Set "This" Variable Easily

Set this variable easily?

There are two methods defined for all functions in JavaScript, call(), and apply(). The function syntax looks like:

call( /* object */, /* arguments... */ );
apply(/* object */, /* arguments[] */);

What these functions do is call the function they were invoked on, assigning the value of the object parameter to this.

var myFunction = function(){
    alert(this.foo_variable);
}
myFunction.call( document.body );

How can I easily set a variable to false with a default of true?

You could use the ternary (conditional) operator like this:

this.listening = config.listening === false ? false : true;

If config.listening is false, this.listening is set to false. If it's any other value, it's set to true.

If you want to check if it's defined, you could use:

this.listening = typeof config.listening !== "undefined"

References:

• https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Operators/Conditional_Operator

How to set a variable by using another variable to name

Javascript variable depend on the lexical scope they are declared/defined in, wether it be a global scope or inside some function, if you're just defining your variable inside a the script tags, you might as well say

window[user_input_var] == true

which is equivalent to (in case user input was "red")

var red = true.

The most important part you should get out of this is variables in javascript are lexically scoped somewhere. Lexically scoped means physically declared somewhere, and based on that declaration we can use them in other places if we have access to this scope.

scopes can open up a lot of discussions here, and i don't think we need to dig down that rabbit whole. my example above should do the trick, if not determine in what scope/location you want your variable to be declared, and define there as in window.

js set x variable to y without using extra variable

1) Only using math operations you can do like this. Keep the sum into the first variable. Then in the second keep the difference of the sum and the current's value, which will give the first value. After this in the first variable keep the difference of the sum and the seconds value, which will give you the first value.

let x = 5;let y = 6;
console.log(`x is ${x}, y is ${y}`);
x = x + y;y = x - y;x = x - y;
console.log(`x is ${x}, y is ${y}`);

How do I change the value of a global variable inside of a function

Just reference the variable inside the function; no magic, just use it's name. If it's been created globally, then you'll be updating the global variable.

You can override this behaviour by declaring it locally using var, but if you don't use var, then a variable name used in a function will be global if that variable has been declared globally.

That's why it's considered best practice to always declare your variables explicitly with var. Because if you forget it, you can start messing with globals by accident. It's an easy mistake to make. But in your case, this turn around and becomes an easy answer to your question.

What is the fastest way to update a variable on a condition?

void reset_if_true(void*& ptr, bool cond)
{
    if (cond)
        ptr = nullptr;
}

The naive solution will undoubtedly be the fastest in the majority of cases. Although it has a branch, which can be slow on modern pipelined processors, it is only slow if the branch is mispredicted. Since branch predictors are very good nowadays, unless the value of cond is extremely unpredictable, it's likely that a simple conditional branch is the fastest way to write the code.

And if it is not, a good compiler should know that and be able to optimize the code to something better, considering the target architecture. Which goes to gnasher729's point: just write the code the simple way and leave optimization in the hands of the optimizer.

While this is good advice in general, sometimes it is taken too far. If you actually care about the speed of this code, you need to check and see what the compiler is actually doing with it. Check the object code that it is generating, and make sure that it is sensible and that the function's code is getting inlined.

Such an examination can be quite revealing. For example, let's consider x86-64, where branches can be quite expensive in cases where branch prediction is foiled (which is really the only time when this is an interesting question, so let's assume that cond is completely unpredictable). Almost all compilers are going to generate the following for the naive implementation:

reset_if_true(void*&, bool):
    test   sil, sil              ; test 'cond'
    je     CondIsFalse
    mov    QWORD PTR [rdi], 0    ; set 'ptr' to nullptr, and fall through
  CondIsFalse:
    ret

This is about as tight of code as you could imagine. But if you put the branch predictor in a pathological case, it might end up being slower than using a conditional move:

reset_if_true(void*&, bool):
    xor    eax, eax              ; pre-zero the register RAX
    test   sil, sil              ; test 'cond'
    cmove  rax, QWORD PTR [rdi]  ; if 'cond' is false, set the register RAX to 'ptr'
    mov    QWORD PTR [rdi], rax  ; set 'ptr' to the value in the register RAX
    ret                          ;  (which is either 'ptr' or 0)

Conditional moves have a relatively high latency, so they are considerably slower than a well-predicted branch, but they might be faster than a completely unpredictable branch. You would expect a compiler to know this when targeting the x86 architecture, but it doesn't (at least in this simple example) have any knowledge about cond's predictability. It assumes the simple case, that branch prediction will be on your side, and generates code A instead of code B.

If you decide that you want to encourage the compiler to generate branchless code because of an unpredictable condition, you might try the following:

void reset_if_true_alt(void*& ptr, bool cond)
{
    ptr = (cond) ? nullptr : ptr;
}

This succeeds in persuading modern versions of Clang to generate branchless code B, but is a complete pessimization in GCC and MSVC. If you hadn't checked the generated assembly, you wouldn't have known that. If you want to force GCC and MSVC to generate branchless code, you will have to work harder. For example, you might use the variation posted in the question:

void reset_if_true(void*& ptr, bool cond)
{
    void* p[] = { ptr, nullptr };
    ptr = p[cond];
}

When targeting x86, all compilers generate branchless code for that, but it is not especially pretty code. In fact, none of them generate conditional moves. Instead, you get multiple accesses to memory in order to build the array:

reset_if_true_alt(void*&, bool):
    mov     rax, QWORD PTR [rdi]
    movzx   esi, sil
    mov     QWORD PTR [rsp-16], 0
    mov     QWORD PTR [rsp-24], rax
    mov     rax, QWORD PTR [rsp-24+rsi*8]
    mov     QWORD PTR [rdi], rax
    ret

Ugly and probably very inefficient. I'd predict that it gives the conditional jump version a run for its money even in the case where the branch is mispredicted. You'd have to benchmark it to be sure, of course, but it is probably not a good choice.

If you were still desperate to eliminate the branch on MSVC or GCC, you'd have to do something uglier involving reinterpreting the pointer bits and twiddling them. Something like:

void reset_if_true_alt(void*& ptr, bool cond)
{
    std::uintptr_t p = reinterpret_cast<std::uintptr_t&>(ptr);
    p &= -(!cond);
    ptr = reinterpret_cast<void*>(p);
}

That will give you the following:

reset_if_true_alt(void*&, bool):
    xor   eax, eax
    test  sil, sil
    sete  al
    neg   eax
    cdqe
    and   QWORD PTR [rdi], rax
    ret

Again, here we've got more instructions than a simple branch, but at least they're relatively low-latency instructions. A benchmark on realistic data will tell you if the tradeoff is worth it. And give you the justification you need to put in a comment if you're going to actually check-in code like this.

Once I went down the bit-twiddling rabbit hole, I was able to force MSVC and GCC to use conditional move instructions. Apparently they weren't doing this optimization because we were dealing with a pointer:

void reset_if_true_alt(void*& ptr, bool cond)
{
    std::uintptr_t p = reinterpret_cast<std::uintptr_t&>(ptr);
    ptr = reinterpret_cast<void*>(cond ? 0 : p);
}

reset_if_true_alt(void*&, bool):
    mov    rax, QWORD PTR [rdi]
    xor    edx, edx
    test   sil, sil
    cmovne rax, rdx
    mov    QWORD PTR [rdi], rax
    ret

Given the latency of CMOVNE and the similar number of instructions, I'm not sure if this would actually be faster than the previous version. The benchmark you ran would tell you if it was.

Similarly, if we bit-twiddle the condition, we save ourselves one memory access:

void reset_if_true_alt(void*& ptr, bool cond)
{
   std::uintptr_t c = (cond ? 0 : -1);
   reinterpret_cast<std::uintptr_t&>(ptr) &= c;
}

reset_if_true_alt(void*&, bool):
     xor    esi, 1
     movzx  esi, sil
     neg    rsi
     and    QWORD PTR [rdi], rsi
     ret

(That's GCC. MSVC does something slightly different, preferring its characteristic sequence of neg, sbb, neg, and dec instructions, but the two are morally equivalent. Clang transforms it into the same conditional move that we saw it generate above.) This may be the best code yet if we need to avoid branches, considering that it generates sane output on all tested compilers while preserving (some degree of) readability in the source code.

Method for asking input easily not assigning input to variable

You need to return your input statement in input method.

static String input(String inputprompt){

  System.out.print(inputprompt);
Scanner userInput;
userInput = new Scanner(System.in);
String input = userInput.nextLine();
return input;/* it is written as a new variable because it generates an
  error otherwise*/}

public static void main(String[] args) {
    String input = "useless for now"; /*this line is because it detects that there is no variable at
  the line where it assigns the value of input to text before the script runs*/
    String text = input("what would you like the following text to be");
    //the line refered to in previous comment
    System.out.print(text);
}

}

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