Difference between usage of pointer to char and char array

Question

In the following code, fizz prints correctly, but buzz does not. What is wrong with using a char array here or in what way am I using it wrong?

#include <stdio.h>

int main() {

    int i;

    char *fizz = "fizz";
    char buzz[4] = "buzz";

    for(i = 1; i <= 100; i++) {
            if(!((i %3 == 0) || (i%5 == 0)))
                    printf("%d", i);
            if(i%3 == 0)
                    printf("%s", fizz);
            if(i%5 == 0)
                    printf("%s", buzz);

            printf("\n");
    }
}

this line: char buzz[4] = "buzz"; will overflow the buzz[] buffer because strings contain a nul termination byte, so "buzz" is actually 5 bytes long. writing past the end of the buffer is undefined behaviour and can lead to a seg fault event — user3629249
– user3629249, Commented Dec 18, 2014 at 3:16

Alexis King · Accepted Answer · 2014-12-18 02:45:16Z

6

Neither string is actually four bytes long. C strings are NUL-terminated, so they're always their length in characters plus one for the '\0' character. Therefore, you should change the declaration of buzz as follows:

char buzz[5] = "buzz";

However, also note that explicitly declaring the variable's length when initializing with a string literal is not required. You can just leave out the number, and it will work fine.

char buzz[] = "buzz";

answered Dec 18, 2014 at 2:45

Alexis King

44.1k17 gold badges144 silver badges218 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

M.M · Accepted Answer · 2014-12-18 02:46:24Z

3

The %s specifier prints a string. A string is a series of characters followed by a null-terminator. To put it another way, you tell printf where your characters start, and printf keeps printing until it encounters a null terminator.

The "buzz" expression is a null-terminated string, however you shoe-horned it into char buzz[4] so the null terminator didn't fit. The result is that the buzz array does not contain a null terminator, and so printf reads off the end of the string, causing undefined behaviour. To fix this write:

char buzz[] = "buzz";

which will include a null terminator and allocate the right amount of space.

answered Dec 18, 2014 at 2:46

M.M

143k26 gold badges223 silver badges400 bronze badges

Comments

sharon · Accepted Answer · 2014-12-18 06:23:38Z

3

You have declared the size of the buzz array as 4 and including \0 the size of the string is 5. so you have to change the size of the buzz array as 5 or don't mention the size.

char buzz[5] = "buzz";

because buzz is not a pointer variable, if the buzz is a pointer variable it will allocate the memory according to the string, so there is no need to specify the size. As buzz is a array you have to mention the size correctly. And while printing we use %s , %s will print the string till the \0 character occurs. In the buzz string there is no \0 character, this is the problem.

edited Dec 18, 2014 at 6:23

answered Dec 18, 2014 at 4:31

sharon

7326 silver badges15 bronze badges

Comments

user3288244 · Accepted Answer · 2014-12-18 03:04:34Z

2

buzz contains 4 character, it has no space reserved for the null character, therefore it cannot be printed correctly with %s. %s specifies that a null terminated string to be printed and it cannot work correctly if the string is not null terminated.

Change your initialization to char buzz[] = "buzz"

answered Dec 18, 2014 at 3:04

user3288244

1215 bronze badges

Collectives™ on Stack Overflow

Difference between usage of pointer to char and char array

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related