#include<stdio.h>
void func1(int n)
{
if(n==0) return;
printf("%d",n);
func2(n-2);
printf("%d",n);
}
void func2(int n)
{
if(n==0) return;
printf("%d",n);
func1(++n);
printf("%d",n);
}
void main()
{
func1(5);
}
输出:53423122233445
我不明白导致上述输出的代码中发生的控制流。有人可以解释一下吗?提前谢谢你:)
通过教学生递归如何与局部变量一起工作,我发现理解它的最简单方法是,如果你完全按照计算机的方式去做,
- 逐步处理它,并记下调用的内容以及变量值何时更改
例如:
main
func1(5)
n=5
printf 5
func2(5-2)
n=3
print 3
++n
n=4
func1(4)
n=4
print 4
func2(4-2)
n=2
print 2
++n
n=3
func1(3)
n=3
print 3
func2(3-2)
n=1
print 1
++n
n=2
func1(2)
n=2
print 2
func2(2-2)
n=0
if n==0 => return
print 2
print 2
print 3
print 3
print 4
print 4
print 5
//done
您还需要了解在每个函数调用中,
函数内对 'n' 的更改不会更改较早的
调用位置的值。
如果你想象计算机正在做这样的事情,你会看得更清楚:
每个函数调用都会在堆栈上创建一组新变量,
当一个函数返回时,它的变量将从堆栈中删除。
stack: (empty)
main
func1(5) ==> push n='5' on stack, then jump to func1()
stack is now { n=5 }
so n is 5
print 5
func2(5-2) ==> push 'n=3' on stack, then jump to func2()
stack is now { n=3 } , { n=5 }
so n is 3
print 3
++n
stack is now { n=4 } , { n=5 }
func1(4) ==> push 'n=4' on stack then jump to func1()
stack is now { n=4} , { n=4 } , { n=5 }
so n is 4
print 4
func2(4-2) ==> push 'n=2' on stack then jump to func()
stack is now {n=2}, {n=4} , { n=4 } , { n=5 }
++n
stack is now {n=3}, {n=4} , { n=4 } , { n=5 }
...etc...
.....
....
stack is eventually {n=0} {n=2}, {n=2}, {n=2} ,{n=1} {n=3}, {n=3}, {n=4} , { n=4 } , { n=5 }
after func(2-2) is called
then:
if n==0 => return;
the return pops one item {n=0} off the stack, so
stack is then {n=2}, {n=2}, {n=2} ,{n=1} {n=3}, {n=3}, {n=4} , { n=4 } , { n=5 }
print 2
return (deletes {n=2})
stack is then {n=2}, {n=2} ,{n=1} {n=3}, {n=3}, {n=4} , { n=4 } , { n=5 }
print 2
return (deletes {n=2})
stack is then {n=2} ,{n=1} {n=3}, {n=3}, {n=4} , { n=4 } , { n=5 }
print 2
return (deletes {n=2})
stack is then {n=1} {n=3}, {n=3}, {n=4} , { n=4 } , { n=5 }
print 1
return (deletes {n=1})
stack is then {n=3}, {n=3}, {n=4} , { n=4 } , { n=5 }
print 3
依此类推,直到完成并打印出最后一个“5”。