Integer operations must result in an integer value within the range of the integer type (that is, the resulting value is the same as the result produced by unlimitedrange integers). Frequently, the range is more restrictive depending on the use of the integer value, for example, as an index. Integer values can be verified by code review or by static analysis.
Integer overflow is undefined behavior, so a compiled program can do anything, including go off to play the Game of Life. Furthermore, a compiler may perform optimizations that assume an overflow will never occur, which can easily yield unexpected results. Compilers can optimize away if
statements that check whether an overflow occurred. See MSC15C. Do not depend on undefined behavior for an example.
Verifiably inrange operations are often preferable to treating outofrange values as an error condition because the handling of these errors has been repeatedly shown to cause denialofservice problems in actual applications. The quintessential example is the failure of the Ariane 5 launcher, which occurred because of an improperly handled conversion error that resulted in the processor being shut down [Lions 1996].
A program that detects an integer overflow to be imminent may do one of two things: (1) signal some sort of error condition or (2) produce an integer result that is within the range of representable integers on that system. Some situations can be handled by an error condition, where an overflow causes a change in control flow (such as the system complaining about bad input and requesting alternative input from the user). Others are better handled by the latter option because it allows the computation to proceed and generate an integer result, thereby avoiding a denialofservice attack. However, when continuing to produce an integer result in the face of overflow, the question of what integer result to return to the user must be considered.
The saturation and modwrap algorithms and the technique of restricted range usage, defined in the following subsections, produce integer results that are always within a defined range. This range is between the integer values MIN
and MAX
(inclusive), where MIN
and MAX
are two representable integers with MIN < MAX
.
Saturation Semantics
For saturation semantics, assume that the mathematical result of the computation is result
. The value actually returned to the user is set out in the following table:
Range of Mathematical Result  Result Returned 







Modwrap Semantics
In modwrap semantics (also called modulo arithmetic), integer values "wrap round." That is, adding 1 to MAX
produces MIN
. This is the defined behavior for unsigned integers in the C Standard, subclause 6.2.5, paragraph 9. It is frequently the behavior of signed integers, as well. However, it is more sensible in many applications to use saturation semantics instead of modwrap semantics. For example, in the computation of a size (using unsigned integers), it is often better for the size to stay at the maximum value in the event of overflow rather than to suddenly become a very small value.
Restricted Range Usage
Another technique for avoiding integer overflow is to use only half the range of signed integers. For example, when using an int
, use only the range [INT_MIN/2
, INT_MAX/2
]. This practice has been a trick of the trade in Fortran for some time, and now that optimizing C compilers are more sophisticated, it can be valuable in C.
Consider subtraction. If the user types the expression a  b
, where both a
and b
are in the range [INT_MIN/2, INT_MAX/2]
, the result will be in the range (INT_MIN, INT_MAX]
for a typical two's complement machine.
Now, if the user types a < b
, an implicit subtraction often occurs. On a machine without condition codes, the compiler may simply issue a subtract instruction and check whether the result is negative. This behavior is allowed because the compiler is allowed to assume there is no overflow. If all explicitly usergenerated values are kept in the range [INT_MIN/2, INT_MAX/2]
, then comparisons will always work even if the compiler performs this optimization on such hardware.
Noncompliant Code Example
In this noncompliant example, i + 1
will overflow on a 16bit machine. The C Standard allows signed integers to overflow and produce incorrect results. Compilers can take advantage of this to produce faster code by assuming an overflow will not occur. As a result, the if
statement that is intended to catch an overflow might be optimized away.
int i = /* Expression that evaluates to the value 32767 */; /* ... */ if (i + 1 <= i) { /* Handle overflow */ } /* Expression involving i + 1 */
Compliant Solution
Using a long
instead of an int
is guaranteed to accommodate the computed value:
long i = /* Expression that evaluates to the value 32767 */; /* ... */ /* No test is necessary; i is known not to overflow */ /* Expression involving i + 1 */
Risk Assessment
Outofrange integer values can result in reading from or writing to arbitrary memory locations and the execution of arbitrary code.
Recommendation  Severity  Likelihood  Remediation Cost  Priority  Level 

INT08C  Medium  Probable  High  P4  L3 
Automated Detection
Tool  Version  Checker  Description 

Astrée  19.04  integeroverflow  Fully checked 
Axivion Bauhaus Suite  6.9.0  CertCINT08  
CodeSonar  5.1p0  ALLOC.SIZE.ADDOFLOW ALLOC.SIZE.IOFLOW ALLOC.SIZE.MULOFLOW ALLOC.SIZE.SUBUFLOW MISC.MEM.SIZE.ADDOFLOW MISC.MEM.SIZE.BAD MISC.MEM.SIZE.MULOFLOW MISC.MEM.SIZE.SUBUFLOW  Addition Overflow of Allocation Size Integer Overflow of Allocation Size Multiplication Overflow of Allocation Size Subtraction Underflow of Allocation Size Addition Overflow of Size Unreasonable Size Argument Multiplication Overflow of Size Subtraction Underflow of Size 
Compass/ROSE  Could detect violations of this recommendation by flagging any comparison expression involving addition that could potentially overflow. For example, instead of comparing  
LDRA tool suite  9.7.1  488 S, 493 S, 493 S  Partially implemented 
Parasoft C/C++test  10.4.2  CERT_CINT08a  Avoid integer overflows 
Polyspace Bug Finder  R2018a  Integer overflow  Overflow from operation between integers 
PRQA QAC  9.5  2800,2801,2802,2803, 2910,2911,2912,2913  Partially implemented 
Related Vulnerabilities
Search for vulnerabilities resulting from the violation of this rule on the CERT website.
Related Guidelines
SEI CERT C++ Coding Standard  VOID INT08CPP. Verify that all integer values are in range 
ISO/IEC TR 24772:2013  Numeric Conversion Errors [FLC] 
13 Comments
Douglas A. Gwyn
Any good application will have an error recovery scheme, and it is much better to use that than to silently substitute *different values* and proceed with the computation. Saturating arithmetic has uses, for example to avoid driving mechanical systems beyond hard limits, but it's not a substitute for detecting errors or malicious input and taking more appropriate measures.
David Svoboda
True, but this begs the question. The recovery scheme could divert control flow to an error recovery mechanism, or it could proceed with the calculation...which to do is a choice for the developer. I have (hopefully) made this more clear in the intro paragraphs.
Also added a bit of background and details on restricted range checking, from email from David Keaton.
Alex Volkovitsky
Regarding ROSE algorithm, who's to say that
bc
can't underflow? Also, am I right to understand that we end up flagging just about every addition inside of a comparison? Seems silly...David Svoboda
The pattern in the 'a + b < c' comparison is that a and b are the same type (or are promoted to the same type), and c > b. Ideally c is the maximal value of its type, at the very least b is known at compile time, and c > b.
The general case may be unenforceable, but there are lots of enforceable subcases. The NCCE for example is: a + b < a, where b > 0. That is automatically a violation, since it presumes modular behavior.
David Svoboda
This rule is still rosepossible, although a checker for INT30C will pretty much catch all violations of this rule.
Masaki Kubo
this sentence doesn't make sense to me:
"eg instead of comparing 'a + b < c', where b and c are compiletime constants and b > c, the code should compare 'a < c  b'."
David Svoboda
Added some background assumptions to the paragraph. The idea is you generally don't know if a+b might overflow but you can guarantee that cb won't underflow, so you do that subtraction instead of the addition.
Masaki Kubo
The 2nd paragraph of "Restricted Range Usage",
(INT_MIN, INT_MAX] should be corrected to [INT_MIN, INT_MAX]? I think INT_MIN is inclusive.
Robert Seacord (Manager)
No, I think this is right. For example, take the 8bit case. 128 / 2 = 64 127 / 2 = 63 r 1
The restricted range is [64, 63] 64  63 = 127 which is exclusive of 128.
Masaki Kubo
Good point!
Shay Green
"Now, if the user types a < b, there is often an implicit subtraction happening. On a machine without condition codes, the compiler may simply issue a subtract instruction and check whether the result is negative. This is allowed, because the compiler is allowed to assume there is no overflow. If all explicitly usergenerated values are kept in the range INT_MIN/2, INT_MAX/2, then comparisons will always work even if the compiler performs this optimization on such hardware."
Is this really claiming that a compiler can legally convert a<b to (ab)<0 (assuming a and b are ints)? If so, then 0<INT_MIN would fail if the machine uses two's complement arithmetic and modwrap semantics, as 0INT_MIN would evaluate to INT_MIN (INT_MIN == INT_MIN on such a machine). Besides, if one were following this (useful) idea of using half the int range, intermediate expressions would still fall outside this half range, and if one of them were compared with another, a compiler performing the above (broken) rewriting would break such code.
Robert Seacord (Manager)
From Tim Wilson:
This is not how comparisons work. The statement a < b is translated into something like the following assembly code:
A jl (jump if less) instruction does not jump simply when the sign flow has been set indicating the subtraction produced a result less than 0.
Instead, a jl instruction jumps when the sign flag does not equal the overflow flag. In normal cases when there is no overflow, the sign flag does indicate whether a is less than b. When there is overflow, the opposite of the sign flag indicates whether a is less than b.
The compiler is completely justified in converting a < b into (a  b) < 0 because such a conversion is always correct.
Robert Seacord (Manager)
The situation you cite only works on machines that have condition codes. For example, the Cray2 had no condition codes, and the compiler took advantage of the standard's allowed assumption of no overflow, saving four instructions per comparison by testing (a  b) < 0 instead of a < b. Therefore, the comparison a < b sometimes got the wrong result, and this was allowed by the standard.