&gt; Why did the logical operators get left out? Is there a good technical reason why it is hard?

**They didn&#39;t**. You can do `&amp;=` or `|=` or `^=` if you want.

    bool b1 = false;
    bool b2 = true;
    b1 |= b2; // means b1 = b1 | b2

The `||` and `&amp;&amp;` operators do not have a compound form because frankly, they&#39;re a bit silly. Under what circumstances would you want to say

    b1 ||= b2;
    b1 &amp;&amp;= b2;

such that the right hand side is not evaluated if the left hand side does not change? It seems like only a few people would actually use this feature, so why put it in?

For more information about the compound operators, see my serious article here:  
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/compound-assignment-part-one

and the follow-up April-Fools article here:  
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/compound-assignment-part-two




maybe just use 

    isAdmin = isAdmin || IsGroupAdmin()


I guess it is partially because `a ||= b` is kind of confusing because there might be two versions of the implementation: `a = a || b`, or `a = b || a`. And they act differently because the right-hand side of the expression is sometimes not evaluated.

I am getting the following error when I use web deploy from visual studio 2010.
Web deployment task failed. (The type initializer for &#39;Microsoft.Web.Deployment.DeploymentManager&#39; threw an exception.)

I suddenly started getting this error! I published my website many many times with all the same settings, but suddenly it started to give me this error. Could it be something in the website properties? I am using asp.net and have Windows 7 and publishing to Windows Server R2. Please help!

Web deployment task failed. (The type initializer for &#39;Microsoft.Web.Deployment.DeploymentManager&#39; threw an exception.)

If you execute this query

    SELECT &#39;test-a1&#39; AS name1, &#39;test-a2&#39; AS name2
the result will be a one row-selection with two columns having these values: 

    test-a1, test-a2

How can I modify the above query to have a selection with several rows, e.g.

    test-a1, test-a2
    test-b1, test-b2
    test-c1, test-c2

I know how to do this with UNION but I feel that there exists a more simple way to do it.

PS. Sorry for such a basic question, it is very hard to google it.

How to select several hardcoded SQL rows?

We have equivalent assignment operators for all Logical operators, Shift operators, Additive operators and all Multiplicative operators.

Why did the logical operators get left out?
Is there a good technical reason why it is hard?

Why are there no ||= or &amp;&amp;= operators in C#?

<p>We have equivalent assignment operators for all Logical operators, Shift operators, Additive operators and all Multiplicative operators.</p>
<p>Why did the logical operators get left out?
Is there a good technical reason why it is hard?</p>


How can I get the date of next Tuesday? 

In PHP, it&#39;s as simple as `strtotime(&#39;next tuesday&#39;);`. 

How can I achieve something similar in .NET

Datetime - Get next tuesday

given this JSON:

    [
      {
        &quot;$id&quot;: &quot;1&quot;,
        &quot;$type&quot;: &quot;MyAssembly.ClassA, MyAssembly&quot;,
        &quot;Email&quot;: &quot;me@here.com&quot;,
      },
      {
        &quot;$id&quot;: &quot;2&quot;,
        &quot;$type&quot;: &quot;MyAssembly.ClassB, MyAssembly&quot;,
        &quot;Email&quot;: &quot;me@here.com&quot;,
      }
    ]

and these classes:

    public abstract class BaseClass
    {
        public string Email;
    }
    public class ClassA : BaseClass
    {
    }
    public class ClassB : BaseClass
    {
    }


How can I deserialize the JSON into:

    IEnumerable&lt;BaseClass&gt; deserialized;

I can&#39;t use `JsonConvert.Deserialize&lt;IEnumerable&lt;BaseClass&gt;&gt;()` because it complains that `BaseClass` is abstract.


how to deserialize JSON into IEnumerable&lt;BaseType&gt; with Newtonsoft JSON.NET

I am currently building a sales module for a clients website. So far I have got the sale price to calculate perfectly but where I have come stuck is formatting the output to 2 decimal places.

I am currently calling this in a variable so that I can data bind the results to a listview.

    Sale = float.Parse(((x.Sale_Price - (x.Sale_Price * (x.Discount_Price / 100))).ToString())),

Can anyone show me how to format the output to 2 decimal places?? Many Thanks!

Formatting a float to 2 decimal places

The goal is to display the information that the application is working. So I&#39;m looking for an intelligent implementation sample of a loading spinner using WPF / MVVM.


WPF loading spinner

I&#39;m trying to understand the binary operators in C# or in general, in particular [^ - exclusive or][1].

For example: 
&gt; Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time and constant space.
&gt;
&gt;This can be done with ^ as follows: Do bitwise XOR of all the elements. Finally we get the number which has odd occurrences. 

How does it work? 

When I do:

&lt;!-- language: java --&gt;

    int res = 2 ^ 3;  
    res = 1;  
    int res = 2 ^ 5;  
    res = 7;  
    int res = 2 ^ 10;  
    res = 8;  

What&#39;s actually happening? What are the other bit magics? Any reference I can look up and learn more about them? 

  [1]: http://msdn.microsoft.com/en-us/library/zkacc7k1.aspx

What does bitwise XOR (exclusive OR) mean?

How come C# doesn&#39;t have a conditional `XOR` operator?

Example:

    true  xor false = true
    true  xor true  = false
    false xor false = false

Conditional XOR?

Yes, I&#39;ve seen [this question](https://stackoverflow.com/questions/1779685/what-is-operator-in-c) and  [this FAQ](http://www.parashift.com/c++-faq-lite/pointers-to-members.html), but I *still* don&#39;t understand what `-&gt;*` and `.*` mean in C++.  
Those pages provide information *about* the operators (such as overloading), but don&#39;t seem to explain well what they *are*.

What are `-&gt;*` and `.*` in C++, and when do you need to use them as compared to `-&gt;` and `.`?

What are the pointer-to-member operators -&gt;* and .* in C++?

I have code:

      def make_all_thumbs(source)
        sizes = [&#39;1000&#39;,&#39;1100&#39;,&#39;1200&#39;,&#39;800&#39;,&#39;600&#39;]
        threads = []
        sizes.each do |s|
          threads &lt;&lt; Thread.new(s) {
            create_thumbnail(source+&#39;.png&#39;, source+&#39;-&#39;+s+&#39;.png&#39;, s)
          }
        end
      end

what does `&lt;&lt;` mean?

What does &lt;&lt; mean in Ruby?

In javascript I have seen `i++` used in many cases, and I understand that it adds one to the preceding value:

&lt;!-- begin snippet: js hide: false console: true babel: false --&gt;

&lt;!-- language: lang-js --&gt;

    for (var i=1; i&lt;=10; i++) {
      console.log(i);
    }

&lt;!-- end snippet --&gt;

But what happens when I do this:

    ++i;

And is it any different using the `--` operator (besides of course that it&#39;s subtraction rather than addition)?

javascript i++ vs ++i

It is universally agreed that a list of n *distinct* symbols has n! permutations. However, when the symbols are not distinct, the most common convention, in mathematics and elsewhere, seems to be to count only distinct permutations. Thus the permutations of the list `[1, 1, 2]` are usually considered to be   
`[1, 1, 2], [1, 2, 1], [2, 1, 1]`. Indeed, the following C++ code prints precisely those three:

&lt;!-- language: lang-cpp --&gt;

    int a[] = {1, 1, 2};
    do {
        cout&lt;&lt;a[0]&lt;&lt;&quot; &quot;&lt;&lt;a[1]&lt;&lt;&quot; &quot;&lt;&lt;a[2]&lt;&lt;endl;
    } while(next_permutation(a,a+3));

On the other hand, Python&#39;s `itertools.permutations` seems to print something else:

&lt;!-- language: lang-py --&gt;

    import itertools
    for a in itertools.permutations([1, 1, 2]):
        print a

This prints

    (1, 1, 2)
    (1, 2, 1)
    (1, 1, 2)
    (1, 2, 1)
    (2, 1, 1)
    (2, 1, 1)

As user Artsiom Rudzenka pointed out in an answer, the [Python documentation](http://docs.python.org/release/2.6.6/library/itertools.html?highlight=itertools#itertools.permutations) says so:

&gt; Elements are treated as unique based on their position, not on their value.

My question: why was this design decision made? 

It seems that following the usual convention would give results that are more useful (and indeed it is usually exactly what I want)... or is there some application of Python&#39;s behaviour that I&#39;m missing? 

[Or is it some implementation issue? The algorithm as in `next_permutation` — for instance explained on StackOverflow [here (by me)](https://stackoverflow.com/questions/352203/generating-permutations-lazily/353248#353248) and [shown here to be O(1) amortised](https://stackoverflow.com/questions/4973077/the-amortized-complexity-of-stdnext-permutation) — seems efficient and implementable in Python, but is Python doing something even more efficient since it doesn&#39;t guarantee lexicographic order based on value? And if so, was the increase in efficiency considered worth it?]

Why does Python&#39;s itertools.permutations contain duplicates? (When the original list has duplicates)

Maybe this question has been answered before, but the word `if` occurs so often it&#39;s hard to find it.

The example doesn&#39;t make sense (the expression is always true), but it illustrates my question.

Why is this code valid:

    StringBuilder sb;
    if ((sb = new StringBuilder(&quot;test&quot;)) != null) {
        Console.WriteLine(sb);
    }

But this code isn&#39;t:

    if ((StringBuilder sb = new StringBuilder(&quot;test&quot;)) != null) {
        Console.WriteLine(sb);
    }

I found a similar question regarding a `while` statement. The accepted answer there says that in a `while` statement, it would mean the variable would be defined in each loop. But for my `if` statement example, that isn&#39;t the case.

So what&#39;s the reason we are not allowed to do this?

Why can&#39;t we define a variable inside an if statement?

I know there&#39;s all sorts of counter-intuitive properties of Java&#39;s generic types. Here&#39;s one in particular that I don&#39;t understand, and which I&#39;m hoping someone can explain to me. When specifying a type parameter for a class or interface, you can bound it so that it must implement multiple interfaces with `public class Foo&lt;T extends InterfaceA &amp; InterfaceB&gt;`. However, if you&#39;re instantiating an actual object, this doesn&#39;t work anymore. `List&lt;? extends InterfaceA&gt;` is fine, but `List&lt;? extends InterfaceA &amp; InterfaceB&gt;` fails to compile. Consider the following complete snippet:

    import java.util.List;
    
    public class Test {
    
      static interface A {
        public int getSomething();
      }
    
      static interface B {
        public int getSomethingElse();
      }
    
      static class AandB implements A, B {
        public int getSomething() { return 1; }
        public int getSomethingElse() { return 2; }
      }

      // Notice the multiple bounds here. This works.
      static class AandBList&lt;T extends A &amp; B&gt; {
        List&lt;T&gt; list;
    
        public List&lt;T&gt; getList() { return list; }
      }
    
      public static void main(String [] args) {
        AandBList&lt;AandB&gt; foo = new AandBList&lt;AandB&gt;(); // This works fine!
        foo.getList().add(new AandB());
        List&lt;? extends A&gt; bar = new LinkedList&lt;AandB&gt;(); // This is fine too
        // This last one fails to compile!
        List&lt;? extends A &amp; B&gt; foobar = new LinkedList&lt;AandB&gt;();
      }
    }

It seems the semantics of `bar` should be well-defined -- I can&#39;t think of any loss of type-safety by allowing an intersection of two types rather than just one. I&#39;m sure there&#39;s an explanation though. Does anyone know what it is?

Why can&#39;t you have multiple interfaces in a bounded wildcard generic?

In *Python* one can do:

    foo = {}
    assert foo.get(&#39;bar&#39;, &#39;baz&#39;) == &#39;baz&#39;

In *PHP* one can go for a trinary operator as in:

    $foo = array();
    assert( (isset($foo[&#39;bar&#39;])) ? $foo[&#39;bar&#39;] : &#39;baz&#39; == &#39;baz&#39; );

I am looking for a golf version. Can I do it shorter/better in PHP?


### UPDATE [March 2020]:

    assert($foo[&#39;bar&#39;] ?? &#39;baz&#39; == &#39;baz&#39;);


It seems that `Null coalescing operator ??` is [worth checking out](https://www.php.net/manual/en/migration70.new-features.php#migration70.new-features.null-coalesce-op) today.


*found in the comments below (+1)*


Is there a better PHP way for getting default value by key from array (dictionary)?

As I only now noticed after commenting on [this answer](https://stackoverflow.com/questions/6900955/python-convert-list-to-dictionary/6900977#6900977), slices in Python 3 return shallow copies of whatever they&#39;re slicing rather than views. Why is this still the case? Even leaving aside numpy&#39;s usage of views rather than copies for slicing, the fact that `dict.keys`, `dict.values`, and `dict.items` all return views in Python 3, and that there are many other aspects of Python 3 geared towards greater use of iterators, makes it seem that there would have been a movement towards slices becoming similar. `itertools` does have an `islice` function that makes iterative slices, but that&#39;s more limited than normal slicing and does not provide view functionality along the lines of `dict.keys` or `dict.values`.

As well, the fact that you can use assignment to slices to modify the original list, but slices are themselves copies and not views, is a contradictory aspect of the language and seems like it violates several of the principles illustrated in [the Zen of Python](http://www.python.org/dev/peps/pep-0020/).

That is, the fact you can do

    &gt;&gt;&gt; a = [1, 2, 3, 4, 5]
    &gt;&gt;&gt; a[::2] = [0, 0, 0]
    &gt;&gt;&gt; a
    [0, 2, 0, 4, 0]

But not

    &gt;&gt;&gt; a = [1, 2, 3, 4, 5]
    &gt;&gt;&gt; a[::2][0] = 0
    &gt;&gt;&gt; a
    [0, 2, 3, 4, 5]

or something like

    &gt;&gt;&gt; a = [1, 2, 3, 4, 5]
    &gt;&gt;&gt; b = a[::2]
    &gt;&gt;&gt; b
    view(a[::2] -&gt; [1, 3, 5])   # numpy doesn&#39;t explicitly state that its slices are views, but it would probably be a good idea to do it in some way for regular Python
    &gt;&gt;&gt; b[0] = 0
    &gt;&gt;&gt; b
    view(a[::2] -&gt; [0, 3, 5])
    &gt;&gt;&gt; a
    [0, 2, 3, 4, 5]

Seems somewhat arbitrary/undesirable.

I&#39;m aware of http://www.python.org/dev/peps/pep-3099/ and the part where it says &quot;Slices and extended slices won&#39;t go away (even if the `__getslice__` and `__setslice__` APIs may be replaced) nor will they return views for the standard object types.&quot;, but the linked discussion provides no mention of why the decision about slicing with views was made; in fact, the majority of the comments on that specific suggestion out of the suggestions listed in the original post seemed to be positive.

What prevented something like this from being implemented in Python 3.0, which was specifically designed to not be strictly backwards-compatible with Python 2.x and thus would have been the best time to implement such a change in design, and is there anything that may prevent it in future versions of Python?

Why are slices in Python 3 still copies and not views?

Until today, I thought that for example:

    i += j;

Was just a shortcut for:

    i = i + j;

But if we try this:

    int i = 5;
    long j = 8;

Then `i = i + j;` will not compile but `i += j;` will compile fine.

Does it mean that in fact `i += j;` is a shortcut for something like this
`i = (type of i) (i + j)`?

    


Why don&#39;t Java&#39;s +=, -=, *=, /= compound assignment operators require casting?

What is the difference between the following Python expressions:

    # First:

    x,y = y,x+y

    # Second:

    x = y
    y = x+y

*First* gives different results than *Second*.

e.g.,

*First:*

    &gt;&gt;&gt; x = 1
    &gt;&gt;&gt; y = 2
    &gt;&gt;&gt; x,y = y,x+y
    &gt;&gt;&gt; x
    2
    &gt;&gt;&gt; y
    3

*Second:*

    &gt;&gt;&gt; x = 1
    &gt;&gt;&gt; y = 2
    &gt;&gt;&gt; x = y
    &gt;&gt;&gt; y = x+y
    &gt;&gt;&gt; x
    2
    &gt;&gt;&gt; y
    4

*y* is 3 in *First* and 4 in *Second*

Multiple assignment and evaluation order in Python

I&#39;m having a little trouble understanding the pass-by-reference properties of `data.table`. Some operations seem to &#39;break&#39; the reference, and I&#39;d like to understand exactly what&#39;s happening.

On creating a `data.table` from another `data.table` (via `&lt;-`, then updating the new table by `:=`, the original table is also altered. This is expected, as per:

`?data.table::copy`
and [stackoverflow: pass-by-reference-the-operator-in-the-data-table-package](&lt;https://stackoverflow.com/questions/8030452/pass-by-reference-the-operator-in-the-data-table-package&gt;)

Here&#39;s an example:

    library(data.table)

    DT &lt;- data.table(a=c(1,2), b=c(11,12))
    print(DT)
    #      a  b
    # [1,] 1 11
    # [2,] 2 12

    newDT &lt;- DT        # reference, not copy
    newDT[1, a := 100] # modify new DT

    print(DT)          # DT is modified too.
    #        a  b
    # [1,] 100 11
    # [2,]   2 12

However, if I insert a non-`:=` based modification between the `&lt;-` assignment and the `:=` lines above, `DT` is now no longer modified:

    DT = data.table(a=c(1,2), b=c(11,12))
    newDT &lt;- DT        
    newDT$b[2] &lt;- 200  # new operation
    newDT[1, a := 100]

    print(DT)
    #      a  b
    # [1,] 1 11
    # [2,] 2 12

So it seems that the `newDT$b[2] &lt;- 200` line somehow &#39;breaks&#39; the reference. I&#39;d guess that this invokes a copy somehow, but I would like to understand fully how R is treating these operations, to ensure I don&#39;t introduce potential bugs in my code.

I&#39;d very much appreciate if someone could explain this to me.

Understanding exactly when a data.table is a reference to (vs a copy of) another data.table

Is there a magic method that can overload the assignment operator, like `__assign__(self, new_value)`?

I&#39;d like to forbid a re-bind for an instance:

    class Protect():
      def __assign__(self, value):
        raise Exception(&quot;This is an ex-parrot&quot;)

    var = Protect()  # once assigned...
    var = 1          # this should raise Exception()

Is it possible? Is it insane? Should I be on medicine?

Is it possible to overload Python assignment?

I was reading through this [nice answer][1] regarding the &quot;Rule-of-five&quot; and I&#39;ve noticed something that I don&#39;t recall seeing before:

    class C {
      ...
      C&amp; operator=(const C&amp;) &amp; = default;
      C&amp; operator=(C&amp;&amp;) &amp; = default;
      ...
    };

What is the purpose of the `&amp;` character placed in front of `= default` for the copy assignment operator and for the move assignment operator? Does anyone have a reference for this?


  [1]: https://stackoverflow.com/a/4782927/1174378

What does an ampersand after this assignment operator mean?

This is a question I&#39;ve been mildly irritated about for some time and just never got around to search the answer to.

However I thought I might at least ask the question and perhaps someone can explain.

Basically many languages I&#39;ve worked in utilize syntactic sugar to write (using syntax from C++):

    int main() {
        int a = 2;
        a += 3; // a=a+3
    }

while in lua the `+=` is not defined, so I would have to write `a=a+3`, which again is all about syntactical sugar. when using a more &quot;meaningful&quot; variable name such as: `bleed_damage_over_time` or something it starts getting tedious to write: 

    bleed_damage_over_time = bleed_damage_over_time + added_bleed_damage_over_time 

instead of: 

    bleed_damage_over_time += added_bleed_damage_over_time

So I would like to know not how to solve this if you don&#39;t have a nice solution, in that case I would of course be interested in hearing it; but rather why lua doesn&#39;t implement this syntactical sugar.


Lua operators, why isn&#39;t +=, -= and so on defined?

When writing code like this in C++:

    bool allTrue = true;
    allTrue = allTrue &amp;&amp; check_foo();
    allTrue = allTrue &amp;&amp; check_bar();

`check_bar()` will not be evaluated if `check_foo()` returned `false`. This is called [short-circuiting or short-circuit evaluation](http://en.wikipedia.org/wiki/Short-circuit_evaluation) and is part of the lazy evaluation principle.

Does this work with the compound assignment operator `&amp;=`?

    bool allTrue = true;
    allTrue &amp;= check_foo();
    allTrue &amp;= check_bar(); //what now?

*For logical `OR` replace all `&amp; ` with `|` and `true` with `false`.*

Do the &amp;= and |= operators for bool short-circuit?

Recently I came across this question: https://stackoverflow.com/questions/50673012/assignment-operator-chain-understanding.

While answering this question I started doubting my own understanding of the behavior of the addition assignment operator `+=` or any other `operator=` (`&amp;=`, `*=`, `/=`, etc.).

My question is, when is the variable `a` in the expressions below updated in place, so that its changed value is reflected in other places in the expression during evaluation, and what is the logic behind it? Please take a look at following two expressions:

**Expression 1**

    a = 1
    b = (a += (a += a))
    //b = 3 is the result, but if a were updated in place then it should&#39;ve been 4

**Expression 2**

    a = 1
    b = (a += a) + (a += a)
    //b = 6 is the result, but if a is not updated in place then it should&#39;ve been 4

In the first expression, when the innermost expression `(a += a)` is evaluated, it seems that it doesn&#39;t update value of `a`, thus the result comes out as `3` instead of `4`.

However, in the second expression, the value of `a` is updated and so the result is 6.

When should we assume that `a`&#39;s value will be reflected in other places in the expression and when should we not?

Addition assignment += behavior in expression

What is the explanation for the result from the following operation?

    k += c += k += c;

I was trying to understand the output result from the following code:

    int k = 10;
    int c = 30;
    k += c += k += c;
    //k=80 instead of 110
    //c=70

and currently I am struggling with understanding why the result for &quot;k&quot; is 80. Why is assigning k=40 not working (actually Visual Studio tells me that that value is not being used elsewhere)?

Why is k 80 and not 110?

If I split the operation to:

    k+=c;
    c+=k;
    k+=c;

the result is k=110.

I was trying to look through the [CIL][1], but I am not so profound in interpreting generated CIL and can not get a few details:

     // [11 13 - 11 24]
    IL_0001: ldc.i4.s     10
    IL_0003: stloc.0      // k

    // [12 13 - 12 24]
    IL_0004: ldc.i4.s     30
    IL_0006: stloc.1      // c

    // [13 13 - 13 30]
    IL_0007: ldloc.0      // k expect to be 10
    IL_0008: ldloc.1      // c
    IL_0009: ldloc.0      // k why do we need the second load?
    IL_000a: ldloc.1      // c
    IL_000b: add          // I expect it to be 40
    IL_000c: dup          // What for?
    IL_000d: stloc.0      // k - expected to be 40
    IL_000e: add
    IL_000f: dup          // I presume the &quot;magic&quot; happens here
    IL_0010: stloc.1      // c = 70
    IL_0011: add
    IL_0012: stloc.0      // k = 80??????

  [1]: http://en.wikipedia.org/wiki/Common_Intermediate_Language


Content Type	Original Author	Original Content on Stackoverflow
Question	George Duckett	View Question on Stackoverflow
Solution 1 - C#	Eric Lippert	View Answer on Stackoverflow
Solution 2 - C#	温靖寰	View Answer on Stackoverflow

Why are there no ||= or &&= operators in C#?

C# Problem Overview

C# Solutions

Solution 1 - C#

Solution 2 - C#

How to select several hardcoded SQL rows?

Web deployment task failed. (The type initializer for 'Microsoft.Web.Deployment.DeploymentManager' threw an exception.)

Attributions