C Program -2 closed

Question

#include<stdio.h>
int f(int a){
 a > 20 ? return 10: return 20;
}
int main(){
int b=fun(20);
return 0;
}

what will be the output of this program ?

Comments

This code also run , if we directly use the printf as expression in the in the body of conditional operator 

#include<stdio.h>
int f(int a){
     a > 20 ? printf("10") : printf("20");
}
 
int main(){
    f(20);
    return 0;
}    

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Check this: https://stackoverflow.com/questions/3566413/ternary-operators-and-return-in-c

Answer 1

It is an erroneous program.

Operators take expressions not statements. Function f(int) makes use of TERNARY OPERATOR, and just like any other operator, its operands (arguments) must be expressions.

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The correct way to achieving the intended behaviour is this

#include<stdio.h>

int f(int a){
    if(a>20)
        return 10;
    else
        return 20;
}

int main(){
    int b=f(20);
    return 0;
}

Comments

@ prateekdwv  I am getting answe-1

Can u tell me why?

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I'm not sure where you are going wrong until you share your idea of executing this code with me. 

However, in case you instinctively feel that using ternary operator in this way should not cause any problem, I can share a code snippet with you which might help you.

int foo()
{
    return 0 + return 1
}

The error caused by the use of ternary operator the way it is used in the question is sort of same error that this code snippet will cause.

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Won't it generate an error that function "fun" is not defined in the program and we are calling that function?

---

Oops! that was certainly a stupid mistake. Fixed it. Thank you for pointing it.

Answer 2

To make this program errorless and execute with conditional operator, following things should be remember:

• operator are defined with expressions not with keywords
• function f should return a value 

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---

---

#include<stdio.h>
int f(int a){
    return a > 20 ? 10 : 20;
}
 
int main(){
    int b = f(20);
    printf("%d ", b);
    return 0;
}

Answer 3

Let's assume your function defined and called are same, either fun(int) or f(int).  exp1 = expression 1 and so on

So z= exp1 ? exp2 : exp3 is calculated as: evaluate exp1 and if it's non-zero then evaluate exp2 and assign z to exp2 value otherwise  evaluate exp3 and assign z to exp3 value. here return are statements; not expressions.

There won't be any output.

Answer 4

Error; there should be no usage of return values in case of ternary operator

Answer 5

Please do not confuse in ternary operators 

you can consider the code like this

int f(int a){
    if(a>20)
        return 10;
    else
        return 20;
}

int main(){
    int b=f(20);
    return 0;
}

this code will execute properly and will return 0 because main function is returning 0; 

int b=f(20); this statement stores value returned by f()

20 will be returned if in main we write return b;