can someone please point out my mistakes , I don't get what is wrong.

Question

/*i want to make a hour to min convertor , but for some reason there are piles of errors showing up ,much bigger than my code
plz only point out my mistakes */
#include <iostream>

void tomin(int *htpr)
{
*htpr*=60;
std::cout << min;
}

int main()
{
int hours;
std::cin>>hours;

int *htpr;
htpr = &hours;

std::cout << tomin(htpr);
}

Answer 1

Yes. And You can use int return type and return calculated result......

Answer 2

// this is my eddited code but still not giving out the exoected output

#include <iostream>

int tomin(int *htpr)
{
*htpr*=60;

}

int main()
{
int hours ;
std::cin>>hours;

int *htpr;
htpr = &hours;

std::cout << tomin(htpr);

std::cout << hours;

return 0;
}

Answer 3

Your function is void type so it don't return anything.. But you expecting to return a value in last statement.
also min is undfined in function. first declars it then assign value and use to print.

Answer 4

you mean the cout function do0es't work in void

Answer 5

what if I add int tomin(*hptr)

Answer 6

if i use
std::cout << hours ;
in function the outcome is the same

Answer 7

#include <iostream>

int tomin(int *htpr)
{
*htpr*=60;
return *htpr; //because return type int, you should return a int value.
}

int main()
{
int hours;
std::cin>>hours;
int *htpr;
htpr = &hours;

std::cout << tomin(htpr);
std::cout << hours;
//both output same
return 0;
}

//since you are doing pass by pointers, this is enough.

#include <iostream>

void tomin(int *htpr)
{
*htpr*=60;
}

int main()
{
int hours;
std::cin>>hours;

int *htpr = &hours;

tomin(htpr);

std::cout << hours;
return 0;
}
// without *htpr, use as tomin(&hour);