..
Read more

Balanced Binary Tree

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of  every  node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true   Constraints: The number of nodes in the tree is in the range  [0, 5000] . -10 4  <= Node.val <= 10 4 Solution: /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }
Read more

Majority Element

Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int repeatLen
Read more

Smallest Range II

Given an array   A   of integers, for each integer   A[i]   we need to choose   either  x = -K  or  x = K , and add   x   to   A[i]  (only once) . After this process, we have some array  B . Return the smallest possible difference between the maximum value of  B  and the minimum value of  B . Example 1: Input: A = [1] , K = 0 Output: 0 Explanation : B = [1] Example 2: Input: A = [0,10] , K = 2 Output: 6 Explanation : B = [2,8] Example 3: Input: A = [1,3,6] , K = 3 Output: 3 Explanation : B = [4,6,3] Note: 1 <= A.length <= 10000 0 <= A[i] <= 10000 0 <= K <= 10000 Solution: class Solution {     public int smallestRangeII(int[] A, int K) {         int i, min=0, max=0, min1, min2, max1, max2, length = A.length, diff
Read more