int j,a[]={1,3,5,7,9,11,13,15,#include main(){ int j,a[]={1,3,5,7,9,11,13,15},*p=a+5;for(j=3; j; j--){ switch(j){ case 1:case 2:printf("%d",*p++); break;case 3:printf("%d",*(--p));}}}答案是9911,是怎样来?

int j,a[]={1,3,5,7,9,11,13,15,#include main(){ int j,a[]={1,3,5,7,9,11,13,15},*p=a+5;for(j=3; j; j--){ switch(j){ case 1:case 2:printf("%d",*p++); break;case 3:printf("%d",*(--p));}}}答案是9911,是怎样来?
int j,a[]={1,3,5,7,9,11,13,15,
#include
main()
{ int j,a[]={1,3,5,7,9,11,13,15},*p=a+5;
for(j=3; j; j--)
{ switch(j)
{ case 1:
case 2:printf("%d",*p++); break;
case 3:printf("%d",*(--p));
}}}
答案是9911,是怎样来?

int j,a[]={1,3,5,7,9,11,13,15,#include main(){ int j,a[]={1,3,5,7,9,11,13,15},*p=a+5;for(j=3; j; j--){ switch(j){ case 1:case 2:printf("%d",*p++); break;case 3:printf("%d",*(--p));}}}答案是9911,是怎样来?
j=3时执行case 3: printf("%d",*(--p));(循环前p指针指向a[5])这时指向a[4];访问a[4]的值为9;