I want to print the largest number that unsigned int can represent, which is 2^32 - 1. I use the following code:

unsigned int n = 2147483647 + 1 + 2147483647;

std::cout << n << std::endl;

However, I get an overflow warning from clangd:

overflow in expression; result is -2147483648 with type ‘int’

According to post here, the expression 2147483647 + 1 overflows because both number are treated as int type. The compiler will choose from int, long, long long in that order to find if a type can fit the number. The maximum positive integer an int type can represent is 2147483647. So the type chosen for 2147483647 is int, not long or long long. Thus we get the overflow warning when we add 1 to it. If you use 2147483648 directly, the compiler will choose long type for this number, which will also not overflow.

The variable type on the left side of the assignment does not impact the choose integer type for the variable on the right side. It only decides whether type conversion will happen when assigning the result to this variable. In this case, although we want to the assign the final result to an unsigned int type, 2147483647 is not converted to unsigned int automatically.

To fix issue, we can explicitly tell the compiler that we want an unsigned int instead of int:

unsigned int n = (unsigned int)2147483647 + 1 + 2147483647;
// or use the following
unsigned int n = 2147483647u + 1 + 2147483647;

By converting one number to unsigned int, the other number will also be elevated to unsigned int and no overflow occurs in this case.

References#

• C++ literal integer type: https://stackoverflow.com/q/42892227/6064933
• Integer literal: https://en.cppreference.com/w/cpp/language/integer_literal
• http://www.c-faq.com/expr/intoverflow1.html