There are three reasons.
First of all, start + (end - start) / 2
works even if you are using pointers, as long as end - start
doesn't overflow1.
int *start = ..., *end = ...;
int *mid = start + (end - start) / 2; // works as expected
int *mid = (start + end) / 2; // type error, won't compile
Second of all, start + (end - start) / 2
won't overflow if start
and end
are large positive numbers. With signed operands, overflow is undefined:
int start = 0x7ffffffe, end = 0x7fffffff;
int mid = start + (end - start) / 2; // works as expected
int mid = (start + end) / 2; // overflow... undefined
(Note that end - start
may overflow, but only if start < 0
or end < 0
.)
Or with unsigned arithmetic, overflow is defined but gives you the wrong answer. However, for unsigned operands, start + (end - start) / 2
will never overflow as long as end >= start
.
unsigned start = 0xfffffffeu, end = 0xffffffffu;
unsigned mid = start + (end - start) / 2; // works as expected
unsigned mid = (start + end) / 2; // mid = 0x7ffffffe
Finally, you often want to round towards the start
element.
int start = -3, end = 0;
int mid = start + (end - start) / 2; // -2, closer to start
int mid = (start + end) / 2; // -1, surprise!
Footnotes
1 According to the C standard, if the result of pointer subtraction is not representable as a ptrdiff_t
, then the behavior is undefined. However, in practice, this requires allocating a char
array using at least half the entire address space.
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