Just had a mini test worth a small percentage of our end of year grade. Wasnt overly happy with how it went the last part of the question was using pointer nonotation only to divide element 0 of an array by element 0 of another array and store them in a third array. And print the value to the screen.what would this look like.
Something like this I would imagine. Depending on your thought processes.Code:(function to take in 3 arrays) *(pointer3+0) = *(pointer1 + 0) / *(pointer2 +0);
Code:int get_random_number(void) { return 4; //chosen by fair dice roll. //guaranteed to be random }
Thanks that's what I had : )
another question about this what would this mean
[code]
*(pointer3+0) = *pointer1 + 0 / *pointer2 +0;
[code/]
Basically you are dereferencing the array ( *pointer1), it is interpreted as a r-value. So what goes into pointer3 will be :Originally Posted by alien1
Also keep in mind where the division operator is.Code:pointer3[0] = pointer1[0] + 0 / pointer2[0] + 0
A page listing operator precedence:Code:pointer3[0] = pointer1[0] + ( 0 / pointer2[0] ) + 0
C++ Operator Precedence - cppreference.com
Last edited by Alpo; 12-02-2014 at 06:21 PM.
WndProc = (2[b] || !(2[b])) ? SufferNobly : TakeArms;
