Imagine you are writing a program to manage a shopping list. Using a vector to hold the shopping list items, write a print function to print out the contents of a vector of strings. Test your print function with a main program that does the following: Create an empty vector. Print it. Append the items, "eggs," "milk," "sugar," "chocolate," and "flour" to the list. Print it. Remove the last element from the vector. Print it. Append the item, "coffee" to the vector. Print it. Write a loop that searches for the item, "sugar" and replace it with "honey." Print the vector. Write a loop that searches for the item, "milk," and then remove it from the vector. Print the vector. Search for the item, "chocolate," and insert the item "tea" before it. Print the vector one more time.

时间: 2024-01-25 20:05:04 浏览: 25
Sure, here's an example program that does what you're asking for: ```cpp #include <iostream> #include <vector> #include <string> void printList(const std::vector<std::string>& list) { for (const auto& item : list) { std::cout << item << std::endl; } } int main() { std::vector<std::string> shoppingList; // Print the empty list std::cout << "Empty list:" << std::endl; printList(shoppingList); // Add items to the list shoppingList.push_back("eggs"); shoppingList.push_back("milk"); shoppingList.push_back("sugar"); shoppingList.push_back("chocolate"); shoppingList.push_back("flour"); // Print the list with items std::cout << std::endl << "List with items:" << std::endl; printList(shoppingList); // Remove the last item from the list shoppingList.pop_back(); // Print the updated list std::cout << std::endl << "List with last item removed:" << std::endl; printList(shoppingList); // Add "coffee" to the list shoppingList.push_back("coffee"); // Print the updated list std::cout << std::endl << "List with coffee added:" << std::endl; printList(shoppingList); // Replace "sugar" with "honey" for (auto& item : shoppingList) { if (item == "sugar") { item = "honey"; } } // Print the updated list std::cout << std::endl << "List with sugar replaced with honey:" << std::endl; printList(shoppingList); // Remove "milk" from the list for (auto it = shoppingList.begin(); it != shoppingList.end(); ) { if (*it == "milk") { it = shoppingList.erase(it); } else { ++it; } } // Print the updated list std::cout << std::endl << "List with milk removed:" << std::endl; printList(shoppingList); // Insert "tea" before "chocolate" for (auto it = shoppingList.begin(); it != shoppingList.end(); ++it) { if (*it == "chocolate") { shoppingList.insert(it, "tea"); break; } } // Print the final list std::cout << std::endl << "Final list with tea added:" << std::endl; printList(shoppingList); return 0; } ``` The output of the program should look like this: ``` Empty list: List with items: eggs milk sugar chocolate flour List with last item removed: eggs milk sugar chocolate List with coffee added: eggs milk sugar chocolate coffee List with sugar replaced with honey: eggs milk honey chocolate coffee List with milk removed: eggs honey chocolate coffee Final list with tea added: eggs honey tea chocolate coffee ```

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3)A digital clock consists of a screen to display the time and a dial for setting in turn the year, month, day, hour and minute. Twisting the dial to the left reduces by one the value being changed but twisting it to the right increases it by one. Pushing the dial alters which value is being adjusted. At first, it is the year but after the dial is pushed once, it is the month, then after the dial is pushed again, it is the day and so on. Imagine the clock is represented by a class with attributes year, month, day etc. The following is what the code for a method rotateDialLeft() might look like. public void rotateDialLeft() { if (mode == YEAR_MODE) { year--; } else if (mode == MONTH_MODE) { month--; } else if (mode == DAY_MODE) { day--; } else if (mode == HOUR_MODE) { hour--; } else if (mode == MINUTE_MODE) { minute--; } } The code for rotateDialRight() is similar. Apply the Open-Closed Principle to explain why the above code is unsatisfactory from the design viewpoint, considering the possibility of future change to the code, giving an example of such a change. 5)Give the code required for the classes introduced in question 3), focusing on the code for a method selectState() which changes the value that is being adjusted from years to months. Make it clear in which classes the code is to be found. Assume the existence of other methods that are needed such as getMonthSetUpState(). 8)Suppose that in a multiplayer role-playing game, a class Client has a dependency to an interface Fighter with public methods attack(), defend() and escape(). The game designer now wishes for Client to use a class Wizard with three different but equivalent public methods castDestructionSpell(), shield() and portal(). Explain how it is possible to do this using an appropriate design pattern.

Another example is the SRIOV_NET_VF resource class, which is provided by SRIOV-enabled network interface cards. In the case of multiple SRIOV-enabled NICs on a compute host, different qualitative traits may be tagged to each NIC. For example, the NIC called enp2s0 might have a trait “CUSTOM_PHYSNET_PUBLIC” indicating that the NIC is attached to a physical network called “public”. The NIC enp2s1 might have a trait “CUSTOM_PHYSNET_INTRANET” that indicates the NIC is attached to the physical network called “Intranet”. We need a way of representing that these NICs each provide SRIOV_NET_VF resources but those virtual functions are associated with different physical networks. In the resource providers data modeling, the entity which is associated with qualitative traits is the resource provider object. Therefore, we require a way of representing that the SRIOV-enabled NICs are themselves resource providers with inventories of SRIOV_NET_VF resources. Those resource providers are contained on a compute host which is a resource provider that has inventory records for other types of resources such as VCPU, MEMORY_MB or DISK_GB. This spec proposes that nested resource providers be created to allow for distinguishing details of complex components of some resource providers. During review the question came up about “rolling up” amounts of these nested providers to the root level. Imagine this scenario: I have a NIC with two PFs, each of which has only 1 VF available, and I get a request for 2 VFs without any traits to distinguish them. Since there is no single resource provider that can satisfy this request, it will not select this root provider, even though the root provider “owns” 2 VFs. This spec does not propose any sort of “rolling up” of inventory, but this may be something to consider in the future. If it is an idea that has support, another BP/spec can be created then to add this behavior.

clc; clear all; [filename,pathname]=uigetfile({'*.jpg;*.tif;*.png;*gif','all imagine files';'*.*','all files'},'select your photo'); path=[pathname,filename]; image=imread(path); % axes(handles.photo); imshow(image);%显示图片 %image processing I=rgb2gray(image); I=rangefilt(I);%滤波 background = imopen(I,strel('disk',11));%使用形态学开来估计背景 I2 = I-background;%从原始图像中减去背景图像 I3 = imadjust(I2);%增强对比度 bw = imbinarize(I3);%阈 值图像 bw = bwareaopen(bw,160);%降噪150,160 bw=edge(bw,'canny'); %边缘检测 %bw=1-bw; % axes(handles.a1); imshow(bw); %se=strel('disk',13);%15 se=strel('square',15);%15 bw1=imclose(bw,se);%闭 bw2=imdilate(bw1,se);%膨胀 bw2=imerode(bw2,se);%腐蚀 bw3=imfill(bw2,'holes'); % axes(handles.a2); imshow(bw3); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %circle detection rmin = 20; rmax = 2500; radiusRange=[rmin rmax]; [center, rad] = imfindcircles(bw3,radiusRange,'EdgeThreshold',0.13);%检测灵敏度(边缘渐变阈值)0.3 display(center); display(rad); % axes(handles.a3); imshow(bw3); viscircles(center, rad,'Color','b'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %initialize the number of coins one=0; half=0; little=0; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %coin recognition [m,n]=size(rad); num=m; i=1; j=num; min=rad(i); max=rad(j); while i<=j if rad(i)<rad(j) if rad(i)<min min=rad(i); else if rad(j)<max max=rad(j); end end else if rad(j)<min min=rad(j); else if rad(i)<max max=rad(i); end end end i=i+1; j=j-1; end sum=0; for i=1:num sum=rad(i)+sum; end % ave=(sum-(min+max))/(num-2); ave = sum/num; for i=1:num if 0.6<(rad(i)/ave)&&(rad(i)/ave)<1.5 if rad(i)>ave one=one+1; else if 0.93<(rad(i)/ave) && rad(i)<=ave half=half+1; else little = little+1; end end end end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %display results sum=half*0.5+one+little*0.1; one half little sum这段代码什么意思

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