CWE Rule 463

This issue occurs when a string does not have enough space to terminate with a null character '\0'.

This defect applies only for projects in C.

A buffer overflow can occur if you copy a string to an array without assuming the implicit null terminator.

If you initialize a character array with a literal, avoid specifying the array bounds.

char three[]  = "THREE";

If the issue occurs after initialization, you might have to increase the size of the array by one to account for the null terminator.

In certain circumstances, you might want to initialize the character array with a sequence of characters instead of a string. In this situation, add comments to your result or code to avoid another review. See:

void countdown(int i)
{
    static char one[5]   = "ONE";
    static char two[5]   = "TWO";
    static char three[5] = "THREE"; //Noncompliant
}

The character array three has a size of 5 and 5 characters 'T', 'H', 'R', 'E', and 'E'. There is no room for the null character at the end because three is only five bytes large. This issue can cause a compilation error in C++.

One possible correction is to change the array size to allow for the five characters plus a null character.

void countdown(int i)
{
    static char one[5]   = "ONE";
    static char two[5]   = "TWO";
    static char three[6] = "THREE";
}

One possible correction is to initialize the string by leaving the array size blank. This initialization method allocates enough memory for the five characters and a terminating-null character.

void countdown(int i)
{
    static char one[5]   = "ONE";
    static char two[5]   = "TWO";
    static char three[]  = "THREE";
}