[英]I'm getting the error: no suitable conversion function from std::basic_istream<char, std::char_traits<char>> to char exists
我試圖使運算符>>重載，但我想從輸入中讀取每個字符，但出現錯誤。 這是代碼：
istream& operator>>(istream& input, Natural_Big_Number number)
{
int x;
input >> x;
number.set_nr_digits(x);
char c;
while ((c = input.get(c)))
{
}
}
如果您希望它返回字符，則無需為.get()指定參數。 https://en.cppreference.com/w/cpp/io/basic_istream/get
#include <iostream>
using namespace std;
istream& operator>>(istream& input, int number)
{
char c;
while ((c = input.get()))
{
}
}
如果要檢查流的布爾值是否false ，則可以執行注釋中提到的操作，而不是：
while (input.get(c))
它將字符存儲在c然后檢查返回的輸入流的bool值。
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[英]I keep getting an error from Xcode that says “ Apple Mach-O Linker (Id) Error ”readIt(std::_1::basic_ifstream<char,std::_1::char_traits<char> >&,"
出於某種原因，我不斷收到此錯誤：“運算符>>”不匹配（操作數類型為“std::istream”{又名“std::basic_istream”<char> '}
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無法綁定'std :: basic_istream <char> '左右'std :: basic_istream <char> &&”
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沒有合適的構造函數來將“ const char [8]”轉換為“ std :: basic_string” <wchar_t, std::char_traits<wchar_t> ，std :: allocator <wchar_t> >”
[英]no suitable constructor exists to convert from “const char [8]” to “std::basic_string<wchar_t, std::char_traits<wchar_t>, std::allocator<wchar_t>>”
ostream和運算符std :: basic_string <char, std::char_traits<char> ，std :: allocator <char> >？
[英]ostream and operator std::basic_string<char, std::char_traits<char>, std::allocator<char>>?
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