unreachable-code

This question already has an answer here: Unreachable code: error or warning? 12 answers class For1 { public static void main(String args[]) { int a = 0; for(;;) { break; System.out.println(a); //Line 1 ++a;//Line 2 } } } I know that Line 1/Line 2 will never be executed. But still I don't understand why a compile time error is thrown. I am getting "unreachable statement" compile error. Does it mean that compiler checks whether it is able to compile for all branches/lines of code ? Does it mean that compiler checks whether it is able to compile for all branches/lines of code ? It means the

I can't seem to find a way to fix this problem. All i'm doing is declaring an integer and it's telling me that the code is unreachable. private class myStack{ Object [] myStack = new Object[50]; private void push(Object a){ int count = 50; while(count>0){ myStack[count]=myStack[count-1]; count--; } myStack[0]=a; } private Object pop(){ return myStack[0]; int count2 = 0; //Unreachable Code } } Once you return from a method, you return to the method that called the method in the first place. Any statements you place after a return would be meaningless, as that is code that you can't reach

Does anyone know why: public void foo() { System.out.println("Hello"); return; System.out.println("World!"); } Would be reported as an "unreachable error" under Eclipse, but public void foo() { System.out.println("Hello"); if(true) return; System.out.println("World!"); } Only triggers a "Dead code" warning? The only explanation I can think of is that the Java compiler only flags the first, and that some extra analysis in Eclipse figures out the second. However, if that is the case, why can't the Java compiler figure out this case at compile time? Wouldn't the Java compiler figure out at