assignment-operator

问题 My expectation is to set the addition assignment operator( +=/-= ) to the transform: translateX() , but have no idea how would I do this. I've tried some ways to do this: $('.inline-grid').css({transform: 'translate(+= 4%, 0)'}) $('.inline-grid').css({transform: 'translate(''+=' + '4' + '%', 0')'}) $('.inline-grid').css({transform: "translate("+=" + "10" + "%", 0)"}) $('.inline-grid').css({transform: '+=' + 'translateX(4%)'}) $('.inline-grid').css({transform: '+=translateX(4%)'}) but none of

问题 Name of an array is pointer to the first element. So Why one character array cannot be assigned another array ? #include<stdio.h> int main() { char str1[]="Hello"; char str2[10]; char *s="Good Morning"; char *q; str2=str1; /* Error */ q=s; /* Works */ return 0; } 回答1: Arrays in expressions automatically converted to pointer pointing to the first element of the array except for operands of sizeof operator and unary & operator, so you cannot assign to arrays. Adding #include <string.h> to the

问题 What's the difference between the following two declarations, assuming I have not specified a copy constructor and operator= in class Beatle ? Beatle john(paul); and Beatle john = paul; Edit: In objects assignment, the operator = implicitly calls the copy constructor unless told otherwise? 回答1: They're different grammatical constructions. The first one is direct initialization , the second is copy initialization . They behave virtually identically, only that the second requires a non-

问题 $a = 1; $a OR $a = 'somthing' echo $a; //1 Why? If = have much precedence then 'OR' then why OR execute first? 回答1: Because if OR has higher precedence then $a OR $a = 'somthing' will be parsed as: ($a OR $a) = 'somthing' that would be technically wrong because you can't assign to an expression (while programmers would like to write expression like this is coding so it should be a valid expression). because precedence of or operator was low hence the expression $a OR $a = 'somthing' pareses

问题 in Does the C/C++ ternary operator actually have the same precedence as assignment operators? Luchian Grigore's answer says that cases like a ? b : c = d will always be inferred as a ? b : ( c = d ) because both = and ?: associate right to left so in c++ k = 21 > 3 ? j = 12 : j = 10; and k = 1 > 3 ? j = 12 : j = 10; both are fine. In C k = 21 > 3 ? 12 : j = 10 returns error invalid lvalue in assignment. Shouldn't above be inferred as (and return no error) k= 21 > 3 ? 12 : ( j = 10 ) I assume

问题 Doubtful result in the following code: public static void main (String[] args) { int i = 2; i = i+=2 + i++; System.out.println(i); } Was expecting 8 as output, as 'i+=2' should update i, but its not behaving so. Output: 6 I infer that the short-hand assignment operator is returning 4 as expected but not updating the same in variable i. Any explanation will be appreciated. 回答1: i++ is a postfix increment - it increments i, then essentially returns the old value of i. The equivalent prefix

问题 I have a class that should hold a reference to some data, without owning that data (i.e. the actual data is guaranteed not to go out of scope). In particular , the class cannot make a copy – the data is easily several gigabytes in size. Now, the usual implementation (I assume) is to have a reference to the data: struct holder_ref { type const& value; holder_ref(type const& value) : value(value) { } }; (Please note that the const ness has absolutely no bearing on the problem). Now, I

问题 I asked the same question before about boost::assign::map_list_of (which didn't get answered), then I thought maybe using brace initialization would help, but it didn't. This works perfectly: std::map<int, char> m = {{1, 'a'}, {3, 'b'}, {5, 'c'}, {7, 'd'}}; But this doesn't: std::map<int, char> m; m = {{1, 'a'}, {3, 'b'}, {5, 'c'}, {7, 'd'}}; Visual Studio 2013 gives the error error C2593: 'operator =' is ambiguous , could be either operator=(std::initalizer_list) or operator=(std::map&&) .

问题 The C++17 standard revised the definitions of the order of operations for the C++ language by a rule stating, to the effect: In every simple assignment expression E1=E2 and every compound assignment expression E1@=E2, every value computation and side-effect of E2 is sequenced before every value computation and side effect of E1 However, when compiling the following code in GCC 8.1 with -std=c++17 and -Wall int v[] { 0,1,2,3,4,5,6,7 }; int *p0 = &v[0]; *p0++ = *p0 + 1; cout << "v[0]: " << v[0]

问题 I'm having a little trouble understanding the pass-by-reference properties of data.table . Some operations seem to 'break' the reference, and I'd like to understand exactly what's happening. On creating a data.table from another data.table (via <- , 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 Here's an example: library(data.table) DT <- data.table