operator-precedence

问题 While I was learning C (I am very new to it), I was playing around with pointers. Here you can see my code: #include <stdio.h> void change(int *i) { *i += 1; } int main() { int num = 3; printf("%d\n", num); change(&num); printf("%d\n", num); return 0; } My aim was to replace incrementing the num value without reassigning it like so: num = change(num); That's why I was passing the memory location of num using the & : so it could be used as a pointer. Before this version everything in the code

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 2 years ago . #include <stdio.h> int main() { int a=-1?2:5 + 8?4:5; printf("%d\n",a); return 0; } The output of above program is 2. But why ? Please explain 回答1: Write human-readable and understandable code. ( Atleast, try to... ) int a=-1?2:5 + 8?4:5; is the same as int a = (-1) ? 2 : ( 5 + ( 8 ? 4 : 5) );

问题 Lets say we have an iterator (iter) over a list of pointers to memory assigned to heap space, if I do that delete (*iter++) am I right that the precedence is first dereference iterator to get memory address then free the space and then increment the iterator to free the next element ? 回答1: The effect is as you write, but it's achieved using a slightly different sequence: The post-increment has highest precedence, so it's evaluated first. However, its return value (which is processed by

问题 I would appreciate if you clarify this for me.Here are two recent questions with their accepted answers: 1) What is the difference between *myptr++ and *(myptr++) in C 2) Yet another sequence point query: how does *p++ = getchar() work? The accepted answer for the first question,concise and easily to understand states that since ++ has higher precedence than * , the increment to the pointer myptr is done first and then it is dereferenced.I even checked that out on the compiler and verified it

问题 I have two CSS rules: .avo-lime-table th, .avo-lime-table td { background-color: grey; } Rule two .avo-lime { background-color: green; } Everything works fine in FireFox, Chrome, Opera and Safari. Obviously Microsoft's browser (as always) has some diffrent ideas about rendering my page... <div class="avo-center-shrink"> <form method="post" action="/someformAction"> <table class="avo-lime-table"> <colgroup> <col> <col> </colgroup> <thead> <tr><th colspan="2" class="avo-lime">Login form heading

问题 If an expression evaluates multiple && operators, and does not evaluate any operators of lower precedence (eg. || , ?: ), will the expression evaluate to 0 as soon as one of the && s return 0, or will it finish evaluating the remaining && s? For example, q=0; w=1; e=1; r=1; if(q && w && r && e) {} Will this if() evaluate to false as soon as q && w evaluates to 0 (since the remaining && must all evaluate to 0 regardless of the right hand operators)? 回答1: Yes, evaluation will terminate early (

问题 Let's consider the following code: <?php class X{ public function test(){ return 'test'; } } //This obviously works: $x = (new X())->test(); //This does not, syntax error $x = new X()->test(); echo $x; Why? BTW: I know it was introduced in php 5.4 It's not about "how", it's about " why " - I know the first syntax is the one documented in the manual. Actually, it's more about where and how to find the answer. From what I've learned so far, while asking my questions elsewhere: The -> is not

问题 I am receiving a lot of conflicting answers about this. But as I always understood it. When we have a pointer in C and use it in a post increment statement, the post increment will always happen after the line of code resolves. int array[6] = {0,1,2,3,4,5}; int* p = array; printf("%d", *p++); // This will output 0 then increment pointer to 1 output : 0 Very simple stuff. Now here's where I am receiving a bit of dissonance in the information people are telling me and my own experience. // Same

问题 When an expression has two operators with the same precedence, the expression is evaluated according to its associativity. I want to know how the following works: i=b + b + ++b i here is 4 So ++b didn't change the first 2 b values, but it executed first, because the execution is from left to right. Here, however: int b=1; i= b+ ++b + ++b ; i is 6 According to associativity, we should execute the 3rd b so it should be: 1+ (++1) + ( ++1 should be done first) . so it becomes: 1 + ++1 + 2 =5

问题 I am confused between precedence of operators and want to know how this statement would be evaluated. # include <stdio.h> int main() { int k=35; printf("%d %d %d",k==35,k=50,k>40); return 0; } Here k is initially have value 35, when I am testing k in printf I think : k>40 should be checked which should result in 0 k==35 should be checked and which should result in 1 Lastly 50 should get assigned to k and which should output 50 So final output should be 1 50 0 , but output is 0 50 1 . 回答1: You