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V697. Number of elements in the allocat…
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V697. Number of elements in the allocated array equals the size of a pointer in bytes.

Jul 24 2014

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: