Warning C26451: Arithmetic overflow: Using operator '%operator%' on a %size1% byte value and then casting the result to a %size2% byte value. Cast the value to the wider type before calling operator '%operator%' to avoid overflow
- 2 minutes to read
-
Contributors
This warning indicates incorrect behavior that results from integral promotion rules and types larger than those in which arithmetic is typically performed.
We detect when a narrow type integral value was shifted left, multiplied, added, or subtracted and the result of that arithmetic operation was cast to a wider type value. If the operation overflowed the narrow type value, then data is lost. You can prevent this loss by casting the value to a wider type before the arithmetic operation.
Example 1
The following code generates this warning:
void leftshift(int i)
{
unsigned __int64 x;
x = i << 31; // C26451 reported here
// code
}
To correct this warning, use the following code:
void leftshift(int i)
{
unsigned __int64 x;
x = (unsigned __int64)i << 31; // OK
// code
}
Example 2
void somefunc(__int64 y);
void callsomefunc(int x)
{
somefunc(x * 2); // C26451 reported here
}
To correct this warning, use the following code:
void callsomefunc(int x)
{
somefunc((__int64)x * 2); // OK
}
Example 3
__int64 add(int x)
{
constexpr auto value = 2;
return x += value; // C26451 reported here
}
To correct this warning, use the following code:
__int64 add(int x)
{
constexpr auto value = 2;
const __int64 y = (__int64)x + value; // OK
return y;
}