How to call a function with an array of strings

Question

I've made a function, but I'm struggling with calling it.

This is the prototype of the function:

char *test(int argc, char **argv);

I've tried calling it this way but it doesn't work :

int main()
{

    char tab[3][3] ={
     "Yo",
     "Hi"};

    test(2, tab);


  return (0);
}

Hello and welcome. "it doesn't work" is not a useful problem statement. Please [edit] your question and be a bit more precise. — Federico klez Culloca, Nov 14 '19 at 15:33
You haven't provided any details of the `test` function, just the prototype. So, what does "It doesn't work"` even mean here? — h0r53, Nov 14 '19 at 15:33
have you tried following something similar to this question? https://stackoverflow.com/questions/8032225/c-passing-char-array-to-function — hurnhu, Nov 14 '19 at 15:34
test expects char**, but he is passing a char*, shouldnt it be &tab — Yamahari, Nov 14 '19 at 15:34
@Yamahari wouldn't work. You cannot pass a `char (*)[3]` to a `char**` parameter. — DarkAtom, Nov 14 '19 at 15:36

score 0 · Answer 1 · answered Nov 14 '19 at 15:44

0

For me this works:

char* test(int index, char** char2Darray)
{
    return char2Darray[index];
}

int main()
{

    char* tab[2] ={
     "Yo",
     "Hi"};

    test(1, tab);


  return (0);
}

I think there are two problems in your code :

the tab what you provided has only two item
your tab is a pointer which pointing to char[3] types and not char*

answered Nov 14 '19 at 15:44

Istvan Nagy

33
1
4

can you explain to me "char *tab[2]" is this a pointer or an array ? – Exdeen Nov 14 '19 at 16:14
Here tab is a char pointer array with 2 element. Another words tab is an array (2 elements) which can store char*. – Istvan Nagy Nov 15 '19 at 07:39

score 0 · Answer 2 · answered Nov 14 '19 at 15:47

When you pass the 2D array

char tab[3][3]

to the function test(), it decays to a pointer of type:

char (*)[3]

which is a pointer to an array of 3 char elements.

This type is not compatible with char** which is a pointer-to-pointer to char. This is the source of your problem. You need to change the function test() to take char (*argv)[3] instead of char **argv.

score 0 · Answer 3 · answered Nov 14 '19 at 15:55

Looking at function prototype: char *test(int argc, char **argv);

char **argv is pointer of pointers and the variable char tab[3][3] is array of arrays so they are incompatible.

You can change function prototype to: char *test(int argc, char argv[][SOMECONSTANT]); or char *test(int argc, char (*argv)[3]);

score -1 · Accepted Answer · answered Nov 14 '19 at 15:50

-1

This way

char * test(int argc,const char **argv);

int main()
 {
  const char * tab[3]=
   {
    "Yo",
    "Hi",
    0
   };

  test(2,tab);

  return 0;
 }

or

char * test(int argc,char **argv);

int main()
 {
  char arg1[]="Yo";
  char arg2[]="Hi";

  char * tab[3]=
   {
    arg1,
    arg2,
    0
   };

  test(2,tab);

  return 0;
 }

answered Nov 14 '19 at 15:50

Sergey Strukov

373
3
9

sorry I'm confused .. "char *tab[3]" what does that mean ? is it an array or a pointer – Exdeen Nov 14 '19 at 16:11
@Exdeen `char *tab[3]` declares `tab` as an array of 3 `char *`. I.e. it is an array of pointers. – Ian Abbott Nov 14 '19 at 16:40

How to call a function with an array of strings

4 Answers4