Question

I have written a function that converts a decimal number to a binary number. I enter my decimal number as a long long int. It works fine with small numbers, but my task is to determine how the computer handles overflow so when I enter (2^63) – 1 the function outputs that the decimal value is 9223372036854775808 and in binary it is equal to -954437177. When I input 2^63 which is a value a 64 bit machine can’t hold, I get warnings that the integer constant is so large that it is unsigned and that the decimal constant is unsigned only in ISO C90 and the output of the decimal value is negative 2^63 and binary number is 0. I’m using gcc as a compiler. Is that outcome correct?

The code is provided below:

#include <iostream>
#include<sstream>
using namespace std;
int main()
{
long long int answer;
long long dec;
string binNum;
stringstream ss;
cout<<"Enter the decimal to be converted:"<< endl;;
cin>>dec;
cout<<"The dec number is: "<<dec<<endl;
while(dec>0)
{
     answer = dec%2;
     dec=dec/2;
     ss<<answer;
     binNum=ss.str();
}
cout<<"The binary of the given number is: ";
for (int i=sizeof(binNum);i>=0;i--){
     cout<<binNum[i];}
return 0;
    }

Answer 1

First, “on a 64-bit computer” is meaningless: long long is guaranteed at least 64 bits regardless of computer. If could press a modern C++ compiler onto a Commodore 64 or a Sinclair ZX80, or for that matter a KIM-1, a long long would still be at least 64 bits. This is a machine-independent guarantee given by the C++ standard.

Secondly, specifying a too large value is not the same as “overflow”.

The only thing that makes this question a little bit interesting is that there is a difference. And that the standard treats these two cases differently. For the case of initialization of a signed integer with an integer value a conversion is performed if necessary, with implementation-defined effect if the value cannot be represented, …

C++11 §4.7/3:
“If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined”

while for the case of e.g. a multiplication that produces a value that cannot be represented by the argument type, the effect is undefined (e.g., might even crash) …

C++11 §5/4:
“If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined.”

Regarding the code I I only discovered it after writing the above, but it does look like it will necessarily produce overflow (i.e. Undefined Behavior) for sufficiently large number. Put your digits in a vector or string. Note that you can also just use a bitset to display the binary digits.

Oh, the KIM-1. Not many are familiar with it, so here’s a photo:

It was, reportedly, very nice, in spite of the somewhat restricted keyboard.