How to read a string array from input in C and pass to function?

Question

I have been stuck on this all day. I want to be able to create a string array in C, and pass that array to the PrintStuff function. I don't want to change my parameters to PrintStuff, but still make it work. Any help please?

void PrintStuff(const char **arr) {
  for (int i = 0; i < 5; ++i) {
        printf ("%s\n", arr[i]);
    }
}

int main ()
{
    //This works
    //char * array[5] = { "this", "is", "a", "test", "megan"};

    //This doesn't work
    char * array[5];
    for (int i=0;i<5;i++)
    {
        //scanf("%9s", array[i]);
        fgets(array[i], 10, stdin);
    }

    Sort(array, 0, 5 - 1);
}

It doesn't do anything and I get this warning that says

passing 'char *[5]' to parameter of type 'const char **' discards qualifiers in nested pointer types [-Wincompatible-pointer-types-discards-qualifiers]

I've got no ideas what that means or how to fix it, HELP ME PLEASE!!!!!!!

You have five pointers. Where do they point? Nowhere. You must allocate some space. — klutt
– klutt, Commented Nov 8, 2020 at 3:29
Have a look at the function malloc but make sure you check the return value for errors. You should do that for fgets too. — klutt
– klutt, Commented Nov 8, 2020 at 3:38

csavvy · Accepted Answer · 2020-11-08 04:30:34Z

2

First Understand what is char* array[5].

that is nothing but array of 5 pointers to char*, so you need memory for each pointer before you can use them.

char* array[5] = NULL;
int noe = sizeof(array) / sizeof (array[0]);

for(int a = 0; a < noe; a++) {
    if( !(array[a] = malloc(BUFFER_SIZE))
        printf("malloc failed at %d entry\n", a);
}

dont forget to free them when you are done using them

for(int a = 0; a < noe; a++) {
    if(array[a]) {
        free(array[a]);
        array[a] = NULL;
    }
}

edited Nov 8, 2020 at 4:30

answered Nov 8, 2020 at 3:46

csavvy

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3 Comments

Daniel Walker Over a year ago

You could also statically allocate some space like char array[5][100];.

csavvy Over a year ago

@DanielWalker, yes we could , since OP using pointers, so suggested that way.

csavvy Over a year ago

@klutt, you are right, i just wrote the array of 5 pointers as 5 pointers to 5 arrays, not sure if OP understands if i say array of pointers, i could correct though.

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