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(#M40158312) IBM Programming question Java Keep an EYE

wap to find desired character in the string and find ascii value in java

Asked In IBM Ashish Nayak (8 years ago)

Unsolved Read Solution (12)

Is this Puzzle helpful?   (14)   (1) Submit Your Solution Program

Read Solution (12) :

VICKY   (8 years ago)

import java.io.IOException;
import java.util.*;
public class ibmchstr {
public static void main(String args[]) throws IOException{
Scanner sc=new Scanner(System.in);
int x=0;
System.out.println("enter the string");
String s=sc.nextLine();
System.out.println("enter the char to search");
char ch=(char)System.in.read();
for(int i=0;i

Is this solution Helpfull? Yes (2) | No (2)  

vinay kumar  (8 years ago)

import java.io.*;
import java.lang.*;
public class CharToASCII{
public static void main(String args[]) throws IOException{
BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the char:");
String str = buff.readLine();
for ( int i = 0; i < str.length(); ++i ){
char c = str.charAt(i);
int j = (int) c;// your work is done here
System.out.println("ASCII OF "+c +" = " + j + ".");
}
}
}

Is this solution Helpfull? Yes (2) | No   

shubham ojha  (7 years ago)

import java.io.*;
import java.lang.*;
public class CharToASCII{
public static void main(String args[]) throws IOException{
BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the char:");
String str = buff.readLine();
for ( int i = 0; i < str.length(); ++i ){
char c = str.charAt(i);
int j = (int) c;// your work is done here
System.out.println("ASCII OF "+c +" = " + j + ".");
}
}
}

Is this solution Helpfull? Yes (1) | No   

Ashish Nayak  (8 years ago)

@Vicky after that

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VICKY   (8 years ago)

@ashish nayak i had posted full solution but i dono why its look lk this

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Ashish Nayak  (8 years ago)

OK can you please mail me ashish.al19@gmail.com

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Ashish Nayak  (8 years ago)

@vicky please send me that code ashish.al19@gmail.com

Is this solution Helpfull? Yes | No   

  (7 years ago)

public JavaApplication() {
}

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);

System.out.println("Enter the String");
String s = sc.nextLine();
int count = (s.length()) + 1;

System.out.println("Enter the character");
char c = sc.next().charAt(0);
//Character c=new Character('a');
for (int i = 0; i < s.length(); i++) {
if (c == s.charAt(i)) {
System.out.println("yes it is present");
int j = (int) c;
System.out.println("Your ASCII Value " + j);
count--;
}
if (count == 0) {
System.out.println("it is not present");
}
}
}
}

Is this solution Helpfull? Yes | No   

Bhavya  (6 years ago)

don't they prefer c language?

Is this solution Helpfull? Yes | No   

  (6 years ago)

import java.util.*;
import java.io.*;
import java.lang.*;

public class DesiredChar
{
public static void main (String[] args)
{
int position=0;
System.out.println("Enter The String: ");
Scanner sc=new Scanner(System.in);
String str=sc.nextLine();
System.out.println("Enter The Character to find: ");
char ch=sc.next().charAt(0);
position=str.indexOf(ch);
if(position>=0)
{
System.out.println("The position of "+ch+" is "+(position));
System.out.println("THe Ascii value of "+ch+" is "+(int)ch);
}
else
System.out.println(ch+" is not Present in the Given string");
}
}

Is this solution Helpfull? Yes | No   

Rayed Ali  (5 years ago)

import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner sc= new Scanner (System.in);
String str=sc.nextLine();
char ch= sc.next(".").charAt(0);
System.out.println(ascii(str,ch));
}
static int ascii (String str,char ch)
{
for (int i=0;i

Is this solution Helpfull? Yes | No   

Ramya Easwaran  (5 years ago)

import java.util.Scanner;
public class Main
{

public static void main(String []args){
Scanner sc=new Scanner(System.in);
String name=sc.nextLine(); //input of a string
char n = sc.next().charAt(0); //input of the string you need to find
int val=name.indexOf(n);
if(val != -1){
System.out.println("present");
System.out.println((int)n);
}else{
System.out.println("no");
}
}
}

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