标签：enter mos speed eterm provides main tom lock section
- Having more than one integer type lets you choose the type that is best suited to a particular need. For example, you could use
shortto conserve space orlongto guarantee storage capacity or to find that a particular type speed up a particular calculation.short ribs = 80; // or short int ribs = 80;
unsigned int q = 42110; // or unsigned q = 42110;
unsigned long ants = 3000000000; // or long long ants = 3000000000;
Note: Don‘t count onintbeing large enough to hold 3,000,000,000. Also if your system supports universal list-initialization, you could use it:
short rbis = {80};
unsigned int q {42110};
long long ants {3000000000};- C++ provides no automatic safeguards to keep you from exceeding integer limits; you can use the
climitsheader file to determine what the limits are.- The constant 33L is type
long, whereas the constant 33 is typeint.- The two statements are not really equivalent, although they have the same effect on some system. Most importantly, the first statement assigns the letter A to
gradeonly on a system using the ASCII code, while the second statement also works for other codes. Second, 65 is a typeintconstant, whereasAis a typecharconstant.- Here are four ways:
char c = 88;
cout << c << endl; // char type prints as character
cout.put(char(88)); // put() prints char as character
cout << char(88) << endl; // new-style type cast value to char
cout << (char) 88 << endl; // old-style type cast value to char- The answer depends on how large the two types are. If
longis 4 bytes, there is no loss. Thats because the largestlongvalue would be about 2 billion, which is 10 digits. Becausedoubleprovides at least 13 significant figures, no rounding would be needed. Thelong longtype, on the other hand, can reach 19 digits, which exceeds the 13 significant figures guaranteed fordouble.a. 8 * 9 + 2 is 72 + 2 is 74
b. 6 * 3 / 4 is 18 / 4 is 4
c. 3 / 4 * 6 is 0 * 6 is 0
d. 6.0 * 3 / 4 is 18.0 / 4 is 4.5
e. 15 % 4 is 3- Either of the following would work for the first task:
int pos = (int) x1 + (int) x2;
int pos = int (x1) + int (x2);
To add them as typedoubleand then convert, you could do either of the following:
int pos = (int) (x1 + x2);
int pos = int (x1 + x2);- a. int b. float c. char d. char32_t e. double
#include <iostream>
const int Factor{12};
int main()
{
using namespace std;
cout << "Input your height(inches): ";
int height, feet, inches;
cin >> height;
feet = height / Factor;
inches = height % Factor;
cout << "Your height is equal to " << feet << " feet," << inches << " inches" << endl;
return 0;
}#include <iostream>
int main()
{
using namespace std;
const double Inches_per_foot = 12;
const double Inches_per_meter{0.0254};
const double Kilo_per_pound{2.2};
cout << "Enter your height in feet and inches(example: 14 17): ";
double feet;
double inches;
cin >> feet;
cin >> inches;
cout << "Enter your weight in pounds: ";
double weight;
cin >> weight;
double i = (feet * Inches_per_foot + inches) * Inches_per_meter;
cout << "Your BMI is: " << weight * Kilo_per_pound / (i * i) << endl;
return 0;
}#include <iostream>
int main()
{
using namespace std;
const double Mid{60.0};
cout << "Enter a latitude in degrees, minutes, and seconds:" << endl;
cout << "First, enter the degrees: ";
int degrees;
cin >> degrees;
cout << "Next, enter the minutes of arc: ";
int minutes;
cin >> minutes;
cout << "Finally, enter the seconds of arc: ";
int seconds;
cin >> seconds;
cout << degrees << " degrees, " << minutes << " minutes, " << seconds << " seconds = "
<< double (degrees) + minutes / Mid + seconds / (Mid * Mid) << endl;
return 0;
}#include <iostream>
int main()
{
using namespace std;
const int H_per_day{24};
const int M_per_hour = {60};
const int S_per_seconds = 60;
cout << "Enter the number of seconds: ";
long long seconds;
cin >> seconds;
long long d = seconds / (H_per_day * M_per_hour * S_per_seconds);
long long h = seconds % (H_per_day * M_per_hour * S_per_seconds) / (M_per_hour * S_per_seconds);
long long m = seconds % (H_per_day * M_per_hour * S_per_seconds) % (M_per_hour * S_per_seconds) / S_per_seconds;
long long s = seconds % (H_per_day * M_per_hour * S_per_seconds) % (M_per_hour * S_per_seconds) % S_per_seconds;
cout << seconds << " seconds = "
<< d << " days, "
<< h << " hours, "
<< m << " minutes, "
<< s << " seconds \n";
return 0;
}#include <iostream>
int main()
{
using namespace std;
cout << "Enter the world's population: ";
long long p_of_world;
cin >> p_of_world;
cout << "Enter the population of the China: ";
long long p_of_nation;
cin >> p_of_nation;
cout << "The population of the China is " << double (p_of_nation) / double (p_of_world) * 100.0
<< "% of the world population. \n";
return 0;
}#include <iostream>
int main()
{
using namespace std;
cout << "Enter you have driven in miles: ";
int mile;
cin >> mile;
cout << "Enter you have costed of gasoline: ";
int gas;
cin >> gas;
cout << double (mile) / double (gas) << " miles per gallon your car has gotten. \n";
return 0;
}#include <iostream>
int main()
{
using namespace std;
const double Miles_per_kilo{6.214e-1};
const double Liters_per_gallon = 3.875;
cout << "Enter European style (liters per 100 kilometers): ";
double eu;
cin >> eu;
int temp = int ((1.0 / eu) * (Miles_per_kilo * 100.0) * Liters_per_gallon);
cout << eu << " l/100 km = " << temp << " mpg \n";
return 0;
}《C++ primer plus 英文版 第六版》Chapter 3
标签：enter mos speed eterm provides main tom lock section
原文地址：https://www.cnblogs.com/narisu/p/8970967.html
