The "error TS7030: Not all code paths return a value." caused by "strict": false in tsconfig.json

At first, I’d like to reproduce my issue.

1. Create a new Angular project

   ng new ng-ts-strict-issue
   cd ng-ts-strict-issue
   

2. Modify compilerOptions in tsconfig.json. Let’s set strict to false.

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   {
     ...
     "compilerOptions": {
       ...
       "strict": false,
       ...
     },
     ...
   }
   

3. Add a method to the app.component.ts.

   test(p: string): string | null | undefined {
     if (p === 'test') {
       return;
     }
     return null;
   }
   

4. Run ng build

   ⠼ Building...✘ [ERROR] TS7030: Not all code paths return a value. [plugin angular-compiler]
   
       src/app/app.component.ts:17:6:
         17 │       return;
           ╵       ~~~~~~
   
   
   Application bundle generation failed. [3.905 seconds]
   

   

Actually, I know this error. The error is related to "noImplicitReturns": true setting in the tsconfig.json file. I actually can turn this option off to avoid this issue.

But my question is why the return; statement will lead to TS7030: Not all code paths return a value. error only when strict been set to false? I actually already returned on every path. Why I violated the noImplicitReturns rule?

>Solution :

When the --strictNullChecks compiler option is disabled, your code is interpreted as:

function test(p: string): string {
  if (p === 'test') {
    return; // error
  }
  return null;
}

That is, string | null | undefined is just string because null and undefined are implicitly contained in every type. According to microsoft/TypeScript#7358, the bare return statement is therefore considered likely to be a mistake because it does not return a string, nor does it explicitly return null or undefined. This behavior was requested in microsoft/TypeScript#5916.

The error message might not be the best, since it looks like it’s saying "not all code paths return" but it really means "not all code paths explicitly return a value".

Anyway, the fix here (assuming you want to keep your compiler options as-is) would be to explicitly return undefined instead:

function test(p: string): string {
  if (p === 'test') {
    return undefined; // okay
  }
  return null;
}

Playground link to code