Difference between int and int received by ParseInt in java

Question

int i = 0;
int k = Integer.parseInt("12");
int j = k;
System.out.println(i+1 + " " + j+1);

Strangely the output received is

1 121

I can not figure out this basic difference. Please help me.

Orison

Answer 1

Use brackets as follows

System.out.println((i+1) + " " + (j+1));

From the docs

The + operator is syntactically left-associative, no matter whether it is later determined by type analysis to represent string concatenation or addition. In some cases care is required to get the desired result. For example, the expression:

a + b + c is always regarded as meaning: (a + b) + c

Extending this to your scenario

i+1 + " " + j+1

it becomes

(((i + 1) + " ") + j)+1

Since i is an int so (i + 1) = 1 , simple addition

" " is a String hence ((i + 1) + " ") = 1 WITH SPACE (String concatenation)

Similarly when j and last 1 is added, its being added to a String hence String concatenation takes place, which justifies the output that you are getting.

See

• JLS 15.18.1 String Concatenation Operator +

Uphemia

Answer 2

that is beacuse of " ".

whenever a String comes, java doesnt do any calculations after that and just append it as string.

So in your case, i+1 is computed to 1, but " " + j+1 has string in it. So, it just appended together to form 121

Bierce

But why it is not appending on the first case ? If it is true it should be like 01 121 – Thundercloud

any primitives/wrappers before the appearance of first String are computed, Anything after that, is just appended together. – Bierce

@AurA: because it basically translates to something equivalent to: " ".concat(j).concat(1) (or a similar version using StringBuilders, depending on your code and JVM version). It can't guess what you want, so it's pretty natural left to right interpretation. – Arguello

@Bierce thanks for editing the answer not it is quite clear. haylem thanks for explaination – Thundercloud

No.. it is not like 01 121... it is conceptually like (0+1) + " ".append(j).append(1) – Bierce

Yes, I got it he meant that only. – Thundercloud

Answer 3

The reason you see this behavior is that the sequence of + operators is evaluated left-to-right. So it is evaluated as if parenthesized:

System.out.println((((i + 1) + " ") + j) + 1);

The first operator adds two int values and produces an int value. The next + adds an int to a String and produces a String. After that, everything is string concatenation. You can introduce your own parentheses to get the result you want.

Dockhand

Yeah +1, this indeed is the best answer – Toor

Answer 4

    int i = 0;
    int k = Integer.parseInt("12");
    int j = k;
    System.out.println(i+1 + " " + (j+1));

basically when you put + " " + after this java just appends values as string.

and when you put (j+1) in brackets then its precedence gets higher and it is executes it first and perform sum operation.

Supervision

Answer 5

When you use " " The expression after that gets evaluated as string.

Using brackets ( and ) around an expression can solve the problem in hand.

System.out.println(i+1 + " " + (j+1));

Thacher

Answer 6

+ operator is overloaded for addition and String concatenation what you're doing is String concatenation and not addition.. Use brackets for performing addition.

Revivalist

Answer 7

parseint will basically return int (Look at Java API), and there is only one int type in Java. in this example you used " ", where java will treat it as string. in any operation make sure you dont mix up strings with calculations. Always use parenthesis to separate String from calculations.

Badoglio

Answer 8

It happens because the + operator has left associativity and has an overloaded function with strings, so when you have this

int i = 0; int k = Integer.parseInt("12"); int j = k; i+1 + " " + j+1

it first sums i + 1 which gives 1 then it sums 1 + " ", which uses the overloading function of it to concatenate 1 and " " so it gives a string with the value of "1 ". After that it sums "1 " + j and since one of the operands is a string, it does the same behavior and so on.

Yardage

Answer 9

Interger.parseInt (String str) is a wrapper class method which is used to convert String obj type to primitive data type (int). this are generally used in collection frame work for converting primitive data type to object and vice versa

Weighin