javascript - Typescript - Inline undefined check not working (Object is possibly 'undefined'.ts(2532)) - Stack O

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I get the following typescript error:

const myFunction = (
  param1: string | undefined,
  param2: { someProp: string } | undefined
) => {
  if (!param1 && !param2) {
    return;
  }

  // Here I get the following Typescript error:
  //  (parameter) param2: { someProp: string } | undefined
  //  Object is possibly 'undefined'.ts(2532)
  const param3 = param1 ? param1 : param2.someProp;
};

the following works:

const param4 = param1 ? param1 : param2 ? param2.someProp : null;

but seems redundant to check for null or undefined twice.

I have to mention that the strictNullChecks option is set to true in the pilerOptions and want to keep it like this.

Any idea why I get this error?

Here's a CodeSandbox with the code:

I get the following typescript error:

const myFunction = (
  param1: string | undefined,
  param2: { someProp: string } | undefined
) => {
  if (!param1 && !param2) {
    return;
  }

  // Here I get the following Typescript error:
  //  (parameter) param2: { someProp: string } | undefined
  //  Object is possibly 'undefined'.ts(2532)
  const param3 = param1 ? param1 : param2.someProp;
};

the following works:

const param4 = param1 ? param1 : param2 ? param2.someProp : null;

but seems redundant to check for null or undefined twice.

I have to mention that the strictNullChecks option is set to true in the pilerOptions and want to keep it like this.

Any idea why I get this error?

Here's a CodeSandbox with the code: https://codesandbox.io/s/jn2mp01q2v

• javascript
• typescript
• tsc

Share Improve this question Follow edited May 3, 2019 at 13:45 Sergiu asked May 3, 2019 at 11:58 SergiuSergiu 1,39611 gold badge1818 silver badges3333 bronze badges 2

• the following works ... it doesn't work if strictNullChecks is set to true – bugs Commented May 3, 2019 at 12:17
• @bugs sorry, my bad. You need to specify the correct type. Please take a look at the CodeSandbox link: codesandbox.io/s/jn2mp01q2v. – Sergiu Commented May 3, 2019 at 13:30

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 6

The sad truth about the TypeScript piler is that it just isn't as smart as a human being (as of TypeScript 3.4 anyway) and so its control flow analysis is but a pale shadow of the sort of analysis you can perform yourself. Of course, it is very consistent about its analysis, whereas mine tends to get worse when I haven't eaten recently.

If you perform a check on a variable of a union type which pletely eliminates one or more of the constituents of that union, the piler will happily narrow the type of the variable for you:

param1.charAt(0); // error, possibly undefined
if (!param1) return;
param1.charAt(0); // okay now

But one thing the piler just doesn't do is keep track of correlated variables outside of discriminated unions. What you've eliminated by checking

if (!param1 && !param2) return;

is the possibility that both param1 and param2 can be undefined at the same time. You've taken two previously independent variables, and made them correlated to each other. Which the piler doesn't keep track of. Since param1 and param2 can both still be undefined (just not at the same time), the piler treats them as still independent and you are left with your problem.

You can do what the other answer suggested and use a type assertion, which is meant for occasions where you're smarter than the piler and don't want to try to lead the piler through the task of understanding what you already know:

const param3 = param1 ? param1 : param2!.someProp; // I'm smarter than the piler 
I get the following typescript error:

const myFunction = (
  param1: string | undefined,
  param2: { someProp: string } | undefined
) => {
  if (!param1 && !param2) {
    return;
  }

  // Here I get the following Typescript error:
  //  (parameter) param2: { someProp: string } | undefined
  //  Object is possibly 'undefined'.ts(2532)
  const param3 = param1 ? param1 : param2.someProp;
};


the following works:

const param4 = param1 ? param1 : param2 ? param2.someProp : null;


but seems redundant to check for null or undefined twice.

I have to mention that the strictNullChecks option is set to true in the pilerOptions and want to keep it like this.

Any idea why I get this error?

Here's a CodeSandbox with the code: 
    
                
I get the following typescript error:

const myFunction = (
  param1: string | undefined,
  param2: { someProp: string } | undefined
) => {
  if (!param1 && !param2) {
    return;
  }

  // Here I get the following Typescript error:
  //  (parameter) param2: { someProp: string } | undefined
  //  Object is possibly 'undefined'.ts(2532)
  const param3 = param1 ? param1 : param2.someProp;
};


the following works:

const param4 = param1 ? param1 : param2 ? param2.someProp : null;


but seems redundant to check for null or undefined twice.

I have to mention that the strictNullChecks option is set to true in the pilerOptions and want to keep it like this.

Any idea why I get this error?

Here's a CodeSandbox with the code: https://codesandbox.io/s/jn2mp01q2v
    

        
            
                
                    
                    
javascript
typescript
tsc
                
            
        

    
        
            
                




    

        
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            edited May 3, 2019 at 13:45
        
        
    
    
        
    
    
        Sergiu
        
            
        
    

                
            
                
    
        
            asked May 3, 2019 at 11:58
        
        
    
    
        
    
    
        SergiuSergiu
        
            1,39611 gold badge1818 silver badges3333 bronze badges
        
    



            
        
    
    





            

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                the following works ... it doesn't work if strictNullChecks is set to true
                
                
– bugs
                
                
                    Commented
                    May 3, 2019 at 12:17
                
            
        
    
    

        
            
            
        
        
            
                
                @bugs sorry, my bad. You need to specify the correct type. Please take a look at the CodeSandbox link: codesandbox.io/s/jn2mp01q2v.
                
                
– Sergiu
                
                
                    Commented
                    May 3, 2019 at 13:30
                
                        
                            
                        
            
        
    

            
	    

        
                        Add a ment
                     | 
            
                 
    

    





    
		


                
                
                    
                    
                        
                            
                                

                                        2 Answers
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The sad truth about the TypeScript piler is that it just isn't as smart as a human being (as of TypeScript 3.4 anyway) and so its control flow analysis is but a pale shadow of the sort of analysis you can perform yourself.  Of course, it is very consistent about its analysis, whereas mine tends to get worse when I haven't eaten recently.  

If you perform a check on a variable of a union type which pletely eliminates one or more of the constituents of that union, the piler will happily narrow the type of the variable for you:

param1.charAt(0); // error, possibly undefined
if (!param1) return;
param1.charAt(0); // okay now


But one thing the piler just doesn't do is keep track of correlated variables outside of discriminated unions.  What you've eliminated by checking 

if (!param1 && !param2) return;


is the possibility that both param1 and param2 can be undefined at the same time.  You've taken two previously independent variables, and made them correlated to each other.  Which the piler doesn't keep track of.  Since param1 and param2 can both still be undefined (just not at the same time), the piler treats them as still independent and you are left with your problem.

You can do what the other answer suggested and use a type assertion, which is meant for occasions where you're smarter than the piler and don't want to try to lead the piler through the task of understanding what you already know:

const param3 = param1 ? param1 : param2!.someProp; // I'm smarter than the piler 

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