Card 04/ 09
if vs switch: Choosing by What the Branches Test
Ranges and mixed tests suit an if ladder; one value against fixed constants suits a switch.
if and switch can both choose between paths, and each can be forced to do the other's job. This card compares them on what they test and what the compiler checks for you, and ends with a rule for choosing.
Forced into the wrong job, each one gets awkward. A switch asked to express a range needs a case for every value in it. An if ladder asked to look up a code becomes twenty rungs that all compare the same variable, one after another.
| Compared on | if / else if | switch |
|---|---|---|
| Tests | A condition per rung | One value against constants |
| Accepts | Anything that produces true or false | int, char, String, an enum, and in Java 21 any object through type patterns |
| Ranges | Natural | Not on an int: every value has to be listed |
| Many values, one outcome | A long chain of || (or) | Comma-separated on one case |
| Every value handled | Nothing checks it | Checked by the compiler for a switch expression over an enum or sealed type |
The decision rule. Ask what the branches are testing. If every rung compares the same variable with a constant, write a switch. If the rungs test different things, or the same thing with > and <, write an if ladder.
The last row matters most in real code. An enum is a type with a fixed list of named values, such as the tiers BRONZE, SILVER and GOLD. A switch expression over it, written without a default, must handle every value, so the day somebody adds PLATINUM the build fails with "the switch expression does not cover all possible input values", instead of giving a wrong answer at run time. A switch statement, and an if ladder, get no such check: they compile, and the new value falls through to whatever the code does by default.
So when an if ladder compares one variable with constants, rewriting it as a switch expression makes it shorter and lets the compiler catch the case you forgot.
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