為了理解該機制，我做了實驗：intmain(void){intd=3;inta=1,b=2,*c=&d;(int*)e=&a,f=&b;return0;}編譯成果返回c.c:Infunction‘main’:c.c:4:11:error:‘e’undeclared(firstuseinthisfunction)(int*)e=&a,f=&b;^c.c:4:11:note:eachundeclaredidentifierisreportedonlyonceforeachfunctionitappearsinc.c:4:19:error:‘f’undeclared(firstuseinthisfunction)(int*)e=&a,f=&b;(int*)e=&a,f=&b;為什么行不通？