我正在撰写一本书的章节评论：在本章的最后，您将完成一些问题/任务。
我决定以程序的格式而不是文本文件来执行它们：
#include <iostream>
int main(int argc, char* argv[]) {
std::cout << "Chapter review\n"
<< "1. Why does C++ have more than one integer type?\n"
<< "\tTo be able to represent more accurate values & save memory by only allocating what is needed for the task at hand.\n"
<< "2. Declare variables matching the following descriptions:\n"
<< "a.\tA short integer with the value 80:\n";
short myVal1 = 80;
std::cout << "\t\t\"short myVal1 = 80;\": " << myVal1 << std::endl
<< "b.\tAn unsigned int integer with the value 42,110:\n";
unsigned int myVal2 = 42110;
std::cout << "\t\t\"unsigned int myVal2 = 42110;\": " << myVal2 << std::endl
<< "c.\tAn integer with the value 3,000,000,000:\n";
float myVal3 = 3E+9;
std::cout << "\t\t\"float myVal3 = 3E+9;\": " << static_cast<unsigned int>(myVal3) << std::endl
<< "3. What safeguards does C++ provide to keep you from exceeding the limits of an integer type?\n"
<< "\tWhen it reaches maximum number it starts from the begging again (lowest point).\n"
<< "4. What is the distinction between 33L and 33?\n"
<< "\t33L is of type long, 33 is of type int.\n"
<< "5. Consider the two C++ statements that follow:\n\tchar grade = 65;\n\tchar grade = 'A';\nAre they equivalent?\n"
<< "\tYes, the ASCII decimal number for 'A' is '65'.\n"
<< "6. How could you use C++ to find out which character the code 88 represents?\nCome up with at least two ways.\n"
<< "\t1: \"static_cast<char>(88);\": " << static_cast<char>(88) << std::endl; // 1.
char myChar = 88;
std::cout << "\t2: \"char myChar = 88;\": " << myChar << std::endl // 2.
<< "\t3: \"std::cout << (char) 88;\" " << (char) 88 << std::endl // 3.
<< "\t4: \"std::cout << char (88);\": " << char (88) << std::endl // 4.
<< "7. Assigning a long value to a float can result in a rounding error. What about assigning long to double? long long to double?\n"
<< "\tlong -> double: Rounding error.\n\tlong long -> double: Significantly incorrect number and/or rounding error.\n"
<< "8. Evaluate the following expressions as C++ would:\n"
<< "a.\t8 * 9 + 2\n"
<< "\t\tMultiplication (8 * 9 = 72) -> addition (72 + 2 = 74).\n"
<< "b.\t6 * 3 / 4\n"
<< "\t\tMultiplication (6 * 3 = 18 -> division (18 / 4 = 4).\n"
<< "c.\t3 / 4 * 6\n"
<< "\t\tDivision (3 / 4 = 0) -> multiplication (0 * 6 = 0).\n"
<< "d.\t6.0 * 3 / 4\n"
<< "\t\tMultiplication (6.0 * 3 -> 18.0) -> division (18.0 / 4 = 4.5).\n"
<< "e.\t 15 % 4\n"
<< "\t\tDivision (15 / 4 = 3.75) Then returns the reminder, basically how many times can 4 go into 15 in this case that is 3 (3*4 = 12).\n"
<< "9. Suppose x1 and x2 are two type of double variables that you want to add as integers and assign to an integer variable. Construct a C++ statement for doing so. What if you wanted to add them as type double and then convert to int?\n"
<< "\t1: \"int myInt = static_cast<double>(doubleVar);\"\n\t2: \"int myInt = int (doubleVar);\".\n"
<< "10. What is the variable type for each of the following declarations?\n"
<< "a.\t\"auto cars = 15;\"\n\t\tint\n"
<< "b.\t\"auto iou = 150.37f;\"\n\t\tfloat\n"
<< "c.\t\"auto level = 'B';\"\n\t\tchar\n"
<< "d.\t\"auto crat = U'/U00002155';\"\n\t\twchar_t ?\n"
<< "e.\t\"auto fract = 8.25f/.25;\"\n\t\tfloat" << std::endl;
return 0;
}
由于移动/其他一些现实生活中的东西，我读了第3章已经有一段时间了。
我在这里不确定的基本上是第3个问题：它以复数形式表示保障措施。
但是我只知道一个：它在达到最大值后再次从头开始？我在这里错过了什么吗？
如果您还看到其他任何错误，请告诉我 - 毕竟我这样做是为了学习：）。
答案 0 :(得分：2)
基本上我不能接受评论作为答案，所以总结一下：
没有任何保障措施，我误解了@ n.m的问题。为我澄清。
10.e错误如@ Jarod42指出的那样，这是正确的。
谢谢！
答案 1 :(得分：0)
至于我＆＃34;声明整数变量，值为3,000,000,000＆＃34;是：
unsigned anInteger = 3000000000;
导致c ++，第11个提供15个整数类型，unsigned int是可以存储如3 000 000 000这样的大整数的最小值。
答案 2 :(得分：0)
C ++将整数溢出归类为“未定义行为” - 任何事情都可能因此而发生。通过以下思考，这本身可称为“保护措施”（虽然它是一个延伸）：
gcc有-ftrapv编译开关，当整数溢出发生时，程序会崩溃。这使您可以轻松调试溢出。这个功能是可行的，因为C ++使得程序在这些情况下崩溃是合法的（本质上是未定义的行为）。我认为C ++委员会在制定C ++标准的这一部分时会考虑到这个确切的情况。
这与例如Java，其中整数溢出导致环绕，并且可能更难调试。