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
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
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。
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