
Type conversions occur explicitly as the result of a cast or implicitly as required by an operation. While conversions are generally required for the correct execution of a program, they can also lead to lost or misinterpreted data.
The C99 standard rules define how C compilers handle conversions. These rules include integer promotions, integer conversion rank, and the usual arithmetic conversions.
Integer Promotions
Integer types smaller than int
are promoted when an operation is performed on them. If all values of the original type can be represented as an int
, the value of the smaller type is converted to an int
; otherwise, it is converted to an unsigned int
.
Integer promotions are applied as part of the usual arithmetic conversions to certain argument expressions, operands of the unary +, --, and ~ operators, and operands of the shift operators. The following code fragment illustrates the use of integer promotions:
char c1, c2; c1 = c1 + c2;
Integer promotions require the promotion of each variable (c1
and c2
) to int
size. The two ints
are added and the sum truncated to fit into the char
type.
Integer promotions are performed to avoid arithmetic errors resulting from the overflow of intermediate values. For example:
char cresult, c1, c2, c3; c1 = 100; c2 = 90; c3 = --120; cresult = c1 + c2 + c3;
In this example, the value of c1
is added to the value of c2
. The sum of these values is then added to the value of c3
(according to operator precedence rules). The addition of c1
and c2
would result in an overflow of the signed char
type because the result of the operation exceeds the maximum size of signed char
. Because of integer promotions, however, c1
, c2
, and c3
are each converted to integers and the overall expression is successfully evaluated. The resulting value is then truncated and stored in cresult
. Because the result is in the range of the signed char
type, the truncation does not result in lost data.
Integer Conversion Rank
Every integer type has an integer conversion rank that determines how conversions are performed. The following rules for determining integer conversion rank are defined in C99.
- No two different signed integer types have the same rank, even if they have the same representation.
- The rank of a signed integer type is greater than the rank of any signed integer type with less precision.
- The rank of long long int is greater than the rank of long int, which is greater than the rank of int, which is greater than the rank of short int, which is greater than the rank of signed char.
- The rank of any unsigned integer type is equal to the rank of the corresponding signed integer type, if any.
- The rank of any standard integer type is greater than the rank of any extended integer type with the same width.
- The rank of char is equal to the rank of signed char and unsigned char.
- The rank of any extended signed integer type relative to another extended signed integer type with the same precision is implementation defined but still subject to the other rules for determining the integer conversion rank.
- For all integer types T1, T2, and T3, if T1 has greater rank than T2 and T2 has greater rank than T3, then T1 has greater rank than T3.
The integer conversion rank is used in the usual arithmetic conversions to determine what conversions need to take place to support an operation on mixed integer types.
Usual Arithmetic Conversions
Many operators that accept arithmetic operands perform conversions using the usual arithmetic conversions. After integer promotions are performed on both operands, the following rules are applied to the promoted operands.
- If both operands have the same type, no further conversion is needed.
- If both operands are of the same integer type (signed or unsigned), the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
- If the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type is converted to the type of the operand with unsigned integer type.
- If the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type is converted to the type of the operand with signed integer type.
- Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type. Specific operations can add to or modify the semantics of the usual arithmetic operations.
Example
In the following example, assume the following code is compiled and executed on IA-32:
signed char sc = SCHAR_MAX; unsigned char uc = UCHAR_MAX; signed long long sll = sc + uc;
Both the signed char sc
and the unsigned char uc
are subject to integer promotions in this example. Because all values of the original types can be represented as int
, both values are automatically converted to int
as part of the integer promotions. Further conversions are possible, if the types of these variables are not equivalent as a result of the "usual arithmetic conversions". The actual addition operation in this case takes place between the two 32-bit int
values. This operation is not influenced by the resulting value is stored in a signed long long integer. The 32-bit value resulting from the addition is simply sign-extended to 64-bits after the addition operation has concluded.
Assuming that the precision of signed char
is 7 bits, and the precision of unsigned char
is 8 bits, this operation is perfectly safe. However, if the compiler represents the signed char
and unsigned char
types using 31 and 32 bit precision (respectively), the variable uc
would need be converted to unsigned int
instead of signed int
. As a result of the usual arithmetic conversions, the signed int
is converted to unsigned and the addition takes place between the two unsigned int
values. Also, because uc
is equal to UCHAR_MAX
which is equal to UINT_MAX
in this example, the addition will result in an overflow. The resulting value is then zero-extended to fit into the 64-bit storage allocated by sll
.
Non-compliant Code Example 1
In the following non-compliant code example, cBlocks is multiplied by 16 and the result is stored in the unsigned long long int alloc. The result of this multiplication can overflow because it is a 32 bit operation and the resulting value stored in alloc invalid.
void* AllocBlocks(size_t cBlocks) { if (cBlocks == 0) return NULL; unsigned long long alloc = cBlocks * 16; return (alloc < UINT_MAX) ? malloc(cBlocks * 16) : NULL; }
Compliant Solution 1
On architectures where unsigned long long int is guaranteed to have 2x the number of bits as size_tupcast the variable used in the multiplication to a 64-bit value. This ensures the multiplication operation is performed
void* AllocBlocks(size_t cBlocks) { if (cBlocks == 0) return NULL; unsigned long long alloc = (unsigned long long)cBlocks*16; return (alloc < UINT_MAX) ? malloc(cBlocks * 16) : NULL; }
The assumption concerning the relationship of unsigned long long int and size_t must be document in the header for each file that depends upon this assumption for correct execution.
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.
Consequences
Improper range checking can lead to buffer overflows and the execution of arbitary code by an attacker.
References
- Seacord 05 Chapter 5 Integers
- ISO/IEC 9899-1999 6.3 Conversions