post-increment

Pre and post increment in a for loop [duplicate]

我的梦境 提交于 2019-12-12 05:27:38
问题 This question already has answers here : Is ++i really faster than i++ in for-loops in java? (11 answers) Closed 6 years ago . Is it more performant to do a pre-increment vs a post-increment in a for loop in java ? Sample code : for (int i=0; i<10; i++) and for (int i=0; i<10; ++i) I notice that when i do a pre-increment, the execution time is lesser that when i do the post-increment. Any suggestions on why this might be the case ? thanks. 回答1: I compiled a minimal example (Oracle jdk1.7.0_07

How post Increment ++ Operator works while initialization

穿精又带淫゛_ 提交于 2019-12-12 05:14:47
问题 I have written following code to reverse the given String: String str = "abcdef"; char[] strToChar = str.toCharArray(); char[] outString = new char[strToChar.length]; for(int j = 0; j < strToChar.length; ) { int len = strToChar.length-j; outString[j++] = strToChar[strToChar.length-j]; } As per my understanding initialization happens from Right to Left . Hence, strToChar[strToChar.length-j] here should throw ArrayIndexOutOfBoundException . But it runs fine. How is that happening? Shouldn't it

MIPS assembly - Label value modification

给你一囗甜甜゛ 提交于 2019-12-12 02:10:39
问题 Is it possible in MIPS to change during execution the value of a label, or to create a label with certain value? I ask this because when using the instruction lw $a0, label($s0) i want to increment the value of label +4 every time we loop, indicating the new memory address of the array. I am aware I can do lw $a0, label+4($s0) but the new value of label will not be stored. Any advise? 回答1: No. In MIPS you must have a constant outside the parentheses when dereferencing (poor wording). If it

Does postfix increment perform increment not on returned value?

只愿长相守 提交于 2019-12-11 18:26:39
问题 Again, a silly question. #include <stdio.h> #include <iostream> using namespace std; int main() { int i = 0; i = i++; cout<<i; return 0; } I get 1 printed as a result of this program though I expected 0: first a temp object created eing equal 0, then i is incremented, then temp object is returned and assigned to i. Just according to: 5.2.6 Increment and decrement [expr.post.incr] 1 The value obtained by applying a postfix ++ is the value that the operand had before applying the operator.

Bash Post Increment

北城余情 提交于 2019-12-11 07:27:32
问题 Just a little question about the right way of doing Post-increment in bash. while true; do VAR=$((CONT++)) echo "CONT: $CONT" sleep 1 done VAR starts at 1 in this case. CONT: 1 CONT: 2 CONT: 3 But if I do this: while true; do echo "CONT: $((CONT++))" sleep 1 done It starts at 0. CONT: 0 CONT: 1 CONT: 2 Seems that the the first case is behaving ok, because ((CONT++)) would evaluate CONT (undefined,¿0?) and add +1. How can I get a behaviour like in echo statement to assign to a variable? EDIT:

Increment, preincrement and postincrement

ぐ巨炮叔叔 提交于 2019-12-11 05:17:08
问题 Help me to resolve this please. The steps that follows that expressions are: //Expression offSpring1[m1++] = temp1; //Steps: 1.- increment m1 2.- assign temp1 to offSpring I have always thought that the expression inside the brackets was the first to be done. But now I am confuse. So if a write this: //Expression offSpring1[++m1] = temp1; //Steps would be: 1.- assign temp1 to offSpring 2.- increment m1 If the steps would be the same as first ones, what is the difference between i++ and ++i?

Why java statement evaluation is happening like these ?

独自空忆成欢 提交于 2019-12-11 01:11:34
问题 int z = 1; System.out.println(z++ == ++z); System.out.println(++z == z++); the output will be: false true and I don't get why, please explain this to me. 回答1: First You Should Clear About Pre Increment & Post Increment Lets Have A Look int i=1; int j=i++; int k=++i; here in the value of j will be 1 and k will be 3 why means in i++ first store the value i into j and after which increments the value of i so j=1,i=2 ok :) now in the case of ++i where first increment the value and after that

Post-Incrementing/decrementing in recursive method calls (Java)

岁酱吖の 提交于 2019-12-10 20:02:17
问题 Say you have a recursive method, and you post-increment/decrement a value in the recursive call. Why will this result in a stack overflow exception when a pre-increment/decrement will not? Ex. numberCount(currentNumber++); //Stack overflow exception numberCount(++currentNumber); //No stack overflow exception Thanks in advance for any clarification. 回答1: The first numberCount(currentNumber++); //Stack overflow exception is equivalent to: numberCount(currentNumber); currentNumber += 1; while

C++ post-increment operator overload in iterators (compiling with -Wall -Werror)

烂漫一生 提交于 2019-12-10 19:07:42
问题 I'm currently creating my own iterator for a b-tree, and I'm stuck on how to implement the post-increment operator without the compiler complaining. The error message is as follows, and is expected (since I am doing exactly what the error message says) cc1plus: warnings being treated as errors error: reference to local variable 'temp' returned I am required to write the function with the -Wall and -Werror tags, so hopefully someone is able to help me with a solution around that. Here is the

implementing a C++ postfix increment operator

允我心安 提交于 2019-12-10 15:13:55
问题 I compiled the following example: #include <iostream> #include <iterator> using namespace std; class myiterator : public iterator<input_iterator_tag, int> { int* p; public: myiterator(int* x) :p(x) {} myiterator(const myiterator& mit) : p(mit.p) {} myiterator& operator++() {++p;return *this;} myiterator& operator++(int) {myiterator tmp(*this); operator++(); return tmp;} bool operator==(const myiterator& rhs) {return p==rhs.p;} bool operator!=(const myiterator& rhs) {return p!=rhs.p;} int&