An expression contains a multiplication or division operation over integer data types. The resulting value is implicitly cast to a floating-point type. When detecting this, the analyzer warns you about a potential error that may cause an overflow or calculation of an incorrect result.
Below are examples of possible errors.
Case one. Overflow.
int LX = 1000; int LY = 1000; int LZ = 1000; int Density = 10; double Mass = LX * LY * LZ * Density;
We want to calculate an object's mass relying on its density and volume. We know that the resulting value may be a large one. That's why we declare the 'Mass' variable as the 'double' type. But this code doesn't take into account that there are variables of the 'int' type which are multiplied. As a result, we'll get an integer overflow in the right part of the expression and the result will be incorrect.
There are two ways to fix the issue. The first way is to change the variables' types:
double LX = 1000.0; double LY = 1000.0; double LZ = 1000.0; double Density = 10.0; double Mass = LX * LY * LZ * Density;
The second way is to use an explicit type conversion:
int LX = 1000; int LY = 1000; int LZ = 1000; int Density = 10; double Mass = (double)(LX) * LY * LZ * Density;
We can cast only the first variable to the 'double' type - that'll be enough. Since the multiplication operation refers to left-associative operators, calculation will be executed in the following way: (((double)(LX) * LY) * LZ) * Density. Consequently, each of the operands will be cast to the 'double' type before multiplication and we will get a correct result.
P.S. Let me remind you that it will be incorrect if you try to solve the issue in the following way: Mass = (double)(ConstMass) + LX * LY * LZ * Density. The expression to the right of the '=' operator will have the 'double' type, but it's still variables of the 'int' type that will be multiplied.
Case two. Loss of accuracy.
int totalTime = 1700; int operationNum = 900; double averageTime = totalTime / operationNum;
The programmer may be expecting that the 'averageTime' variable will have value '1.888(8)', but the result will equal '1.0' when executing the program. It happens because the division operation is performed over integer types and only then is cast to the floating-point type.
Like in the previous case, we may fix the error in two ways.
The first way is to change the variables' types:
double totalTime = 1700; double operationNum = 900; double averageTime = totalTime / operationNum;
The second way is to use an implicit type conversion.
int totalTime = 1700; int operationNum = 900; double averageTime = (double)(totalTime) / operationNum;
Certainly, in some cases it's exactly division of integers that you need to execute. In such cases you can use the following comment to suppress false positives:
This diagnostic is classified as:
You can look at examples of errors detected by the V636 diagnostic.