Evaluating lazily 'f<-' ?

Hello,


I can include code for "padding<-"as well, but the error is before that,
namely in 'right<-':

right = function(x, val) {print("Right");};
# more options:
padding = function(x, right, left, top, bottom) {print("Padding");};
'padding<-' = function(x, ...) {print("Padding = ");};
df = data.frame(x=1:5, y = sample(1:5, 5));


### Does NOT work
'right<-' = function(x, val) {
  ? ??? print("Already evaluated and also does not use 'val'");
  ? ??? x = substitute(x); # x was evaluated before
}

right(padding(df)) = 1;

I want to capture the assignment event inside "right<-" and then call
the function padding() properly.

I haven't thought yet if I should use:

padding(x, right, left, ... other parameters);

or

padding(x, parameter) <- value;


It also depends if I can properly capture the unevaluated expression
inside "right<-":

'right<-' = function(x, val) {

# x is automatically evaluated when using 'f<-'!

# but not when implementing as '%f%' = function(x, y);

}


Many thanks,


Leonard


On 9/13/2021 4:11 PM, Duncan Murdoch wrote:

On 12/09/2021 10:33 a.m., Leonard Mada via R-help wrote:

How can I avoid evaluation?

right = function(x, val) {print("Right");};
padding = function(x) {print("Padding");};
df = data.frame(x=1:5, y = sample(1:5, 5));

### OK
'%=%' = function(x, val) {
 ? ??? x = substitute(x);
}
right(padding(df)) %=% 1; # but ugly

### Does NOT work
'right<-' = function(x, val) {
 ? ??? print("Already evaluated and also does not use 'val'");
 ? ??? x = substitute(x); # is evaluated before
}

right(padding(df)) = 1

That doesn't make sense.? You don't have a `padding<-` function, and
yet you are trying to call right<- to assign something to padding(df).

I'm not sure about your real intention, but assignment functions by
their nature need to evaluate the thing they are assigning to, since
they are designed to modify objects, not create new ones.

To create a new object, just use regular assignment.

Duncan Murdoch