模式匹配，循环重复次数超出预期

Question

I've never used string or string functions until today and I'm running into a problem that I don't understand. This program as is, should just accept a command line argument, load the file and display it to memory. However it displays it multiple times. I'm pretty sure the for loop is the problem, but it is the same technique as what is used in the programming reference I am using.

#include <iostream>
#include <fstream>
#include <vector>
#include <string>

char* getFile( char* fileName ){
    std::fstream inFile( fileName );
    if( !inFile ) std::cout << "Could not open " << fileName << ".\n";
    else{
        inFile.seekg(0,inFile.end);
        int len = inFile.tellg();
        inFile.seekg(0,inFile.beg);
        char* buffer = new char[len];
        inFile.read( buffer, len);
        inFile.close();
        std::cout.write(buffer,len);
        return buffer;
        }

    }
int main(int argc, char** argv){
    if(argc != 2) std::cout << "Parameter required\n";
    else{
        std::string f = getFile( argv[1] );
        for( size_t i = f.find( 0x0A, 0 ); i != std::string::npos ; i = f.find( 0x0A, i) ){
            std::cout << f.substr(0,i)<<std::endl;
            i++;
            }
    }
}

I see at least one of the problems with my code. I re-wrote the loop as a while loop because it was easier to follow and paid a little more attention to where I am starting and stopping. However it still seems to be printing twice.

int main(int argc, char** argv){
    if(argc != 2) std::cout << "Parameter required\n";
    else{
        std::string f = getFile( argv[1] );
        size_t start = 0;
        size_t end = 1;
        while( end != std::string::npos ){
            end = f.find( 0x0A, start );
            std::cout << f.substr(start,end)<<std::endl;
            start = ( end + 1 );
            }

Answer 1

This is because you have two printing statements that are displaying the contents of the file.

The first print statement is this one:

std::cout.write(buffer,len);

The second one is this:

std::cout << f.substr(0,i)<<std::endl;