variable-assignment

This question already has an answer here: Scope of Default function parameters in javascript 1 answer Assigning a default value using a variable with the same name throw a reference error: var a = 'adef'; var x = (a=a) => console.log(a); x(); => "ReferenceError: a is not defined" But this is fine: var other = 'otherdef'; var x = (a=other) => console.log(a); x(); => "otherdef" My assumption was that the value of a in the outer scope would be assigned to the new scope. I have tried using const instead of var , and class / function instead of an arrow-function, but the result is always the same

In the Go web server example here: http://golang.org/doc/effective_go.html#web_server The following line of code works var addr = flag.String("addr", ":1718", "http service address") but changing it to addr := flag.String("addr", ":1718", "http service address") is a compilation error. Why? Does it have anything to do with the face that the return type of the function is *string instead of string ? What difference does that make? UPDATE : Thanks for pointing out that := is not allowed at the top level. Any idea why this inconsistency is in the spec? I don't see any reason for the behaviour to

In Swift, why does var x: Int? = nil if let y: Int? = x { ... } behave differently from if let y: Int? = nil { ... } My understanding of why the first case succeeds suggests that the second should as well, so I must not really be understanding. The latter is not failing because of an invalid assignment, nor because of optional chaining; and otherwise it seems the same as the former. Why does the latter fail, and how is it different from the former. Exactly at what point, and for what reason, is the second optional binding abandoned? rintaro if let y: Int? = nil { ... } is equivalent to if let

As the title says I found such a sentence in some C lecture notes. I can't invent any example proving that sentence. In my opinion every of assignment operations is evaluated once, because when we want it to be evaluated more than once we put in in a loop. What am I missing then? I've searched but couldn't find an answer here on SO. ouah C says: (C99, 6.5.16.2p3) "A compound assignment of the form E1 op= E2 differs from the simple assignment expression E1 = E1 op (E2) only in that the lvalue E1 is evaluated only once." Below are some examples of why it matters: Example 1: a[i++] += 1; is the

I know I can use the comma operator like this for (int i = 1, j = 15; j>10; i++, j--) { // do something neat } but some articles seem to suggest that the comma operator can be used outside of the for loop declaration, for example int j = 2, k = 4 ; int x ; // Assignment statement with comma operator x = j + 1, k ; source: http://www.cs.umd.edu/~clin/MoreJava/ControlFlow/comma.html or int x = (expression) ? (i++,2) : 3; source: https://stackoverflow.com/a/12047433/1084813 This would be a neat trick for a code obfuscation contest or to confuse my colleagues, but neither of the examples will

Possible Duplicate: How can I change the values of multiple points in a matrix? I have a matrix A and three vectors of the same length, r , holding the indexes of the rows to assign to, c , holding the indexes of the columns to assign to, and v containing the actual values to assign. What I want to get is A(r(i),c(i))==v(i) for all i . But doing A(r,c)=v; Doesn't yield the correct result as matlab interprets it as choosing every possible combination of r and c and assigning values to it, for instance n=5; A=zeros(n); r=1:n; c=1:n; A(r,c)=1; Yields a matrix of ones, where I would like to get