Print out elements of an array of strings in c

Question

I've created a function that takes a pointer to the first element in a C-String array. This is the array, and how I went about making it:

char element1[60] = "ubunz";
char element2[60] = "uasdasffdnz";
char* array[10] = {element1,element2};

Then I created a pointer to the first element in the array:

char *pointertoarray = &array[0];

Then I passed the pointer to the function I made:

void printoutarray(char *pointertoarray){
  int i = 0;
  while (i < 2){
    printf("\n Element is: %s \n", *pointertoarray);
    pointertoarray = pointertoarray + 1;
    i = i+1;
  }
}

When I run the program the array never prints out.

I've done this program before in C++ but I used the STL string type, and made a pointer to an array of strings. I think my problem lies with the way I'm going about making an array of strings in C, and making a pointer to it.

Ask yourself what type `*pointertoarray` is and what type `%s` expects. — chris, Jan 13 '13 at 18:45
If you enable warning messages in your compiler, it will tell you that there's a problem here... — Oliver Charlesworth, Jan 13 '13 at 18:46

score 10 · Accepted Answer · edited Nov 19 '18 at 20:11

When printf() is called with %s it will more or less do the following

for(i = 0; pointer[i] != '\0'; i++)
{
   printf("%c", pointer[i]);
}

What you want do is something like

for(i = 0; pointer[i] != '\0'; i++)
{
   printf("\n Element is %c", pointer[i]);
}

When you are referencing specific elements in an array you must use a pointer AND an index i.e. pointer[0] in this case 0 is the index and 'pointer' is the pointer. The above for loops will move through the entire array one index at a time because 'i' is the index and 'i' increases at the end of every rotation of loop and the loop will continue to rotate until it reaches the terminating NULL character in the array.

So you might want to try something along the lines of this.

void printoutarray(char *pointertoarray)
{
  for(i = 0; i <= 2; i++)
  {
    printf("\n Element is: %s \n", pointertoarray[i]);
  }
}

score 7 · Answer 2 · answered Jan 13 '13 at 18:46

7

Does this:

char *pointertoarray = &array[0];

compile without warnings on your compiler? If so, your compiler is broken. Please read the errors and warnings it prints out.

&array[0] is a char**, not a char*

answered Jan 13 '13 at 18:46

gustaf r

1,224
9
14

Thank you. I'm still fairly new to C-Strings. – turnt Jan 13 '13 at 18:49

score 2 · Answer 3 · answered Jan 13 '13 at 18:45

2

Remove the asterisk

printf("\n Element is: %s \n", pointertoarray);

answered Jan 13 '13 at 18:45

hmatar

2,437
2
17
27

This works to, but I'd rather have the pointertoarray point back to an element in the array. – turnt Jan 13 '13 at 18:54

Print out elements of an array of strings in c

3 Answers3