How to pass a variable reference to a function in JavaScript?

Question

Suppose I have the following react-native code, in which I want the "press" function to be different based on the value of bProps.type:

const press = bProps.type > 0 
    ? props.function1(arg1, arg2) 
    : props.function2(arg1, arg2);
return <Button onPress={press}></Button>;

But the problem is that both function1 and function2 seem to be called before pressing the buttons, and pressing the buttons doesn't seem to call these functions. Is there a way I can set the value of "press" so that pressing the buttons calls the right functions?

Quentin · Accepted Answer · 2020-01-14 16:27:10Z

3

Currently you are calling the function and assigning its return value to press.

You need to create a new function (which will call the function you want to call with arguments when it itself is called by the event being triggered).

const press = bProps.type > 0 
    ? function() { props.function1(arg1, arg2) }
    : function() { props.function2(arg1, arg2) };
return <Button onPress={press}></Button>;

or

const press = bProps.type > 0 
    ? props.function1.bind(null, arg1, arg2) }
    : props.function2.bind(null, arg1, arg2) };
return <Button onPress={press}></Button>;

or

const press = bProps.type > 0 
    ? () => props.function1(arg1, arg2)
    : () => props.function2(arg1, arg2);
return <Button onPress={press}></Button>;

answered Jan 14, 2020 at 16:27

Quentin

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Comments

user47589 · Accepted Answer · 2020-01-14 16:27:27Z

0

You can try this:

const press = bProps.type > 0 
    ? () => props.function1(arg1, arg2) 
    : () => props.function2(arg1, arg2);
return <Button onPress={press}></Button>;

The original code had to evaluate function1 and function2 before it could evaluate the ternary operator. Wrapping them in lambdas means the lambdas need to be evaluated, but that doesn't mean they are immediately called.

answered Jan 14, 2020 at 16:27

user47589

Comments

Max · Accepted Answer · 2020-01-14 16:35:42Z

0

You can try this:

const press = bProps.type > 0 ? props.function1 : props.function2;
return <Button onPress={() => press(arg1,arg2)}></Button>;

answered Jan 14, 2020 at 16:35

Max

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