variable-assignment | 易学教程

How to append a string value during assignment in Javascript?

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问题 Hey i dont know if that is possible but i want to set a given variable in js by reference . What i want to do is, that each time i pass a string to the function addstring that the value of the textfield is added like += function addstring(string) { document.getElementById("string").value = string; // its a textfield } How can i do that? 回答1: += works fine. var str = "stack"; str += "overflow"; console.log(str); //alert(str); Use firebug!! stackoverflow 回答2: Javascript does not support passing

Counting Rooms While Knowing Where Walls Are

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问题 This question is on code for C++ builder 6. The bounty is interested in a standard C++ algorithm to solve the problem given a standardized input (see this for more information.) The txt file which also represents the data I have in an array: 1101 0110 1101 0110 1100 0101 0110 1110 1001 0110 1011 1010 1111 1010 1000 0101 0011 1110 1011 1110 1010 1011 1101 0101 0001 0101 0011 1011 Explanation of the txt: The numbers from the txt file are a 4-bit representation of the walls to a room, with a set

Is variable assignment and reading atomic operation?

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问题 I was unable to find any reference to this in the documentations... Is assigning to a double (or any other simple type, including boolean) an atomic operation viewed from the perspective of threads? double value = 0; public void First() { while(true) { value = (new Random()).NextDouble(); } } public void Second() { while(true) { Console.WriteLine(value); } } In this code sample, first method is called in one thread, and the second in another. Can the second method get a messed up value if it

Are multiple variable assignments done by value or reference?

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问题 $a = $b = 0; In the above code, are both $a and $b assigned the value of 0 , or is $a just referencing $b ? 回答1: With raw types this is a copy. test.php $a = $b = 0; $b = 3; var_dump($a); var_dump($b); Output : int(0) int(3) With objects though, that is another story (PHP 5) test.php class Obj { public $_name; } $a = $b = new Obj(); $b->_name = 'steve'; var_dump($a); var_dump($b); Output object(Obj)#1 (1) { ["_name"]=> string(5) "steve" } object(Obj)#1 (1) { ["_name"]=> string(5) "steve" }

SET a variable in SELECT statement - MySQL

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问题 I'm using this code which has an error: SET @rejects = ''; SELECT * FROM list WHERE maker = 1 AND by_ids IN ('10','11') AND country LIKE '%I%' AND ( src IS NULL || src NOT IN (@rejects) AND checkSrc(src) = 'yes' AND SET @rejects = CONCAT(@rejects,',',src) ); What's causing the issue? 回答1: The issue is that you cannot mix select and set in one statement, there'll surely be syntax error: select*from t where 1 and set@a=1; ERROR 1064 (42000): You have an error in your SQL syntax; check the

Pointer initialisation gives segmentation fault

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问题 I wrote a C program as follows: CASE 1 int *a; /* pointer variable declaration */ int b; /* actual variable declaration */ *a=11; a=&b;/* store address of b in pointer variable*/ It gives a segmentation fault when running the program. I changed the code as follows: CASE 2 int *a; /* pointer variable declaration */ int b; /* actual variable declaration */ a=&b;/* store address of b in pointer variable*/ *a=11; Now it's working fine. If anyone knows please explain why it is giving a

Manipulating matrix elements in tensorflow

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问题 How can I do the following in tensorflow? mat = [4,2,6,2,3] # mat[2] = 0 # simple zero the 3rd element I can't use the [] brackets because it only works on constants and not on variables. I cant use the slice function either because that returns a tensor and you can't assign to a tensor. import tensorflow as tf sess = tf.Session() var1 = tf.Variable(initial_value=[2, 5, -4, 0]) assignZerosOP = (var1[2] = 0) # < ------ This is what I want to do sess.run(tf.initialize_all_variables()) print

How to assign hash['a']['b']= 'c' if hash['a'] doesn't exist?

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Is there any way simpler than if hash.key?('a') hash['a']['b'] = 'c' else hash['a'] = {} hash['a']['b'] = 'c' end The easiest way is to construct your Hash with a block argument: hash = Hash.new { |h, k| h[k] = { } } hash['a']['b'] = 1 hash['a']['c'] = 1 hash['b']['c'] = 1 puts hash.inspect # "{"a"=>{"b"=>1, "c"=>1}, "b"=>{"c"=>1}}" This form for new creates a new empty Hash as the default value. You don't want this: hash = Hash.new({ }) as that will use the exact same hash for all default entries. Also, as Phrogz notes, you can make the auto-vivified hashes auto-vivify using default_proc :

What is the “pin” operator for, and are Elixir variables mutable?

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问题 Currently trying to understand the "^" operator in Elixir. From the website: The pin operator ^ can be used when there is no interest in rebinding a variable but rather in matching against its value prior to the match: Source - http://elixir-lang.org/getting_started/4.html With this in mind, you can attach a new value to a symbol like so: iex> x = 1 # Outputs "1" iex> x = 2 # Outputs "2" I can also do: iex> x = x + 1 # Outputs "3"! So my first question is; Are Elixir variables mutable? It

R: += (plus equals) and ++ (plus plus) equivalent from c++/c#/java, etc.?

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Does R have a concept of += (plus equals) or ++ (plus plus) as c++/c#/others do? Patrick Cuff No, it doesn't, see: R Language Definition: Operators Following @GregaKešpret you can make an infix operator: `%+=%` = function(e1,e2) eval.parent(substitute(e1 <- e1 + e2)) x = 1 x %+=% 2 ; x R doesn't have a concept of increment operator (as for example ++ in C). However, it is not difficult to implement one yourself, for example: inc <- function(x) { eval.parent(substitute(x <- x + 1)) } In that case you would call x <- 10 inc(x) However, it introduces function call overhead, so it's slower than