Unexpected !.Why ?

Question

#include<stdio.h>

int main()
{
int a=1;
printf(" %d %d %d \n",a,a++,++a);

return 0;
}

Answer 1

It starts by evaluating the last parameter of a function, then the second last and so on until the first.
Example :
int a = 5;
printf("%d %d %d\n",a,++a,a++);
->
printf(..., a, ++a, 5); // a == 6
printf(..., a, 7, 5); // a == 7
printf(..., 7, 7, 5);

Answer 2

I'm not sure it work exactly that wat @Baptiste, for your answer doesn't explain the output of the following code... Apparently the compiler evaluate the term "++a" prior the call of the function printf :

#include<stdio.h>

int main()
{
int a=1;
printf(" %d %d %d \n",a,++a,a);

return 0;
}

Answer 3

finally u got the point man :)

Answer 4

The answer is simple : GCC evaluate function's parameters from right to left.
I said GCC as it is SoloLearn's compiler and as this behavior is not defined in C norms. It is compiler dependant so you might have another behavior with mingw or clang for example

@Infinity, cout can't be used as a test as it is different than printf, it is like you were doing :
<<(<<(<<(cout, a++), a++) a);
where you have to replace << by operator<<
(Sorry, I was lazy :p)

Answer 5

nope

Answer 6

absolutely correct @Infinity. Was a Mistake in the code which i've corrected right now. Have a Look at it and run the code again and tell why is it giving abnormal output

Answer 7

i've corrected (edited) this code. look at the question again !

Answer 8

i'd love to hear the reason why this happens.

Answer 9

yeah bro ! the abnormal output ..m still finding the reason.

Answer 10

@Prunier i am still not able to understand this fact... sorry m beginner. can u please explain it little further

Answer 11

Of course ! But, which fact ? Being compiler dependent or the way GCC evaluate functions' parameters ?

Answer 12

the way GCC works

Answer 13

Prior to it expression, but the expression is used as a parameter. Not prior to the line (which is different)
So my explication is still correct ^^