R lpsolve see all possible solutions of an integral LP - r

Question

Is there a way to make lpSolve return multiple solutions? In below case i want (5,0) and (0,5) both.
If lpSolve cannot do that then is there any other R package which will return all possible solutions of an integral linear optimization program?
library("lpSolve")
A=matrix (c(1, 1), nrow=1, byrow=TRUE)
b=(5)
signs='=='
c_=c(1,1)
res = lpSolve::lp('max', c_, A, signs, b, all.int = TRUE)
res$solution
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I would also like to know why lpSolve package provides all possible solutions if all decision variables are binary. Why cannot it repeat the same when all variables are integer...

Answer 1

Code:
library(lpSolveAPI)
vBiv_of_v <- function (nbits,v){
taillev<-length(v)
taillevBivalent<-nbits*taillev
vBivalent<-rep(0,taillevBivalent)
for(iLg in seq(1,taillev)) {
iCoef<-1
for(iDelta in seq(1,nbits)){
vBivalent[(iLg-1)*nbits+iDelta]<- iCoef*v[iLg]
iCoef<-iCoef*2
}
}
vBivalent
}
vBiv_to_v <- function (nbits,vBivalent) {
taillevBivalent<-length(vBivalent)
taillev<-taillevBivalent/nbits
v<-rep(0,taillev)
for(iLg in seq(1,taillev)) {
for(iDelta in seq(1,nbits)){
v[iLg]<-v[iLg]+2^(iDelta-1)*vBivalent[(iLg-1)*nbits+iDelta]
}
}
v
}
nbVariable<-2
nbBits=3
nbVariableBivalentes<-nbVariable*nbBits
f.obj<-rep(0,nbVariableBivalentes)
mylp <- make.lp(0, nbVariableBivalentes)
set.objfn(mylp,f.obj)
add.constraint(mylp, vBiv_of_v(nbBits,c(1,1)), "=", 5)
set.type(mylp, 1:nbVariableBivalentes , type = "binary")
repeat {
status<-solve(mylp)
if(status == 0) {
last_sol<-get.variables(mylp)
vRes<-vBiv_to_v(nbBits,last_sol)
cat(vRes[1],vRes[2],"\n")
#add cutting
new_rhs <- 0;
f.condSup<-rep(0,nbVariableBivalentes)
for (iCol in 1:nbVariableBivalentes) {
f.condSup[iCol] <- 2 * last_sol[iCol] - 1
new_rhs <- new_rhs + last_sol[iCol];
}
add.constraint(mylp, f.condSup, "<=", new_rhs - 1)
}
else if(status == 2) {
cat("No more solution.\n")
break
}
}
Result:
5 0
4 1
3 2
1 4
2 3
0 5
No more solution.