Review and Reinforcement: Operators

Tom Kelliher, CS18

February 7, 1996

Nested Loops

(Finishing from last time)

Arbitrary nesting possible

Example:

i = j = 0;
for (i = 0; i < 10; i++) {
   cout << setw(10) << i << j << endl;   // first 12 values printed?

   for (j = 9; j >= 0; j--)
      cout << setw(10) << i << j << endl;
}

goto Considered Bad

How could we break out of both loops in last example?

Operators

Includes:

• Arithmetic
• Relational
• Equality, inequality
• Assignment
• Logical

Two Somewhat Obscure Operators

Comma ( ,) binary operator:

for (sum = 0, i = 0; i < MAX; i++)
   sum += data[i];

cout << sum << endl;

Not the comma separating items in a list (arguments, identifiers in a declaration)

Conditional ( ? :) trinary operator:

max = (i > j) ? i : j;   // condition parenthesized by convention

if (i > j)
   max = i;
else
   max = j;

Another example:

cout << i << "item" << ((i > 0) ? "s" : "") << endl;

Why the parentheses around the entire conditional operator?

There is No Assignment Statement

Assignment, being an operator, yields a value

All sequential statements (excepting void functions) have/return a value, usually discarded

Consider:

i = j = 0;

i = (j = 5) + 6;

if ((objectPointer = (object*) malloc(sizeof(object)) == NULL)
   fatal("Memory allocation failure");

Order of Operator Evaluation

Affected by:

1. Operator precedence
2. Operator associativity

Assignment, unary operators right associate; all others left associate

Parentheses change the order of evaluation

Expressions Types and Conversions

2 / 5 is 0. Why??? Overloading.

How can we fix it?

What is the type of 2.5 + 3?

How could we fix this:

int i = 2;
int j = 5;
float x;

x = 2 / 5;   // x == 0.0

In a mixed-type expression, how should we convert types?