Unable to understand the output

Question

1.00/5 (1 vote)

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C++

main()
{
int k=25;
printf("\n %d %d %d",k==25,k=50,k>10);
return 0;
}

my question is why is the output 0 50 and 1
i have problem with the 0 as output should be 1 as the value of k has been defined as 25
plez help

What I have tried:

and its my first question here so plez ignore my faults in asking .....though future directions are politely welcome :)

Posted 5-Mar-17 15:36pm

yexxx786

Updated 5-Mar-17 17:05pm

PIEBALDconsult

v2

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Comments

CPallini 6-Mar-17 3:30am

Have also a look at:
https://en.wikipedia.org/wiki/Sequence_point

2 solutions

Solution 1

The first argument, k==25, is a comparison expression. The result being either true (1) or false (0). So, the first number you got was false (0).

The second argument, k=50, is an assignment expression. You set the value of k to be 50. The second number is just saying what the value of k is, 50.

The last argument is another comparison expression. Since k is 50 and 50 is greater than 10, you get true (1).

This reeks of a homework question. I certainly hope you get these concepts because you're in serious jeopardy of failing the class without them.

Posted 5-Mar-17 15:43pm

Dave Kreskowiak

Updated 5-Mar-17 15:44pm

v2

Comments

Patrice T 6-Mar-17 1:50am

Hi Dave,
I think the OP already know these concepts.
He is asking explanation on the k==25 being False, which is rather surprising when one read the code.
See my solution.

Dave Kreskowiak 6-Mar-17 8:18am

Whoops! My bad. I misread the question.

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Patrice T · Accepted Answer · 2017-03-05T16:02:00

Solution 2

You are in a gray zone, this code is unpredictable and will depend on your compiler.
A C compiler id allowed to rewrite your code as it see fit.
Depending on your compiler, your code

C++

int k=25;
printf("\n %d %d %d",k==25,k=50,k>10);

can translate to (I have rewritten the code with 1 operation per line to illustrate the compiler work)

C++

int k=25;
printf("\n %d",k==25); // 1
k=50;
printf(" %d",k); // 50
printf(" %d",k>10); // 1

or in your case

C++

int k=25;
k=50;
printf("\n %d",k==25); // 0
printf(" %d",k); // 50
printf(" %d",k>10); // 1

In C, never use a variable more than once in a line if you also change its value.

Posted 5-Mar-17 16:02pm

Patrice T

Updated 5-Mar-17 20:04pm

v5

Comments

Richard MacCutchan 6-Mar-17 3:18am

It is usually because the arguments are pushed onto the stack in reverse order. So we get the following expressions pushed:

k>10 ...  1
k=50 ... 50, and k now has the value 50
k==25 ... 0

Patrice T 6-Mar-17 3:33am

I know it.
Just wanted to push an easy to understand picture.

CPallini 6-Mar-17 3:28am

I don't quite understand why someone voted two. Have my 5.

Patrice T 6-Mar-17 3:35am

Me too.
Thank you.

nv3 6-Mar-17 4:17am

My 5. Good explanation.

Patrice T 6-Mar-17 4:34am

Thank you.