Convert binary to hex value (32 bits) - c++

Question

i am trying to convert binary value as hex value and i got the following code and it is working well up to 28 bit but not for 32 bit.
The code is as follows.
int main()
{
long int longint=0;
string buf;
cin>>buf;
int len=buf.size();
for(int i=0;i<len;i++)
{
longint+=( buf[len-i-1]-48) * pow((double)2,i);
}
cout<<setbase(16);
cout<<longint;
return 0;
}
If I input 28 '1' (111111111111111111111111) then the output is fffffff
but if i input 32 '1' (11111111111111111111111111111111) then the output is 80000000.
Can anyone please explain why this is happenning and also in the above code why 48 is subtracted .

Answer 1

The problem seems to be with the use of pow, which uses floating-point math, if I recall correctly.. You may be running into issues with overflow.
A more elegant way to calculate powers of two is by using bit-shifts:
2^0 = 1 << 0 = 1
2^1 = 1 << 1 = 2
2^2 = 1 << 2 = 4
2^n = 1 << n

Answer 2

You are using int32 and it is getting out of range when you use it for 32 bytes,try using int64 i.e long long
unsigned long long longint=0; //Change Here
string buf;
cin>>buf;
int len=buf.length();
for(int i=0;i<len;i++)
{
longint+=( buf[len-i-1]-48) * pow((double)2,i);
}
cout<<setbase(16);
cout<<longint;

Answer 3

As Nathan's post, it'll display correct when changing your code like this,
longint += (buf[len-i-1]-'0') << i;

Answer 4

This is because you have forced ( buf[len-i-1]-48) * pow((double)2,i) to be converted to double, and double is 8 bytes long, but it has to store extra information, it can not be full charged to represent 0x80000000, you can find more information here and your last ( buf[len-i-1]-48) * pow((double)2,i) (when i is 31) expression already overflow.
But something weard happend when converting from 4294967295.0000000(which is 0xffffffff) to int , it just come out 0x80000000 , but I am very sorry I don't know why（Please reference the comment from TonyK）.
You can change it to
longint+=(long int)(( buf[len-i-1]-48) * pow(2,i));
Why minus 48 ? because ascii for '0' is 48, you want to convert from literal '0' to numeric 0, you have to do it.

Answer 5

The reason is that float-value is with limit precision and pow() computer use numerical calculation approximation which isn't definitely precise. To get a precise value, you should use bit-wise ">>" instead.
You can see how pow() function works as below.
I changed your code
longint+= ( buf[len-i-1]-48) * pow((double)2,i)
to the code below, which is equal, becase pow() return a double value.
double temp = ( buf[len-i-1]-48) * pow((double)2,i);
longint+= temp;
cout<<setbase(16);
cout<<"temp"<<endl;
cout<<temp<<endl;
cout<<longint<<endl;
the output is as below
temp
1.34218e+08
fffffff
temp
2.68435e+08
1fffffff
temp
5.36871e+08
3fffffff
temp
1.07374e+09
7fffffff
temp
2.14748e+09
80000000
final
80000000
Which shows clearly pow() is with limited precision. 2.14748e+09 is not equal to (2^31).
You should use ">>" which is the best or just use conversion to integer which isn't 100 percently correclty, either.
You can see conversion as below.
when I change
double temp = ( buf[len-i-1]-48) * pow((double)2,i);
to
int temp = ( buf[len-i-1]-48) * pow((double)2,i);
the result is
temp
8000000
fffffff
temp
10000000
1fffffff
temp
20000000
3fffffff
temp
40000000
7fffffff
temp
80000000
ffffffff
final
ffffffff
Which works correctly.
substract 48
You got a char from standard input, for example, you got '1' from terminal instead of 1. In order to get 1. You should use '1'-'0'.
The reason : computer store '0'～'9' as a byte with value（48～57）. As a result， '1' - '0' equals '1' - 48.