Ok. Why did you not say that? :)
x=[58.61111111 67.32055556 70.56194444 74.22694444 78.39388889 85.11555556];
y=[18.93000031 19.42000008 19.51000023 19.67000008 19.68000031 19.71999931];
S = spline(x,y)
breaks: [58.611 67.321 70.562 74.227 78.394 85.116]
coefs: [5x4 double]
0.00053154 -0.013366 0.13235 18.93
0.00053154 0.00052218 0.020489 19.42
-0.0013274 0.005691 0.040628 19.51
0.00061302 -0.0089032 0.028855 19.67
0.00061302 -0.00124 -0.013411 19.68
Each row of S.coefs is the set of coefficients of one cubic polynomial segment. It can be evaluated by polyval, but be CAREFUL here. Those cubic polynomials are defined to be evaluated relative to the break point at the beginning of that interval.
So if I wanted to evaluate the first cubic segment at the point xhat = 59, I would do this:
xhat = 59;
polyval(S.coefs(1,:),xhat - S.breaks(1))
As you can see, it yields the same prediction as ppval did. The reason for this offset is it makes the polynomial evaluation more stable with respect to numerical problems.
As part of my SLM toolbox (found on the File Exchange), I do provide a tool that allows you to extract the polynomials in a symbolic form.
polys = slmpar(S,'symabs')
[1x2 double] [1x2 double] [1x2 double] [1x2 double] [1x2 double]
[1x1 sym ] [1x1 sym ] [1x1 sym ] [1x1 sym ] [1x1 sym ]
So in symbolic form I removed the break point offset that was built into the polynomial, then return it as a symbolic "function" of x.
0.00053154425392522449377030735462313*x^3 - 0.1068293870770472518768171460159*x^2 + 7.1771491067641370730246290387668*x - 141.76724752093537873401498012911
And if I now substitute 59 into that polynomial, you also get the value you should expect.
I'm still not positive exactly what you intend as a goal, so if you need more help, please add further clarification. For example, my own SLM toolbox gives you a tool that does regression spline modeling.