函数在PHP 7中返回void

Question

PHP 7增加了对返回类型声明的支持：

    ##########
    # Manual renaming with partial matches
    ##########

    # a) Take a look at the desired column of factor variables
    sort(unique(MYDATA$names))   # take a look

    # ****
    Sensthreshold <- 0.2   # sensitivity of agrep, usually 0.1-0.2 get it right
    Searchstring <- "Invesstment LLC"   # what should I search?
    # ****

    # User-defined function: returns similar string on query in column
    Searcher <- function(input, similarity = 0.1) {
      unique(agrep(input, 
                   MYDATA$names,   # <-- define your column here
                   ignore.case = TRUE, value = TRUE,
                   max.distance = similarity))
    }

    # b) Make a search of desired string
    Searcher(Searchstring, Sensthreshold)   # using user-def function 
    ### PLEASE INSPECT THE OUTPUT OF THE SEARCH
    ### Did it get it right?

 =============================================================================#
    ## ACTION! This changes your dataframe!
    ## Please make backup before proceeding
    ## Please execute this code as a whole to avoid errors

    # c) Make a vector of cells indexes after checking output
    vector_of_cells <- agrep(Searchstring, 
                       MYDATA$names, ignore.case = TRUE,
                       max.distance = Sensthreshold)
    # d) Apply the changes
    MYDATA$names[vector_of_cells] <- Searchstring # <--- CHANGING STRING
    # e) Check result
    unique(agrep(Searchstring, MYDATA$names, 
                 ignore.case = TRUE, value = TRUE, max.distance = Sensthreshold))
=============================================================================#

无论如何申报程序？ （没有回报的功能）

function sum($a, $b): float {
    return $a + $b;
}