I find myself dealing with a complex method in Typescript that contains multiple if else if else constructs. It's a void method, and I'm wondering how I can exit the method based on a specific if condition without having to execute the remaining conditions. This would allow me to skip over the else block entirely; essentially replacing it with more if statements.
In Java, achieving this is simple by using the return; statement within the void method.
In Javascript, Typescript, and Java, the process is identical: you simply utilize a straightforward return;.
if (true) {
return;
}
console.log('Never printed')
This procedure remains consistent across multiple languages such as C, C++, C#, with minor differences in syntax.
To terminate the function execution, you can employ a blank return statement:
function example() {
if(true) {
return;
} else if (true) {
console.log('The message will not be displayed as "return" is used within the first "if" condition');
}
}
example();Perhaps the issue lies in the type of value being returned as undefined. When working with TypeScript, I usually compare values to null, but for certain types such as Date, I also need to consider the possibility of undefined, like so:
if (this.creationDate === null || this.creationDate === undefined) { return; }
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