CODEWITHSUNDEEP
c++filefile-io
calccrypto
calccrypto

Reputation: 8991

I can't open variable file name

Is there any reason why Dev C++ is not letting me do file.open(file_name_variable)? I don't understand why it's not letting me open anything but a hardcoded name like file.open("abc.txt") how do get around this? Don't use Dev C++?

here's basically what I have:

int open_file(string file_name){
    ifstream file;
    file.open(file_name);
    if (!file.is_open()){
        return 0;       
    }
    return 1;
}

Upvotes: 0

Views: 4522

Answers (3)

Benjamin Lindley
Benjamin Lindley

Reputation: 103703

You need to pass it a c-string. Use:

file.open( file_name.c_str() );

In C++11, this is no longer necessary. A signature that takes std::string was added.

Upvotes: 6

log0
log0

Reputation: 10917

fstream::open requires a const char * as first argument.

void open ( const char * filename,
        ios_base::openmode mode = ios_base::in | ios_base::out );

It does not take a std::string (which is a shame by the way).
You need to convert your std::string to a const char *

file.open(file_name.c_str())

Upvotes: 1

Stuart Golodetz
Stuart Golodetz

Reputation: 20616

Alexandre's right, I think. The signature of open is:

void open(const char *filename, ios_base::openmode mode);

You're trying to pass a std::string in as a const char *, but std::string doesn't have an operator const char * (for safety reasons). Instead, you have to use the c_str() method. The issue is not that the implicit conversion to const char * should be available, but that basic_ifstream should have an open overload that accepts a string -- to the best of my knowledge, this is being added in C++0x, but I don't have a reference for that offhand.

Upvotes: 0

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