![]() At this point, j is set to i - 1 = 0 and the difference is found. (lowestStrengthDifference) Įverything is fine up until when i = 1 on line 20. If we decrease the number of instructions in an inner loop then the running time of a program may be improved even if we increase the amount of code outside. Loop optimization, type sharpening, dead-code elimination, and intrinsics are just some of the other. LowestStrengthDifference = tempStrengthDifference A primer on JIT compilation in Java HotSpot VM. If (tempStrengthDifference < lowestStrengthDifference) //store the lowest found thus far in lowestSD TempStrengthDifference = Math.abs(strengthArray - strengthArray) //finding the difference between two values Dead code is code that can never be executed. Int tempStrengthDifference=99999 //junk difference set arbitrarily high - bad practice I knowįor (int i = 0 i 0) //do not execute the next for loop if i = 0 as cannot evaluate sAįor (int j = i - 1 j < 1 j-) // **this is line 20** from wherever the array has populated up to, work backwards to compare the numbers which have been fed in thus far Loop distribution is the opposite of loop fusion, that is, decomposing a single loop into multiple loops. Int strengthArray = new int //Initialise the array to be of length = N, the first input line A loop with at most one iteration is equivalent to the use of an if statement to conditionally execute one piece of code. Also, if a return is placed in the middle of the. Int N = in.nextInt() //How many data points there will be Most of them when they see and infinite loop, will not compile anything in the function that follows that loop. Whatever we can do for a while we can do it with a Java for loop too (and of course with a do-while too). Obviously I could separate those concerns, but I am experimenting. Java For Loop, is probably the most used one out of the three loops. I am trying to build the program in a way in which the data points are populated into an array at the same time as the comparisons are made. There are 3 data points (5 8 9), and the smallest difference is between 8 and 9, so the program should return 1. When the test is moved to the bottom of the loop, current has been assigned a new value which the compiler recognizes as potentially null. The solution was not nor will ever be posted since it is not evaluated. If current.key has not already thrown a NullPointerException through the attempt to access the key member, current cannot possibly be null at the beginning of the while loop. ![]() So, for instance, a sample input would be (on separate lines): 3 5 8 9 Dead code on a nested for loop that returns a boolean Ask Question Asked 4 years, 6 months ago Modified 4 years, 6 months ago Viewed 582 times 0 I am trying to complete an exercise that we received in class. ![]() ![]() The program should then print the smallest difference between all of the data points. Fljava work cosc 3304 wordnet Shorte estorjava cOSC File Edit Tools. Line 1 - How many data points there are following (N) having trouble with dead code error in for for loops any suggestion on fixing it. I have debugged this code and it appears to run a for loop even though the termination condition is met. ![]()
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