Compound assignment operators, what happens if the value is modified (in the meanwhile)?

Question

Consider the following pseudocode (language agnostic):

int f(reference int y) {
   y++;

   return 2;
}

int v = 1;

v += f(v);

When the function f changes y (that is v) while evaluating v += f(v), is the original value of v "frozen" and changes to v "lost"?

v += f(v); // Compute the address of v (l-value)
           // Evaluate v (1)
           // Execute f(v), which returns 2
           // Store 1 + 2 
printf(v); // 3

score 3 · Accepted Answer · answered Jun 08 '13 at 20:13

3

In most languages += operator (as well as any other compound assignment operator, as well as simple assignment operator) has right-to-left associativity. That means f(v) value will be evaluated first, then its result will be added to the current value of v.

So in your example it should be 4, not 3:

C++: (demo)

int f(int& v) {
  v++;
  return 2;
}

int main() {
  int v = 1;
  v += f(v);
  cout << v; // 4
}

Perl: (demo)

sub f {
  $_[0]++;
  return 2;
}

my $v = 1;
$v += f($v);

print $v; # 4

answered Jun 08 '13 at 20:13

raina77ow

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That is, the same as _just increment the value of v by the quantity on the right side_, right? – gremo Jun 08 '13 at 20:17
I'd say 'evaluate the right side, then increment v by its result'. The key point here is the order of evaluation. – raina77ow Jun 08 '13 at 20:18
Good, got the point. Thanks for your answer and live examples. – gremo Jun 08 '13 at 20:20

Compound assignment operators, what happens if the value is modified (in the meanwhile)?

1 Answers1