Why Java Can't Swap Primitive Values Inside a Method – The Correct Way

When you pass an object (or primitive) as a parameter to a Java method, the method receives only a copy of the value. Modifying that copy does not change the original variable in the caller's scope. The article illustrates this with a simple int‑swap example.

Faulty implementation

int a = 1, b = 2;
output("a:" + a, "b:" + b);
changeNum(a, b);
output("a:" + a, "b:" + b);

public void changeNum(int a, int b) {
    int i = a;
    a = b;
    b = i;
}

The console output shows that the values of a and b remain unchanged after the method call:

Before call: a:1 b:2
After call:  a:1 b:2

Correct implementation using a temporary variable

int a = 1, b = 2;
output("a:" + a, "b:" + b);
int i = a;
a = b;
b = i;
output("a:" + a, "b:" + b);

Now the output correctly shows the values swapped:

Before swap: a:1 b:2
After swap:  a:2 b:1

Correct implementation using a return value

public int[] changeNum(int a, int b) {
    int[] abc = { b, a };
    return abc;
}

int a = 1, b = 2;
output("a:" + a, "b:" + b);
int[] abc = changeNum(a, b);
a = abc[0];
b = abc[1];
output("a:" + a, "b:" + b);

This version returns an array containing the swapped values, which the caller then assigns back to a and b, achieving the desired effect.

The key takeaway is that any method that needs to modify primitive data must either operate on a mutable wrapper object or return the new values, because Java's pass‑by‑value semantics prevent direct modification of the caller's primitives.