Conversions can occur explicitly as the result of a cast or implicitly as required by an operation. Although conversions are generally required for the correct execution of a program, they can also lead to lost or misinterpreted data. Conversion of an operand value to a compatible type causes no change to the value or the representation.
The C integer conversion rules define how C compilers handle conversions. These rules include integer promotions, integer conversion rank, and the usual arithmetic conversions. The intent of the rules is to ensure that the conversions result in the same numerical values and that these values minimize surprises in the rest of the computation. Prestandard C usually preferred to preserve signedness of the type.
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 shows the application 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 int
values are added and the sum is truncated to fit into the char
type. Integer promotions are performed to avoid arithmetic errors resulting from the overflow of intermediate values:
signed char cresult, c1, c2, c3; c1 = 100; c2 = 3; c3 = 4; cresult = c1 * c2 / c3;
In this example, the value of c1
is multiplied by c2
. The product of these values is then divided by the value of c3
(according to operator precedence rules). Assuming that signed char
is represented as an 8-bit value, the product of c1
and c2
(300) cannot be represented. Because of integer promotions, however, c1
, c2
, and c3
are each converted to int
, and the overall expression is successfully evaluated. The resulting value is truncated and stored in cresult
. Because the final result (75) is in the range of the signed char
type, the conversion from int
back to signed char
does not result in lost data.
Integer Conversion Rank
Every integer type has an integer conversion rank that determines how conversions are performed. The ranking is based on the concept that each integer type contains at least as many bits as the types ranked below it. The following rules for determining integer conversion rank are defined in the C Standard, Section 6.3.1.1 [ISO/IEC 9899:2011]:
- No two signed integer types shall have the same rank, even if they have the same representation.
- The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.
- The rank of
long long int
shall be greater than the rank oflong int
, which shall be greater than the rank ofint
, which shall be greater than the rank ofshort int
, which shall be greater than the rank ofsigned char
. - The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.
- The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.
- The rank of
char
shall equal the rank ofsigned char
andunsigned char
. - The rank of
_Bool
shall be less than the rank of all other standard integer types. - The rank of any enumerated type shall equal the rank of the compatible integer type.
- 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
, andT3
, ifT1
has greater rank thanT2
andT2
has greater rank thanT3
, thenT1
has greater rank thanT3
.
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
The usual arithmetic conversions are rules that provide a mechanism to yield a common type when both operands of a binary operator are balanced to a common type or the second and third operands of the conditional operator ( ? :
) are balanced to a common type. Conversions involve two operands of different types, and one or both operands may be converted. 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 code is compiled using an implementation with 8-bit char
, 32-bit int
, and 64-bit long long
:
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 being 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 to 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
, the addition results in an overflow in this example. The resulting value is then zero-extended to fit into the 64-bit storage allocated by sll
.
Noncompliant Code Example (Comparison)
The programmer must be careful when performing operations on mixed types. This noncompliant code example shows an idiosyncrasy of integer promotions.
int si = -1; unsigned int ui = 1; printf("%d\n", si < ui);
In this example, the comparison operator operates on a signed int
and an unsigned int
. By the conversion rules, si
is converted to an unsigned int
. Because −1 cannot be represented as an unsigned int
value, the −1 is converted to UINT_MAX
in accordance with C, Section 6.3.1.3, paragraph 2 [ISO/IEC 9899:2011]:
Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.
Consequently, the program prints 0 because UINT_MAX
is not less than 1.
Compliant Solution
The noncompliant code example can be modified to produce the intuitive result by forcing the comparison to be performed using signed int
values.
int si = -1; unsigned ui = 1; printf("%d\n", si < (int)ui);
This program prints 1 as expected. Note that (int)ui
is correct in this case only because the value of ui
is known to be representable as an int
. If it were not known, the compliant solution would need to be written as
int si = /* some signed value */; unsigned ui = /* some unsigned value */; printf("%d\n", (si < 0 || (unsigned)si < ui));
Noncompliant Code Example
This noncompliant code example demonstrates how performing bitwise operations on integer types smaller than int
may have unexpected results.
uint8_t port = 0x5a; uint8_t result_8 = ( ~port ) >> 4;
In this example, a bitwise complement of port
is first computed and then shifted 4 bits to the right. If both of these operations are performed on an 8-bit unsigned integer, then result_8
will have the value 0x0a
. However, port
is first promoted to a signed int
, with the following results (on a typical architecture where type int
is 32 bits wide):
Expression | Type | Value | Notes |
---|---|---|---|
|
|
|
|
|
|
|
|
|
|
| Whether or not value is negative is implementation-defined. |
|
|
|
|
Compliant Solution
In this compliant solution, the bitwise complement of port
is converted back to 8 bits. Consequently, result_8
is assigned the expected value of 0x0aU
.
uint8_t port = 0x5a; uint8_t result_8 = (uint8_t) (~port) >> 4;
Risk Assessment
Misunderstanding integer conversion rules can lead to errors, which in turn can lead to exploitable vulnerabilities. The major risks occur when narrowing the type (which requires a specific cast or assignment), converting from unsigned to signed, or converting from negative to unsigned.
Recommendation | Severity | Likelihood | Remediation Cost | Priority | Level |
---|---|---|---|---|---|
INT02-C | medium | probable | medium | P8 | L2 |
Automated Detection
Tool | Version | Checker | Description |
---|---|---|---|
9.7.1 | 52 S | Fully implemented | |
PRQA QA-C | Unable to render {include} The included page could not be found. | 0290 | Fully implemented |
Related Vulnerabilities
This vulnerability in Adobe Flash arises because Flash passes a signed integer to calloc()
. An attacker has control over this integer and can send negative numbers. Because calloc()
takes size_t
, which is unsigned, the negative number is converted to a very large number, which is generally too big to allocate, and as a result, calloc()
returns NULL
, causing the vulnerability to exist.
Search for vulnerabilities resulting from the violation of this rule on the CERT website.
Related Guidelines
CERT C++ Secure Coding Standard | INT02-CPP. Understand integer conversion rules |
ISO/IEC TR 24772:2013 | Numeric Conversion Errors [FLC] |
MISRA-C | Rule 10.1 Rule 10.3 |
MITRE CWE | CWE-192, Integer coercion error CWE-197, Numeric truncation error |
Bibliography
[Dowd 2006] | Chapter 6, "C Language Issues" ("Type Conversions," pp. 223–270) |
[ISO/IEC 9899:2011] | Section 6.3.1.1, "Boolean, Characters, and Integers" |
[Seacord 2013] | Chapter 5, "Integer Security" |