Removing String (java)

Question

Is there any possible way to remove the string? For example:

String name = "1 + 1";
If( nam == 2){ //I want this like to act like this 👉 1+1 == 2;
System.out.println("hello");
}

Is there any way to change the data type from String to void?

___________________________________
I tried like this this but failed lol...
👇
void num = System.out.print("1+1");

And

void num = System.out.print("1 " + " 1");

______________________________________

Is that even possible by anyway? If so please tell me :)

Thank you programmers! :)

Answer 1

Mr. 🍊range ,
As Avinesh said you can use switch but that'll not be enough to evaluate expressions in proper order.

In order to maintain proper evaluation order you need stack.

For example in 5+3*6 should be 23 not 48. Stack helps you to evaluate according to operator precedence.

I have used stack to evaluate postfix expression (oparands are digit only 0-9) . //but you need infix.

https://code.sololearn.com/cG9uKJ8aT7so/?ref=app

Answer 2

👑 Prometheus 🇸🇬

To ask the user to type their question for example "3 + (8*3)" then calc their question...

I don't find any other way to solve my problem 😂 I know there a way to solve it... but I don't know how😂

Answer 3

Avinesh

Yes that will work but...how do u convert arithmetic operations?

For example if they says
"40 - 3" how do u get that minus from the string?

Answer 4

Mr. 🍊range , see this example by John sir. It does exactly what you asked.

https://code.sololearn.com/ctEXGtkw5ahO/?ref=app

Answer 5

🇮🇳Omkar🕉 and Avinesh thank you both for ur help :)

Answer 6

But why would you store a System.out.print()?

Answer 7

Mr. 🍊range you can probably make use of switch case which can contain all the operators that can be mapped.

Answer 8

🇮🇳Omkar🕉 since he mentioned only two operands separated by an operator I suggested him to go with switch directly.

But for more number of operands and operators, precedence comes into picture just like you said.

Nice example there👍

Answer 9

Apart from using eval you can also use stack.

Just search for `infix expression evaluation using stack` on internet.

Answer 10

For example
Scanner sc = new Scanner(System.in);
String name=sc.nextLine().replaceAll(“ “, “”);
int res=0;
if (name.contains(“+”)){
String[] arr = name.split(“+”);
for (int i=0; i<arr.length; i++{
int res+=Integer.parseInt(arr[i]);
}
}
System.out.println(res);

If user enter 32 + 6
you will replace all whitespaces
String name = “32+6”
arr[0]=32
arr[1]=6

res=38;

Answer 11

Take string input and convert them to integers for performing any operation.

Answer 12

I dont know what you are creating but you can try this code see if the format works for you.

int a = 1;
int b = 1;

String Name = Integer.toString(a+b);
int name = Integer.parseInt(Name);

if(name==2){
System.out.println("hello");

Answer 13

No removing

Answer 14

Int a=1;
Int c=1;

String name= int.toStrinh(a+b);
int name=int.parselnt(name);
if (name==2){
System.out.println("hello");