E.g: in your function Class_Grade() you're want to access the variable grade1, grade2, grade3 which you previously declared in the main-function, but this isn't allowed in C, you should pass these values as an argument, this is an example of a modified Class_Grade() function:
int Class_Grade()
{
int g1, g2, g3, gradeavg;
printf("Please enter Class grade");
scanf("%d%d%d",&g1,&g2,&g3);
gradeavg=(g1+g2+g3)/3;
/* Put g1+g2+g3 between brackets because of the algebraic rules of math */
return gradeavg;
}
Your main() function misses a '}' at the end, your functions haven't specified their return value ... If your function isn't returning a value then you should declare it as void: void [I]yourfunction[/I] Your main function doesn't return a value, declare it as int main() :
int main(void)
{
/* The code of your main function here */
return 0;
}
It's looking for a compiler called g++, but comes up with an error because it can't find it. I think the right one that really does exist is the gcc compiler. If so how do I tell it to use that one?
Did you set the MINGDIR and PATH environment variables correctly ?
That is not intended for a 'perfect' system, it's only a disclaimer. If you want to bypass it, it's your problem, you're violating the law. Look the little legend on every drug, that doesn't avoid to swallow a pill by a minor.
Agreed, but I only wanted to mention that this system isn't going to prevent anyone to view your site's contents ...
What's so confusing to you? I would recommend to create a design of what your application has to be able to do and how you're going to achieve this before going onto the real coding part ...
For legal purposes? You can quickly bypass this mechanisms, if you really want to be sure you should use credit card validation or something (but keep in mind that this will discourage people to continue) ...
That's doable with simple conditionals. The most important field is year. We assume if is not > 1991 or 91. You wan to check boundaries, also use strtotime('-18 years')
I know, but it isn't pretty for the user to each time select his day of birth, then the month and thereafter the year ...
Yup, first read the forum rules and by the way people here usually don't respond to questions like: 'Please I want someone to help me'
You need to show us what you've done so far (forum rules), it's a nice assignment and it isn't too difficult, so give it a shot, ask specific questions about what exactly you don't understand and we'll help you :) to get a step nearer to the solution ...
In C++ you don't have to supply the 'struct' word each time before you declare a struct-variable, though it's still allowed to do so, you can save some typing :) ...
Edit:: I don't get what you mean, do you have compiler errors ? (please post them if so)
Yup, fstream's open-method only accepts a c-string as it's argument for the filename, so you'll have to convert your C++ string to a c-string first, you can easily do this by using the c_str() method of the string class :) ...
void trim(string & s)
{
int strlen = s.length();
string tmp;
for(int i = 0; i < strlen; i++)
if(s[i] != ' ') tmp += s[i];
s = tmp;
}
> Why are you using a seperate counter 'k', you can just add something to a string by using the '+' operator ...
> Also avoid the use of system("pause"); (look here for the reasons), it's better to use cin.get(); instead ...
> Edit:: Your mistake is on the line where you say: if (copy[i] = ' ') , it has to be if (copy[i] == ' ') , if you write if (copy[i] = ' ') , copy[i] will get the value ' ' , so the condition is always true, then you're running the continue statement everytime which is just skipping to the next i of the loop ...
But because the string copy is in the beginning (of your function) just an exact copy of the string s, the string s' value will remain the same :) ...
> Edit:: You're just overwriting string s, that causes a bad result, the last part of the 'trimmed' string will be the same as the original, but it still contains the whole (trimmed) expression ... ==> If someone can explain this in a better way please do so, as I'm not very good at it :P ...
I tested this program and can confirm it works, but try running it with the wrong number of arguments (1 instead of 2) directly causes the program to crash, it's very 'crash-sensitive' ...
The program is also using 'old C++' code, for example old file handling using the FILE-pointer, etc. ...
I think Microsoft backup is able to do that, but it will be a more pleasant experience if you just use Acronis TrueImage which is designed for that purpose ...
> Get the input (the age) from the user > Use some conditional statements ( if(condition) { /... } ) to check whether the script has to grant access or not > Now you only need a command which redirects the user to a page in case the age validation has failed ...
For the first case: If your salary is 1000.23 it has to be 1000.25: We multiply 1000.23 by 20, round it off to the nearest integer and divide it back by 20 :) ...
Edit:: In the second case (if your salary is 1000.26 and it has to rounded off to 1000.30): the Math.round function should work fine here !
As part of an assignment I am creating a class to handle rationals with numerators and denominators, overloading the operators for cin and cout, as well as +-*/. So these are the functions for those operators, now where exactly should I include these in my class?
Rational addition(Rational a, Rational b){
Rational result = Rational((a.numerator()*b.denominator() + b.numerator()*a.denominator()), a.denominator()*b.denominator());
return result;
}
Rational subtraction(Rational a, Rational b){
Rational result = Rational((a.numerator()*b.denominator() - b.numerator()*a.denominator()), a.denominator()*b.denominator());
return result;
}
Rational multiplication((Rational a, Rational b){
Rational result = Rational(a.numerator()*b.numerator(), a.denominator()*b.denominator());
return result;
}
Rational division((Rational a, Rational b){
Rational result = Rational(a.numerator()*b.denominator(), a.denominator()*b.numerator());
return result;
}
You aren't overloading any operators in this way ...
