How to avoid integer overflow?

Question

In the following C++ code, 32767 +たす 1 =わ -ひく32768.

#include <iostream>
int main(){
short var = 32767;
var++;
std::cout << var;
std::cin.get();
}

Is there any way to just leave "var" as 32767, without errors?

• blu: Do you get whats going on here? You've hit the ceiling for an integer (short), so adding one more flips the signs to the maximum negative amount for an integer.
• dmazzoni: If you want a different mode where integers don't automatically wrap around, it doesn't exist, sorry.
• James McNellis: This reminds me of why I can't sleep. :-)
• MSalters: This is called saturation arithmetic. It can be highly illogical, so be warned. What's 32767+1-1?

Answer 1

Yes, there is:

if (var < 32767) var++;

By the way, you shouldn't hardcode the constant, use numeric_limits<short>::max() defined in <limits> header file instead.

You can encapsulate this functionality in a function template:

template <class T>
void increment_without_wraparound(T& value) {
 if (value < numeric_limits<T>::max())
 value++;
}

and use it like:

short var = 32767;
increment_without_wraparound(var); // pick a shorter name!

Answer 2

#include <iostream> 
int main(){ 
unsigned short var = 32767; 
var++; 
std::cout << var; 
std::cin.get(); 
} 

• i_am_jorf: This answers the question that was asked, but not, I suspect, the question that was intended.

Answer 3

use 'unsigned short int' or 'long int'

#include <iostream>
int main(){
long int var = 32767;
var++;
std::cout << var;
std::cin.get();
}

• Wooble: Then x = 32676; x++ will result in x == 32677, not 32676.
• lornova: OK, so how do you avoid the overflow on an unsigned 64 bit integer?