Problems with arrays of pointers

Question

I'm having an entry level C pointer problem... Let's say I have two strings and I want to print them. What am I misunderstanding in the code below?

void print_array(char **array[]) {
    int i = 0;
    while((*array)[i] != NULL) {
        printf("%s\n", (*array)[i++]);
    }
    return;
}

int main(int argc, char** argv) {
    char str1[] = "hello";
    char str2[] = "world";
    char **array = {str1, str2};

    print_array(&array);
    return (EXIT_SUCCESS);
}

For my code to work I need to access the array like it is in print_array

Sorry I should have said the data in array isn't correct. And at runtime it says Seg. Fault — Eric Fossum
– Eric Fossum, Commented Sep 12, 2014 at 3:41
I assume you might want something like print_array(char** array) and char** array = {str1, str2, NULL}; print_array(array); — SleuthEye
– SleuthEye, Commented Sep 12, 2014 at 3:43
You know, both your return-statements are superfluous... Also, there's no percentage in normalizing a value in a boolean context. — Deduplicator
– Deduplicator, Commented Sep 12, 2014 at 3:49

Deduplicator · Accepted Answer · 2014-09-12 04:47:57Z

Another take on it (Much more refactoring).

Bugfixes:

Corrected confusion what should be dereferenced and how in print_array (Don't aspire to 3-star-programming, unless you must)
Added the missing sentinel-0 in to the array passed from main to print_array

Other changes:

Removed superfluous return-statements (main has an implicit return 0; at the end)
Removed superfluous checks for unequality to 0 / NULL
Removed one level of indirection from print_array
Use of const in print_array where appropriate
Eliminated counter-variable in print_array
Used a constant compound-literal in main (Needs C99)

#include <stdio.h>

void print_array(const char *const array[]) {
    while(*array)
        printf("%s\n", *array++);
}

int main() {
    print_array((const char*[]){"hello", "world", 0});
}

See here on coliru

Undoing the cleanup steps changing the signature of print_array:

#include <stdio.h>

void print_array(char **array[]) {
    for(char** p = *array; *p; p++)
        printf("%s\n", *p);
}

int main() {
    print_array(&(char**){(char*[]){"hello", "world", 0}});
}

Live on coliru

John Kugelman · Accepted Answer · 2014-09-12 04:00:37Z

There's one too many *'s in print_array. It ought to be this:

void print_array(char *array[]) {
    int i = 0;
    while(array[i] != NULL) {
        printf("%s\n", array[i++]);
    }
}

That makes calling it straightforward. Change char **array to char *array[]. And don't forget to include a NULL entry at the end of the array.

int main(int argc, char** argv) {
    char str1[] = "hello";
    char str2[] = "world";
    char *array[] = {str1, str2, NULL};

    print_array(array);
    return EXIT_SUCCESS;
}

On the other hand, if you're not supposed to modify print_array, then that's annoying. You'll need another variable to hold array's address temporarily.

char *array[] = {str1, str2, NULL};
char **p = array;

print_array(&p);

That's why I say there's one too many *'s in print_array.

Aniket Inge · Accepted Answer · 2014-09-12 03:46:04Z

1

since you're dealing with arrays, please pass in the number of array members to print_array. The code above is UB.

To fix your code:

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

int main(int argc, char** argv) {
    char str1[] = "hello";
    char str2[] = "world";
    char *arr[] = {str1, str2};

    print_array(arr, 2);
    return (0);
}

answered Sep 12, 2014 at 3:46

Aniket Inge

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2 Comments

M.M Over a year ago

Being more specific, the original code is UB because it iterates off the end of the array. OP perhaps expected that reading off the end of an array would give 0 or equivalent; of course it doesn't.

M.M Over a year ago

the extra level of indirection was not needed in order to just read the array, but it wasn't a mistake either

Nipun Talukdar · Accepted Answer · 2014-09-12 03:49:27Z

0

Another way:

#include <stdio.h>
void print_array(char *array[]) {
   int i = 0;
   while (array[i] != 0) {
   printf("%s\n", array[i++]);
}
return;
}

int main(int argc, char** argv) {
    char str1[] = "hello";
    char str2[] = "world";
    char *array[] = {str1, str2, 0};

    print_array(array);
    return 0;
}

answered Sep 12, 2014 at 3:49

Nipun Talukdar

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7 Comments

John Kugelman Over a year ago

@Aniket That's silly. Comparing against 0 is a perfectly fine way of checking for null pointers.

Aniket Inge Over a year ago

@Deduplicator it IS 0, but comparing it directly AS 0.. seems odd to me. Eyes hurt

Nipun Talukdar Over a year ago

@Aniket I think you know NULL and 0 means the same thing in C/C++. Eg. if you have a char pointer or pointer something else ptr, you can do "if (ptr != 0)" or "if (ptr != NULL)" .

M.M Over a year ago

@Aniket there is implicit conversion from integer constant expression with value 0 to null pointer

Aniket Inge Over a year ago

@MattMcNabb I just think NULL is more readable. Just personal preference I guess.

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