📄 subject_14773.htm
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序号:14773 发表者:流浪 发表日期:2002-09-11 12:59:15
<br>主题:高手帮帮忙啊!
<br>内容:enum color{red,yellow,blue,white,black}; <BR>enum color i,j,k,pri;<BR>.<BR>.<BR>.<BR>for(i=red;i<=black;i++) //用变量i和j来做循环变量<BR>.<BR>.<BR>.<BR><BR>编译器报:<BR>'enum main::color' does not define this operator or a conversion to a type acceptable to the predefined operator<BR><BR>但是我想要对enum color类型的变量i和j做自增运算,我该怎么做啊?<BR>
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回复者:dongfanghong 回复日期:2002-09-11 15:08:54
<br>内容:enum color{red,yellow,blue,white,black}; <BR>//enum color j,k,pri;<BR>for(int i=red;i<=black;i++) //用变量i和j来做循环变量<BR>{<BR>}
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回复者:流浪 回复日期:2002-09-11 16:00:14
<br>内容:我的程序是这样的:输出五种颜色的组合方案,后面还要用到i和j变量。int main(int argc, char* argv[])<BR>{<BR> enum color{red,yellow,blue,white,black}; enum color i,j,k,pri;<BR> int n,loop;<BR> n=0;<BR> for(i=red;i<=black;i++) <BR> for(j=red;j<black;j++)<BR> if(i!=j)<BR> {<BR> for(k=red;k<=black;k++)<BR> if((k!=i)&&(k!=j))<BR> {<BR> n=n+1;<BR> printf("%d-4d",n);<BR> for(loop=1;loop<=3;loop++)<BR> {<BR> switch(loop)<BR> {<BR> case 1:pri = i;break;<BR> case 2:pri = j;break;<BR> case 3:pri = k;break;<BR> default :break;<BR> }<BR> switch(pri)<BR> {<BR> case red:printf("$-10s","red");break;<BR> case yellow:printf("$-10s","yellow");break;<BR> case blue:printf("$-10s","blue");break;<BR> case white:printf("$-10s","white");break;<BR> case black:printf("$-10s","black");break;<BR> default :break;<BR> }<BR> printf("\n");<BR> }<BR> }<BR> printf("\ntotal:%5d\n",n);<BR> }<BR> return 0;<BR>}
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<font color=red>答案被接受</font><br>回复者:dongfanghong 回复日期:2002-09-11 16:52:55
<br>内容:enum color{red,yellow,blue,white,black}; <BR>enum color pri;<BR> int n,loop;<BR> n=0;<BR> for(int i=red;i<=black;i++) <BR> for(int j=red;j<black;j++)<BR> if(i!=j)<BR> {<BR> for(int k=red;k<=black;k++)<BR> if((k!=i)&&(k!=j))<BR> {<BR> n=n+1;<BR> printf("%d-4d",n);<BR> for(loop=1;loop<=3;loop++)<BR> {<BR> switch(loop)<BR> {<BR> case 1:pri = (enum color)i;break;<BR> case 2:pri = (enum color)j;break;<BR> case 3:pri = (enum color)k;break;<BR> default :break;<BR> }<BR> switch(pri)<BR> {<BR> case red:printf("$-10s","red");break;<BR> case yellow:printf("$-10s","yellow");break;<BR> case blue:printf("$-10s","blue");break;<BR> case white:printf("$-10s","white");break;<BR> case black:printf("$-10s","black");break;<BR> default :break;<BR> }<BR> printf("\n");<BR> }<BR> }<BR> printf("\ntotal:%5d\n",n);<BR> }<BR>
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回复者:流浪 回复日期:2002-09-11 17:22:47
<br>内容:谢谢您了。高明!!!
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回复者:绝对烟头 回复日期:2002-10-06 01:04:30
<br>内容:不好意思,我真不知道这两段程序有什么区别.
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回复者:流浪 回复日期:2002-10-08 09:02:36
<br>内容:首先自定义联合体变量不能进行自加或自减操作,在定义循环控制变量的时候,原来的程序将循环变量定义为联合体,这样做是不正确的;后来的程序是将循环变量修改为整形变量,然后将他们进行强制转换,得到想要的结果,例如 for(int i=red;i<=black;i++)... case 1:pri = (enum color)i;break;<BR>
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