C programming increment and decrement doubt

[shariefbe]

Hello friends,
I have an big doubt in increment and decrement in c programming
i am trying to compile the below code in gcc

#include<stdio.h>
main()
{
int a=10,b=12;

printf("\n%d\n%d\n%d\n%d\n%d\n",a++,++a,a--,--a,a);
}

and i am getting the below output

9
10
9
10
10

i am very much confused. how it is parsing and how i will get this output?please explain me

[gerard4143]

Here's your program with a few corrections, plus when its compiled with strict settings it produced a few warnings.

#include<stdio.h>

int main()
{
	int a = 10;

	printf("\n%d\n%d\n%d\n%d\n%d\n",a++,++a,a--,--a,a);

	return 0;
}

Now this is what the compiler had to say...

warning: operation on ‘a’ may be undefined
warning: operation on ‘a’ may be undefined
warning: operation on ‘a’ may be undefined
warning: operation on ‘a’ may be undefined

Undefined behavior - Wikipedia, the free encyclopedia

By the way here's how I compiled the above code:

gcc testit.c -Wall -ansi -pedantic -o testit

[shariefbe]

No man. again i am getting the same output which i got from my program

[gerard4143]

No man. again i am getting the same output which i got from my program

That's because the operation(s) may produce undefined behaviour.

[blinky]

Out of interest tried a similar thing in java

public class test
{
   public static void main(String[] args)
   {
      int a = 10;
      print(a++, ++a, a--, --a, a);
   }
   
   public static void print(int a, int b, int c, int d, int e)
   {
      System.out.println(a + "\n" + b + "\n" +  c + "\n" +  d + "\n" +  e);
   }
}

It produces what I would have expected

10
12
12
10
10

Don't have a windows/other compiler to see what we get.
Could'nt find a compiler switch to make it do operations in strict order.

[markcole]

No man. again i am getting the same output which i got from my program

There is no way to tell what the state of your variable "a" is at the time of execution, that is why the behavior is undefined.

You don't know which order the variable will be processed.

[Cabhan]

markcole is correct. When you have multiple operations on a single line, you don't know the order that they will occur. This is why decrements seem to be happening before increments, even though the increments appear leftmost.

Java may have different rules, which is why you are seeing different behaviour there.

To get what you want, you would have to separate it into multiple lines:

printf("\n%d", a++);
printf("\n%d", ++a);
...

[Rubberman]

First, the order of evaluation of arguments is compiler-defined. NEVER, NEVER, NEVER do what you did here! I can't even begin to guess how many applications that other engineers wrote that I had to debug because of this sort of cruft!

Second, in an expression, prefix increment/decrement operators are applied before the variable is used, and postfix operators are applied after the variable is used. So, in this case "a = ++b" the value of a will be set to b AFTER b has been incremented and after the expression a and b will be equal, whereas in the case "a = b--", a will be set before b has been decremented, so after the expression, b will be one less than a.

[Monni]

Hello friends,
I have an big doubt in increment and decrement in c programming
i am trying to compile the below code in gcc

#include<stdio.h>
main()
{
int a=10,b=12;

printf("\n%d\n%d\n%d\n%d\n%d\n",a++,++a,a--,--a,a);
}

and i am getting the below output

9
10
9
10
10

i am very much confused. how it is parsing and how i will get this output?please explain me

The reason for such an output is nothing but, LIFO concept. That is "Last In First Out" . Try evaluating it from the right hand side. You will get it.
The last is a whose value is 10, that is displayed at the last line. Then come to the next and so on . This is how that output is generated.

[Cabhan]

This thread is 8 months old. Closing.