variable-assignment

问题 class phone { public: phone(int x) { num = x; } int number(void) { return num; } void number(int x) { num = x; } private: int num; }; int main(void) { phone p1(10); p1 = 20; // here! return 0; } Hi, guys Just I declared a simple class like above one. After that I assigned int value to the object that class, then it worked! (I printed its value. It was stored properly) If there is not a construct with int parameter, a compile error occurred. So, I think it's related with a constructor. Is that

问题 class phone { public: phone(int x) { num = x; } int number(void) { return num; } void number(int x) { num = x; } private: int num; }; int main(void) { phone p1(10); p1 = 20; // here! return 0; } Hi, guys Just I declared a simple class like above one. After that I assigned int value to the object that class, then it worked! (I printed its value. It was stored properly) If there is not a construct with int parameter, a compile error occurred. So, I think it's related with a constructor. Is that

问题 How can i substitute this code with a loop? m1 <- ggplot(foo) m2 <- ggplot(foo) ... m9 <- ggplot(foo) I guess i need something like get() to replace magic(), But get did not work here. for (i in 1:9){ magic(i) <- ggplot(foo) } 回答1: Use a list: out <- list() for (i in 1:9){ out[[i]] <- ggplot(foo) } 回答2: This will work: p <- qplot(data = mtcars, wt, mpg) for(i in 1:9) assign(paste0("m",i),p) ls() [1] "i" "m1" "m2" "m3" "m4" "m5" "m6" "m7" "m8" "m9" "p" 回答3: Something like this should work: for

问题 This is a practice I've seen before, but not very often: A variable is assigned to a value at the same time the value itself is evaluated (or is it the expression itself that is evaluated?). Example: // Outputs "The value is 1" $value = 1; if ($var = $value) { echo "The value is $var"; } Seems to be the same as: $value = 1; $var = $value; if ($var) { echo "The value is $var"; } Another example: // Outputs "The value is 1" $value = 1; echo "The value is ".$var = $value; I've been using this a

问题 I was doing a simple calculator with the following code. Right now it executes perfectly. When I tried to change things around, however, it doesn't work. I used BOOL program to check whether to continue asking for input from the person or finish the program. If I change the expression of while statement to just (program) and change YES / NO in the program statements, why does the code fail to do what is inside the while ? // A simple printing calculator { NSAutoreleasePool * pool = [

问题 I have a small issue with understanding why I'm getting this output. var arr = ["a", "b", "c", "d", "e", "f"]; arr.splice(2,0,"1"); console.log(arr); var arr2 = ["a", "b", "c", "d", "e", "f"]; arr2 = arr2.splice(2,0,"2"); console.log(arr2); ouput is: [ 'a', 'b', '1', 'c', 'd', 'e', 'f' ] [] Why is the second line of output not: [ 'a', 'b', '2', 'c', 'd', 'e', 'f' ] Is it an issue of assignment or what? 回答1: Reading about splice method. It returns An array containing the removed elements. If

问题 I was wondering if it's possible to make my class Time { public: Time(); explicit Time( const double& d); Time& operator=( const Time& time); Time& operator=( const double& d); }; assignable to the primitive double? I'm using Time as an IV a lot and need to do a lot of scalar operations on it, so it needs to "mingle" with DV's which are usually ordinary doubles. Adding a second assignment operator did the trick the other way around. A lot of operations still aren't possible with just this

问题 What is the difference in these two statements in python? var = foo.bar and var = [foo.bar] I think it is making var into a list containing foo.bar but I am unsure. Also if this is the behavior and foo.bar is already a list what do you get in each case? For example: if foo.bar = [1, 2] would I get this? var = foo.bar #[1, 2] and var = [foo.bar] #[[1,2]] where [1,2] is the first element in a multidimensional list 回答1: [] is an empty list. [foo.bar] is creating a new list ( [] ) with foo.bar as

问题 error: invalid conversion from 'int' to 'int*' int *q = 8; Works fine. *q = 6; Why can't I directly assign an int to an int pointer like this: int *q = 6 ; and I can assign it safely in the next line? 回答1: Because they're different things at all. The 1st one is definition of variable with initializer expression, i.e an initialization (of the pointer itself): int * q = 8; ~~~~~ ~ ~~~ type is int*; name of variable is q; initialized with 8 The 2nd one is assignment (of the object pointed by the

问题 error: invalid conversion from 'int' to 'int*' int *q = 8; Works fine. *q = 6; Why can't I directly assign an int to an int pointer like this: int *q = 6 ; and I can assign it safely in the next line? 回答1: Because they're different things at all. The 1st one is definition of variable with initializer expression, i.e an initialization (of the pointer itself): int * q = 8; ~~~~~ ~ ~~~ type is int*; name of variable is q; initialized with 8 The 2nd one is assignment (of the object pointed by the