#include "stdafx.h"
long fact(int n);
long rfact(int n);

int main()
{
    int num;
    printf("This program calculates factorials.\n");
    printf("Enter a value in the range 0-12(q to quit):\n");
    while (scanf_s("%d", &num) == 1)
    {
        if (num < 0)
            printf("No negative numbers,please.\n");
        else if (num > 12)
            printf("Keep input under 13.\n");
        else
        {
            printf("loop:%d factorial=%ld\n",num,fact(num));
            printf("recursion:%d factorial=%ld\n",num,rfact(num));
        }
        printf("Enter a value in the range 0-12(q ti quit):\n");
    }
        return 0；
}
    long fact(int n)
    {
        long ans;
        for (ans = 1; n > 1; n--)
            ans *= n;
        return ans;
    }
    long rfact(int n)
    {
        long ans;
        if (n > 0)
            ans = n*rfact(n - 1);
        else
            ans = 1;
        return ans;
    }

[此贴子已经被作者于2016-10-29 20:19编辑过]

#include "stdafx.h" //  有这个头文件的吗？我没见过。找到的话告诉我一声。
long fact(int n);
long rfact(int n);

int main()
{
    int num;
    printf("This program calculates factorials.\n");
    printf("Enter a value in the range 0-12(q to quit):\n");
    while (scanf_s("%d", &num) == 1) //scanf有scanf_s 在c语言中这样子的吗？能不能告诉我这个啥。
    {
        if (num < 0)
            printf("No negative numbers,please.\n");
        else if (num > 12)
            printf("Keep input under 13.\n");
        else
        {
            printf("loop:%d factorial=%ld\n",num,fact(num));
            printf("recursion:%d factorial=%ld\n",num,rfact(num));
        }
        printf("Enter a value in the range 0-12(q ti quit):\n");
    }
        return 0；// 你确定这里用了英文输入？
}
    long fact(int n)
    {
        long ans;
        for (ans = 1; n > 1; n--)
            ans *= n;
        return ans;
    }
    long rfact(int n)
    {
        long ans;
        if (n > 0)
            ans = n*rfact(n - 1);
        else
            ans = 1;
        return ans;
   

@题主： ，至于题主说的函数非法定义，我没发现，如果可以截个图我看看哪里出错了。最好带上完整的编译信息

@百里：那个头文件是有的，不过好像是MFC的还是VC的才有的吧。