int main(){
int a[5]= {5, 1, 15, 20, 25};
int i, j, k=1, m;
i = ++a[1]; // i becomes a[1] + 1 which is 2 but a[1] changes to 2 even though I'm assigning a value to i?
j = a[1]++; // j becomes a[1]+1 which is 2+1 but stay as 2?
m = a[i++]; // because i is 2, shouldn't the output be a[3]?
printf("\n%d %d %d", i, j, m);
return 0;
}
In this code, the output comes out as 3, 2, 15.
How does the increments work specifically? Thank you.
The pre-increment operator (++x) increments the value and returns the new value. The post-increment operator (x++) increments the value and returns the old value.
It's easy to see what happens if you decompose the pre-increment and post-increment operators into two different lines.
int a[5]= {5, 1, 15, 20, 25};
int i, j, k=1, m;
// i = ++a[1]; becomes:
++a[1]; // a = {5, 2, 15, 20, 25}
i = a[1]; // i = 2
// j = a[1]++; becomes:
j = a[1]; // j = 2
a[1]++; // a = {5, 3, 15, 20, 25}
// m = a[i++]; becomes:
m = a[i]; // m = a[2] = 15
i++; // i = 3
So, in the end:
i = 3
j = 2
m = 15
In particular, in this case:
m = a[i++];
The variable i previously had a value of 2, so the post-increment operator increments i (which becomes 3), but returns 2, so that line is equivalent to:
m = a[i]; // Use old value here
i++; // Increment after
i = ++a[1];
//i becomes a[1] + 1 which is 2 but a[1] changes to 2 even though i am assigning a value to i?
Correct, i becomes the value of a[1] + 1 and a[1] do so too. Both, a[1] and j have the value of 2 after the expression.
j = a[1]++;
// j becomes a[1]+1 which is 2+1 but stay as 2?
No, j do not get to the value evaluated by a[1] + 1.
After this expression, The value inside of j is still a[1] or better said 2, but the value of a[i] will change to 3 after the expression.
The Postincrement-Operator (variable)++, as opposed to the Preincrement-Operator, ++(variable), increases the value of the operand only after the respective operation:
C: What is the difference between ++i and i++?
m = a[i++];
// because i is 2, shouldn't the output be a[3]?
No, because like I said above, The Postincrement-Operator only increments the value of the operand after the operation, so i in m = a[i++]; is still i, not i + 1.
Pre-increment first increments the value, then returns the incremented value to you. Post-increment returns the value before it gets incremented.
The first statement initially sets i=2, but the third statement i++ increments it again. That's why i=3 in the result. j=2 because at the time this was the value of a[1]. m=15 because at the time the statement is executed, i=2 and this is the (zero-based) value of that element in the array.
The code as written does many things which are difficult to decipher ... hence the question. It's not a good idea to write code in this way except as an exercise.
