C Programming

Question

#include <stdio.h>

int main()
{
    int a=9;
   printf("%d %d %d",++a,a++,++a);

    return 0;
}

what is the output?

According to me it should be     12 10 10

Comments

with in a sequence point, a updated more than once, therefore it is undefined behavior in C.

Moreover coming to GATE, there are no such questions ( compiler dependent, implementation dependent and undefined behavior etc ) asked

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In printf  precedence  of printing  is followed  from left to right or from right to left

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order of evaluation of function parameters are not standard in C language.

as per my knowledge, that is the main reason for showing above line is UNDEFINED behavior

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@Shaik Order of evaluation of arguments is not defined in C. But even if it was defined, the given code produces undefined behaviour.

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@Arjun sir, if really order of evaluation of arguments defined, then why not comma acts as operator?

if comma acts as operator, it is sequence point only, right?

 

sir moreover i need confirmation on these

https://gateoverflow.in/234888/normalization

https://gateoverflow.in/235632/%23doubt

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In argument list, "," is not acting as comma operator but only as a separator (operator overloading) and is not forming a sequence point.

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Yes, output is correct

As we know that In c the execution of a function is done from "Right to left "
and print from from "left to Right"

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@Magma

read the comments clearly.

 

Answer 1

The program exhibits undefined behaviour. Don't solve such questions for GATE.

Answer 2

When parameters are passed to a function, the value of every parameter is evaluated before being passed to the function. What is the order of evaluation of parameters - left-to-right or right-to-left?  Unfortunately, there is no fixed order defined by the C standard. A compiler may choose to evaluate either from left-to-right. So the output is compiler dependent.