C++ wordlist generator

[]=-antr4xx-=[]

Mein selbst programmierter Wordlist-Generator in C++.
Ist das C++-pendant zu diesem Thread: http://free-hack.com/perl/52676-perl...generator.html

Hier noch einmal im Pastebin:
C++ | #include <iostream> #include - C++ wordlist generator - yiJ6jffj - Pastebin.com

#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
#include <cstdlib>

int number_of_chars;
char chars_to_use[200];

using namespace std;

ofstream out;

void outputcout(string x, int number_of_chars_left, string separator) {

    char c;
        
    if(number_of_chars_left==0) {
        cout << x << separator;    
        return;        
    }
    
    else {    
            for(c=0; c<number_of_chars; c++) {
                outputcout(x+chars_to_use[c], number_of_chars_left-1, separator);
            }        
    }
}

void output(string x, int number_of_chars_left, string separator) {

    char c;
        
    if(number_of_chars_left==0) {
        out << x << separator;    
        return;        
    }
    
    else {        
            for(c=0; c<number_of_chars; c++) {
                output(x+chars_to_use[c], number_of_chars_left-1, separator);
            }        
    }
}

void usage() {
    cerr << "usage: wlg (-l|-w) (no_chars) (chars) [output file]" << endl;
    exit(1);
}


int main(int argc, char *argv[]) {

    int how_many;
    string separator;    
        
    if(argc<4)
            usage();
            
    if(!strcmp(argv[1], "-l"))
        separator="\n";
    else if(!strcmp(argv[1], "-w"))
        separator="\r\n";
                
    how_many=atoi(argv[2]);
    strcpy(chars_to_use, argv[3]);
    number_of_chars=strlen(chars_to_use);
            
    if(argc==4)                
            outputcout("", how_many, separator);    
    
    if(argc==5) {    
            out.open(argv[4]);            
            output("", how_many, separator);
            out.close();    
    }
            
    
    
}

[blackberry]

Nach Sourcecodes verschoben.

[cTPu1bjo]

Congratulations for coding a Buffer Overflow...

[]=-antr4xx-=[]

Meinst du hier:

strcpy(chars_to_use, argv[3]);

?

Wenn man das durch

strncpy(chars_to_use, argv[3], 199);

ersetzt, ist imo der Overflow weg?

[luxury11]

Damit der BufferOverflow nicht ensteht:

strncpy((char*)chars_to_use,(const char*)argv[3],(size_t)strlen((const char*)argv[3])-1)

Zusätzlich:

char chars_to_use[200] = {0};