C ++类型的参数ifstream :: open() [英] C++ type of argument to ifstream::open()
本文介绍了C ++类型的参数ifstream :: open()的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
必须使用什么类型的文件名作为 ifstream.open()
?
What type must I make my file name to use it as an argument to ifstream.open()
?
int main(int argc, char *argv[]) {
string x,y,file;
string file = argv[1];
ifstream in;
in.open(file);
in >> x;
in >> y;
...
使用此代码,我会收到以下错误:
With this code, I get the following error:
main.cpp|20|error: no matching function for call to 'std::basic_ifstream<char,
std::char_traits<char> >::open(std::string&)'|
gcc\mingw32\4.4.1\include\c++\fstream|525|note: candidates are: void std::basic_ifstream<_CharT,
_Traits>::open(const char*, std::_Ios_Openmode) [with _CharT = char, _Traits = std::char_traits<char>]|
UPDATE:
推荐答案
构造函数使用一个const char *(http://www.cplusplus.com/reference/iostream/ifstream/ifstream/),因此您应该这样做:
The constructor takes a const char* (http://www.cplusplus.com/reference/iostream/ifstream/ifstream/) so you should do it like this:
in.open(argv[1]);
或者如果您真的要使用文件字符串变量,
or if you really want to use the file string variable, then
in.open(file.c_str());
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