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Hey, I'm revising C++ for a new job I'm trying to get. I'm having some trouble with some very basic stuff:
If I enter a value outside of those ranges it completely ignores the if statement and prints "You have entered 0". In other words it never reaches the else statement no matter what values I input!Code:if((aRegularInt > -2147483647) && (aRegularInt < 2147483646){
cout << "You have entered " << aRegularInt;
}
else{
cout << "That is not a valid Integer!";
}
Umm . . . (almost) no possible (standard) values that you could enter would cause the if statement to be false. It's true if the number is more than the lowest number possible and the number is less that the highest possible.
I'll give you some advice to avoid wasting your time: revising C++ for a job application is a waste of time and rather useless for any job. Instead focus on making an actual application that does something using the language you have available already.Quote:
Originally Posted by rickyoswaldiow
There is an error in your code too. You are missing a bracket on the end of your logical if statement
The problem is that an int can't store numbers larger or smaller than that on your platform, evidently. Try something like this instead:Quote:
If I enter a value outside of those ranges it completely ignores the if statement and prints "You have entered 0". In other words it never reaches the else statement no matter what values I input!
Code:#include <iostream>
int main(void) {
int x;
std::cout << "Enter the percentage that you got: ";
std::cin >> x;
if(x < 0 || x > 100) {
std::cout << "Wow, really?\n";
}
else {
std::cout << "Good for you.\n";
}
}
jverkoey : The job is actually for a games developer, the company use C++. I have a portfolio of apps I've made in other languages and I'm making a game in Dark Basic Pro from scratch to show I can create all the media and compile it. I'm also making a few pointless C++ apps to show that I understand the basic concepts of programming and show how fast (or slow) I can pick up the language. Thanks for the advice though :D
swgh : It's just a typo on my post, the syntax on the other computer is sound :)
So I can't check on values outside of the range of the integers maximum ranges?
The idea of the program is to show the range that an integer value can hold. The app explains the range of numbers that an integer can hold and then asks the user to input any valid integer value. If the value is outside the range of an integer then the "else" message is shown. Is it not possible to check if the user is entering a value within the range of that variable type with an IF statement?
I'm guessing that it's the 'cin << aRegularInt' that's causing this since if a value outside of the int ranges is entered it will simply stay at 0? If this is the case, how can I check if a value is within the range of an integer?
<climits> should have macros for INT_MAX and INT_MAX, as well as ranges for other types.
As for checking user input, one thing is clear, you can't take the input as int, because you can't put more in it than it can hold. You should take it as a string and then in you can validate in any way you want.Code:#include <iostream>
#include <climits>
int main()
{
std::cout << "Max int = " << INT_MAX << std::endl;
std::cout << "Min int = " << INT_MIN << std::endl;
std::cin.get();
}
If you just want to check if user input was valid, you can use something like that. (Note that it can't tell exactly what the exact problem with the input was.)
Code:#include <iostream>
int main()
{
int num = 0;
std::cout << "Enter an integer: ";
if (!(std::cin >> num)) {
std::cout << "Invalid input" << std::endl;
std::cin.clear();
}
else
std::cout << "Ok: " << num << std::endl;
std::cin.ignore();
std::cin.get();
}
>The idea of the program is to show the range that an integer value can hold.
Try this.
Code:#include <iostream>
#include <limits>
using namespace std;
int main()
{
long long num;
cout << "Please enter a integer between " << std::numeric_limits<int>::min()
<< " and " << std::numeric_limits<int>::max() << ": ";
cin >> num;
if (num < std::numeric_limits<int>::min() || num > std::numeric_limits<int>::max())
cout << "The number entered (" << num << ") is outside the range of an int" << endl;
else
cout << num << " is a valid int" << endl;
return 0;
}
Eh, Dev-C++ says: ISO C++ does not support `long long'. And if it did, the user could still enter a number that is out of the "long long" range, and you couldn't tell it from just invalid input.
Apart from that, this code tells me always that I have entered a valid integer - even if I type random characters. I guess if cin fails, num will just have an uninitialized value.
Ah brilliant :)
That's exactly what I needed to know, I'll have to go try it out now on the other computer. Unfortunately this is my flatmates laptop, I don't have net access :/
One question, you've included "using namespace std" yet you prefix some of your commands with std::. Just a typo or is there a reason for this?
See, when you read in a value beyond the max value it wraps around.
Say you have 4 bits for your number, and it is unsigned.
1111
is the highest number, which is 15.
one more and you have
10000
but since it is limited to 4 bits, you get
0000
But don't count on the wrap around value of a unsigned number, because how they are represented in memory is undefined.
I understand.
Anon: The code checks the length of a regular int against the user inputed long int.
A long int can't hold all values. A string could any value a user would put in without bashing the keyboard for several hours to get enough characters to fill up memory and swap.But, then you would have to manually parse through it to check to see how large it is. But that prevents wrap around saying it was a valid number but when it just wraped around so much it fell back in normal bounds.
>Eh, Dev-C++ says: ISO C++ does not support `long long'.
What switches are you using? I'm using Dev-C++, and it compiles fine.
More of a typo than anything. "using namespace std" was necessary since I used cout vs std::cout, and endl vs std::endl. But since numeric_limits is less commonly used, I added the std for emphasis.Quote:
One question, you've included "using namespace std" yet you prefix some of your commands with std::. Just a typo or is there a reason for this?
so to declare a long int in Borland C++ 5.5 I use the syntax 'long int <name>'? and is a short 'short int <name>'?
>A long int can't hold all values.
Yeah, that's true. If you enter a 20 digit integer, it's not going to work. A universal solution would probably require reading it as a string, then using a series of if statements.
>so to declare a long int in Borland C++ 5.5 I use the syntax 'long int <name>'? and is a short 'short int <name>'?
I believe it would be __int64 (two underscores).
It doesn't seem to be so with cin. If cin >> num fails, it seems to leave num as it is.Quote:
See, when you read in a value beyond the max value it wraps around.
Anyway, here's what I would try:Code:int num = 42;
cout << "Please don't enter a valid integer: ";
cin >> num;
cout << num; //prints 42
Code:#include <iostream>
#include <string>
#include <sstream>
enum {NOT_INTEGER = 0, OUT_OF_RANGE, OK};
int CheckRange(const std::string & a)
{
//If anybody knows a simpler way, please tell so!
//test 1: is it fully convertible
int num;
std::stringstream test(a);
if (test >> num && test.eof()) return OK;
/*test 2a: minus sign should be at pos 0 if anywhere*/
size_t pos = a.rfind('-');
if (pos && pos != std::string::npos) return NOT_INTEGER;
/*test 2b: if minus sign is at pos 0, string must be
longer than 1 character*/
if (pos == 0 && a.size() == 1) return NOT_INTEGER;
/*test 3: find other non-digits*/
if (a.find_first_not_of("-1234567890") == std::string::npos)
return OUT_OF_RANGE;
else
return NOT_INTEGER;
}
int main()
{
std::string s;
while (1) {
std::cout << "Enter an integer: ";
std::cin >> s;
switch (CheckRange(s)){
case OK:
std::cout << "That's a valid integer." << std::endl;
break;
case OUT_OF_RANGE:
std::cout << "Numeric but out of range." << std::endl;
break;
case NOT_INTEGER:
std::cout << "That's not an integer." << std::endl;
break;
}
}
}
Here's a crude method. It doesn't check for nonnumeric input though.
Code:#include <iostream>
#include <limits>
#include <cstring>
#include <sstream>
using namespace std;
int main()
{
string s_num;
cout << "Please enter a integer between " << std::numeric_limits<int>::min()
<< " and " << std::numeric_limits<int>::max() << ": ";
cin >> s_num;
ostringstream oss;
oss << std::numeric_limits<int>::max();
string int_max = oss.str();
if (s_num[0] == '-')
{
s_num = s_num.substr(1);
}
if (s_num.length() == 10)
{
if (s_num > int_max)
cout << "Outside range." << endl;
else if (s_num <= int_max)
cout << "Valid integer." << endl;
}
else if (s_num.length() < 10)
cout << "Valid integer." << endl;
else
cout << "Outside range." << endl;
return 0;
}
Okay great, I get it. I'll just stick with using a long int for the input and checking it against numeric_limits<int>. The program doesn't need to be totally foolproof and I've only been learning c++ for two days :)
Maybe I'll make it foolproof after I've worked on arrays and functions, thanks for the help!
If by a long int you mean long, be warned: very often it would be just as large as a regular int, and you'd gain nothing.Quote:
Originally Posted by rickyoswaldiow
If you mean __int64 and your compiler accepts it: why bother to check if the input is in range of int? I mean, you'd still have to validate the input otherwise (the user might enter non-numbers anyway). But why would you be interested to restrict your user to 32 bits, if you can use and are using 64-bit integers?
You could just as easily demonstrate your understanding of checking ranges with other examples (e.g precentages as suggested above). It might also be more practical to learn what to do if user enters "fourty-two" where an int is expected.
anon: it's "forty-two" ;)
Try this instead, if you want to hold larger numbers. Now if the user happens to enter a number which just won't fit into the RAM...