Question

我有以下代码：

abc <- function() {
    def <- function() { }
    def_enter <- function() { print("def_enter") }
    def_exit <- function() { print("def_exit") }
    trace(def, def_enter, exit = def_exit)
    def()
}

abc_enter <- function() { print("abc_enter") }
abc_exit <- function() { print("abc_exit") }

trace(abc, abc_enter, exit = abc_exit)

abc()

我希望输出像这样：

> abc()
Tracing abc() on entry 
[1] "abc_enter"
Tracing def() on entry
[1] "def_enter"
Tracing def() on exit
[1] "def_exit"
Tracing abc() on exit 
[1] "abc_exit"
NULL

但得到这样的输出：

> abc()
Tracing abc() on entry 
[1] "abc_enter"
Tracing function "def" in package "base"
Tracing abc() on exit 
[1] "abc_exit"
NULL

是否有可能获得我期望的输出？我如何修复上面的代码？

Answer 1

您需要指定调用trace中函数的环境。只需在where的定义中的跟踪abc中提及它：

abc <- function() {
    def <- function() { }
    def_enter <- function() { print("def_enter") }
    def_exit <- function() { print("def_exit") }
    trace(def, def_enter, exit = def_exit, where=environment())
    def()
}

通过致电abc，您将获得预期的结果：

#>abc()
#Tracing abc() on entry 
#[1] "abc_enter"
#Tracing function "def" in package ".GlobalEnv"
#Tracing def() on entry 
#[1] "def_enter"
#Tracing def() on exit 
#[1] "def_exit"
#Tracing abc() on exit 
#[1] "abc_exit"
#NULL

在另一个函数内声明的跟踪函数

1 个答案: