j
k
j a
j l
Anders Sundman writes:
(_1*_1)(_1 - _2) computes the (i1[i] - i2[i])^2 part
OK; I didn't know that's what you meant by "sum of squared differences". I'd call that "the difference squared". -- Dave Abrahams Boost Consulting www.boost-consulting.com
Back to the thread
Back to the list