C++ Undefined Behavior


What is undefined behavior (UB)? According to the ISO C++ Standard (§1.3.24, N4296), it is "behavior for which this International Standard imposes no requirements."

This means that when a program encounters UB, it is allowed to do whatever it wants. This often means a crash, but it may simply do nothing, make demons fly out of your nose, or even appear to work properly!

Needless to say, you should avoid writing code that invokes UB.


If a program contains undefined behavior, the C++ standard places no constraints on its behavior.

  • It may appear to work as the developer intended, but it may also crash or produce strange results.
  • The behavior may vary between runs of the same program.
  • Any part of the program may malfunction, including lines that come before the line that contains undefined behavior.
  • The implementation is not required to document the result of undefined behavior.

An implementation may document the result of an operation that produces undefined behavior according to the standard, but a program that depends on such documented behavior is not portable.

Why undefined behavior exists

Intuitively, undefined behavior is considered a bad thing as such errors can't be handled graciously through, say, exception handlers.

But leaving some behavior undefined is actually an integral part of C++'s promise "you don't pay for what you don't use". Undefined behavior allows a compiler to assume the developer knows what he's doing and not introduce code to check for the mistakes highlighted in the above examples.

Finding and avoiding undefined behavior

Some tools can be used to discover undefined behavior during development:

  • Most compilers have warning flags to warn about some cases of undefined behavior at compile time.
  • Newer versions of gcc and clang include a so-called "Undefined Behavior Sanitizer" flag (-fsanitize=undefined) that will check for undefined behavior at runtime, at a performance cost.
  • lint-like tools may perform more thorough undefined behavior analysis.

Undefined, unspecified and implementation-defined behavior

From C++14 standard (ISO/IEC 14882:2014) section 1.9 (Program Execution):

  1. The semantic descriptions in this International Standard define a parameterized nondeterministic abstract machine. [CUT]

  2. Certain aspects and operations of the abstract machine are described in this International Standard as implementation-defined (for example, sizeof(int)). These constitute the parameters of the abstract machine. Each implementation shall include documentation describing its characteristics and behavior in these respects. [CUT]

  3. Certain other aspects and operations of the abstract machine are described in this International Standard as unspecified (for example, evaluation of expressions in a new-initializer if the allocation function fails to allocate memory). Where possible, this International Standard defines a set of allowable behaviors. These define the nondeterministic aspects of the abstract machine. An instance of the abstract machine can thus have more than one possible execution for a given program and a given input.

  4. Certain other operations are described in this International Standard as undefined (or example, the effect of attempting to modify a const object). [ Note: this International Standard imposes no requirements on the behavior of programs that contain undefined behavior. —end note ]

Reading or writing through a null pointer

int *ptr = nullptr;
*ptr = 1; // Undefined behavior

This is undefined behavior, because a null pointer does not point to any valid object, so there is no object at *ptr to write to.

Although this most often causes a segmentation fault, it is undefined and anything can happen.

No return statement for a function with a non-void return type

Omitting the return statement in a function which is has a return type that is not void is undefined behavior.

int function() {  
    // Missing return statement

int main() {
    function(); //Undefined Behavior

Most modern day compilers emit a warning at compile time for this kind of undefined behavior.

Note: main is the only exception to the rule. If main doesn't have a return statement, the compiler automatically inserts return 0; for you, so it can be safely left out.

Modifying a string literal

char *str = "hello world";
str[0] = 'H';

"hello world" is a string literal, so modifying it gives undefined behaviour.

The initialisation of str in the above example was formally deprecated (scheduled for removal from a future version of the standard) in C++03. A number of compilers before 2003 might issue a warning about this (e.g. a suspicious conversion). After 2003, compilers typically warn about a deprecated conversion.


The above example is illegal, and results in a compiler diagnostic, in C++11 and later. A similar example may be constructed to exhibit undefined behaviour by explicitly permitting the type conversion, such as:

char *str = const_cast<char *>("hello world");
str[0] = 'H'; 

Accessing an out-of-bounds index

It is undefined behavior to access an index that is out of bounds for an array (or standard library container for that matter, as they are all implemented using a raw array):

 int array[] = {1, 2, 3, 4, 5};
 array[5] = 0;  // Undefined behavior

It is allowed to have a pointer pointing to the end of the array (in this case array + 5), you just can't dereference it, as it is not a valid element.

 const int *end = array + 5;  // Pointer to one past the last index
 for (int *p = array; p != end; ++p)
   // Do something with `p`

In general, you're not allowed to create an out-of-bounds pointer. A pointer must point to an element within the array, or one past the end.

Integer division by zero

int x = 5 / 0;    // Undefined behavior

Division by 0 is mathematically undefined, and as such it makes sense that this is undefined behavior.


float x = 5.0f / 0.0f;   // x is +infinity

Most implementation implement IEEE-754, which defines floating point division by zero to return NaN (if numerator is 0.0f), infinity (if numerator is positive) or -infinity (if numerator is negative).

Signed Integer Overflow

int x = INT_MAX + 1;

// x can be anything -> Undefined behavior

If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined.

(C++11 Standard paragraph 5/4)

This is one of the more nasty ones, as it usually yields reproducible, non-crashing behavior so developers may be tempted to rely heavily on the observed behavior.

On the other hand:

unsigned int x = UINT_MAX + 1;

// x is 0

is well defined since:

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer.

(C++11 Standard paragraph 3.9.1/4)

Sometimes compilers may exploit an undefined behavior and optimize

signed int x ;
if(x > x + 1)
    //do something

Here since a signed integer overflow is not defined, compiler is free to assume that it may never happen and hence it can optimize away the "if" block

Using an uninitialized local variable

int a;
std::cout << a; // Undefined behavior!

This results in undefined behavior, because a is uninitialised.

It is often, incorrectly, claimed that this is because the value is "indeterminate", or "whatever value was in that memory location before". However, it is the act of accessing the value of a in the above example that gives undefined behaviour. In practice, printing a "garbage value" is a common symptom in this case, but that is only one possible form of undefined behaviour.

Although highly unlikely in practice (since it is reliant on specific hardware support) the compiler could equally well electrocute the programmer when compiling the code sample above. With such a compiler and hardware support, such a response to undefined behaviour would markedly increase average (living) programmer understanding of the true meaning of undefined behaviour - which is that the standard places no constraint on the resultant behaviour.


Using an indeterminate value of unsigned char type does not produce undefined behavior if the value is used as:

  • the second or third operand of the ternary conditional operator;
  • the right operand of the built-in comma operator;
  • the operand of a conversion to unsigned char;
  • the right operand of the assignment operator, if the left operand is also of type unsigned char;
  • the initializer for an unsigned char object;

or if the value is discarded. In such cases, the indeterminate value simply propagates to the result of the expression, if applicable.

Note that a static variable is always zero-initialized (if possible):

static int a;
std::cout << a; // Defined behavior, 'a' is 0

Multiple non-identical definitions (the One Definition Rule)

If a class, enum, inline function, template, or member of a template has external linkage and is defined in multiple translation units, all definitions must be identical or the behavior is undefined according to the One Definition Rule (ODR).


class Foo {
    double x;
    int y;

Foo get_foo();


#include "foo.h"
Foo get_foo() { /* implementation */ }


// I want access to the private member, so I am going to replace Foo with my own type
class Foo {
    double x;
    int y;
Foo get_foo(); // declare this function ourselves since we aren't including foo.h
int main() {
    Foo foo = get_foo();
    // do something with foo.y

The above program exhibits undefined behavior because it contains two definitions of the class ::Foo, which has external linkage, in different translation units, but the two definitions are not identical. Unlike redefinition of a class within the same translation unit, this problem is not required to be diagnosed by the compiler.

Incorrect pairing of memory allocation and deallocation

An object can only be deallocated by delete if it was allocated by new and is not an array. If the argument to delete was not returned by new or is an array, the behavior is undefined.

An object can only be deallocated by delete[] if it was allocated by new and is an array. If the argument to delete[] was not returned by new or is not an array, the behavior is undefined.

If the argument to free was not returned by malloc, the behavior is undefined.

int* p1 = new int;
delete p1;      // correct
// delete[] p1; // undefined
// free(p1);    // undefined

int* p2 = new int[10];
delete[] p2;    // correct
// delete p2;   // undefined
// free(p2);    // undefined

int* p3 = static_cast<int*>(malloc(sizeof(int)));
free(p3);       // correct
// delete p3;   // undefined
// delete[] p3; // undefined

Such issues can be avoided by completely avoiding malloc and free in C++ programs, preferring the standard library smart pointers over raw new and delete, and preferring std::vector and std::string over raw new and delete[].

Accessing an object as the wrong type

In most cases, it is illegal to access an object of one type as though it were a different type (disregarding cv-qualifiers). Example:

float x = 42;
int y = reinterpret_cast<int&>(x);

The result is undefined behavior.

There are some exceptions to this strict aliasing rule:

  • An object of class type can be accessed as though it were of a type that is a base class of the actual class type.
  • Any type can be accessed as a char or unsigned char, but the reverse is not true: a char array cannot be accessed as though it were an arbitrary type.
  • A signed integer type can be accessed as the corresponding unsigned type and vice versa.

A related rule is that if a non-static member function is called on an object that does not actually have the same type as the defining class of the function, or a derived class, then undefined behavior occurs. This is true even if the function does not access the object.

struct Base {
struct Derived : Base {
    void f() {}
struct Unrelated {};
Unrelated u;
Derived& r1 = reinterpret_cast<Derived&>(u); // ok
r1.f();                                      // UB
Base b;
Derived& r2 = reinterpret_cast<Derived&>(b); // ok
r2.f();                                      // UB

Floating point overflow

If an arithmetic operation that yields a floating point type produces a value that is not in the range of representable values of the result type, the behavior is undefined according to the C++ standard, but may be defined by other standards the machine might conform to, such as IEEE 754.

float x = 1.0;
for (int i = 0; i < 10000; i++) {
    x *= 10.0; // will probably overflow eventually; undefined behavior

Calling (Pure) Virtual Members From Constructor Or Destructor

The Standard (10.4) states:

Member functions can be called from a constructor (or destructor) of an abstract class; the effect of making a virtual call (10.3) to a pure virtual function directly or indirectly for the object being created (or destroyed) from such a constructor (or destructor) is undefined.

More generally, some C++ authorities, e.g. Scott Meyers, suggest never calling virtual functions (even non-pure ones) from constructors and dstructors.

Consider the following example, modified from the above link:

class transaction
    transaction(){ log_it(); }
    virtual void log_it() const = 0;

class sell_transaction : public transaction
    virtual void log_it() const { /* Do something */ }

Suppose we create a sell_transaction object:

sell_transaction s;

This implicitly calls the constructor of sell_transaction, which first calls the constructor of transaction. When the constructor of transaction is called though, the object is not yet of the type sell_transaction, but rather only of the type transaction.

Consequently, the call in transaction::transaction() to log_it, won't do what might seem to be the intuitive thing - namely call sell_transaction::log_it.

  • If log_it is pure virtual, as in this example, the behaviour is undefined.

  • If log_it is non-pure virtual, transaction::log_it will be called.

Deleting a derived object via a pointer to a base class that doesn't have a virtual destructor.

class base { };
class derived: public base { }; 

int main() {
    base* p = new derived();
    delete p; // The is undefined behavior!

In section [expr.delete] §5.3.5/3 the standard says that if delete is called on an object whose static type does not have a virtual destructor:

if the static type of the object to be deleted is different from its dynamic type, the static type shall be a base class of the dynamic type of the object to be deleted and the static type shall have a virtual destructor or the behavior is undefined.

This is the case regardless of the question whether the derived class added any data members to the base class.

Accessing a dangling reference

It is illegal to access a reference to an object that has gone out of scope or been otherwise destroyed. Such a reference is said to be dangling since it no longer refers to a valid object.

#include <iostream>
int& getX() {
    int x = 42;
    return x;
int main() {
    int& r = getX();
    std::cout << r << "\n";

In this example, the local variable x goes out of scope when getX returns. (Note that lifetime extension cannot extend the lifetime of a local variable past the scope of the block in which it is defined.) Therefore r is a dangling reference. This program has undefined behavior, although it may appear to work and print 42 in some cases.

Extending the `std` or `posix` Namespace

The standard ( generally forbids extending the std namespace:

The behavior of a C++ program is undefined if it adds declarations or definitions to namespace std or to a namespace within namespace std unless otherwise specified.

The same goes for posix (

The behavior of a C++ program is undefined if it adds declarations or definitions to namespace posix or to a namespace within namespace posix unless otherwise specified.

Consider the following:

#include <algorithm>

namespace std
    int foo(){}

Nothing in the standard forbids algorithm (or one of the headers it includes) defining the same definition, and so this code would violate the One Definition Rule.

So, in general, this is forbidden. There are specific exceptions allowed, though. Perhaps most usefully, it is allowed to add specializations for user defined types. So, for example, suppose your code has

class foo
    // Stuff

Then the following is fine

namespace std
    struct hash<foo>
        size_t operator()(const foo &f) const;

Overflow during conversion to or from floating point type

If, during the conversion of:

  • an integer type to a floating point type,
  • a floating point type to an integer type, or
  • a floating point type to a shorter floating point type,

the source value is outside the range of values that can be represented in the destination type, the result is undefined behavior. Example:

double x = 1e100;
int y = x; // int probably cannot hold numbers that large, so this is UB

Invalid base-to-derived static cast

If static_cast is used to convert a pointer (resp. reference) to base class to a pointer (resp. reference) to derived class, but the operand does not point (resp. refer) to an object of the derived class type, the behavior is undefined. See Base to derived conversion.

Function call through mismatched function pointer type

In order to call a function through a function pointer, the function pointer's type must exactly match the function's type. Otherwise, the behaviour is undefined. Example:

int f();
void (*p)() = reinterpret_cast<void(*)()>(f);
p(); // undefined

Modifying a const object

Any attempt to modify a const object results in undefined behavior. This applies to const variables, members of const objects, and class members declared const. (However, a mutable member of a const object is not const.)

Such an attempt can be made through const_cast:

const int x = 123;
const_cast<int&>(x) = 456;
std::cout << x << '\n';

A compiler will usually inline the value of a const int object, so it's possible that this code compiles and prints 123. Compilers can also place const objects' values in read-only memory, so a segmentation fault may occur. In any case, the behavior is undefined and the program might do anything.

The following program conceals a far more subtle error:

#include <iostream>

class Foo* instance;

class Foo {
    int get_x() const { return m_x; }
    void set_x(int x) { m_x = x; }
    Foo(int x, Foo*& this_ref): m_x(x) {
        this_ref = this;
    int m_x;
    friend const Foo& getFoo();

const Foo& getFoo() {
    static const Foo foo(123, instance);
    return foo;

void do_evil(int x) {

int main() {
    const Foo& foo = getFoo();
    std::cout << foo.get_x() << '\n';

In this code, getFoo creates a singleton of type const Foo and its member m_x is initialized to 123. Then do_evil is called and the value of foo.m_x is apparently changed to 456. What went wrong?

Despite its name, do_evil does nothing particularly evil; all it does is call a setter through a Foo*. But that pointer points to a const Foo object even though const_cast was not used. This pointer was obtained through Foo's constructor. A const object does not become const until its initialization is complete, so this has type Foo*, not const Foo*, within the constructor.

Therefore, undefined behavior occurs even though there are no obviously dangerous constructs in this program.

Access to nonexistent member through pointer to member

When accessing a non-static member of an object through a pointer to member, if the object does not actually contain the member denoted by the pointer, the behavior is undefined. (Such a pointer to member can be obtained through static_cast.)

struct Base { int x; };
struct Derived : Base { int y; };
int Derived::*pdy = &Derived::y;
int Base::*pby = static_cast<int Base::*>(pdy);

Base* b1 = new Derived;
b1->*pby = 42; // ok; sets y in Derived object to 42
Base* b2 = new Base;
b2->*pby = 42; // undefined; there is no y member in Base

Invalid derived-to-base conversion for pointers to members

When static_cast is used to convert T D::* to T B::*, the member pointed to must belong to a class that is a base class or derived class of B. Otherwise the behavior is undefined. See Derived to base conversion for pointers to members

Invalid pointer arithmetic

The following uses of pointer arithmetic cause undefined behavior:

  • Addition or subtraction of an integer, if the result does not belong to the same array object as the pointer operand. (Here, the element one past the end is considered to still belong to the array.)

    int a[10];
    int* p1 = &a[5];
    int* p2 = p1 + 4; // ok; p2 points to a[9]
    int* p3 = p1 + 5; // ok; p2 points to one past the end of a
    int* p4 = p1 + 6; // UB
    int* p5 = p1 - 5; // ok; p2 points to a[0]
    int* p6 = p1 - 6; // UB
    int* p7 = p3 - 5; // ok; p7 points to a[5]
  • Subtraction of two pointers if they do not both belong to the same array object. (Again, the element one past the end is considered to belong to the array.) The exception is that two null pointers may be subtracted, yielding 0.

    int a[10];
    int b[10];
    int *p1 = &a[8], *p2 = &a[3];
    int d1 = p1 - p2; // yields 5
    int *p3 = p1 + 2; // ok; p3 points to one past the end of a
    int d2 = p3 - p2; // yields 7
    int *p4 = &b[0];
    int d3 = p4 - p1; // UB
  • Subtraction of two pointers if the result overflows std::ptrdiff_t.

  • Any pointer arithmetic where either operand's pointee type does not match the dynamic type of the object pointed to (ignoring cv-qualification). According to the standard, "[in] particular, a pointer to a base class cannot be used for pointer arithmetic when the array contains objects of a derived class type."

    struct Base { int x; };
    struct Derived : Base { int y; };
    Derived a[10];
    Base* p1 = &a[1];           // ok
    Base* p2 = p1 + 1;          // UB; p1 points to Derived
    Base* p3 = p1 - 1;          // likewise
    Base* p4 = &a[2];           // ok
    auto p5 = p4 - p1;          // UB; p4 and p1 point to Derived
    const Derived* p6 = &a[1];
    const Derived* p7 = p6 + 1; // ok; cv-qualifiers don't matter

Shifting by an invalid number of positions

For the built-in shift operator, the right operand must be nonnegative and strictly less than the bit width of the promoted left operand. Otherwise, the behavior is undefined.

const int a = 42;
const int b = a << -1; // UB
const int c = a << 0;  // ok
const int d = a << 32; // UB if int is 32 bits or less
const int e = a >> 32; // also UB if int is 32 bits or less
const signed char f = 'x';
const int g = f << 10; // ok even if signed char is 10 bits or less;
                       // int must be at least 16 bits

Returning from a [[noreturn]] function


Example from the Standard, [dcl.attr.noreturn]:

[[ noreturn ]] void f() {
  throw "error"; // OK
[[ noreturn ]] void q(int i) { // behavior is undefined if called with an argument <= 0
  if (i > 0)
    throw "positive";

Destroying an object that has already been destroyed

In this example, a destructor is explicitly invoked for an object that will later be automatically destroyed.

struct S {
    ~S() { std::cout << "destroying S\n"; }
int main() {
    S s;
} // UB: s destroyed a second time here

A similar issue occurs when a std::unique_ptr<T> is made to point at a T with automatic or static storage duration.

void f(std::unique_ptr<S> p);
int main() {
    S s;
    std::unique_ptr<S> p(&s);
    f(std::move(p)); // s destroyed upon return from f
}                    // UB: s destroyed

Another way to destroy an object twice is by having two shared_ptrs both manage the object without sharing ownership with each other.

void f(std::shared_ptr<S> p1, std::shared_ptr<S> p2);
int main() {
    S* p = new S;
    // I want to pass the same object twice...
    std::shared_ptr<S> sp1(p);
    std::shared_ptr<S> sp2(p);
    f(sp1, sp2);
} // UB: both sp1 and sp2 will destroy s separately
// NB: this is correct:
// std::shared_ptr<S> sp(p);
// f(sp, sp);

Infinite template recursion

Example from the Standard, [temp.inst]/17:

template<class T> class X {
    X<T>* p; // OK
    X<T*> a; // implicit generation of X<T> requires
             // the implicit instantiation of X<T*> which requires
             // the implicit instantiation of X<T**> which ...