how this printf() function output these values?

Question

#include<stdio.h>
void main()
{
int i=10;
printf("%d %d %d\n",a,--a,++a); // output  10 10 10

}

how this code gives same output? what is the precedence of these increment and decrement operators?

score 0 · Answer 1 · answered Apr 10 '16 at 20:07

0

This mean the argument of printf are evaluated before the printf function is called.

For the compiler, your code looks like

#include<stdio.h>
void main()
{
  int a=10;
  a;
  --a;
  ++a;
  printf("%d %d %d\n",a,a,a); // output  10 10 10
}

answered Apr 10 '16 at 20:07

ced_c

2

Incorrect. The evaluation order is unspecified and the program has undefined behaviour. It's not equivalent to anything. – Lightness Races in Orbit Apr 10 '16 at 20:18

score -1 · Answer 2 · answered Apr 10 '16 at 18:36

-1

It gives you the same output because - - a print the original value of a and then It became 9 (or 11 with ++a)

answered Apr 10 '16 at 18:36

Hoffman

What is the difference between a++ and ++a ? – Vladimir Tsykunov Apr 10 '16 at 18:41
can you explain it more? – Promod Sampath Elvitigala Apr 10 '16 at 18:43
@VladimirTsykunov ++a will increment the value of a, and then return the incremented value. a++ will increment the value of a but It returns the original value. – Hoffman Apr 10 '16 at 19:00
No, the behaviour is undefined. – Lightness Races in Orbit Apr 10 '16 at 20:18

2 Answers2