Missing null in string array

String does not terminate with null character

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Description

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

This defect applies only for projects in C.

Risk

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

Fix

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

char three[]  = "THREE";

The compiler automatically allocates space for a null terminator. In the preceding example, the compiler allocates sufficient space for five characters and a null terminator.

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:

Address Results in Polyspace User Interface Through Bug Fixes or Justifications if you review results in the Polyspace user interface.
Address Results in Polyspace Access Through Bug Fixes or Justifications (Polyspace Access) if you review results in a web browser.
Annotate Code and Hide Known or Acceptable Results if you review results in an IDE.

Examples

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Array size is too small

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

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.

Correction — Increase Array Size

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";
}

Correction — Change Initialization Method

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";
}

Result Information

Group: Programming

Language: C

Default: On for handwritten code, off for generated code

Command-Line Syntax: MISSING_NULL_CHAR

Impact: Low

Version History

Introduced in R2013b