在命令提示符下忽略以'＃'开头的行

Question

我在写一段代码，要求我忽略注释行（即以＃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 

所以注释行可以在任何地方。它只需要忽略整条线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

从你所显示的，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

请问任意数量根据你给reqd值数。 如果一行中的第一个字符本身是#，它也会起作用 - 将再次询问该行。如果`＃之前没有数字，还会读取另一行。

#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; 
}