Question

我正在编写一个代码，要求我忽略注释行（即以＃符号开头的行）直到行尾。我正在使用linux来编写c ++代码。例如：在添加两个数字的情况下。

xxx@ubuntu:~ $ ./add
Enter the two numbers to be added
1 #this is the first number
2 #this is the second number
result: 3

因此注释行可以在任何地方。它只需要忽略整行，并将下一个值作为输入。

#include <iostream>
using namespace std;
int main()
{
int a,b;


cout<< "Enter the two numbers to be added:\n";
while(cin >>a >>b)
{
if (a == '#'|| b == '#')
continue;
cout << "Result: "<<a+b;
}
return 0;
}

Answer 1

根据你的表现，我认为这可能是你想要的。

int main()
{
    string comment;
    int nr1,nr2;
    // Read the first number. It should be the first one always. No comment before number!
    cin >> nr1;            

    // See if we can read the second number Successfully. Which means it is an integer.
    if(cin >> nr2) {
    } 
    // Otherwise clear cin and read the rest of the comment line                        
    else {
        cin.clear();           
        getline(cin,comment);
        // Now read the second number from the second line
        cin >> nr2;           
    }
    // Read the rest of second the line.
    getline(cin,comment);   

    cout << "result: " << nr1 + nr2 << endl;
    return 0;
}

Answer 2

根据您提供的值reqd，是否会有任意数量的数字。如果一行中的第一个字符是#，也会有效 - 将再次询问该行。如果在＃。

之前没有数字，也会读另一行

#include <iostream>
#include <sstream>
#include <ctype.h>

using namespace std;

int main ()
{
    const int reqd = 2;
    string sno[reqd];
    int no[reqd];
    int got = 0;
    size_t pos;
    istringstream is;

    cout<< "Enter "<<reqd<<" numbers to be added:\n";
    while(got < reqd)
    {
        getline(cin, sno[got]);
        if((pos = sno[got].find('#')) && isdigit(sno[got][0]))
        {
            is.str(sno[got]);
            is>>no[got];
            ++got;
        }
    }

    int sum = 0;
    for(int i = 0; i < reqd; ++i)
        sum+=no[i];

    cout<<"Result : "<<sum;
    return 0;
}

忽略命令提示符中以“＃”开头的行

2 个答案: