Warning C26832
Allocation size is the result of a narrowing conversion that could result in overflow
Remarks
This warning reports that the size specified for an allocation may be the result of a narrowing conversion that results in a numerical overflow. For example:
void* SmallAlloc(int);
void foo(unsigned short i, unsigned short j)
{
unsigned short size = i + j;
int* p = (int*)SmallAlloc(size); // Warning: C26832
p[i] = 5;
}
In the expression i + j
, both i
and j
are promoted to integers, and the result of the addition is stored in a temporary integer. Then, the temporary integer is implicitly cast to an unsigned short
before the value is stored in size
. The cast to unsigned short
might overflow, in which case SmallAlloc
may return a smaller buffer than expected. That will likely lead to out of bounds attempts to access the buffer later on. This code pattern can result in remote code execution vulnerabilities
This check applies to common allocation functions like new
, malloc
, and VirtualAlloc
. The check also applies to custom allocator functions that have alloc
(case insensitive) in the function name.
This check sometimes fails to recognize that certain checks can prevent overflows because the check is conservative.
This warning is available in Visual Studio 2022 version 17.7 and later versions.
Example
To fix the previous code example in which i+j
might overflow, introduce a check to make sure it won't. For example:
void *SmallAlloc(int);
void foo(unsigned short i, unsigned short j)
{
if (i > 100 || j > 100)
return;
unsigned short size = i + j;
int* p = (int*)SmallAlloc(size);
p[i] = 5;
}