string array with garbage character at end

Question

I have a char array buffer that I am using to store characters that the user will input one by one. My code below works but has a few glitches that I can't figure out:

1. when I execute a printf to see what's in Buffer, it does fill up but I get garbage characters at the end
2. it won't stop at 8 characters despite being declared as char Buffer[8];

Can somebody please explain to me what is going on and perhaps how I could fix this? Thanks.

char Buffer[8]; //holds the byte stream
int i=0;

if (/* user input event has occurred */) 
{
        Buffer[i] = charInput;
        i++;

        // Display a response to input
        printf("Buffer is %s!\n", Buffer);

}

Output:

tagBuffer is 1┬┬w!
tagBuffer is 12┬w!
tagBuffer is 123w!
tagBuffer is 1234!
tagBuffer is 12345!
tagBuffer is 123456=!
tagBuffer is 1234567!
tagBuffer is 12345678!

tagBuffer is 123456789!

Answer 1

You have to end the string with a \0 character. That's why they are called zero terminated strings.

It is also wise to allocate 1 extra char to hold the \0.

Answer 2

The only thing you are passing to the printf() function is a pointer to the first character of your string. printf() has no way of knowing the size of your array. (It doesn't even know if it's an actual array, since a pointer is just a memory address.)

printf() and all the standard c string functions assume that there is a 0 at the end of your string. printf() for example will keep printing characters in memory, starting at the char that you pass to the function, until it hits a 0.

Therefore you should change your code to something like this:

char Buffer[9]; //holds the byte stream
int i=0;

if( //user input event has occured ) 
{
        Buffer[i] = charInput;
        i++;

        Buffer[i] = 0; // You can also assign the char '\0' to it to get the same result.

        // Display a response to input
        printf("Buffer is %s!\n", Buffer);

}