INT_MAX + 1 = INT_MIN in integers?

Question

for (i = 0; i <= N; ++i) { ... }

This particular operator will cause an infinite loop if N is equal INT_MAX. It is known that Unsigned Overflows are overflows the assumption iand Nfor unsigned, the compiler can be assumed that the loop will iterate exactly N+1once, if ithere is an overflow undefined. The following should be noted here: if I create loops like,

for (i = 0; i < N; ++i) { ... }

Will it still be undefined behav?

Why is it INT_MAX + 1not necessarily equal INT_MINin the case of signed integers?

Answer 1

INT_MAX + 1

this operation causes undefined behavior. Signed integer overflow is undefined behavior in C.

INT_MIN, . .

Answer 2

Why INT_MAX + 1 is not surely equal to INT_MIN in case of signed integers? 

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