Parameter Passing Method Exercise
Consider the following program in some unidentified language which supports procedures with parameters.
function joe(int a, int b, int c)
begin
a := b + c;
b := c + 1;
print a, b, c;
end

function main
begin
int i := 5;
int j := 10;
ink k := 15;
joe(i, j, j + k);
print i, j, k;
end
What values are printed by the program under each of the following assumptions?
1. All parameters are passed by value.
35 26 25
5 10 15
Parameters are independent variables initialized to the values of the argument expressions. Changes to them do not effect the arguments.
2. Pass a and b by reference, and c by value.
35 26 25
35 26 15
This is very much the same, except the changes to a and b are also made to i and j since these parameters are aliases.
3. Pass a and b by value-result, and c by value.
35 26 25
35 26 15
This has the same effect as pass-by-reference, since the values of a and b are returned to i and j when the function returns.
4. All parameters are passed by name.
35 26 41
35 26 15
This is again similar to the last two, since changes to a and b also change i and j. The difference is the value of 41 printed for c, since c is an alias for the expression j + k. The assignment b := c + 1 changes b to 26, which changes j, which changes j + k, now 26 + 15 = 41.
And now for something completely different.
function phil(int a, int b, int c)
begin
b := b + 5;
b := a + c + 4;
print a, b, c;
end

function main
begin
int j := 10;
ink k := 15;
phil(j, j, j + k);
print j, k;
end
What values are printed by the program under each of the following assumptions?
1. All parameters are passed by value.
10 39 25
10 15
Okay, so this one isn't all that different.
2. Pass a and b by reference, and c by value.
44 44 25
44 15
Here both a and b are aliases for j. So the first assignment to b also changes a to 15. The second adds 29 more to b (as well as a and j), making everyone 44.
3. Pass a and b by value-result, and c by value.
10 39 25
39 15
Here value-result differs from reference. To make this happens usually requires sending the same parameter in two places, or sending a global as a parameter. So it's either something a professor makes up, or an unexpected accident that can be very hard to figure out.

The first line is the same as pass-by-value since the process begins the same way. While the function is running, a and b are different values, so the first assignment to b is completely destroyed by the second one. Then the values are copied back. The second line given here depends on the assumption that the values are copied back in left-to-right order so that the 10 is copied to j, then the 39, so the later becomes the final value.

4. All parameters are passed by name.
49 49 64
49 15
This is like pass-by-reference, except now the first assignment to b changes j, and hence both a and c also, since a is j, and c is j + k. Thus the second assignment to b totals up to 49, which becomes the new value of a, b and j. That also changes c, since it's still j + k, and so prints out as 64.