减少赋值右侧的 size_t

Question

Why is idx1 -1 in this case?

long long idx = (size_t)0 - 1;

cout << (size_t)0 - 1; This prints ULLONG_MAX

cout << idx; This prints -1

What would change if idx was integer instead?

Answer 1

In this statement

cout << (size_t)0 - 1;

the type of the expression (size_t)0 - 1 is the unsigned integer type size_t due to the usual arithmetic conversions. Because it can't be negative, it has the value ULLONG_MAX (it seems that in the used system the type size_t is defined as an alias for the type unsigned long long int ).

In this declaration

long long idx = (size_t)0 - 1;

the initializer is converted to the signed type long long int that is to the type of the declared object. The UNIT_MAX value cannot fit into a signed long long, so it gives -1.

As the width of the type size_t in the used system is the same as the width of the type long long int. So the internal representation of the initializer of the type size_t is interpreted as a value of the type long long That is the most significant bit of the unsigned initializer is considered as the sign bit.int.