## C++ and Object-oriented Numeric Computating For Scientists by Yang, D.

By Yang, D.

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Extra resources for C++ and Object-oriented Numeric Computating For Scientists And Engineers

Example text

It is always executed at least once. For example, 44 2. Expressions and Statements int x = 10 do { cout << x++ << "\n" } while (x > 100) It will stop after the rst iteration and thus print out only 10: The continue statement can also be used in a while or do-while loop to cause the current iteration to stop and the next iteration to begin immediately, and the break statement to exit the loop. Note that continue can only be used in a loop while break can also be used in switch statements. For example, the code segment int x = 10 do { cout << x++ << '\n' if (x == 20) break } while (true) // print x, then increment x // exit loop when x is 20 should print out all integers from 10 up to 19 and the code segment int x = 10 do { if (x++ == 15) continue cout << x << '\n' if (x == 20) break } while (true) // test equality, then increase x should print out all integers from 11 up to 20 except for 16: However, the code int x = 10 while (true) { if (x == 15) continue cout << x++ << '\n' if (x == 20) break } rst prints out integers 10 11 12 13 14 and then goes into an in nite loop (since x will stay equal to 15 forever).

That is, nd the sum n2 + (n + 1)2 + + m2 for two given integers n and m with n < m. 1, the variable sum should be declared as double or long double in order to handle large values of n and m. In this exercise, try to compute sum in two ways: as a long double and as an int, and compare the results. On a computer with 4 bytes for storing int, the second way calculates the sum 12 +22 +32 + +50002 as ;1270505460. Why could a negative number as the output be possible? 2. , less than or equal to 12) integers.

In the pre x form ++n the value of n is rst incremented by 1 and then the new value of n is used. A similar explanation holds for the decrement operator ;; : For example, int i++ ++j int int m = n = i = 4, j = 4 m = i++ n = ++j --i j-- // // // // // // // i and j are both initialized to 4 i = i+ 1. So i= 5 j = j+ 1. So j= 5 first m=i, then i=i+1. So m=5, i=6 first j=j+1, then n=j. So n=6, j=6 first i=i-1, then m=i. So m=5, i=5 first n=j, then j=j-1. So n=6, j=5 In the statement m = ;; i the value of i is decremented by 1 rst and the new value of i is assigned to m, while in the statement n = j ;; the value of j is rst assigned to n and then the value of j is decremented by 1: Compound operators may sometimes cause side e ects.