Yes, the output is correct: // This line increments the variable then prints its value. Console.WriteLine(“{0}”, ++ Cash); // This prints the value of the (incremented variable) Console.WriteLine(“{0}”, Cash); // The prints the value of the variable *then* increments its value Console.WriteLine(“{0}”, Cash ++); 1 solved C# help declaring variable i initialize it to
Why does this code print 10 times “Hello world.”? No, this will be an infinite loop because the following statement resets the value of k to 1: k=+1; Also, k=k++ will not change the value of k because it is processed something like int temp = k; k++; k = temp; You can try the … Read more
The problem is because you call that function in a while, and because of for(counter = 1; counter < 51; counter++) the counter will become again 51 (everytime the function is called) and then will enter again in that if (counter == 51). To resolve that you could declare a global variable int ok = … Read more
The order in which the arguments passed to a function are evaluated is not specified, and the order of evaluating the operants of + is unspecified, too. So in printf(“%d %d”, i+1, i-1), for example, you cannot rely on the order of evaluation of the arguments; i+1 might be evaluated after i-1, actually; You will … Read more
The output will be 1. Your expression ++x will be x = x+1; In both the printf() you get 1 So the value of x is modified with the pre-increment operator here and in printf() in the second line prints the new value of x which is 1 printf() just prints the value of x … Read more
By the time p3 equals 10, it will never re-enter the while (p3 < 10) loop (because once you’ve incremented it to 10, it will never be less than 10 and will never re-enter that loop). So yes, p and p2 do increment, but they never get printed because p3 == 10 at that point … Read more
You need to understand the concepts of post increment(decrement) and pre increment(decrement). Post increment cout << x++<<endl; You can understand this line as “Return the value of x” + “increment the value of x”. I.e The return value is before the increment. So return 0 and increase the value of x to 1. Pre increment … Read more
I would prefer i++ over the other two because it expresses a common idiom. Those lines of code all end up doing exactly the same thing. But if I’m scanning code for an increment, the ++ just sort of jumps off the page. Also, the i = i++ is actually incorrect. You meant i = … Read more
A correct program can look for example the following way #include <iostream> int main() { int num, a = 15; num = ( —-a )–; std::cout << num << std::endl; std::cout << a << std::endl; return 0; } Its output is 13 12 The post-decrement operator has a higher priority than the pre-decrement operator and … Read more
I do not know whether this solution is more efficient, but maybe it is more readable and easier to code: #include <iostream> #include <cstdlib> int main() { const int magicNumber = 32640; const int maxOffset = 1920; int n; std::cin >> n; int y = 20; const std::div_t divresult = std::div(n, magicNumber); if (divresult.rem > … Read more
<?php $counter = 0; $value = 50; // Intial position $numbers = 230 // Lets say you have total 230 iterations. for ($i = 0 ; $i <= $numbers ; i++) { if($i == $value) // if 50 counter is increased and we are setting the value to 100 { $counter += 1; $value = … Read more
You are incrementing i at two places. Inside for loop for (i = 1; i <= 16; i++) and $tile.length + i++ Because of this it is incrementing two times solved Why my function incrementing plus 2?
You can’t apply the ++ operator to a string, but you can implement this logic yourself. I’d go over the string from its end and handle each character individually until I hit a character that can just be incremented simply: public static String increment(String s) { StringBuilder sb = new StringBuilder(s); boolean done = false; … Read more
Here’s 3 solutions: the first two are somewhat arithmetic incrementations while the third is more a character manipulations. The 3 implementations all pass the same unit tests: assertEquals(“1DDA01A”, MyClass.increment(“1DDA00Z”)); assertEquals(“1A9AV00”, MyClass.increment(“1A9AU99”)); assertEquals(“AFH00”, MyClass.increment(“AFG99”)); assertEquals(“A2GF24”, MyClass.increment(“A2GF23”)); assertEquals(“ABAA0000”, MyClass.increment(“AAZZ9999”)); assertEquals(“11AB0A”, MyClass.increment(“11AA9Z”)); First: public static String increment(String number) { Pattern compile = Pattern.compile(“^(.*?)([9Z]*)$”); Matcher matcher = compile.matcher(number); String … Read more
Part one should print i = 2. This is because: public class Sample { public static void main(String[] args) throws Exception { //part 1 int i=1; // i++ means i is returned (which is i = 1), then incremented, // therefore i = 1 because i is incremented to 2, but then reset // to … Read more