Conditional Expression

Question

What is the output of given program?

#include<stdio.h>
int main()
{
    int x;
    x = 4 > 8 ? 5!= 1<5 == 0 ? 1:2:3;
    printf("%d",x);
    return 0;
}
option a. 1 b.2 c.3 d.compilation error

Comments

@Shaik masthan will you please explain the order of evaluation of your code I am not getting for multiple ternary operation?

---

@Ravi Dubey

h=a++?b++?c++?d++:e++:f++:g++;

note that ternary operator is right associative.

 

step 1:-

h = a++ ? b++ ? ( (c++) ? (d++) : (e++) ) : f++ : g++ ;

 

step 2:-

h = a++ ? ( (b++) ? ( (c++) ? (d++) : (e++) ) : (f++) ): g++ ;

 

step 3:-

h = ( (a++) ? ( (b++) ? ( (c++) ? (d++) : (e++) ) : (f++) ): (g++) ) ;

 

finally it look like as,

$h = (\, \, \, \underbrace{ (a++) }\, \, \, ? \, \, \, \underbrace{ (\, \, \, \overbrace{ (b++)} \, \, \, ?\, \, \, \overbrace{ ( \, \, \, \underbrace{(c++)} \, \, \, ? \, \, \, \underbrace{(d++)} :\, \, \, \underbrace{(e++)} \, \, \, )} : \overbrace{(f++)} \, \, \, )} : \underbrace{ (g++) }\, \, \,) ;$

 

after parenthesize, i hope it is simple,

first evaluate expr₁ (a++) ==> return true ( return 10 which is non-zero but note that after sequence point a=11 )

evaluate expr₂ of first ternary operator ==> ( (b++) ? (c++?d++:e++) : (f++) )

          evaluate b++ ===> return false ====> evaluates f++ and return that value to h

---

@shaik masthan really thanks I understood.

Answer 1

Answer: 3

Refer this: http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm