What happens when you attempt to compile and run the following code?
#include
#include
#include
#include
#include
using namespace std;
int main(){
int t[] ={ 3, 4, 2, 1, 6, 5, 7, 9, 8, 0 };
vector
map
for(vector
stringstream s; s<<*i<<*i; m.insert(pair
}
for(map
cout<<*i<<" ";
}
return 0;
}
What happens when you attempt to compile and run the following code?
#include
using namespace std;
template
void f(A &a)
{
cout<<1< } void f(int &a) { cout<<2< } int main() { int a = 1; f(a); return 0; }
What will happen when you attempt to compile and run the code below, assuming that file test.in contains the following sequence: 1 2 3?
#include
#include
#include
#include
#include
using namespace std;
template
ostream & out;
Out(ostream & o): out(o){}
void operator() (const T & val ) {out< int main () { ifstream f("test.in"); list for( ; f.good() ; ) { int i; f>>i; l.push_back(i); } f.close(); for_each(l.begin(), l.end(), Out return 0; } Program will output:
What happens when you attempt to compile and run the following code?
#include
using namespace std;
int main()
{
cout.setf(ios::hex, ios::basefield);
cout<<100.33<<" ";
cout.setf(ios::showbase);
cout<<100.33<<" ";
return 0;
}
Program outputs:
What happens when you attempt to compile and run the following code?
#include
#include
#include
using namespace std;
class B { int val;
public:
B(int v):val(v){}
int getV() const {return val;} bool operator < (const B & v) const { return val ostream & operator <<(ostream & out, const B & v) { out< template ostream & out; Out(ostream & o): out(o){} void operator() (const T & val ) { out< int main() { int t[]={8, 10, 5, 1, 4, 6, 2, 7, 9, 3}; set s1(t, t+10); sort(s1.begin(), s1.end()); for_each(s1.begin(), s1.end(), Out(cout));cout< return 0; } Program outputs: