What causes the error "this condition will always return 'true'" when using an else if clause?

Exploring the concepts outlined in the handbook basics, it is noted that TypeScript is able to identify logic errors. In an attempt to experiment with this, I modified the else if section of the if...else statement as demonstrated in the handbook's example to the following:

const value = Math.random() < 0.5 ? "a" : "b";
if (value !== "a") {

} else if (value !== "b") {
    
}

However, I encountered an error message:

"This condition will always return 'true' since the types '"a"' and '"b"' have no overlap."

I am puzzled by this. Furthermore, it is worth noting that this approach worked in JavaScript but seems to be ineffective in TypeScript. What could be causing this inconsistency?

Answer №1

Exploring control flow analysis and the concept of type narrowing.

Initially, TypeScript assigns the type value as "a" | "b". The conditional check value !== "a" results in value being narrowed down to "b" in the true branch and "a" in the false branch. This ensures that TypeScript accurately infers the type of value based on the condition.

By analyzing this, it becomes evident that the condition value !== b will always evaluate to true since value is initially "a". TypeScript detects this and provides a warning to alert the developer of this logical certainty.

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