let choice = Math.random() < 0.5 ? "a" : "b";
if (choice !== "a") {
// ...
} else if (choice === "b") {
This situation will always be false because the values 'a' and 'b' are completely distinct.
// Oops, unreachable
}
Came across this snippet in a coding guide. I'm curious about the message:
"types '"a"' and '"b"' have no overlap."What does this actually mean?
When it comes to TypeScript, "a" is classified as a literal type, meaning it has only one value, which is "a". The same goes for "b".
Based on how you initialize the variable value, TypeScript recognizes its type as "a" | "b", a union type indicating that the value must be either "a" or "b".
If the type (and hence the value) of value is not "a", then the first if block will be executed.
This implies that the else if condition will certainly work with a value of type and value "a". Consequently, you are comparing "a" with "b".
However, in TypeScript, no value can simultaneously belong to both "a" and "b" types because these types do not overlap (meaning no value can be of both types at once).
For example, consider the types string and "a" have some overlap since the value "a" can be classified under both types.
In this scenario, the variable value is restricted to either the string 'a' or 'b', which means:
if (value !== 'a') {
// value can ONLY be 'b' here
} else {
// value can ONLY be 'a' here
}
Therefore, the condition else if (value === 'b') will never be fulfilled and TypeScript alerts you that 'a' !== 'b'.
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