CODEWITHSUNDEEP
java
Lion
Lion

Reputation: 19027

Array expressions in Java

Consider the following expressions in Java.

int[] x={1, 2, 3, 4};
int[] y={5, 6, 7, 0};

x=y;
System.out.println(x[0]);

Would display 5 on the console because x is made to point to y through the expression x=y and x[0] would obviously be evaluated to 5 which is actually the value of y[0].

When I replace the above two statements with the following statement combining both of them into a single statement,

System.out.println(x[(x=y)[3]]);

it displays 1 which is the value of x[0] even though it seems to be equivalent to those above two statements. How?

Upvotes: 4

Views: 2328

Answers (5)

Ömer
Ömer

Reputation: 188

It's all about the precedence here

lets take a look at this

we have following expression

x= 2 + 5 *10

so on running the about expression it first multiply 5 * 10 then add it with 2 and assign it to the x because the precedence of "=" is least

x[(x=y)[3]] in this case what confuses the most is that (x = y ) term, and most of programmers think that it assigns the reference of y to x first then proceed with the remaining part but here assignment operator is resolved least

that's why when you try to print the x[0] after the above expression is run, then i will give value of 5 because the zeroth index of y contains 5

Upvotes: 0

Jeff
Jeff

Reputation: 36573

x[(x=y)[3]]

breaks down to

int[] z = y; // new "temp" array (x=y) in your expression above
int i = z[3]; // index 3 in y is 0
System.out.println(x[i]); // prints x[0] based on the value of i...which is 1

Upvotes: 1

Joe
Joe

Reputation: 82614

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

...println(x[y[3]]); // 1

Upvotes: 1

Tudor
Tudor

Reputation: 62439

It's because x = y produces the new x inside the index. So now x[3] = y[3] = 0 and x[0] = 1, because it still uses the old x array on the outside.

Upvotes: 2

hvgotcodes
hvgotcodes

Reputation: 120198

the third index in the array y points to is 0, so 1 is the correct output.

So you have

x[(x=y)[3]]

which is x[y[3]], but y[3] is 0, because arrays are 0-indexed, and x[0] is 1.

Upvotes: 6

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