Increments: From There to Here and Some Things I Learned Along the Way

While there may be a few use cases for it, it should typically be avoided. In the event of an error in a component, getDerivedStateFromError runs and you can update state to reflect that an error occurred. Use this method copiously. The following CodePen snippet shows the steps in the mounting phase:. You may have used higher-order components, or HOCs, already. But what exactly is a HOC? When we call connect , we get a HOC back that we can use to wrap a component. What HOCs allow us to do is abstract shared logic between components into a single overarching component.

A good use case for an HOC is authorization. You could write your authentication code in every single component that needs it. It would quickly and unnecessarily bloat your code. Using HOCs, you might do something like so:.

The AuthWrapper component lifts all authentication logic into a unifying component. All it does is take a prop called isLoggedIn and returns the WrappedComponent or a paragraph tag based on whether or not that prop is true or false. As you can see, HOCs are extremely useful because they let us reuse code and remove bloat.

Most of you have probably used React state, we even used it in our HOC example. The only way you should change state is via the setState method. This method takes an object and merges it into the current state.

On the past, present, and future of COBOL – Increment: Programming Languages

On top of this, there are a few things you should also know about it. First, setState is asynchronous. Looking at the above image, you can see that we call setState and then console. So what if we want to access the new state after setState actually updates state? This brings us to the next piece of knowledge that we should know about setState and that is it can take a callback function. Because setState is asynchronous, relying on it to create our new value will have some pitfalls. For example, by the time setState runs, another setState could have mutated state. Passing setState a function gives us two benefits.

The first is it allows us to take a static copy of our state that will never change on its own. The second is that it will queue the setState calls so they run in order. Just take a look at the following example where we try to increment the counter by 2 using two consecutive setState calls:. The above is what we saw earlier while we have the fix below.

CodePen for above code. In the first picture, both setState functions directly use this. Thus, we get 1 instead of 2 because both setState functions are setting counter to 1. In the second picture, we pass setState a function which will guarantee both setState functions run in order. On top of this, it takes a snapshot of state, rather than using the current, un-updated state.

A snapshot of programming language history

Now we get our expected result of 2. This brings us now to React context which is just global state for components. The React context API allows you to create global context objects that can be given to any component you make. This allows you to share data without having to pass props down all the way through the DOM tree.

The React docs describe setting context in a component like so:. However, in CodePen React What we are doing is wrapping our component with the Context. Consumer component and passing in context as a prop. Now we can write something like the following:. How do we change context you might ask. We see this in the following example:.

There's only one side effect applied to x in this statement, but it's still dangerous because we're not sure when the side effect will be applied. Perhaps an alternative wording could be, "if you modify a variable's value in a statement, don't use that variable again in the same statement, otherwise undefined behavior may result". That makes it a lot more clear to me, at least I will know to avoid using variables in that way.

I partially get your explanation about side effects but it's still a little too abstract to me and I struggle to see it in the actual code. Please correct me if I'm wrong, I know I'm probably simplifying it: A side effect is always a value. Not any value, but one that is a result of an operation and is stored in a variable.

So if we use that variable in the same statement as the side effect we can't know in which order the values will be evaluated. A side effect isn't a value, a side effect is anything that modifies the permanent state of something else. But the remainder is true: Well from my understanding "else if" will only be executed if your previous "if" statement was false.

The push function is not a problem, because you're not assigning a new value to stack, you're only modifying it's members. Think of it like this.

Why learn web development with Rails in 2018

So this is a define situation. Pre-increment should be resolve first like this: With side effects, this gets a bit weird. I think i got it, operator precedence only say us how to associate operator with expresion but nothing about when to resolve side effect. Only Sequence points define when side effect have to be resolve if I got it right. This would presumably perform much better than having to load 2 registers and using an ADD operation. But this is highly dependent on the compiler and the CPU instruction set. The default increment and decrement operators are fixed at 1.

Later on you'll learn about defining own data types and own operators which you could use to implement such behavior.

3.3 — Increment/decrement operators, and side effects

Hi Alex, I wanted to check whether my compiler evaluates the left or right argument first, using the code you provided in this lesson. In theory, one of the two values should have been 11 and the other 12, no? In fact both return Am I understanding things incorrectly here? Don't use a variable with side effects applied more than once in an expression.

By looking whether left is 8 and right is 9 or vice-versa, we can determine which operand got evaluated first. And apologies for forcing you to repeat the same thing again. I thought I was not using it twice in an argument actually as I was using the operator only once.

JavaScript Increment ++ and Decrement --

I guess it's the variable that I should count and not how many times I am using the operator. I'm failing to understand your program. The way I see it, the function "gen " is called two times with its own local scope each time, so it would return 9 every time. Getting called once wouldn't have any effect on any subsequent calls. This is especially true considering "x" is a static integer. Well that's what I had initially thought, anyway.

What am I understanding incorrectly?

How do the 2 calls to gen provide different results? You can treat x as a global variable with limited scope. So it's because of the static variable, which comes in the future.

• What is Ruby on Rails.
• Madame Salvatins Hemmelighed (Danish Edition).
• 9.7 — Overloading the increment and decrement operators.
• It’s COBOL all the way down;
• Babylon on a Thin Wire;

I was thinking "static" was another way of having a constant variable but that's where the English language messed me up, haha. Hi Alex, The post increment operator has a higher precedence than assignment operator. Doesn't it mean that. The actual increment part is not guaranteed to happen immediately think of post-increment as "increment at some point in the future".

This was probably originally done for performance reasons, but is one of those annoying quirks of the language that you have to learn not to do. You should feel free to use the increment and decrement operators. Just don't use the variables you're using those operators on more than once in a given statement. Then 'Evaluate x' means 'assign value of temp copy 5 to y', discard temp copy, incrementing x from 5 to 6. In this case, "First, a memory location x with value 5 is made a temporary copy with temp copy which has value 5 now has two memory locations with value 5.

Somehow, my understanding lacks of incorporation of r-value and l-value concept inside and so it seems compiler should compile though I think something is amiss. Don't worry about r-values and l-values in this context. Hi Alex, What do you mean by te outputting of x modifying the console? Outputting the value x to the console changes the state of the console such that it now displays the value of x, whereas before it did not. Can you please explain this program. This is the same program given in example in this chapter but I made a change in the line where the add function is called.

I added pre-decrement operator. When I executed your example without any change, the output was 12, which means that my compiler code blocks first operates on the right argument. If it is so, then this program should have given 11 as output but the output given is Don't use a variable that has a side effect applied to it more than once in a given statement. If you do, the result is undefined. I have a bit of a question I wrote a code that finds the prime numbers up to It works fine, but I can't figure it out how to deal with the first prime number, witch is 2. I put there a separate if statement, just to output 2, and after that it goes normal for the rest of the numbers.

But I would like to make the code more generic, without working separately just for 2. I have tried to change the initial values of dividend and divisor but without any success. That way your while loop logic will work for the number 2. I can't think of a way to not special-case it. I edited your code slightly and used it as a means of practice to give the user the ability to set their own limit.

In the example below, does this mean that it is not guaranteed that line 11, i. Your program doesn't call any functions with multiple arguments, so the order of evaluation issue isn't applicable here. From the Table in section 3. How is the rule applied now? I guess I'm doing it wrong in some part, but I don't know how. I've updated the lesson with a little more detail about how these expressions evaluate.

Have a re-read and let me know if you have additional questions. I think I might know the cause of his confusion cause as I've scratched my head for a few seconds as well. In the first paragraph, I think it should be prefix operator instead of postfix. Thanks for the awesome tutorials! I have problem with this code , I am not understanding the output result that is 9 8 10 6 10 rather according to me it should 5 6 8 8 10 please reply.

Not quite sure what you're asking. Question 3 is fine though, because you're not violating this rule. What value does this program print?

Unfortunately, the answer is somewhat complicated, and has to do with something called sequence points. A sequence point is a point in the program where it's guaranteed that all side effects from previous evaluations have been performed. Between sequence points, the order of side effects resolution is indeterminate. That 1 is assigned to x.

The value of 1 from the previous evaluation is then assigned to x, leaving x with the final value of 1. Why This code is producing the result 23? I think it should produce the result When you use a variable with a side effect applied more than in a single statement, the result is undefined. On your compiler, it produces On another compiler, it may produce On a third compiler, it may produce something else. Printing takes place from left to right. There's no point in explaining it. You should never use a variable with side effects applied more than once in a statement.

Email will not be published required. Enter your search terms Submit search form. It depends on what order your compiler evaluates the function arguments in. Facebook Twitter Google Pinterest. October 23, at 1: October 25, at October 24, at 1: October 5, at 7: