V123. Allocation of memory by the pattern "(X*)malloc(sizeof(Y))" where the sizes of X and Y types are not equal.

The analyzer found a potential error related to the operation of memory allocation. When calculating the amount of memory to be allocated, the sizeof(X) operator is used. The result returned by the memory allocation function is converted to a different type, "(Y *)", instead of "(X *)". It may indicate allocation of insufficient or excessive amount of memory.

Consider the first example:

int **ArrayOfPointers = (int **)malloc(n * sizeof(int));

The misprint in the 64-bit program here will cause allocation of memory twice less than necessary. In the 32-bit program, the sizes of the "int" type and "pointer to int" coincide and the program works correctly despite the misprint.

This is the correct version of the code:

int **ArrayOfPointers = (int **)malloc(n * sizeof(int *));

Consider another example where more memory is allocated than needed:

unsigned *p = (unsigned *)malloc(len * sizeof(size_t));

A program with such code will most probably work correctly both in the 32-bit and 64-bit versions. But in the 64-bit version, it will allocate more memory than it needs. This is the correct code:

unsigned *p = (unsigned *)malloc(len * sizeof(unsigned));

In some cases the analyzer does not generate a warning although the types X and Y do not coincide. Here is an example of such correct code:

BYTE *simpleBuf = (BYTE *)malloc(n * sizeof(float));

Bugs Found

Checked Projects
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12 668