operator-precedence | 易学教程

How to force matlab to call a regular function rather than class method when they are overloaded?

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问题 Assume I have an object X of class MyClass . MyClass has a method compute , and when I call U = compute(X,...) , matlab automatically calls the class method. However, what I actually want is to call another function also called compute whose parameters start with a MyClass object though. How do I force matlab to call this regular function rather than go into the class method? 回答1: There is no way to do this without making some changes either to the function's name or location. If you check

Difference between i = ++i and ++i [duplicate]

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问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…) What is the difference between i = ++i; and ++i; where i is an integer with value 10 ? According to me both do the same job of incrementing i i.e after completion of both the expressions i =11 . 回答1: i = ++i; invokes Undefined Behaviour whereas ++i; does not. C++03 [Section 5/4] says Between the previous and next sequence point a

Post increment and Pre increment in C

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问题 I have a question about these two C statements: x = y++; t = *ptr++; With statement 1, the initial value of y is copied into x then y is incremented. With statement 2, We look into the value pointed at by *ptr, putting that into variable t, then sometime later increment ptr. For statement 1, the suffix increment operator has higher precedence than the assignment operator. So shouldn't y be incremented first and then x is assigned to the incremented value of y? I'm not understanding operator

Precedence of ++ and — operators in Java

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问题 I read from the official tutorial of Java that prefix and postfix ++ -- have different precedences: postfix: expr++ expr-- unary: ++expr --expr +expr -expr ~ ! Operators According to the tutorial, shouldn't this d = 1; System.out.println(d++ + ++d); print out 6 ( d++ makes d 2, ++d makes it 3) instead of 4? I know the explanation of ++d being evaluated beforehand, but if d++ has higher precedence then ++d , why isn't d++ being first evaluated? And what is more, in what case should d++ shows

Precedence of dereference and postfix

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问题 When I read the TCPL by K&R, I just couldn't understand two expressions: *p++ = val; /*push val onto stack */ Here is my idea: dereference and postfix has the same precedence, and associativity is right to left,so *p++ = val maybe the same with *(p++) = val , because the pointer usually is the next position to the top , so in this code, p increase 1 first because of the parenthesis, so the p is the two units above the current top ,but not the one unit above the current top ,where the val

Behaviour and Order of evaluation in C# [duplicate]

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Possible Duplicate: C#: Order of function evaluation (vs C) Code snippet: i += ++i; a[++i] = i; int result = fun() - gun(); //statement of similar kind Are their behavior well-defined in C#? In C++, such code invokes undefined/unspecified behavior . Please also quote the relevant sections from the language specification in your response! The key here is the table in "1.4 Expressions", and "7.3.1 Operator precedence and associativity". I won't duplicate the table from 1.4, but to quote 7.3.1: Except for the assignment operators, all binary operators are left-associative, meaning that operations

Java: pre-,postfix operator precedences

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I have two similar questions about operator precedences in Java. First one: int X = 10; System.out.println(X++ * ++X * X++); //it prints 1440 According to Oracle tutorial : postfix (expr++, expr--) operators have higher precedence than prefix (++expr, --expr) So, I suppose that evaluation order: 1) first postfix operator: X++ 1.a) X++ "replaced" by 10 1.b) X incremented by one: 10+1=11 At this step it should look like: System.out.println(10 * ++X * X++), X = 11; 2) second POSTfix operator: X++ 2.a) X++ "replaced" by 11 2.b) X incremented by one: 11+1=12 At this step it should look like: System

The output of cout << 1 && 0; [closed]

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I don't understand why the below code prints 1 . 1 && 0 is not the same as true && false -> false ? Why doesn't this print 0 ? #include <iostream> using namespace std; int main(){ cout << 1 && 0; return 0; } Smit Ycyken It's all about Operator Precedence . The Overloaded Bitwise Left Shift Operator operator<<(std::basic_ostream) has a higher priority than the Logical AND Operator && . #include <iostream> int main() { std::cout << (1 && 0); return 0; } If you are not 146% sure about the priority of an operator, do not hesitate to use brackets. Most modern IDEs will tell you if you don't need to

Why is `return a or b` a void value expression error in Ruby?

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This is just fine: def foo a or b end This is also fine: def foo return a || b end This returns void value expression : def foo return a or b end Why? It doesn't even get executed; it fails the syntax check. What does void value expression mean? return a or b is interpreted as (return a) or b , and so the value of return a is necessary to calculate the value of (return a) or b , but since return never leaves a value in place (because it escapes from that position), it is not designed to return a valid value in the original position. And hence the whole expression is left with (some_void_value)

Is indexing a new map element and having something that reads it assigned to it undefined behaviour, or just unspecified?

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问题 After answering this question, there was a long discussion over whether the code in question was undefined behaviour or not. Here's the code: std::map<string, size_t> word_count; word_count["a"] = word_count.count("a") == 0 ? 1 : 2; First of all, it was well-established that this was at least unspecified. The result differs based on which side of the assignment is evaluated first. In my answer, I followed through each of the four resulting cases, with factors of which side is evaluated first