示例

int main(int argc, char* argv[]) { if(argc != 2){ cout << "argc  was not 2.. "; }        END: cout << "done"; return 0; }

解决方案

可以除去不使用的标签。 int main(int argc, char* argv[]) { if(argc != 2){ cout << "argc  was not 2.. "; }        cout << "done"; return 0; }