The analyzer found a possible error inside an arithmetic expression related to the implicit type conversion to memsize type. An overflow error may be caused by the changing of the permissible interval of the values of the variables included into the expression. This warning is almost equivalent to warning V104 with the exception that the implicit type conversion occurs due to the use of '?:' operation.
Let's give an example of the implicit type conversion while using operation:
int i32; float f = b != 1 ? sizeof(int) : i32;
In the arithmetic expression the ternary operation '?:' is used which has three operands:
The result of the expression "b != 1 ? sizeof(int) : i32" is the value of type 'size_t' which is then converted into type 'float' value. Thus, the implicit type conversion realized for the 3rd operand of '?:' operation.
Let's examine an example of the incorrect code:
bool useDefaultVolume; size_t defaultVolume; unsigned width, height, depth; ... size_t volume = useDefaultVolume ? defaultVolume : width * height * depth;
Let's suppose, we're developing an application of computational modeling which requires three-dimensional calculation area. The number of calculating elements which are used is determined according to the variable 'useDefaultSize' value and is assigned on default or by multiplication of length, height and depth of the calculating area. On the 32-bit platform the size of memory which was already allocated, cannot excess 2-3 Gb (depending on the kind of OS Windows) and as consequence the result of the expression "width * height * depth" will be always correct. On the 64-bit platform, using the opportunity to deal with a larger memory size, the number of calculating elements may excess the value 'UINT_MAX' (4 Gb). In this case an overflow will occur while determining the expression "width * height * depth" because the result of this expression had type 'unsigned'.
Correction of the code may consist in the changing of the type of the variables 'width', 'height' and 'depth' to memsize type as follows:
... size_t width, height, depth; ... size_t volume = useDefaultVolume ? defaultVolume : width * height * depth;
Or in use of the explicit type conversion:
unsigned width, height, depth; ... size_t volume = useDefaultVolume ? defaultVolume : size_t(width) * size_t(height) * size_t(depth);
In addition, we advise to read the description of a similar warning V104, where one can learn about other effects of the implicit type conversion to memsize type.
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