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//假设题没有出错是自己理解偏了,那么强行理解
int*min(int x,int y)
{
return *((int*)x)>*((int*)y)?(int*)x:(int*)y;
}
#include <stdlib.h>
#include <stdio.h>
int result;
int*
mymin(int x,int y){
result = x < y ? x : y;
return &result;
}
int
main(int argc,char **argv)
{
int z;
z = *(mymin(3,6));
printf("z = %d\n",z);
exit(0);
}