c++递归函数调用小常识

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1，float和double类型转化在数值很大的时候慎用，比如6423.32563255（double）强制转化float类型时，精确度只能6423.325；应用场景举例：a,b两个double类型，通过求中点求离a小于0.0001的数，在求中点过程中，若有类型转化就会出现无限递归下去

2，我们经常会用到递归函数，但是如果递归深度太大时，往往导致栈溢出。而递归深度往往不太容易把握，所以比较安全一点的做法就是：用循环代替递归，方法如下：

int SomeFuncLoop(int n, int &retIdx)
18 {
19      // (First rule)
20     struct SnapShotStruct {
21        int n;        // - parameter input
22        int test;     // - local variable that will be used 
23                      //     after returning from the function call
24                      // - retIdx can be ignored since it is a reference.
25        int stage;    // - Since there is process needed to be done 
26                      //     after recursive call. (Sixth rule)
27     };
28     // (Second rule)
29     int retVal = 0;  // initialize with default returning value
30     // (Third rule)
31     stack<SnapShotStruct> snapshotStack;
32     // (Fourth rule)
33     SnapShotStruct currentSnapshot;
34     currentSnapshot.n= n;          // set the value as parameter value
35     currentSnapshot.test=0;        // set the value as default value
36     currentSnapshot.stage=0;       // set the value as initial stage
37     snapshotStack.push(currentSnapshot);
38     // (Fifth rule)
39     while(!snapshotStack.empty())
40     {
41        currentSnapshot=snapshotStack.top();
42        snapshotStack.pop();
43        // (Sixth rule)
44        switch( currentSnapshot.stage)
45        {
46        case 0:
47           // (Seventh rule)
48           if( currentSnapshot.n>0 )
49           {
50              // (Tenth rule)
51              currentSnapshot.stage = 1;            // - current snapshot need to process after
52                                                    //     returning from the recursive call
53              snapshotStack.push(currentSnapshot);  // - this MUST pushed into stack before 
54                                                    //     new snapshot!
55              // Create a new snapshot for calling itself
56              SnapShotStruct newSnapshot;
57              newSnapshot.n= currentSnapshot.n-1;   // - give parameter as parameter given
58                                                    //     when calling itself
59                                                    //     ( SomeFunc(n-1, retIdx) )
60              newSnapshot.test=0;                   // - set the value as initial value
61              newSnapshot.stage=0;                  // - since it will start from the 
62                                                    //     beginning of the function, 
63                                                    //     give the initial stage
64              snapshotStack.push(newSnapshot);
65              continue;
66           }
67           ...
68           // (Eighth rule)
69           retVal = 0 ;
70           
71           // (Ninth rule)
72           continue;
73           break; 
74        case 1: 
75           // (Seventh rule)
76           currentSnapshot.test = retVal;
77           currentSnapshot.test--;
78           ...
79           // (Eighth rule)
80           retVal = currentSnapshot.test;
81           // (Ninth rule)
82           continue;
83           break;
84        }
85     }
86     // (Second rule)
87     return retVal;
88 }

c++递归函数调用小常识

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原文地址：https://www.cnblogs.com/yinuohome/p/14898790.html