assignment-operator

问题 What is this idiom and when should it be used? Which problems does it solve? Does the idiom change when C++11 is used? Although it\'s been mentioned in many places, we didn\'t have any singular \"what is it\" question and answer, so here it is. Here is a partial list of places where it was previously mentioned: What are your favorite C++ Coding Style idioms: Copy-swap Copy constructor and = operator overload in C++: is a common function possible? What is copy elision and how it optimizes copy

问题 What are the differences between the assignment operators = and <- in R? I know that operators are slightly different, as this example shows x <- y <- 5 x = y = 5 x = y <- 5 x <- y = 5 # Error in (x <- y) = 5 : could not find function \"<-<-\" But is this the only difference? 回答1: What are the differences between the assignment operators = and <- in R? As your example shows, = and <- have slightly different operator precedence (which determines the order of evaluation when they are mixed in

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

问题 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) ? 回答1: As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract: A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T