给定这样一个函数boolRequestStatus(){...if(code==myCode){returntrue;}else{returnfalse;}}为什么编译器会提示“函数应该返回值”。除非我遗漏了什么，否则它怎么可能不返回true或false？是不是因为myCode的值依赖于运行时，所以编译器不确定逻辑路径？ 最佳答案 如果您编写return(code==myCode);，您将节省行数，让编译器满意，并且通常以更像C++的风格编写。 关于c++-函数应该返回值问题，我们在St

我试图在以下非常简单的程序中引起缓冲区溢出:#include#includevoidbadf(intn,charc,char*buffer){charmycode[]={0xeb,0x0f,0xb8,0x0b,0x00,0x00,0x00,0x8b,0x1c,0x24,0x8d,0x0c,0x24,0x31,0xd2,0xcd,0x80,0xe8,0xec,0xff,0xff,0xff,0x2f,0x62,0x69,0x6e,0x2f,0x6c,0x73,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};//37bytes//OverwriteBaseP

我试图在以下非常简单的程序中引起缓冲区溢出:#include#includevoidbadf(intn,charc,char*buffer){charmycode[]={0xeb,0x0f,0xb8,0x0b,0x00,0x00,0x00,0x8b,0x1c,0x24,0x8d,0x0c,0x24,0x31,0xd2,0xcd,0x80,0xe8,0xec,0xff,0xff,0xff,0x2f,0x62,0x69,0x6e,0x2f,0x6c,0x73,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};//37bytes//OverwriteBaseP