为什么这个代码会报这个错？(语言-c语言)

Question

#include<stdio.h>
int main()
{
    enum Color {red, yellow, blue, white, black};
    enum Color i, j, k, pri;
    int n, loop;
    n = 0;
    for(i = red; i <= black; i++)
        for(j = red; j <= black; j++)
        if(i != j)
    {
        for(k = red; k <= black; k++)
            if((k != i) && (k != j))
        {
            n = n + 1;
            printf("%-4d", n);
            for(loop = 1; loop <= 3; loop++)
            {
                switch(loop)
                {
                    case 1: pri = i; break;
                    case 2: pri = j; break;
                    case 3: pri = k; break;
                    default: break;
                }
                switch(pri)
                {
                    case red:printf("%-10s", "red"); break;
                    case yellow:printf("%-10s", "yellow"); break;
                    case blue:printf("%-10s", "blue"); break;
                    case white:printf("%-10s", "white"); break;
                    case black:printf("%-10s", "black"); break;
                    default: break;
                }
            }
            printf("\n");
        }
    }
    printf("\ntotal: %5d\n", n);
    return 0;
}

Answer 1

基于Monster 组和GPT的调写：
是因为在C语言中，枚举类型是不支持自增或自减运算的，但在该代码的第15行、第16行和第19行中，枚举变量 i、j 和 k 都进行了自增运算，导致编译器报错。想要遍历枚举类型的值,用循环和枚举类型中的变量来实现，而不是使用自增或自减运算，改成：

for(i = red; i <= black; i = (enum Color)(i + 1))
    for(j = red; j <= black; j = (enum Color)(j + 1))
        if(i != j)
{
    for(k = red; k <= black; k = (enum Color)(k + 1))
        if((k != i) && (k != j))
    {
        // 循环体
    }
}

完整代码：

#include<stdio.h>
int main()
{
    enum Color {red, yellow, blue, white, black};
    enum Color i, j, k, pri;
    int n, loop;
    n = 0;
    for(i = red; i <= black; i = (enum Color)(i + 1))
    for(j = red; j <= black; j = (enum Color)(j + 1))
        if(i != j)
{
    for(k = red; k <= black; k = (enum Color)(k + 1))
        if((k != i) && (k != j))
    {
            n = n + 1;
            printf("%-4d", n);
            for(loop = 1; loop <= 3; loop++)
            {
                switch(loop)
                {
                    case 1: pri = i; break;
                    case 2: pri = j; break;
                    case 3: pri = k; break;
                    default: break;
                }
                switch(pri)
                {
                    case red:printf("%-10s", "red"); break;
                    case yellow:printf("%-10s", "yellow"); break;
                    case blue:printf("%-10s", "blue"); break;
                    case white:printf("%-10s", "white"); break;
                    case black:printf("%-10s", "black"); break;
                    default: break;
                }
            }
            printf("\n");
        }
    }
    printf("\ntotal: %5d\n", n);
    return 0;
}