Question

我在尝试理解运行时遇到了麻烦。任何帮助将不胜感激！

int foo(int x) {
  if (x <= 0) return x; 
     cout << x;
     return foo (x-1);
}

void bar(int n) {
  if (n <= 0) return;
  cout << foo (n) << endl;
  bar (n-1);
  cout << n;
}

int main() {
 int n;
 cin >> n;
 bar(n);
 return 0;
}

Answer 1

Foo正在打印您传递的数字，它将执行x--直到为0，如果您传递5将打印543210，因此您可以将其称为n减1并打印结果

Bar正在做同样的事情而不打印，只是递减并调用foo

这是一个3递归级别，尝试使用一个小数字并按照流程，如4

-Bar得到4，大小写为0,4大于0所以它会继续，它会调用foo（4）（请记住，foo是一个递归调用，将打印4到0 =＆gt; 43210 ，每次递减1） - 再次调用条形图，这次使用n-1 = 4 -1 = 3，值为3，它将调用foo（3）并且相同

当您在bar中遇到案例库时，n == 0您将停止调用其他函数并且此函数将获得返回的值，您可以在screnn上打印但它并不意味着函数结束，它正在等待要返回的其他值，当它返回时会打印出调用它的n，因此它将是1234，即值1,2,3和4是一次输入一个条形的值并打印出值foo的结果（对于4,3,2,1），因为它是你得到的

int foo(int x) { //Decrementing by one and printing the value
  // If x reaches 0 exit (you need an exit in a recursion)
  if (x <= 0) return x; 
  // Print the value x (the first time x will be the n, the second n-1)
  cout << x;
  // Call the same function in we are but with x-1 value
  return foo (x-1);
}
// It will produce foo(4)=>foo(3)=>foo(2)=>foo(1) and foo(0)
// foo(0) will break the recursion and finish here (only prints)

 // It is where main calls and enters n is the value you type
 void bar(int n) {
   // Case base to break recursion when it reaches 0
   if (n <= 0) return;
   // Here is the code that prints the line that foo will make
   cout << foo (n) << endl;
   // Here you will call the same function but with n-1, like foo does
   bar (n-1);
   // When all the recursion above is over you will get here
   // In the stack you will have done n calls into here with n-x
   // So every one will print the value of n that in the paremeter 
   // when the call was make 
   cout << n;
 }

递归函数的大Theta（Θ）运行时

1 个答案: