"std：：cin >> 中"运算符>>"的不明确重载

ambiguous overload for ‘operator>>’ in ‘std::cin >>

本文关键字：gt 不明确 重载 std cin 运算符      更新时间：2023-10-16

我正试图制作一些东西，这样你就可以输入一个文件名，然后输入文本写入文件，但当我试图编译它时，它会说：

files.cc: In function ‘int main()’:
files.cc:11: error: ambiguous overload for ‘operator>>’ in ‘std::cin >> filetoopen’
/usr/include/c++/4.2.1/istream:131: note: candidates are: std::basic_istream<_CharT, _Traits>& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>& (*)(std::basic_istream<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>] <near match>
/usr/include/c++/4.2.1/istream:135: note:                 std::basic_istream<_CharT, _Traits>& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_ios<_CharT, _Traits>& (*)(std::basic_ios<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>] <near match>
/usr/include/c++/4.2.1/istream:142: note:                 std::basic_istream<_CharT, _Traits>& std::basic_istream<_CharT, _Traits>::operator>>(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>] <near match>
/usr/include/c++/4.2.1/istream:250: note:                 std::basic_istream<_CharT, _Traits>& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_streambuf<_CharT, _Traits>*) [with _CharT = char, _Traits = std::char_traits<char>] <near match>
files.cc:14: error: no match for ‘operator>>’ in ‘std::cout >> text’
files.cc:16: error: conversion from ‘std::fstream’ to non-scalar type ‘std::ofstream’ requested
files.cc: In function ‘char* openfile(std::fstream, char*)’:
files.cc:21: error: no matching function for call to ‘std::basic_fstream<char, std::char_traits<char> >::open()’
/usr/include/c++/4.2.1/fstream:780: note: candidates are: void std::basic_fstream<_CharT, _Traits>::open(const char*, std::_Ios_Openmode) [with _CharT = char, _Traits = std::char_traits<char>]


#include <iostream>
#include <fstream>
using namespace std;
char* openfile(ofstream file, char* words);
int main()
{
  fstream filetoopen;
  char* text;
  cout << "Enter the name of a file to write to." << endl;
  cin >> filetoopen;
  cout << "Now write somthing to the file." << endl;
  cin >> text;
  openfile(filetoopen, text);
}
char * openfile (fstream file, char* words)
{
  file.open();
  file << words << endl;
  file.close();
  return words;
}

我对C++很陌生，不知道这意味着什么。此外，我不确定如何使函数返回数组，所以我有点猜测。有人能帮忙吗？

可能你写了这样的东西：

fstream filetoopen;
std::cin>>filetoopen;

以打开用户指定的文件上的CCD_ 1。这不是它的工作方式：您必须读取包含文件名的字符串，然后使用fstream的构造函数或其open方法打开文件：

std::string fileName;
std::getline(cin, fileName);
std::fstream fileStream(fileName.c_str());

fstream filetoopen;
cin >> filetoopen;

这是错误的，您无法从标准输入中读取流。您可以将文件的名称读取为字符串，然后使用该名称打开文件流。

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