Usage of '+=' in c

Question

I don't understand why, on this code, 'b+=' return 6 instead of 5. The operation on right-end of operator '+=', should be 0.

i/2 = 4
a-4= 0

So Operator '+=' should just add: 0.

#include<stdio.h>

int main(){
  int a=4, i=8;
  int b;
  
  b=++a; 
  printf("%i\n",b);

  b+=a-i/2;
  printf("%i\n",b);

}

Just using theory of sintax

That's C, which is correctly tagged. And there's no `=+` operator. `b=++a` is really `b = ++a`. — Some programmer dude, Mar 26 '23 at 18:35
As for `b+=a-i/2` it's the same as `b += a - i / 2`, which is the same as `b = b + (a - (i / 2))`. — Some programmer dude, Mar 26 '23 at 18:36
You might be more enlightened if you print the value of `a` too. — pmacfarlane, Mar 26 '23 at 18:41
a++ and ++a is the shortcut for a = a + 1 so the value actually changes, the difference between them is post-increment changes the value of the variable a after statement executed and pre-increment changes the value of the variable a before assigning it to variable b. — ayseguldmrblk, Mar 26 '23 at 18:44

Vlad from Moscow · Accepted Answer · 2023-03-27T08:06:17.267

After this statement

b=++a;

b and a will be equal to 5 due to the prefix increment operator ++.

From the C Standard (6.5.3.1 Prefix increment and decrement operators)

2 The value of the operand of the prefix ++ operator is incremented. The result is the new value of the operand after incrementation. The expression ++E is equivalent to (E+=1)

So in this compound assignment statement

b+=a-i/2;

that may be rewritten like

b = b + a - i/2;

you have

b = 5 + 5 - 4

So as a result you have b is equal to 6.

You could get the expected by you result if to initialize the variable b by zero

int b = 0;

and if to remove this statement

b=++a;

score 1 · Answer 2 · answered Mar 26 '23 at 18:42

1

When you execute b=++a;, a is incremented. Its value becomes 5. Then afterward, a-i/2 is 5-8/2 then 1.

answered Mar 26 '23 at 18:42

Frédéric LOYER

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Usage of '+=' in c

2 Answers2