Hi I was giving this following program:
http://euclid.cs.qc.cuny.edu/13-page-handout.pdf
Example 2 on first page
class FinalExam {
static int e = 1;
static int a[] = {0,1,2};
public static void main(String args[])
}
test(a[e], a[e-1]);
System.out.println(a[0] + " " + a[1]
+ " " + a[2] + " " + e);
{
static void test (int x, int y)
}
a[1] = 6;
e = 2;
x += 3;
y--;
System.out.print(x + " " + y + " ");
{
{
class FinalExam
the output from the answer key was: Output for each parameter passing mode: x y a[0] a[1] a[2] e
Pass By Name: x=5 y=5 a[0]=0 a[1]=5 a[2]=5 e=2
Can you tell me how they arrive开发者_运维技巧 at that answer? I don't get where 5 comes from
Looking at your handout, I'm guessing you are learning the differences between the various evaluation strategies a programming language can implement (call-by-value, call-by-reference, call-by-name, etc...). The code snippet for #2 is in java, but they are asking you to evaluate the output AS IF a given evaluation strategy was being used, rather than provide a code snippet in multiple languages.
In Java there is only call by value, so the output of this java program will match their answer marked call-by-value (4 -1 0 6 2 2).
Regarding the call-by-name answer: the parameters to a function are actually substituted into the function, and are re-evaluated each time they are used.
Before test(a[e], a[e-1])
: e = 1, a = {0, 1, 2}
test(x = a[e], y = a[e-1]) { // inside the function x becomes a[e], y becomes a[e-1]
a[1] = 6; // simple -> e = 1, a = {0, 6, 2}
e = 2; // simple -> e = 2, a = {0, 6, 2}
x += 3; // a[e] += 3 -> e = 2, a = {0, 6, 5}
y--; // a[e-1]--; -> e = 2, a = {0, 5, 5}
System.out.print(x + " " + y + " "); // print(a[e] ... a[e-1]) -> 5, 5
After test(a[e], a[e-1])
: e = 2, a = {0, 5, 5}
System.out.println(a[0] + " " + a[1] + " " + a[2] + " " + e); // 0, 5, 5, 2
Final output: 5, 5, 0, 5, 5, 2
They weren't asking what this java program would output, but what a similar program using call-by-name would output.
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