这个回调函数示例是如何工作的？

Question

(I ask understanding from fellow programmers that I am re-studying nodeJS after I got out of the army. I am a beginner and question maybe too simple but please help with my understanding of the example code bellow)

function add(a, b, callback) {
    var result = a + b;
    callback(result);
    var count = 0;

    var history = function() {
        count += 1;
        return count +  ' : ' + a + ' + ' + b + ' = ' + result;  
    };
    return history;

}

var add_history = add(20, 20, function(result) {
    console.log('addition result : ' + result);
});


console.log('execute callback function: ' + add_history());
console.log('execute callback function: ' + add_history());

I expect the result to be following:

addition result : 40
execute callback function: 1 : 20 + 20 = 40
addition result : 40
execute callback function: 2 : 20 + 20 = 40

However, the result says:

addition result : 40
execute callback function: 1 : 20 + 20 = 40
execute callback function: 2 : 20 + 20 = 40

Why is console.log('addition result : ' + result); not repeated each time add_history() is called from the last two statements?

Answer 1

Note that you are returning a function from the add function.

when this block runs:

var add_history = add(20, 20, function(result) {
    console.log('addition result : ' + result);
});

The callback gets executed, the addition result : 40 logs to the console, and add_history becomes a reference to the history function you have defined in the add function.

Now, when you call add_history you are calling a reference to the history function which has nothing to do with the callback. But since history was created in the scope of add , you have access to the count variable defined in that scope.

You could call your callback by defining the history function inside the add function like this:

var history = function() {
   callback(result)
   count += 1;
   return count +  ' : ' + a + ' + ' + b + ' = ' + result;  
}

This would cause the callback to run each time you call add_history .