CODEWITHSUNDEEP
carrayspointersgccchar
Jack Walter
Jack Walter

Reputation: 11

Difference between usage of pointer to char and char array

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");
    }
}

Upvotes: 0

Views: 91

Answers (4)

sharon
sharon

Reputation: 734

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.

Upvotes: 3

user3288244
user3288244

Reputation: 121

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"

Upvotes: 2

M.M
M.M

Reputation: 141574

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.

Upvotes: 3

Alexis King
Alexis King

Reputation: 43842

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";

Upvotes: 6

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