我试图在 C++ 中的类上重载 << 运算符。每当我将一个普通字符串(如“”)插入输出流时,我都会遇到无法理解的编译错误。我以前做过一次没有问题,所以我很困惑。
friend std::ostream& operator<<(std::ostream& out, Variable v);
std::ostream& operator<<(std::ostream& out, Variable v) {
out << v.type;
out << " ";
out << v.name;
return out;
}
这是输出:
src/Variable.cpp: In function 'std::ostream& operator<<(std::ostream&, Variable)':
src/Variable.cpp:35:9: error: no match for 'operator<<' in 'out << " "'
src/Variable.cpp:35:9: note: candidates are:
src/Variable.cpp:33:15: note: std::ostream& operator<<(std::ostream&, Variable)
src/Variable.cpp:33:15: note: no known conversion for argument 2 from 'const char [2]' to 'Variable'
In file included from /usr/local/Cellar/gcc/4.7.0/gcc/lib/gcc/x86_64-apple-darwin10.8.0/4.7.0/../../../../include/c++/4.7.0/string:54:0,
from src/../inc/Variable.h:4,
from src/Variable.cpp:1:
/usr/local/Cellar/gcc/4.7.0/gcc/lib/gcc/x86_64-apple-darwin10.8.0/4.7.0/../../../../include/c++/4.7.0/bits/basic_string.h:2750:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/local/Cellar/gcc/4.7.0/gcc/lib/gcc/x86_64-apple-darwin10.8.0/4.7.0/../../../../include/c++/4.7.0/bits/basic_string.h:2750:5: note: template argument deduction/substitution failed:
src/Variable.cpp:35:9: note: mismatched types 'const std::basic_string<_CharT, _Traits, _Alloc>' and 'const char [2]'
make: *** [bin/Variable.o] Error 1
最佳答案
德普。我没有包括 iostream。但是,这对我来说没有多大意义......因为只要我没有向 ostream 添加字符串,它就会起作用。我会认为编译器根本无法找到 ostream,并且会对此提示
关于c++ - 插入字符串 << 重载 C++ 时出错,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11514375/