V121. Implicit conversion of the type of 'new' operator's argument to size_t type

The analyzer detected a potential error related to calling the operator new. A value of a non-memsize type is passed to the operator "new" as an argument. The operator new takes values of the type size_t, and passing a 32-bit actual argument may signal a potential overflow that may occur when calculating the memory amount being allocated. Here is an example: 

unsigned a = 5;
unsigned b = 1024;
unsigned c = 1024;
unsigned d = 1024;
char *ptr = new char[a*b*c*d]; //V121

Here you may see an overflow occurring when calculating the expression "a*b*c*d". As a result, the program allocates less memory than it should. To correct the code, use the type size_t:

size_t a = 5;
size_t b = 1024;
size_t c = 1024;
size_t d = 1024;
char *ptr = new char[a*b*c*d]; //Ok

The following code also contains an error:

#define x 1024
#define y 1024
#define z 1024
#ifdef _M_X64
#define n 5
#else
#define n 1
#endif
char *p = new char[x*y*z*n]; //V121

This error is diagnosed only when the option MoreThan2Gb in the analyzer's settings is set to "true". The error will not be diagnosed if the value of the argument is defined as a safe 32-bit constant value. Here is an example of safe code:

char *ptr = new char[100]; 
const int size = 3*3;
char *p2 = new char[size];

This warning message is similar to the warning V106.