Background, writing a function will take a list of {x,y} pairs and check that the first and last x-values of the list are 0 and 1. If not, then the list is fixed by adding points to the list. Then an interpolating function for the list is created. One constraint: would be best not to modify the original list.

Have tried a few iterations and settled on the following. My question is, how many copies of the list are being created with the code below. (I think 2) and is there a better alternative.

Thanks for your comments and suggestions.

Current implementation:

ClearAll[makeInterp, x1, x2, f1, f2];
makeInterp[ptsIn_List, intOrder_: 1] := Module[{ptsAll},
  ptsAll = ptsIn;
  If[ First[ptsAll][[1]] > 0, PrependTo[ptsAll, {0, First[ptsIn][[2]] }] ];
  If[ Last[ptsAll][[1]] < 1, AppendTo[ptsAll, {1, Last[ptsIn][[2]] }] ];
  Interpolation[ptsAll, InterpolationOrder -> intOrder]
  ]
x1 = {{0.1, 3}, {0.5, 5}, {0.7, 6} };
f1 = makeInterp[x1];

x2 = {{0, 0}, {0.1, 2}, {0.5, 8}, {0.7, 9}, {1, 8} };
f2 = makeInterp[x2];
Plot[{f1[x], f2[x]}, {x, 0, 1}] 

Is this code making two copies of the list? One copy when calling the function, which provides values for ptsIn; and one copy for ptsAll. I considered a few alternatives.

Alternative #1 -- operate directly on ptsIn. But this results in errors with PrependTo and AppendTo

makeInterp[ptsIn_, intOrder_: 1] := Module[{},
  If[ First[ptsIn][[1]] > 0, PrependTo[ptsIn, {0, First[ptList][[2]] }] ];
  If[ Last[ptsIn][[1]] < 1, AppendTo[ptsIn, {1, Last[ptList][[2]] }] ];
  Interpolation[ptsIn, InterpolationOrder -> intOrder]
  ]

Alternative #2 -- operate directly on ptsIn and use HoldFirst. This works, but it modifies the original list, which isn't a good idea.

ClearAll[makeInterp, x, f];
SetAttributes[makeInterp, {HoldFirst}] 
makeInterp[ptsIn_, intOrder_: 1] := Module[{},
  If[ First[ptsIn][[1]] > 0, PrependTo[ptsIn, {0, First[ptList][[2]] }] ];
  If[ Last[ptsIn][[1]] < 1, AppendTo[ptsIn, {1, Last[ptList][[2]] }] ];
  Interpolation[ptsIn, InterpolationOrder -> intOrder]
  ]
x = {{0.1, 3}, {0.5, 5}, {0.7, 6} };
f = makeInterp[x]
Plot[f[x], {x, 0, 1}]
x

Alternative #2A -- not sure why the following fails. Seems that cannot provide a pattern for the first argument, when using HoldFirst?

ClearAll[makeInterp, x, f];
SetAttributes[makeInterp, {HoldFirst}] 
makeInterp[ptsIn_List, intOrder_: 1] := Module[{},
  If[ First[ptsIn][[1]] > 0, PrependTo[ptsIn, {0, First[ptList][[2]] }] ];
  If[ Last[ptsIn][[1]] < 1, AppendTo[ptsIn, {1, Last[ptList][[2]] }] ];
  Interpolation[ptsIn, InterpolationOrder -> intOrder]
  ]
x = {{0.1, 3}, {0.5, 5}, {0.7, 6} };
f = makeInterp[x]
(*output:  makeInterp[x]*)