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class Popup1 { constructor( ) { this.name &#61; 'Popup1'; } printName( ) {g

동생 · Answer

자바스크립트 다시 공부 하셔야할듯

mirheeoj · Answer

chatgpt의 답변입니다. In fact, not binding the instance in this case results in unexpected behavior because the printName method relies on the correct this value to access the instance's properties. Without binding, this inside printName will refer to the global object (e.g., window in a browser environment), or undefined in strict mode, which will not have the name property defined.이렇게 수정하면 값이 나옵니다. 이게 글쓰신 분의 의도라면요. let popup1 = new Popup1(); callFunction(popup1.printName.bind(popup1));

gasdff

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