int x=10,y=3,z; cout

int x=10,y=3,z; cout int w=1,x=2,y=3,z=4; cout int x=1,y=2,z=3; x=x^z; y=y^z; z=x^y; cout int x=-5,y;cout '++' needs l-value int main(){int x=2,y=3;char ch='a';float z=1.1;y=(y+x)++;cout int c=7,d=8,z=3,a,x,y; a=c^d; x=c^d&-z; y=a&-z; cout #include using namespace std; int main() { int a,b,c; a=3; int f(int x,int y,int z);#include using namespace std; int main() {int a,b,c;a=3;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout #include void mian() { int x,y; x=3; y=5*(++x); cout int a=3；cout x=1; y=2; z=3; x+=y+=z; cout int max(int x,int y) { int z; if(x>y) z=x; else z=y; return(z); struct st{int x,int*y;}*p; int s[]={5,6,7,8} st a[]={10,&s[0],20,&s[1]30,&[2],40,&s[3]} main( ) {p=a;cout Ｃ＋＋两个数相除保留两位小数怎么回事? #include using namespace std; int main() { double x,y,z; cin>>x>>y; z=x/y; cout 3、以下正确的函数定义是( ).A.double fun(int x,int y); {int z; z=x+y; return z；}3、以下正确的函数定义是( ).A.double fun(int x,int y); {int z; z=x+y; return z；}B.fun(int x,y) {int z; return z；}C.double fun( c/c++类的问题 下面有详细才程序 B(int i,int j):A(i){y=j;}这一句 #includeclass A{int x;int z;public:A(int i,int j){x=i;z=j;}void dispa(){cout 若有宏定义# define s(x) x*x-x,设int k=3; 问cout 求高手解释一下这段程序#include class Test {int x,y; public: Test(int i,int j=0) {x=i;y=j;} int get(int i,int j) {return i+j;} }; void main() {Test t1(2),t2(4,6); int (Test::*p)(int,int=10); p=Test::get; cout int x=0;y=0;z=0;x=(y=3;z=5*y,y=10);x=?y=?z=?