In this section we will learn about Operator Precedence and Operator Associativity.
Operator Precedence
Precedence decides which operator will be evaluated first in a case where more than one operator is present in the same calculation.
Operators | Precedence(High to Low) |
---|---|
postfix | expr expr-- |
unary | expr expr expr expr ~ ! |
multiplicative | * / % |
additive | + - |
shift | << >> >>> |
relational | < > <= >= instanceof |
equality | == != |
bitwise AND | & |
bitwise exclusive OR | ^ |
bitwise inclusive OR | | |
logical AND | && |
logical OR | || |
ternary | ? : |
assignment | = += -= *= /= %= &= ^= |= <<= >>= >>>= |
Example of Precedence
/* * Here we will see the effect of precedence in operators life */ class OperatorPrecedenceExample { public static void main(String args[]) { int i = 40; int j = 80; int k = 40; int l = i + j / k; /* * In above calculation we are not using any bracket. So which operator * will be evaluated first is decided by Precedence. As precedence of * divison(/) is higher then plus(+) as per above table so divison will * be evaluated first and then plus. * * So the output will be 42. */ System.out.println("value of L :" + l); int m = (i + j) / k; /* * In above calculation brackets are used so precedence will not come in * picture and plus(+) will be evaluated first and then divison()/. So * output will be 3 */ System.out.println("Value of M:" + m); } }
Operator Associativity
If two operators have the same precedence in the calculation then Associativity of the operators will be used to decide which operator will be executed first.
Example of Associativity
package jbt.bean; /* * Here we will see the effect of precedence in operators life */ public class OperatorAssociativityExample { public static void main(String args[]) { int i = 40; int j = 80; int k = 40; int l = i / k * 2 + j; /* * In above calculation we are not using any bracket. And there are two * operator of same precedence(divion and multiplication) so which * operator(/ or *) will be evaluated first is decided by association. * Associativity of * & / is left to right. So divison will be evaluated * first then multiplication. * * So the output will be 82. */ System.out.println("value of L :" + l); int m = i / (k * 2) + j; /* * In above calculation brackets are used so associativity will not come * in picture and multiply(*) will be evaluated first and then * divison()/. So output will be 80 */ System.out.println("Value of M:" + m); } }
Operators in Java
Let us discuss each operator individually.
Assignment (=) and Arithmetic operators(+, -, *, /) work the same way as they do in other programming languages, so we will not discuss them here. The precedence for ‘/‘ and ‘*‘ operators is higher than sum(+) or minus(–) or modular division(%)
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Could you please explain to me the following expression
int m=i/(k*2)+j;
(precedence doesn’t come into action and that is understood)
While evaluating the expression this is what happens(40/(40*2)+80)=(40/(80)+80)==(1+80)==81
Always the (/) is considered to give us the quotient.
Please do explain the same.
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m = i / (k * 2) + j;
there m=80 is incorrect.
Value will be 80 only. Please check once again.
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