So, with the given code, we can say that c[0]=0x10 and c[1]=0x11.Code:#include <iostream> using namespace std; int main() { char * c = new char[1]; *(c+0) = 0x10; *(c+1) = 0x11; unsigned int i = *(c); cout << i; return 0; }
My question is, how could I combine both values into a single integer(2bytes)?
What I want is my program to output 4113 (0x1011)
Don't create a new thread with the same question because you aren't getting an answer. That is a very sure way of getting both threads locked. Now the size of int, short, ect. are implementatino specific so be sure to verify your implementation. Note my use of short instead of unsigned int.
Code:#include <iostream> using namespace std; int main() { char * c = new char[1]; short result=0; *(c+0) = 0x10; *(c+1) = 0x11; for(short i = 0; i < sizeof(i); i++) result = (result << 8) + c[i]; short j = 0x1011; cout << result << " " <<j; cin.get(); }
Last edited by AndrewHunter; 07-02-2011 at 06:01 PM. Reason: Fixed spacing
Originally Posted by anduril462
Warning: Some or all of my posted code may be non-standard and as such should not be used and in no case looked at.
This question is actually slightly different to the other.
If we ignore the problem of writing two characters of a one character array, the value of i in this case will always be 0x10
Last edited by grumpy; 07-02-2011 at 06:11 PM.
