Question

I was given task to generate random 80 byte keys and i have decided following strateges

in my computer sizeof(char)=1 so i have created array of english alphabetical letters

char *p=" ";
char a[0..26] and in cycle 


for (int  i=0;i<=80;i++){
   *(p+i)+= a[(rand()+100) % 26];
}

but it does not work it stops execution please help sorry if my code is stupid but i can’t think at this time otherwise
thanks
code

#include <iostream>
#include <string.h>
#include <cstdlib>
using   namespace std;
int main(){

    char *p=" ";
    char  a[]= { 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
 for (int i=0;i<=80;i++){
        *(p+i)+=(a[(rand()+100)%26]);
    }

     cout<<p<<endl;


     return 0;

}

Answer 1

Try this out:

#include <iostream>
#include <string.h>
#include <cstdlib>
using   namespace std;
int main(){
    // ensure the target has enough memory for the key and a null terminator
    char p[81];

    // this string will do as nicely as the character array
    char a[] = "abcdefghijklmnopqrstuvwxyz";

    // no += here. I assign the random character directly to the target buffer
    for (int i=0;i<=80;i++)
        p[i] = a[rand()%26];

    // alternately, you can calculate a random English character with:
    // p[i] = rand()%26 + 'a';
    // which removes the need for the a[] buffer at all

    // don't forget to null-terminate
    p[80] = '\0'

    // output results
    cout<<p<<endl;
    return 0;
}