post-increment | 易学教程

Incrementing in C++ - When to use x++ or ++x?

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I'm currently learning C++ and I've learned about the incrementation a while ago. I know that you can use "++x" to make the incrementation before and "x++" to do it after. Still, I really don't know when to use either of the two... I've never really used "++x" and things always worked fine so far - so, when should I use it? Example: In a for loop, when is it preferable to use "++x"? Also, could someone explain exactly how the different incrementations (or decrementations) work? I would really appreciate it. It's not a question of preference, but of logic. x++ increments the value of variable x

Is there a performance difference between i++ and ++i in C?

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问题 Is there a performance difference between i++ and ++i if the resulting value is not used? 回答1: Executive summary: No. i++ could potentially be slower than ++i , since the old value of i might need to be saved for later use, but in practice all modern compilers will optimize this away. We can demonstrate this by looking at the code for this function, both with ++i and i++ . $ cat i++.c extern void g(int i); void f() { int i; for (i = 0; i < 100; i++) g(i); } The files are the same, except for

Java: Prefix/postfix of increment/decrement operators?

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问题 From the program below or here, why does the last call to System.out.println(i) print the value 7 ? class PrePostDemo { public static void main(String[] args){ int i = 3; i++; System.out.println(i); // \"4\" ++i; System.out.println(i); // \"5\" System.out.println(++i); // \"6\" System.out.println(i++); // \"6\" System.out.println(i); // \"7\" } } 回答1: i = 5; System.out.println(++i); //6 This prints out "6" because it takes i adds one to it and returns the value. 5+1=6; This is prefixing,

Is there a performance difference between i++ and ++i in C++?

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问题 We have the question is there a performance difference between i++ and ++i in C? What\'s the answer for C++? 回答1: [Executive Summary: Use ++i if you don't have a specific reason to use i++ .] For C++, the answer is a bit more complicated. If i is a simple type (not an instance of a C++ class), then the answer given for C ("No there is no performance difference") holds, since the compiler is generating the code. However, if i is an instance of a C++ class, then i++ and ++i are making calls to

C: What is the difference between ++i and i++?

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问题 In C, what is the difference between using ++i and i++ , and which should be used in the incrementation block of a for loop? 回答1: ++i will increment the value of i , and then return the incremented value. i = 1; j = ++i; (i is 2, j is 2) i++ will increment the value of i , but return the original value that i held before being incremented. i = 1; j = i++; (i is 2, j is 1) For a for loop, either works. ++i seems more common, perhaps because that is what is used in K&R. In any case, follow the

Post-increment and Pre-increment concept?

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问题 I don\'t understand the concept of postfix and prefix increment or decrement. Can anyone give a better explanation? 回答1: All four answers so far are incorrect , in that they assert a specific order of events. Believing that "urban legend" has led many a novice (and professional) astray, to wit, the endless stream of questions about Undefined Behavior in expressions. So. For the built-in C++ prefix operator, ++x increments x and produces (as the expression's result) x as an lvalue, while x++

What is x after “x = x++”?

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问题 What happens (behind the curtains) when this is executed? int x = 7; x = x++; That is, when a variable is post incremented and assigned to itself in one statement? I compiled and executed this. x is still 7 even after the entire statement . In my book, it says that x is incremented! 回答1: x does get incremented. But you are assigning the old value of x back into itself. x = x++; x++ increments x and returns its old value. x = assigns the old value back to itself. So in the end, x gets assigned

How do the post increment (i++) and pre increment (++i) operators work in Java?

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问题 Can you explain to me the output of this Java code? int a=5,i; i=++a + ++a + a++; i=a++ + ++a + ++a; a=++a + ++a + a++; System.out.println(a); System.out.println(i); The output is 20 in both cases 回答1: Does this help? a = 5; i=++a + ++a + a++; => i=6 + 7 + 7; (a=8) a = 5; i=a++ + ++a + ++a; => i=5 + 7 + 8; (a=8) The main point is that ++a increments the value and immediately returns it. a++ also increments the value (in the background) but returns unchanged value of the variable - what looks