Randomizing an array of C strings

Question

I'm having a problem with a function meant to insert random C-strings into an array. I've read a few other questions about this on stackexchange, but none seem to work for me.

void rand(const char *tab[], int n){
    int i, j;
    char c[10];
    for(i=0; i<n; i++){
        for(j=0; j<10; j++){
            c[j]= rand()%26 + 97;
        }
        tab[i]=c;
    }
}

When trying to print it out, I get a blank screen, as if the array is empty. I declared the array as const char *tab[] and use the function rand(tab, 5). What could be wrong?

`c` is a local variable. It has no meaning outside the function, and dereferencing its copied value outside the function results in undefined behaviour. — Kerrek SB, Jan 28 '15 at 21:57
your question is very confusing. we can't know exactly what you want. what does `n` mean? — Wagner Patriota, Jan 28 '15 at 22:00
I'm sorry about that. `n` is the ammount of string that are supposed to be generated. — photons3432, Jan 28 '15 at 22:02

pm100 · Accepted Answer · 2015-01-28T22:38:31.527

0

you need to make a copy of the generated string. and you need a string terminator

void rand(const char *tab[], int n){
    int i, j;
    char c[10];
    for(i=0; i<n; i++){
        for(j=0; j<9; j++){
            c[j]= rand()%26 + 97;
        }
      c[9] = '\0';
        tab[i]=strdup(c) <<<<=====
    }
}

the strings are allocated on the heap - free them when you are finised

for(int i=0; i < NO_STRINGS; i++)
{
   free(tab[i]);
}

edited Jan 28 '15 at 22:38

answered Jan 28 '15 at 22:13

pm100

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remember to free the strings after – pm100 Jan 28 '15 at 22:22
I'm sorry, what do you mean by "free the strings"? – photons3432 Jan 28 '15 at 22:34

Randomizing an array of C strings

1 Answers1