在 C++ 中访问具有多种不同类型的数组时会发生什么：int16_t、int 和 int64_t？

Question

Suppose I have a dynamically allocated array as in

int64_t * arr; arr = new int64_t[M];

where M is of type int64_t and is large enough to handle the accessed value I show below in the subscript operator.

Question : What happens to each kind of integer and integer-multiplication in my subscript operator of `arr' when I access it with the following:

int16_t A = 12;
int B = INT_MAX;
int64_t N = int64_t(B)*int64_t(B) // This will store the correct value of 800^3 into N and will not cause wrap-around 

int64_t my_value = arr[N + B + A]

int64_t my_value_2 = arr[A + B + N]

int64_t my_value_3 = arr[B]

int64_t my_value_4 = arr[12L*(Nx+Nx*Ny+1)]

?

I am unsure of the rules for these different integer types are handled in the subscripting operator. I am familiar with working with subscripting operators where all variables used in the operator are all of type `int' and are small enough that they will not overflow. I'm unsure of the behavior when there are multiple int types all called in the same operator, especially for large arrays where int64_t is required.

Answer 1

The addition operator ( + ) is left-to-right associative. As a result, given

int16_t A;
int B;
int64_t N;

then N + B + A and B + A + N both have the same final result type, but the former uses int64_t (or int , if it is larger) to compute the subexpression N + B and the latter uses int (which cannot be smaller than int16_t ) to compute the subexpression B + A .

If B + A overflows int , B + A + N will cause you problems that don't exist with N + B + A .

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Note that INT_MAX can be 2 ¹⁵ -1, 2 ³¹ -1, 2 ⁶³ -1 (among other legal values) and therefore it is not portable to assume that int64_t(INT_MAX)*int64_t(INT_MAX) does not overflow.