CODEWITHSUNDEEP
java
Bhavesh Jain
Bhavesh Jain

Reputation: 53

Type mismatch: cannot convert from long to int

I had the following lines of code

long longnum = 555L;
int intnum = 5;
intnum+=longnum;
intnum= intnum+longnum; //Type mismatch: cannot convert from long to int
System.out.println("value of intnum is: "+intnum);

I think line-3 and line-4 do same task, then why compiler showing error on line-4 "Type mismatch: cannot convert from long to int"

please help.

Upvotes: 4

Views: 8174

Answers (3)

Ramya Ravi
Ramya Ravi

Reputation: 21

Long id = 50;
Integer sum = Integer.valueOf(Long.toString(id));

Upvotes: 0

Jon Skeet
Jon Skeet

Reputation: 1503629

I think line-3 and line-4 do same task, then why compiler showing error on line-4 "Type mismatch: cannot convert from long to int"

Because they don't do the same thing. Compound assignment operators have an implicit cast in them.

From section 15.26.2 of the JLS:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

So your third line is more like:

intnum = (int) (intnum + longnum);

The cast is required because in the expression intnum + longnum, binary numeric promotion is applied before addition is performed in long arithemtic, with a result of long. There's no implicit conversion from long to int, hence the cast.

Upvotes: 11

Rohit Jain
Rohit Jain

Reputation: 213391

That's because the compound assignment operator does implicit casting.

From JLS Compound Assignment Operator:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

While in case of binary + operator, you have to do casting explicitly. Make your 4th assignment:

intnum = (int)(intnum+longnum);

and it would work. This is what your compound assignment expression is evaluated to.

Upvotes: 12

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