What is 0[p] doing?

2018-07-01 02:12:09

This question already has an answer here:

With arrays, why is it the case that a[5] == 5[a]? 17 answers

The C Standard defined the operator [] this way:

Whatever a and b are a[b] is considred as *((a)+(b))

And that's why 0[p] == *(0 + p) == *(p + 0) == p[0] which is the first element of the array.

0[p] is equivalent to p[0] . Both are converted as

0[p] = *(0+p) and p[0] = *(p+0)

From above statements both are equal.

0[p]

in 0[p] = 42;

is equivalent to p[0]

+ operation is commutative and we have:

p[0] == *(p + 0) == *(0 + p) == 0[p]

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