programming doubt closed

Question

#include<stdio.h>
int main()
{
    int i = 1;
    printf("%d %d %d\n", i++, i++, i);
    return 0;
}
 on running i m getting answer : 2 1 3 why answer cant be 1 2 3

Comments

printf will evaluate it from right to left , in turbo c i got        2 1  1

value of i =1    

printf  (  i++       ,                i++       ,             i      )               

               2                             1                    i=1                     ( go right to left )        i =1

 

.//post increment //  i++ ( use then change ) so middle one will have value 1  as soon as  we come to leftmost       i++     value of i will become 2

and again leftmost is i++ (use then change )     we used i  value as 2 and printf will print it   2  1   1     , after this  i final value will be 3

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which compiler yoou used ?  //

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Code block

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It's an ide

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it is undefined behavior in C

updating same operand more than once in a single statement leads to  undefined behavior

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yes ..printf execute from R to L but like this ->

printf  (  i++       ,                i++       ,             i      )               

               2                             1                                         ( go right to left )        i =1 

and i++ will be replaced with 2,  i++ will be replaced with 1 (similar to immediate mode ) ...but i will be evaluated from memory(bcoz i is a variable not expression...) o/p:-2,1,3

ex-1..let consider this one...

 printf  (  i++       ,                i++       ,             a=i      )    in this case 

value of i would assigned to 'a' and at the time of evaluation 'a' it will be fetched from memory (@location 'a' is present)..o/p:- 2,1,1

ex-2..

printf  ( c= i++       ,             b=   i++       ,             a=i      )  in this case at the time of evalution a,b,c would be fetched from it's original location and then printed...o/p;- 2,1,1

Answer 1

  printf("%d %d %d\n", i++, i++, i);

In the above statement, the arguments i++ & i++ are the increment operators and the argument i is simply printing the value of i

For the increment/decrement operator the associativity is from right to left.

So, the second i++ will be evaluated first & will print 1 & increment i to 2

The first i++ will be evaluated next & will print 2 & increment i to 3

Lastly, i will be printed i.e. 3

So, answer is 2 1 3

Comments

Tysm

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thank you sir,for nice explanation ,but different compilers are giving different outputs.what would be your answer to this problem sir,

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in case

#include <stdio.h>

int main()

{

    int i = 1;

    printf("%d %d %d\n", i++, ++i, i);

    return 0;

}

 

the output is coming out to be

2 3 3

from your logic it should be 2 2 3