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:

Answer 1

Since Buffer is not initialized, it starts with all 9 garbage values. From the observed output, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th and 2 immediate next memory location(outside the array) elements are clearly 'T', 'T', 'W', '\0', '\0', '=', '\0', '\0', '\0'.

Strings consume all the characters up until they see NULL character. That is why, in every iteration, as the array elements are assigned one by one, buffer is printed up to the part where a garbage NULL is present.

That is to say, string has an undefined behavior if the character array doesn't end with '\0'. You can avoid this by having an extra space for '\0' at the end of the buffer.

Answer 2

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 3

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);

}

Answer 4

If you are programming in C or C++, you have to remember that: 1) the strings are finished with a \0 character. 2) C does not have boundary check at strings, they are just character arrays.

Answer 5

You might also want to look into using a stringstream.

Answer 6

In addition to the previous comments about zero termination, you also have to accept responsibility for not overflowing your own buffer. It doesn't stop at 8 characters because your code is not stopping! You need something like the following (piggy-backing onto Jeremy's suggestion):

#define DATA_LENGTH 8
#define BUFFER_LENGTH (DATA_LENGTH + 1)

char Buffer[BUFFER_LENGTH]; //holds the byte stream
int charPos=0;  //index to next character position to fill

while (charPos <= DATA_LENGTH  ) { //user input event has occured
    Buffer[i] = charInput;

    Buffer[i+1] = '\0';

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

    i++; 

}

In other words, make sure to stop accepting data when the maximum length has been reached, regardless of what the environment tries to push at you.