Borg said:
...
return 0;
}Don't worry about the calcuation or value of s. It's just an example. According to the initial description, s may have changed before the method is called.CAF123 said:
The method requires you to return an int (i.e. public int someRandomMethod()). If it makes it all the way to the end, it has to return an int. I just hardcoded a 0 as a default.CAF123 said:
Yes, correct!CAF123 said:
Borg said:
Console myConsole = System.console();
String S = myConsole.readLine("Enter an integer value for s:");
int s = Integer.parseInt("S");
boolean condition1;
int u = 2/s;
condition1 = s > 100;
int p = 10*t - s; //All of this in the class, outwith the method.
public int someRandomMethod() {
int t = 2;
if (condition1) {
return u;
}else {
return p;
}
return 0;
}I'm not sure what you're asking is OK. There is always a different way to write any program. My goal is just to describe how variables are scoped.CAF123 said:
public static int s = 10; // Some class level variable
// private int u = 0;
public int someRandomMethod() {
int t = 2;
// int u = 0;
if (s > 100) {
int u = 2 / s;
if (u > 250) {
return u; // Leave the method and return the value of u
}
// If u <= 250 the code will leave the if/else block and continue at the EXIT POINT
}
else { // If s was greater than 100, it won't go in here
int u = (10 * t) - s;
if (u < 25) {
return u;
}
// If u <= 25 the code will leave the if/else block and continue at the EXIT POINT
// int u = 0;
}
// EXIT POINT
// The compiler knows about s and t
// The compiler does not know about u at this point because it can't look up and into brackets
// int u = 0;
return 0;
}Just noticed that I wrote four.Borg said:
public class Random1 {
public static int s = 10; // Some class level variable
// private int u = 0; //will compile if placed here - while in method, we don't care there exists a class level varriable called u- if want to refer to class level, use this. construct.
public int someRandomMethod() {
int t = 2;
// int u = 0; //u cannot go here since it is again defined intu=2/s and intu=10t-s within the method.
if (s > 100) {
int u = 2 / s;
if (u > 250) {
return u; // Leave the method and return the value of u
}
// If u <= 250 the code will leave the if/else block and continue at the EXIT POINT
}
else { // If s was greater than 100, it won't go in here
int u = (10 * t) - s;
if (u < 25) {
return u;
}
// If u <= 25 the code will leave the if/else block and continue at the EXIT POINT
// int u = 0; //u cannot go here since it is already defined inside method.(i.e int u = 2/s above or int u = 10t-s)
}
// EXIT POINT
// The compiler knows about s and t
// The compiler does not know about u at this point because it can't look up and into brackets
// int u = 0; //ok to have here since u (private int u) defined outside method and intu=2/s and intu=10t-s defined inside brackets so compiler does not know about them.
return 0;
} //close method
}Borg said:
Yes, that's all correct - very good. Does it all make better sense now how variables are visible to the compiler throughout a class?CAF123 said:
I misread your code. I see now that one of the variables is a capital S. By normal conventions, variables are Camel Case where the first word starts with a non-capitalized letter. I guess that caused me to see it wrong. The convention for class names is Camel Case where the first word starts with a capitalized letter. It makes it easier to read declarations like this:CAF123 said:
CAF123 said:
import java.io.Console;
public class Random {
int s = s;
boolean condition1 = s > 100;
int u = 2/s;
int t = 2;
int p = 10*t - s; //All of this in the class, outwith the method.
public int someRandomMethod() {
if (condition1) {
return u;
}else {
return p;
}
}
public static void main(String args[]) {
String S = myConsole.readLine("Enter an integer value for s:");
int s = Integer.parseInt(S);
}
}