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An example of a list that proves this point is the list (2,3,4,5,1), which would only need to go through one pass of cocktail sort to become sorted, but if using an ascending bubble sort would take four passes. However one cocktail sort pass should be counted as two bubble sort passes.
For example, in a program, two variables may be defined thus (in pseudocode): data_item x := 1 data_item y := 0 swap (x, y); After swap() is performed, x will contain the value 0 and y will contain 1; their values have been exchanged.
Using the XOR swap algorithm to exchange nibbles between variables without the use of temporary storage. In computer programming, the exclusive or swap (sometimes shortened to XOR swap) is an algorithm that uses the exclusive or bitwise operation to swap the values of two variables without using the temporary variable which is normally required.
Bubble sort, sometimes referred to as sinking sort, is a simple sorting algorithm that repeatedly steps through the input list element by element, comparing the current element with the one after it, swapping their values if needed.
Is a generalisation of normal compare-and-swap. It can be used to atomically swap an arbitrary number of arbitrarily located memory locations. Usually, multi-word compare-and-swap is implemented in software using normal double-wide compare-and-swap operations. [16] The drawback of this approach is a lack of scalability. Persistent compare-and-swap
for i from 1 to 52 j := i + randomInt(53 - i) - 1 a.swapEntries(i, j) where a is an array object, the function randomInt(x) chooses a random integer between 1 and x, inclusive, and swapEntries(i, j) swaps the ith and jth entries in the array. In the preceding example, 52 and 53 are magic numbers, also not clearly related to each other. It is ...
Since the sorting algorithm only involves comparison-swap operations and is oblivious (the order of comparison-swap operations does not depend on the data), by Knuth's 0–1 sorting principle, [7] [8] it suffices to check correctness when each is either 0 or 1. Assume that there are 1s.
→ 1.0 push the constant 1.0 (a double) onto the stack ddiv 6f 0110 1111 value1, value2 → result divide two doubles dload 18 0001 1000 1: index → value load a double value from a local variable #index: dload_0 26 0010 0110 → value load a double from local variable 0 dload_1 27 0010 0111 → value load a double from local variable 1 dload ...