int *p,m=5,n; scanf("%d",&n); *p=n; 上述程序段为何不正确环境VC++6.0

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int *p,m=5,n; scanf(

int *p,m=5,n; scanf("%d",&n); *p=n; 上述程序段为何不正确环境VC++6.0
int *p,m=5,n; scanf("%d",&n); *p=n; 上述程序段为何不正确
环境VC++6.0

int *p,m=5,n; scanf("%d",&n); *p=n; 上述程序段为何不正确环境VC++6.0
*p=n;操作之前,p没有指向有效的地址,所以出错.
如果前面有p=&m;操作,p指向了m的地址,*p=n;就没问题了.
如果这里是要求p指向n的地址,那应该这样p=&n;

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