Are you getting the error “Yield expression implicitly results in an ‘any’ type” in TypeScript and don’t know how to fix it? Don’t worry because in this tutorial, we will show you the cause of this problem and how to solve it properly.
What causes this error?
Generator has been a very important feature in TypeScript since it was released. Yield is a keyword that is used to pause and resume the generator function.
Syntax:
yield [expression]
Parameter:
- expression: the value to be returned from the generated function.
Example 1:
function* lineGenerator() {
console.log('This is line 1!');
yield 1; //value=1
console.log('This is line 2!');
yield 2; //value=2
console.log('This is line 3!');
yield 3; //value=3
console.log('This is line 4');
yield 4; //value=4
}
var line = lineGenerator();
line.next();
line.next();
line.next();
line.next();Output:
This is line 1!
This is line 2!
This is line 3!
This is line 4!Reproduce the “Yield expression implicitly results in an ‘any’ type” error in TypeScript
Let’s take a look at example 2:
function* myGenerator() {
var myStr = yield 'LearnShareIT.com';
};
var result = myGenerator();
console.log(result.next());Output:
'yield' expression implicitly results in an 'any' type because its containing generator lacks a return-type annotation.
'myStr' is declared but its value is never read.As we run the program, we will have this error. This is because TypeScript cannot identify the type to be received from the expression. So, here are some solutions for this issue:
How to fix the error “Yield expression implicitly results in an ‘any’ type” in TypeScript?
Set the type of the generator function to ‘any’
One quick way to fix this is to set the type of our generator function to any.
function* myGenerator() : any {
var myStr = yield 'LearnShareIT.com';
};
var result = myGenerator();
console.log(result.next());Output:
{
"value": "LearnShareIT.com",
"done": false
} However, this method is not highly recommended since what it does is turning off the type-checking, or we can say that, it tricks the compiler.
Set the return-type annotation for the generator
Instead of setting the type of the generator to any, we can set the return-type annotation.
function* myGenerator(): Generator<string> {
var myStr = yield 'LearnShareIT.com';
};
var result = myGenerator();
console.log(result.next());Output:
{
"value": "LearnShareIT.com",
"done": false
}The type that we provide to the generator is the type that we want to yield. So in our example, it will be string.
Not using the result
As we can see in example 1, we are not using the results of yield, so there is no error. Therefore, if we don’t use the results, we don’t have to worry much about their type.
Summary
In this tutorial, we have shown you some of the solutions to fix the error “Yield expression implicitly results in an ‘any’ type” in TypeScript. The idea is that we can set the type of the generator to any or the exact return-type annotation. Let’s try it.
Maybe you are interested:
- Catch clause variable type annotation must be any or unknown if specified
- This expression is not callable. Type ‘String’ has no call signatures
- This expression is not callable. Type ‘#’ no call signatures in TypeScript
Hello. My name is Khanh Hai Ngo. I graduated in Information Technology at VinUni. My advanced programming languages include C, C++, Python, Java, JavaScript, TypeScript, and R, which I would like to share with you. You will benefit from my content.
Name of the university: VinUni
Major: EE
Programming Languages: C, C++, Python, Java, JavaScript, TypeScript, R