in printf (in c) what is the order of displaying variable and push into stack

Question

Suppose i have the following code:

#include <stdio.h>    
main()
{
   int a,b,c;
    b=1;
    c=2;

    printf("%d\n",10,b=20,b=30,c=50,c=100);
    printf("%d\n",b);
    printf("%d\n",c);
}

o/p-10,20,50 how did the value of b became 20,not 30 ..and also the same for c?

you have both undefined and implementation dependent behavior. i.e. it won't work — Bryan Chen, Sep 14 '14 at 09:29
It is not related to printf. The order of evaluation of function arguments is implementation dependent. It can be from left to right or right to left or any other order. — Marian, Sep 14 '14 at 09:29
it will start to push in stack from right to left(generally) — Anshuman Singh, Sep 14 '14 at 09:30

score 2 · Accepted Answer · answered Sep 14 '14 at 09:32

2

The order of evaluation of argument expressions and their pushing on the stack are different things.

The order of evaluation of argument expressions are unspecified in C. So it might be that at first b = 20 will be evaluated and then b = 30 or vice versa.

The order of placing arguments in the stack is the following: the right most argument is placed first.

answered Sep 14 '14 at 09:32

Vlad from Moscow

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+1'd,To clarify more--->shouldn't it be `The order of placing arguments in the stack is the following: the right most argument is placed at the bottom.` – Am_I_Helpful Sep 14 '14 at 09:37
It depends on the ABI - some don't even use the stack, or use a mixture of registers and stack. – Paul R Sep 14 '14 at 09:45
Even worse, this is multiple assignments to one location without intervening sequence point, so it is undefined behaviour. – Vatine Sep 14 '14 at 10:06

in printf (in c) what is the order of displaying variable and push into stack

1 Answers1