C++ basic_string::_S_construct null 无效
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basic_string::_S_construct null not valid
提问by starrystar
I am reading an input file from the command line.
我正在从命令行读取输入文件。
int main(int argc, char **argv)
{
Scene myScene;
string filename = argv[1];
myScene = Parser(filename);
...
}
from another file I use the parser function which is declerated like that;
从另一个文件中，我使用了像这样声明的解析器函数；
Scene Parser(string filename)
{
string line;
ifstream myfile (filename.c_str());
...
return scene;
}
I am getting the error; terminate called after throwing an instance of 'std::logic_error' what(): basic_string::_S_construct null not valid
我收到错误；在抛出“std::logic_error”what() 实例后调用终止：basic_string::_S_construct null 无效
Program received signal SIGABRT, Aborted.
程序收到信号 SIGABRT，中止。
I have searched the error. I think it is because of these lines. But I cannot find the actual reason. Can anybody help me?
我已经搜索了错误。我想是因为这些线。但我找不到真正的原因。有谁能够帮助我？
回答by Halim Qarroum
This means that filenameis NULLin Parser, probably because you are not passing any arguments to your program's command line.
这意味着filename是空的Parser，可能是因为你没有传递任何参数到你的程序的命令行。
Make sure to always check whether the expected number of arguments are passed to your program. For instance, you could do :
确保始终检查是否将预期数量的参数传递给您的程序。例如，你可以这样做：
int main(int argc, char *argv[]) {
if (argc != NUMBER_OF_EXPECTED_ARGUMENTS) {
exit(EXIT_FAILURE);
}
// ...
string filename(argv[1]);
Scene myScene = Parser(filename);
// ...
}
回答by Vlad from Moscow
It is possible that you forgot to specify command line arguments and as the result argv[1] is equal to NULL. You should check whether the user entered command line arguments. For example
您可能忘记指定命令行参数，结果 argv[1] 等于 NULL。您应该检查用户是否输入了命令行参数。例如
int main(int argc, char **argv)
{
Scene myScene;
string filename;
if ( 1 < argc ) filename.assign( argv[1] );
