operator-precedence

问题 I cant seem to wrap my head around the first part of this code ( += ) in combination with the ternary operator. h.className += h.className ? \' error\' : \'error\' The way i think this code works is as following: h.className = h.className + h.className ? \' error\' : \'error\' But that isn\'t correct because that gives a error in my console. So my question is how should i interpet this code correctly? 回答1: h.className = h.className + (h.className ? ' error' : 'error') You want the operator to

问题 I like Scala\'s propose of operator precedence but in some rare cases, unmodified rules may be inconvenient, because you have restrictions in naming your methods. Are there ways to define another rules for a class/file, etc. in Scala? If not, would it be resolved in the future? 回答1: Operator precedence is fixed in the Scala Reference - 6.12.3 Infix Operations by the first character in the operator. Listed in increasing order of precedence: (all letters) | ^ & = ! < > : + - * / % (all other

问题 Trying to understand the behaviour of pointers in C, I was a little surprised by the following (example code below): #include <stdio.h> void add_one_v1(int *our_var_ptr) { *our_var_ptr = *our_var_ptr +1; } void add_one_v2(int *our_var_ptr) { *our_var_ptr++; } int main() { int testvar; testvar = 63; add_one_v1(&(testvar)); /* Try first version of the function */ printf(\"%d\\n\", testvar); /* Prints out 64 */ printf(\"@ %p\\n\\n\", &(testvar)); testvar = 63; add_one_v2(&(testvar)); /* Try

问题 Why does ++x || ++y && ++z calculate ++x first, even though the precedence of operator && is higher than || ? 回答1: Unwind, R and others have explained what really happens. So let me just add: The premise of your question is faulty. The fact that the && has higher precedence doesn't mean that operands that surround it must be evaluated before any operands in the expression with lower precedence. Even where the special case short-circuiting of || and && this wouldn't necessarily be so. For

问题 Okay, I\'m aware that the standard dictates that a C++ implementation may choose in which order arguments of a function are evaluated, but are there any implementations that actually \'take advantage\' of this in a scenario where it would actually affect the program? Classic Example: int i = 0; foo(i++, i++); Note: I\'m not looking for someone to tell me that the order of evaluation can\'t be relied on, I\'m well aware of that. I\'m only interested in whether any compilers actually do

问题 This question already has an answer here: Why does the expression 0 < 0 == 0 return False in Python? 9 answers When I was looking at answers to this question, I found I didn\'t understand my own answer. I don\'t really understand how this is being parsed. Why does the second example return False? >>> 1 in [1,0] # This is expected True >>> 1 in [1,0] == True # This is strange False >>> (1 in [1,0]) == True # This is what I wanted it to be True >>> 1 in ([1,0] == True) # But it\'s not just a

问题 This question already has an answer here: Difference between pre-increment and post-increment in a loop? 21 answers The following for loops produce identical results even though one uses post increment and the other pre-increment. Here is the code: for(i=0; i<5; i++) { printf(\"%d\", i); } for(i=0; i<5; ++i) { printf(\"%d\", i); } I get the same output for both \'for\' loops. Am I missing something? 回答1: After evaluating i++ or ++i , the new value of i will be the same in both cases. The

问题 The terms \'operator precedence\' and \'order of evaluation\' are very commonly used terms in programming and extremely important for a programmer to know. And, as far as I understand them, the two concepts are tightly bound; one cannot do without the other when talking about expressions. Let us take a simple example: int a=1; // Line 1 a = a++ + ++a; // Line 2 printf(\"%d\",a); // Line 3 Now, it is evident that Line 2 leads to Undefined Behavior, since Sequence points in C and C++ include:

问题 What are the implications of the voted in C++17 evaluation order guarantees (P0145) on typical C++ code? What does it change about things like i=1; f(i++, i) and std::cout << f() << f() << f() ; or f(g(),h(),j()); 回答1: Some common cases where the evaluation order has so far been unspecified , are specified and valid with C++17 . Some undefined behaviour is now instead unspecified. What about things like i=1; f(i++, i) was undefined but is now unspecified. Specifically, what is not specified

问题 Are the two statements below equivalent? SELECT [...] FROM [...] WHERE some_col in (1,2,3,4,5) AND some_other_expr and SELECT [...] FROM [...] WHERE some_col in (1,2,3) or some_col in (4,5) AND some_other_expr Is there some sort of truth table I could use to verify this? 回答1: And has precedence over Or , so, even if a <=> a1 Or a2 Where a And b is not the same as Where a1 Or a2 And b, because that would be Executed as Where a1 Or (a2 And b) and what you want, to make them the same, is the