int max(int x, int y)
{
    return(x > y ? x : y);
}

int min(int x, int y)
{
    return(x > y ? y : x);
}

#include<stdio.h>
int main()
{
    int max(int x, int y);
    int min(int x, int y);
    int(*p)(int x, int y);
    int a = 1, b = 2, n, c;
    int flag = 0;
    while (flag == 0)
    {
        printf("choose 1 or 2 in your select:\n");
        scanf_s("%d", &n);
        if (n == 1 || n == 2) flag = 1;
        else
        {
            printf("There is an error in your enter,please try again!\n");
            flag = 0;
        }
    }
    if (n == 1)
        p = max;
    else if (n == 2)
        p = min;
    c = (*p)(a, b);
    printf("a=%d,b=%d\n", a, b);
    if (n == 1) printf("max=%d\n", c);
    else if (n == 2) printf("min=%d\n", c);
    return 0;
}

[此贴子已经被作者于2017-1-14 09:39编辑过]

好了

int max(int x, int y)
{
    return(x > y ? x : y);
}

int min(int x, int y)
{
    return(x > y ? y : x);
}

#include<stdio.h>
int main()
{
    int max(int x, int y);
    int min(int x, int y);
    int(*p)(int x, int y) = NULL;
    int a = 1, b = 2, n, c;
    int flag = 0;
    while (flag == 0)
    {
        printf("choose 1 or 2 in your select:\n");
        scanf_s("%d", &n);
        if (n == 1 || n == 2) flag = 1;
        else
        {
            printf("There is an error in your enter,please try again!\n");
            flag = 0;
        }
    }
    if (n == 1) p = max;
    else if (n == 2) p = min;
    c = (*p)(a, b);
    printf("a=%d,b=%d\n", a, b);
    if (n == 1) printf("max=%d\n", c);
    else if (n == 2) printf("min=%d\n", c);
    return 0;
}

[此贴子已经被作者于2017-1-14 09:38编辑过]

怎么啦~

VS2017rc控制内存溢出

回复2楼见注释

int max(int x, int y)
{
    return(x > y ? x : y);
}

int min(int x, int y)
{
    return(x > y ? y : x);
}

#include<stdio.h>
int main()
{
    int max(int x, int y);
    int min(int x, int y);
    int(*p)(int x, int y) = NULL;//首先将没有确定指向的函数指针指向一个确定的、安全的“空”位置，否则p将指向一个无法预料的位置
    int a = 1, b = 2, n, c;
    int flag = 0;
    while (flag == 0)
    {
        printf("choose 1 or 2 in your select:\n");
        scanf_s("%d", &n);
        if (n == 1 || n == 2) flag = 1;
        else
        {
            printf("There is an error in your enter,please try again!\n");
            flag = 0;
        }
    }
    if (n == 1) p = max;
    else if (n == 2) p = min;
    c = (*p)(a, b);
    printf("a=%d,b=%d\n", a, b);
    if (n == 1) printf("max=%d\n", c);
    else if (n == 2) printf("min=%d\n", c);
    return 0;
}

int max(int x, int y)
{
    return(x > y ? x : y);
}

int min(int x, int y)
{
    return(x > y ? y : x);
}

#include<stdio.h>
int main()
{
    int max(int x, int y);    //这里声明局部函数，max。
    int min(int x, int y);    //以上类同。
    int(*p)(int x, int y) = NULL;    //声明函数的指针，也按要求先赋NULL。
    int a = 1, b = 2, n, c;    //声明变量。
    int flag = 0;    
    while (flag == 0)    
    {
        printf("choose 1 or 2 in your select:\n");
        scanf_s("%d", &n);
        if (n == 1 || n == 2) flag = 1;
        else
        {
            printf("There is an error in your enter,please try again!\n");
            flag = 0;
        }
    }            //flag控制输入的n只能是 1或 2
    if (n == 1) p = max;    //是 1 则把max函数的地址复制给p函数指针
    else if (n == 2) p = min;    //是 2 则把min赋给p，上同
    c = (*p)(a, b);    //然后就开始使用p函数指针，使用方法与原函数一样，返回结果给c
    printf("a=%d,b=%d\n", a, b);
    if (n == 1) printf("max=%d\n", c);    //最后输出结果c，并打印max
    else if (n == 2) printf("min=%d\n", c);    //上同，不过打印min
    return 0;
}

这段是演示函数指针的代码，这段代码是没问题，没错误的。

vs2017有免费版本吗

有，社区版

都是大侠啊