decrement

问题 This question already has answers here : Unexpected order of evaluation (compiler bug?) [duplicate] (3 answers) Closed 5 years ago . I'm new to C++ and study the increment and decrement operators. So I tried this example: int x = 4; cout << ++x << " " << x++ << " " << x++ << endl << endl; cout << x << endl; It returns this weird output on C++ .NET and QtCreator and 5 online C++ compilers: 7 5 4 7 The weird thing is that I expect something like this: 5 5 6 7 Can you explain what happens? 回答1:

问题 I am a beginner in JS and working at a shopping cart. I have several products which are rendered in the page with ES6 template strings. Everything is working so far, you can add items to the basket and the basket and total update correctly. The only part I am having trouble with is the increase/decrease buttons: they only work once, if you click again the amount printed in the console stays the same. I did find other SO post related to increment/decrement functions but the button keeps

问题 This question already has answers here : Unexpected order of evaluation (compiler bug?) [duplicate] (3 answers) Closed 4 years ago . I'm new to C++ and study the increment and decrement operators. So I tried this example: int x = 4; cout << ++x << " " << x++ << " " << x++ << endl << endl; cout << x << endl; It returns this weird output on C++ .NET and QtCreator and 5 online C++ compilers: 7 5 4 7 The weird thing is that I expect something like this: 5 5 6 7 Can you explain what happens? 回答1:

问题 What is the idiomatic Common Lisp way to increment/decrement numbers and/or numeric variables? 回答1: Use the built-in "+" or "-" functions, or their shorthand "1+" or "1-", if you just want to use the result, without modifying the original number (the argument). If you do want to modify the original place (containing a number), then use the built-in "incf" or "decf" functions. Using the addition operator: (setf num 41) (+ 1 num) ; returns 42, does not modify num (+ num 1) ; returns 42, does

问题 What is the idiomatic Common Lisp way to increment/decrement numbers and/or numeric variables? 回答1: Use the built-in "+" or "-" functions, or their shorthand "1+" or "1-", if you just want to use the result, without modifying the original number (the argument). If you do want to modify the original place (containing a number), then use the built-in "incf" or "decf" functions. Using the addition operator: (setf num 41) (+ 1 num) ; returns 42, does not modify num (+ num 1) ; returns 42, does

问题 I had questions about incremental and decremental operators.I couldn't understand why java gave these outputs. x = 5; y = 10; System.out.println(z = y *= x++); // output is 50 x = 2; y = 3; z = 4; System.out.println("Result = "+ z + y++ * x); // output is Result = 46 x = 5; System.out.println( x++*x); // output is 30 x = 5; System.out.println( x*x++); // output is 25 For example, in 2nd println function y is multiplicated without increasing 1 and in 3rd function x is multiplicated with x+1.

问题 I have this text with numbers: My numbers are 04, and 0005 My numbers are 05, and 0006 My numbers are 06, and 0035 My numbers are 07, and 0007 My numbers are 08, and 0009 This is the code I always used to increment or decrement numbers in a selection/block selection/column: p.e. increment the last 4 numbers in above text with 8: '<,'>s/\%V\<\d\{4}\>/\=submatch(0)+8/g but I noted today that it does strange things. This is the output: My numbers are 04, and 13 My numbers are 05, and 14 My

问题 What I have: I have a readonly textbox and two links. The first link increments the value of the textbox by 1 whereas the second link decrements the value by 1. What I need: I need the textbox value to stop decrementing at zero (I don't want negative numbers). My code: jQuery: jQuery(".increment").on('click',function(){ jQuery(this).next(".amount input").val(parseInt(jQuery(this).next(".amount input").val())+1); }); jQuery(".decrement").on('click',function(){ jQuery(this).prev(".amount input"

问题 In coding with C++ iterators if you wanted to get the previous value to what an iterator points to would you write: *--Iter or would you think it better to add parentheses like so: *(--Iter) ? 回答1: It doesn't matter as far as program correctness is concerned. But I always express this as *(--Iter) because this is more self-documenting and easier to understand to human beings than *--Iter . 回答2: I would use the latter for clarity, but both are completely valid and equivalent. Note: In case

问题 Confusion with ++ and -- operator int a = 10; printf("%d\n", -(--a) ); // valid output: -9 But, problem occurs when following is used: printf("%d\n", --(-a)); // error, invalid Why? 回答1: The ++ and -- operator works on only lvalue, not value. An lvalue is something that can stand on the left side of an assignment. printf("%d\n", -(--a) ); Here, -- operator works on variable a , so this is valid. But, printf("%d\n", --(-a)); Here, (-a) returns a value. -- is applied to a value, which is not