int main(int argc, char* argv[])
{
char * buffer = new char[512];
int *p;
int a = 0;
p = &a;
cout << (void*)buffer << endl;
return 0;
}
...全文
683打赏收藏
请教一个void*的问题
用new分配一块buffer空间后,为何输出其地址需要将其转换成(void*)? #include using namespace std; int main(int argc, char* argv[]) { char * buffer = new char[512]; int *p; int a = 0; p = &a; cout << (void*)buffer << endl; return 0; }