Bitwise shifts include left shift operations of the form shift-expression << additive-expression and right shift operations of the form shift-expression >> additive-expression. The integer promotions are performed on the operands, each of which has integer type. The type of the result is that of the promoted left operand. If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.
Non-Compliant Code Example (left shift, signed type)
The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has a signed type and nonnegative value, and E1 * 2 E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.
The following code can result in undefined behavior because there is no check to ensure that left and right operands have nonnegative values, and that the right operand is greater than or equal to the width of the promoted left operand.
int si1, si2, sresult; sresult = si1 << si2;
Compliant Solution (left shift, signed type)
This compliant solution eliminates the possibility of undefined behavior resulting from a left shift operation on signed and unsigned integers. Smaller sized integers are promoted according to the integer promotion rules [[INT02-A]].
int si1, si2, sresult;
if ( (si1 < 0) || (si2 < 0) || (si2 >= sizeof(int)*CHAR_BIT) || si1 > (INT_MAX >> si2) ) {
/* handle error condition */
}
else {
sresult = si1 << si2;
}
In C99, the CHAR_BIT macro defines the number of bits for smallest object that is not a bit-field (byte). A byte, therefore, contains CHAR_BIT bits.
Non-Compliant Code Example (left shift, unsigned type)
The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. According to C99, if E1 has an unsigned type, the value of the result is E1 * 2 E2, reduced modulo one more than the maximum value representable in the result type. Although C99 specifies modulo behavior for unsigned integers, unsigned integer overflow frequently results in unexpected values and resultant security vulnerabilities (see [[INT32-C]]). Consequently, unsigned overflow is generally non-compliant and E1 * 2 E2 must be representable in the result type. Modulo behavior is allowed if the conditions in the exception section are met.
The following code can result in undefined behavior because there is no check to ensure that the right operand is greater than or equal to the width of the promoted left operand.
unsigned int ui1, ui2, uresult; uresult = ui1 << ui2;
Compliant Solution (left shift, unsigned type)
This compliant solution eliminates the possibility of undefined behavior resulting from a left shift operation on unsigned integers. Example solutions are provided for the fully compliant case (unsigned overflow is prohibited) and the exceptional case (modulo behavior is allowed).
unsigned int ui1, ui2, uresult;
unsigned int mod1, mod2; /* modulo behavior is allowed on mod1 and mod2 by exception */
if ( (ui2 >= sizeof(unsigned int)*CHAR_BIT) || (ui1 > (UINT_MAX / (1 << ui2))) ) {
/* handle error condition */
}
else {
uresult = ui1 << ui2;
}
if (mod2 >= sizeof(unsigned int)*CHAR_BIT) {
/* handle error condition */
}
else {
uresult = mod1 << mod2; /* modulo behavior is allowed by exception */
}
Non-Compliant Code Example (right shift)
The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2 E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined and may be either an arithmetic (signed) shift as shown below or a logical (unsigned) shift.
This non-compliant code example fails to test if the right operand negative or is greater than or equal to the width of the promoted left operand, allowing undefined behavior.
int si1, si2, sresult; unsigned int ui1, ui2, uresult; sresult = si1 >> si2; uresult = ui1 >> ui2;
Compliant Solution (rightshift)
This compliant solution tests the suspect shift operation to guarantee there is no possibility of unsigned overflow.
int si1, si2, sresult;
unsigned int ui1, ui2, result;
if ( (si2 < 0) || (si2 >= sizeof(int)*CHAR_BIT) ) {
/* handle error condition */
}
else {
sresult = si1 >> si2;
}
if (ui2 >= sizeof(unsigned int)*CHAR_BIT) {
/* handle error condition */
}
else {
uresult = ui1 >> ui2;
}
As already noted, if the left operand has a signed type and a negative value, the resulting right shift could be either an arithmetic (signed) shift or a logical (unsigned) shift. For implementations in which an arithmetic shift is performed, and the sign bit can be propagated as the number is shifted.
int stringify;
char buf[sizeof("256")];
sprintf(buf, "%u", stringify >> 24);
If stringify has the value 0x80000000, stringify>> 24 evaluates to 0xFFFFFF80 and the subsequent call to sprintf() results in a buffer overflow.
For bit extraction, one remediation is to use the idiom ((number >> 24) & 0xff).
Exceptions
Unsigned integers can be allowed to exhibit modulo behavior if and only if
- the variable declaration is clearly commented as supporting modulo behavior
- each operation on that integer is also clearly commented as supporting modulo behavior
If the integer exhibiting modulo behavior contributes to the value of an integer not marked as exhibiting modulo behavior, the resulting integer must obey this rule.
References
A test program for this rule is available.
[[Dowd 06]] Chapter 6, "C Language Issues"
[[ISO/IEC 9899-1999]] Section 6.5.7, "Bitwise shift operators"
[[Seacord 05]] Chapter 5, "Integers"
[[Viega 05]] Section 5.2.7, "Integer overflow"
[[ISO/IEC 03]] Section 6.5.7, "Bitwise shift operators"