This is a sample code I have from someone and it runs giving the answers
3, 2, 15
Can someone please explain how this piece of code works and how it got to those outputs?
Code:
int a[5] = { 5, 1, 15, 20, 25 };
int i, j, m;
i = ++a[1];
j = a[1]++;
m = a[i++];
printf("%d, %d, %d", i, j, m);
_getch();
You should know about pre-increment (++var) and post-increment (var++).
Let's break down the code line by line, shall we?
i = ++a[1]; Here, a[1] is 1, (2nd element of array a), is pre-incremented to 2 and that value is stored into i.
j = a[1]++; same, the 2 is stored into j and then a[1] is post-incremented to the value 3.
m = a[i++]; a[2], i.e., 15 is stored into m, and i is post-incremented to 3.
printf("%d, %d, %d", i, j, m); -- Surprise!!
Exactly so:
**Pre- ** Increment: increment the value in the variable, then use the incremented value.
** Post- ** Increment: grab the value and use it. After grabbing the not-yet-incremented value, increment the value that's stored in the variable. Proceed, using the value as it was before.
"Operator precedence" rules mean that, for example, ++a[1] means that the meaning of a[1] will be resolved first, then the ++ operator will be applied to that memory-location. The value of the number-one element (which is actually the second element ...) will be incremented, and the incremented value will be returned for use by the statement.
a[1] refers to the second element which is 1i=++a[1],a[1] is incremented to 2 and assigned to ia[1] is incremented again and the value of a[1] is incremented again to 3m=a[i++], value of i is 2 from step 1 , but it is a post increment .so, m will remain m=a[2]
3, 2, 15. He doesn't expect output, he would like to understand where it came from.