Question

我正在尝试使用getLine读取带空格的字符串，但是我收到错误＆＃34;错误：函数＆＃34; getLine＆＃34;必须有原型。＆＃34; 。即使在使用namespace std;

之后，我也收到此错误

   void buildAhardCodedSQL4() 
    {
       cout << "Enter source : ";
       getLine(cin,source);
    }

Answer 1

将getLine(cin,source);替换为getline(cin,source);

有关如何使用getline()的详细信息，请参阅以下内容：http://www.cplusplus.com/reference/string/string/getline/

Answer 2

阅读手册：

http://en.cppreference.com/w/cpp/string/basic_string/getline

#include <string>
#include <iostream>
#include <sstream>

int main()
{
  // greet the user
  std::string name;
  std::cout << "What is your name? ";
  std::getline(std::cin, name);
  std::cout << "Hello " << name << ", nice to meet you.\n";

  // read file line by line
  std::istringstream input;
  input.str("1\n2\n3\n4\n5\n6\n7\n");
  int sum = 0;
  for (std::string line; std::getline(input, line); ) {
    sum += std::stoi(line);
  }
  std::cout << "\nThe sum is: " << sum << "\n";
}

错误：函数＆＃34; getLine＆＃34;必须有原型

2 个答案: