Are you able to turn off the TypeScript rule TS(1345)?

Question

Are you able to turn off the TypeScript rule TS(1345)?

When using valid debugging code like the example below, TypeScript may throw an error ("An expression of type 'void' cannot be tested for truthiness.ts(1345)"):

const getFooPlusTwo = (foo) => console.log(foo) || foo + 2;

There are various hacks that can be used to work around this issue, such as adding +, ,, etc. However, these workarounds may not be effective in all circumstances and can add unnecessary complexity without any real benefit, as JavaScript functions fine without them.

Fortunately, most TypeScript rules can be disabled. Are there any TypeScript (or ESLint) rules that I can turn off to prevent TypeScript from throwing these errors?

typescript

Answer 1

Answer №1

If you're looking to bypass this typescript rule, keep in mind that console.log yields a void result which cannot be used for a truthiness comparison.

One workaround could involve utilizing the comma operator to divide the log action and the return value. The comma operator appraises each operand (starting from the left) and produces the last operand's value.

export const getFooPlusTwo = (foo: number) => (console.log(foo), foo + 2);

Answer 2

If you're looking to bypass this typescript rule, keep in mind that console.log yields a void result which cannot be used for a truthiness comparison.

One workaround could involve utilizing the comma operator to divide the log action and the return value. The comma operator appraises each operand (starting from the left) and produces the last operand's value.

export const getFooPlusTwo = (foo: number) => (console.log(foo), foo + 2);

Answer 3

Answer №2

If you want to customize the global console object so that its methods always return undefined instead of void, you can do so safely by overwriting the method definitions. For example, you can modify the console.log method like this:

declare global { 
   interface Console { 
      log(msg: any, ...args: any[]): undefined; 
   } 
}

const calculateSum = (num1: number, num2: number) => console.log(num1 + num2) || num1 + num2;

You can also check out the Playground Link for a live demonstration and further explanation on how to declare global variables in TypeScript.

This modification will prevent unintended use of the void keyword in conditional statements, ensuring better code safety. For instance, if you mistakenly pass a callback that returns 1 to a function expecting void, it could lead to unexpected behavior. By forcing all console methods to return undefined, such errors can be minimized.

function processResult(callback: () => void) {
    return callback() || 'default';
}
processResult(() => 1)

The compiler will raise an error stating "An expression of type 'void' cannot be tested for truthiness," highlighting the potential issue with using void in certain scenarios.

Answer 4

If you want to customize the global console object so that its methods always return undefined instead of void, you can do so safely by overwriting the method definitions. For example, you can modify the console.log method like this:

declare global { 
   interface Console { 
      log(msg: any, ...args: any[]): undefined; 
   } 
}

const calculateSum = (num1: number, num2: number) => console.log(num1 + num2) || num1 + num2;

You can also check out the Playground Link for a live demonstration and further explanation on how to declare global variables in TypeScript.

This modification will prevent unintended use of the void keyword in conditional statements, ensuring better code safety. For instance, if you mistakenly pass a callback that returns 1 to a function expecting void, it could lead to unexpected behavior. By forcing all console methods to return undefined, such errors can be minimized.

function processResult(callback: () => void) {
    return callback() || 'default';
}
processResult(() => 1)

The compiler will raise an error stating "An expression of type 'void' cannot be tested for truthiness," highlighting the potential issue with using void in certain scenarios.

Are you able to turn off the TypeScript rule TS(1345)?

Answer №1

Answer №2

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