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V697. Number of elements in the allocated array equals the size of a pointer in bytes.
The number of items in an array allocated by the 'new' operator equals the pointer size in bytes, which makes this code fragment very suspicious.
Take a look at an example demonstrating how such a fragment is introduced into the code. At first, the program contained a fixed array consisting of bytes. We needed to create an array of the same size but consisting of float items. As a result, we wrote the following code:
void Foo()
{
char A[10];
....
float *B = new float[sizeof(A)];
....
}We won't discuss the quality of this code now; what we are interested in is that the 'A' array has become dynamic too as a result of refactoring. The fragment where the 'B' array is created was forgotten to be changed. Because of that, we get the following incorrect code:
void Foo(size_t n)
{
char *A = new char[n];
....
float *B = new float[sizeof(A)];
....
}The number of items in the 'B' array is 4 or 8, depending on the platform bitness. It is this problem that the analyzer detects.
The fixed code:
void Foo(size_t n)
{
char *A = new char[n];
....
float *B = new float[n];
....
}This diagnostic is classified as:
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