I encountered an interesting question, I defined a sub-function (it’s a model) with a variable vector of `9×1`

dimension, and the variables are as follows

```
@variable(model, varFlow[1:9], Bin)
```

The problem can be solved, and I returned the solution by

```
solFlow = value.(varFlow)
return solFlow
```

While I want to concatenate `solFlow`

with another `9×1`

dimension vector to formulate a `9×2`

dimension matrix as follows

```
NewFlow = theFunc() # return the solFlow
hcat(LastFlow, NewFlow)
```

However, the `LastFlow`

is not changed (NewFlow is not concatenated horizontally), so why? I’m confused, because Matlab can do this, while Julia cannot?

Furthermore, I tried `push!()`

function, while it gives me the type of the parameter is `Vector{Any} with 1 element`

, and it cannot be further executed as I want, the codes are like

```
AllFlow = []
push!(AllFlow, NewFlow) # Vector{Any} with 1 elements
```