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 8bit 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, subclause 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 implementationdefined 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.
Example
In the following example, assume the code is compiled using an implementation with 8bit char
, 32bit int
, and 64bit 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 32bit int
values. This operation is not influenced by the resulting value being stored in a signed long long
integer. The 32bit value resulting from the addition is simply signextended 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 32bit 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 zeroextended to fit into the 64bit 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 the C Standard, subclause 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 8bit 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 implementationdefined 



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 

INT02C  Medium  Probable  Medium  P8  L2 
Automated Detection
Tool  Version  Checker  Description 

CodeSonar  5.1p0  ALLOC.SIZE.TRUNC LANG.CAST.COERCE LANG.CAST.VALUE MISC.MEM.SIZE.TRUNC  Truncation of Allocation Size Coercion Alters Value Cast Alters Value Truncation of Size 
1.2  CC2.INT02  Fully implemented  
Klocwork  2018  MISRA.CAST.INT MISRA.CAST.UNSIGNED_BITS MISRA.CONV.INT.SIGN MISRA.CVALUE.IMPL.CAST MISRA.UMINUS.UNSIGNED PRECISION.LOSS  
LDRA tool suite  9.7.1  52 S, 93 S, 96 S, 101 S, 107 S, 332 S, 334 S, 433 S, 434 S, 446 S, 452 S, 457 S, 458 S  Fully implemented 
Parasoft C/C++test  10.4.2  CERT_CINT02a  Implicit conversions from wider to narrower integral type which may result in a loss of information shall not be used 
Polyspace Bug Finder  R2019b  Checks for sign change integer conversion overflow (rec. fully supported)  
PRQA QAC  9.5  1250, 1251, 1252, 1253, 1256, 1257, 1260, 1263, 1266, 1274, 1290, 1291, 1292, 1293, 1294, 1295, 1296, 1297, 1298, 1299, 1800, 1802, 1803, 1804, 1810, 1811, 1812, 1813, 1820, 1821, 1822, 1823, 1824, 1830, 1831, 1832, 1833, 1834, 1840, 1841, 1842, 1843, 1844, 1850, 1851, 1852, 1853, 1854, 1860, 1861, 1862, 1863, 1864, 1880, 1881, 1882, 2100, 2101, 2102, 2103, 2104, 2105, 2106, 2107, 2109, 2110, 2111, 2112, 2113, 2114, 2115, 2116, 2117, 2118, 2119, 2120, 2122, 2124, 2130, 2132, 2134, 4401, 4402, 4403, 4404, 4405, 4410, 4412, 4413, 4414, 4415, 4420, 4421, 4422, 4423, 4424, 4425, 4430, 4431, 4432, 4434, 4435, 4436, 4437, 4440, 4441, 4442, 4443, 4445, 4446, 4447, 4460, 4461, 4463, 4464, 4470, 4471, 4480, 4481  Fully implemented 
PVSStudio  6.23  V555, V605, V673 
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
SEI CERT C++ Coding Standard  VOID INT02CPP. Understand integer conversion rules 
ISO/IEC TR 24772:2013  Numeric Conversion Errors [FLC] 
MISRA C:2012  Rule 10.1 (required) Rule 10.3 (required) Rule 10.4 (required) Rule 10.6 (required) Rule 10.7 (required) Rule 10.8 (required) 
MITRE CWE  CWE192, Integer coercion error CWE197, Numeric truncation error 
Bibliography
[Dowd 2006]  Chapter 6, "C Language Issues" ("Type Conversions," pp. 223–270) 
[Seacord 2013]  Chapter 5, "Integer Security" 
13 Comments
Masaki Kubo
In NCCE, why do you need to say "unsigned int is treated modularly" since unsigend in ui = 1 and there's no possibility of wraparound?
David Svoboda
Well, because there IS possibility of wraparound; it's just not mandated by the standard. On many platforms, including x86, signed ints do wrap. But the standard leaves this as undefined behavior, allowing other platforms to treat signed int overflow differently.
I changed the NCCE language you cite to reference the standard, and added the relevant quote from C99. It boils down to modular arithmetic, but is not defined in those terms, and we try to prefer the language used by the standard. HTH.
Simon Nicolussi
The explicit cast in the last line of the first compliant solution seems to be redundant:
The automatic conversion in such a case is mentioned above and confirmed by C99, Section 6.3.1.8, paragraph 1.
David Svoboda
Yes it is technically redundant. It's still worthwhile IMHO as the conversion is not obvious to most programmers.
Boy van Duuren
This doesn't seem to be correct. IIUC the 1 is converted to an unsigned integer by adding UINT_MAX+1.
Outputs 4294967286, which is not the same as UINT_MAX.
Aaron Ballman
Did you mean to use 10 instead of 1?
A quick test of:
Yields:
Which is what I would expect given the section quoted in the standard. Of special note, the footnote in that paragraph is not replicated. It reads:
which helps explain how this conversion process works.
Boy van Duuren
I understood the part I quoted to mean that every signed value that can't be represented as an unsigned value is converted to UINT_MAX. I used 10 to demonstrate that that is not correct.
For 1 the conversion will actually yield UINT_MAX, because 1+UINT_MAX+1 equals UINT_MAX. For every other value that cannot be represented by an unsigned this is not the case.
Am I misunderstanding the quoted part, or is it incorrect?
Aaron Ballman
I think you're misunderstanding the quoted part. It's saying that the signed value 10 cannot be represented as an unsigned int, so the value is converted to an unsigned int as described by the standard. The way the standard phrases that is rather obtuse. In this example, the new type is
unsigned int
, and the maximum value for that isUINT_MAX
. The way the conversion works is to take the original value and add/subtract UINT_MAX + 1 until you get a value that is in the range allowed by unsigned int. Eg)unsigned int converted_value = 10 + UINT_MAX + 1;
which is how you arrive at the value 4294967286.Yozo TODA
there is a strange sentence at the end of the usual arithmetic conversion rules:
(added since revision 43)
I think it should be removed. Or, any meaning for this sentence I didn't notice?
C11 and C++11 both have nothing like this around the usual arithmetic conversion rules...
Aaron Ballman
I don't think that text was meant to be part of a bullet point, but was meant to be a new paragraph after. I think the intent was to call out operations where the usual arithmetic conversions normally apply, but with slightly different semantics (the wording for
?:
used to have something for that, IIRC). However, in a quick look through the C11 spec, I could not find any cases where the usual arithmetic conversions apply with different semantics, and so I've removed that text entirely.Yozo TODA
well, thanks!
Filip Szymański
Is that recommendation is also applicable for comparisons?
Let's look into code:
uint8 a = 1;
uint8 b = 2;
if (a == b)
{
*/do something */
}
Do you think that in line if(a == b) INT02 is applicable? We compare the same types. I have met several indication about INT02 in similar cases and I am confused how to solve it. For me if((uint32) a == (uint32) b) looks strange, so may be you have better solution.
best regards!
David Svoboda
By the usual arithmetic conversions, a and b have the same type, so no conversion is necessary, and their comparison will produce the value you expect. (Both a and b will be promoted to int or unsigned int, but their types will still be identical).