EXP05-A. Do not cast away a const qualification

Do not attempt to modify a variable, array, or pointer declared as const. The specification of const in variables and parameters implies to a maintainer or caller that, despite knowing some memory location, the code will not modify its content. Although C allows you to remove the specifier using typecasts, doing so violates the implication of the specifier.

Non-Compliant Code Example

In this example, the function f is passed a const char pointer. It then typecasts the const specification away, and proceeds to modify the contents.

void f(const char *str, int slen) {
   char *p = (char*)str;
   int i;
   for (i = 0; i < slen && str[i]; i++) {
      if (str[i] != ' ') *p++ = str[i];
   }
}

Compliant Solution

In this compliant solution the function f is passed a non-const char pointer. The calling function must ensure that the null-terminated byte string passed to the function is not const by making a copy of the string or by other means.

void f(char *str, int slen) {
   char *p = str;
   int i;
   for (i = 0; i < slen && str[i]; i++) {
      if (str[i] != ' ') *p++ = str[i];
   }
}

Non-Compliant Code Example

In this example, a const int array vals is declared and its content modified by memset() with the function, clearing the contents of the vals array.

const int vals[] = {3, 4, 5};
memset((int *)vals, 0, sizeof(vals));

Compliant Solution

If the intention is to allow the array values to be modified, do not declare the array as const.

int vals[] = {3, 4, 5};
memset((int *)vals, 0, sizeof(vals));

Otherwise, do not attempt to modify the contents of the array.

Priority: P6 Level: L2

If the object really is constant, the compiler may have put it in ROM or write-protected memory. Trying to modify such an object may lead to a program crash. This could allow an attacker to mount a denial of service attack.

References

• ISO/IEC 9899-1999 Section 6.7.3 Type qualifiers