Question

c99 standard says that result of modulo operation has same sign as first operand. So -9 % 7 = -2 and 9 % -7 = 2.

I read in one book that c89 standard depends on implementation. So -9 % 7 could yield -2 or 5??? How remainder of -9 / 7 could be 5?

Answer 1

Consider two numbers a and b.

The quotient q=a/b and remainder r=a%b satisfy the equation a == q*b + r.

An (hypothetical) implementation of C89 in which -9 % 7 produces 5 is an implementation in which -9 / 7 is computed as -2.

The mathematical (Euclidian) division constrains r to be positive and smaller than b. C99 constrains it to be of the same sign as a and strictly between -b and b.
It is all only a matter of convention.