infix-notation

Convert from an infix expression to postfix (C++) using Stacks

那年仲夏 提交于 2019-12-05 03:15:24
My lecturer gave me an assignment to create a program to convert and infix expression to postfix using Stacks. I've made the stack classes and some functions to read the infix expression. But this one function, called convertToPostfix(char * const inFix, char * const postFix) which is responsible to convert the inFix expression in the array inFix to the post fix expression in the array postFix using stacks, is not doing what it suppose to do. Can you guys help me out and tell me what I'm doing wrong? The following is code where the functions to convert from inFix to postFix is and

Scala: How can I create a function that allows me to use dot notation when calling it?

≡放荡痞女 提交于 2019-12-04 14:53:49
I have been confused about this for a while, even despite reading the Scala Style Guide - Method Invocation several times. I want to be able to call this method def foldRVL[A,B](l: List[A], z: B)(f: (A, B) => B) = //"Right-Via-Left" l.reverse.foldLeft(z)((a, b) => f(b, a)) using dot notation like this List(1,2,3).foldRVL(0)(_ + _) . And not like this: foldRVL(List(1,2,3), 0)(_ + _) . Also, sometimes I've seen code that shows methods that actually either takes zero parameters in their signature, or one fewer parameters than I would expect them to take, and still properly take a parameter using

Why does Scala evaluate the argument for a call-by-name parameter if the method is infix and right-associative?

我只是一个虾纸丫 提交于 2019-12-04 02:57:34
问题 As I understood call-by-name parameters of a method, the corresponding argument expression will not be evaluated when passing it to the method, but only when (and if) the value of the parameter is used in the method body. In the following example, however, this is only true in the first two method calls, but not in the third one, although it should be a merely syntactical variation of the second case!? Why is the argument expression evaluated in the third method call? (I tested this code

Haskell infix function application precedence

眉间皱痕 提交于 2019-12-03 14:46:42
问题 Let f x y = x * y . We can apply this function in two ways: f 5 6 , or, using infix notation, 5 `f` 6 . Do the operator rules apply to this last expression? What precedence will this application have? Is it just another form of function application, and so will it also have the highest precedence? I suppose that the compiler sees this special form (due to `` and/or the name starting with a letter(?)), and actually treats this as ordinary function application, instead of considering it an

Any reason I couldn't create a language supporting infix, postfix, and prefix functions, and more?

北城余情 提交于 2019-12-03 08:06:45
I've been mulling over creating a language that would be extremely well suited to creation of DSLs, by allowing definitions of functions that are infix, postfix, prefix, or even consist of multiple words. For example, you could define an infix multiplication operator as follows (where multiply(X,Y) is already defined): a * b => multiply(a,b) Or a postfix "squared" operator: a squared => a * a Or a C or Java-style ternary operator, which involves two keywords interspersed with variables: a ? b : c => if a==true then b else c Clearly there is plenty of scope for ambiguities in such a language,

What do Push and Pop mean for Stacks?

北城余情 提交于 2019-12-03 05:10:39
问题 long story short my lecturer is crap, and was showing us infix to prefix stacks via an overhead projector and his bigass shadow was blocking everything so i missed the important stuff he was referring to push and pop, push = 0 pop = x he gave an example but i cant see how he gets his answer at all, 2*3/(2-1)+5*(4-1) step 1 Reverse : )1-4(*5+)1-2(/3*2 ok i can see that he then went on writing x's and o's operations and i got totally lost answer 14-5*12-32*/+ then reversed again to get +/*23-21

Haskell infix function application precedence

走远了吗. 提交于 2019-12-03 04:34:10
Let f x y = x * y . We can apply this function in two ways: f 5 6 , or, using infix notation, 5 `f` 6 . Do the operator rules apply to this last expression? What precedence will this application have? Is it just another form of function application, and so will it also have the highest precedence? I suppose that the compiler sees this special form (due to `` and/or the name starting with a letter(?)), and actually treats this as ordinary function application, instead of considering it an operator. The Haskell 98 Report has a section on Operator Applications that clears it up: An operator is

Understanding infix method call and cons operator(::) in Scala

◇◆丶佛笑我妖孽 提交于 2019-12-03 03:47:48
问题 I'm quite new to Scala programming language, and was trying something out stucked in my mind while I was following the lecture notes at here. I think I couldn't really understand how cons operator works, here are some things I tried: I've created a pseudo-random number generator, then tried to create a list of one random value: scala> val gen = new java.util.Random gen: java.util.Random = java.util.Random@1b27332 scala> gen nextInt 3 :: Nil <console>:7: error: type mismatch; found : List[Int]

What do Push and Pop mean for Stacks?

≯℡__Kan透↙ 提交于 2019-12-02 17:33:52
long story short my lecturer is crap, and was showing us infix to prefix stacks via an overhead projector and his bigass shadow was blocking everything so i missed the important stuff he was referring to push and pop, push = 0 pop = x he gave an example but i cant see how he gets his answer at all, 2*3/(2-1)+5*(4-1) step 1 Reverse : )1-4(*5+)1-2(/3*2 ok i can see that he then went on writing x's and o's operations and i got totally lost answer 14-5*12-32*/+ then reversed again to get +/*23-21*5-41 if some one could explain to me the push pop so i could understand i would be very greatful, i

Understanding infix method call and cons operator(::) in Scala

巧了我就是萌 提交于 2019-12-02 17:14:28
I'm quite new to Scala programming language, and was trying something out stucked in my mind while I was following the lecture notes at here . I think I couldn't really understand how cons operator works, here are some things I tried: I've created a pseudo-random number generator, then tried to create a list of one random value: scala> val gen = new java.util.Random gen: java.util.Random = java.util.Random@1b27332 scala> gen nextInt 3 :: Nil <console>:7: error: type mismatch; found : List[Int] required: Int gen nextInt 3 :: Nil ^ But it tried to pass List(3) to nextnt method. When i used