Any Idea why this if statment isn't working? [closed] - c++

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I know it's probably a simply mistake, but I am rather new at this. Any idea why this might not be working.
#include <iostream.h>
#include <conio.h>
int main()
{
float x, y;
cout << "Please enter the miles traveled:", x;
cin >> x;
if (x <= 0)
cout << "Invalid number please try again.";
if (x > 1) {
y = ((x * .75) + 2);
cprintf("Your total amount owed is $%.2f"), y;
}
getch();
return 0;
}

Answer 1

First, you have some minor mistakes in your program.
In the line
cprintf("Your total amount owed is $%.2f"), y;
you should put the y inside the function call, i.e.
cprintf("Your total amount owed is $%.2f", y);
Also, the , x in the line
cout<< "Please enter the miles traveled:", x;
should be removed. Both , x and , y in the above quoted lines have no effect at all.
Then, you should add the return type int to your main() function:
int main() {
...
}
Finally, since you are using cin and cout from the namespace std, you should either write std::cin and std::cout respectively, or put a using namespace std; after your includes, or put using std::cin; and using std::cout; after the includes (the last option is recommended).
You also shouldn't #include <iostream.h> (I don't know that one, it is non-standard) but rather <iostream>, and I guess if you change that, explicitly mentioning the namespace, like in the above point, is then required.
So far, the program should "work", but I guess not in the way you want it to behave. You probably want the program to repeat after the user made an input error. For this, you should use a loop. Something like this should do the job:
cin >> x;
while (x <= 0)
{
cout << "Invalid number please try again.";
cin >> x;
}
Also, what should your program do if the user enters a number between 0 and 1? I guess instead of x > 1 you want x > 0, i.e. just the opposite of the "error case" x <= 0. This is usually expressed with an else block. But if you write a loop repeating until the user entered a positive x, you can be sure that it is really positive after the loop.

Answer 2

When declaring or defining a function it's required to have a type, be it void or auto while auto is, strictly-speaking, no type. A strictly-conforming main function is int main() or int main(int, char**).
This line
cout<< "Please enter the miles traveled:", x;
looks dubious. What's the x supposed to do here? Remove the , x part.
This line
cprintf("Your total amount owed is $%.2f"), y;
is bad. Pass y inside the parantheses, like
cprintf("Your total amount owed is $%.2f", y);
cout etc. are from the std namespace. Either add a std:: before all of them or put a using namespace std; before main.
This line
y=((x*.75)+2);
isn't necessarily wrong but it multiplies a float with a double, killing off precision. Use .75f instead.
float y is only used in one if-clause, so narrow down its scope to that.
<iostream.h> isn't guaranteed to exist by the C++ standard. It's <iostream>.
Remove that <conio.h> inclusion. That's wicked because non-standard! Read this instead.
Proceeding with the 8th point, don't use all that non-standard C-ish crap. C++ is not C!

Answer 3

int main()
you haven't mentioned return type of main()
cprintf("Your total amount owed is $%.2f", y);
remove
getch()
and
cout<< "Please enter the miles traveled: " << x;
EDIT
Also with gcc you cannot use conio.h

Answer 4

main, being a function, needs to return type int (void too will work in some systems). The x and the y following the cout/cprintf statements, should be removed. You may also want to enclose the content of the first if statement into a while loop, until the desired condition becomes true.
#include <iostream>
using namespace std;
int main()
{
float x, y;
cout << "Please enter the miles traveled: ";
cin >> x;
while (x <= 0.f) {
cout << "Invalid number please try again.";
cin >> x;
}
y = ((x*.75) + 2);
cout << "Your total amount owed is : " << y;
return 0;
}