如何在 ANSI C 中退出递归 [关闭]

Question

我正在编写一个简单的程序来计算 ANSI C 中的阶乘。该程序可以正常工作到 12，因为超过 13 的数字对于 INT 来说太大了。我想要做的是打印前 12 个，然后以某种方式 BREAK 并发送第 13 个数字的错误。但是，问题是当我到达第 13 个数字并发送错误时，前 12 个不打印。是否有可能返回错误并仍然获得前 12 个？

我的代码：

#include <math.h>
#include<stdio.h>

int getIntFactorial(int x){

printf("Processing factorial of (%d) \n", x);
if (x <=1){
    printf("\n Reached base case, returning %d \n", x);
    printf("Now returning total value \n");
    return 1;
}else if (x < 13){
    printf("\n Doing recursion by calling factorial (%d - 1) \n", x);

    int counter = x * getIntFactorial(x - 1);

    printf("Receiving results of factorial (%d) = %d * %d = %d \n", x, x, (x-1), counter);
    return counter;
}else {
///number is greater than 13
    printf("Sorry, we cannot do the factorial for %d, only goes up to 12 \n", x);
    return;
}



}


main()
{
    getIntFactorial(13);


}

13 的输出：

Processing factorial of (13)
Sorry, we cannot do the factorial for 13, only goes up to 12

Process returned 62 (0x3E)   execution time : 0.009 s
Press any key to continue.

12 的输出：

Processing factorial of (12)

 Doing recursion by calling factorial (12 - 1)
Processing factorial of (11)

 Doing recursion by calling factorial (11 - 1)
Processing factorial of (10)

 Doing recursion by calling factorial (10 - 1)
Processing factorial of (9)

 Doing recursion by calling factorial (9 - 1)
Processing factorial of (8)

 Doing recursion by calling factorial (8 - 1)
Processing factorial of (7)

 Doing recursion by calling factorial (7 - 1)
Processing factorial of (6)

 Doing recursion by calling factorial (6 - 1)
Processing factorial of (5)

 Doing recursion by calling factorial (5 - 1)
Processing factorial of (4)

 Doing recursion by calling factorial (4 - 1)
Processing factorial of (3)

 Doing recursion by calling factorial (3 - 1)
Processing factorial of (2)

 Doing recursion by calling factorial (2 - 1)
Processing factorial of (1)

 Reached base case, returning 1
Now returning total value
Receiving results of factorial (2) = 2 * 1 = 2
Receiving results of factorial (3) = 3 * 2 = 6
Receiving results of factorial (4) = 4 * 3 = 24
Receiving results of factorial (5) = 5 * 4 = 120
Receiving results of factorial (6) = 6 * 5 = 720
Receiving results of factorial (7) = 7 * 6 = 5040
Receiving results of factorial (8) = 8 * 7 = 40320
Receiving results of factorial (9) = 9 * 8 = 362880
Receiving results of factorial (10) = 10 * 9 = 3628800
Receiving results of factorial (11) = 11 * 10 = 39916800
Receiving results of factorial (12) = 12 * 11 = 479001600

Process returned 479001600 (0x1C8CFC00)   execution time : 0.015 s
Press any key to continue.

Answer 1

您可以先测试一下您的参数是否大于 12，如果是，则打印您的错误消息，然后以 12 作为参数调用您的递归函数。

Answer 2

您在这里要求做什么并不完全清楚。很明显，鉴于您的代码，您观察到的行为是正确的：当然，当您尝试调用 getIntFactorial(13) 时，前十二个数字的阶乘不会打印出来。您的代码就是这样做的。

getIntFactorial 的伪代码结构是这样的：

if the number is less than or equal to 1:
    return 1;
if the number is less than 13:
    return the number times the factorial of the number minus one
if the number is greater than or equal to 13:
    print an error and return nothing

注意，顺便说一下，如果您的输入大于 13，您调用 return 而不返回任何内容。您的函数声称返回一个int，它是something。您返回的 nothing 与 something 相比非常不同。你的编译器应该抱怨，而且应该大声抱怨。如果没有，你应该买一个新的。

Answer 3

根据我对需求的理解，将最后一个else 块中的return; 更改为return getIntFactorial(x-1) 会产生您想要的结果吗？