cin.getline()和cin.get()都是对输入的面向行的读取,即一次读取整行而不是单个数字或字符,但是二者有一定的区别。 cin.get()每次读取一整行并把由Enter键生成的换行符留在输入队列中,比如: #include <iostream> using std::cin; using std::cout; const int SIZE = 15; int main( ){ cout << "Enter your name:"; char name[SIZE]; cin.get(name,SIZE); cout << "name:" << name; cout << "\nEnter your address:"; char address[SIZE]; cin.get(address,SIZE); cout << "address:" << address; } 输出: Enter your name:jimmyi shi name:jimmyi shi Enter your address:address: 在这个例子中,cin.get()将输入的名字读取到了name中,并将由Enter生成的换行符'/n'留在了输入队列(即输入缓冲区)中,因此下一次的cin.get()便在缓冲区中发现了'/n'并把它读取了,最后造成第二次的无法对地址的输入并读取。解决之道是在第一次调用完cin.get()以后再调用一次cin.get()把'/n'符给读取了,可以组合式地写为cin.get(name,SIZE).get();。 cin.getline()每次读取一整行并把由Enter键生成的换行符抛弃,如: #include <iostream> using std::cin; using std::cout; const int SIZE = 15; int main( ){ cout << "Enter your name:"; char name[SIZE]; cin.getline(name,SIZE); cout << "name:" << name; cout << "/nEnter your address:"; char address[SIZE]; cin.get(address,SIZE); cout << "address:" << address; } 输出: Enter your name:jimmyi shi name:jimmyi shi Enter your address:YN QJ address:YN QJ 由于由Enter生成的换行符被抛弃了,所以不会影响下一次cin.get()对地址的读取。 |
|