CODEWITHSUNDEEP
c
John
John

Reputation: 151

please explain the output

#include<stdio.h>
#include<conio.h>

int t=8;

int dok(int);
int doky(int);

int main()
{
    int clrscr();
    int x,y;
    int s=2;
    s*=3;
    x=dok(s);
    y=doky(s);
    printf("%d%d%d",s,y,x);
    getch();
    return 0;
}

int dok(int a)
{
    a+=-5;
    t-=4;
    return(a+t);
}

int doky(int a)
{
    a=1;
    t+=a;
    return(a+t);
}

Answer to above code: 665

I understand why s=6, x=1+4=5 (a=6-5=1,t=8-4=4)... Please tell me how y comes as 6, I thought y would be 1+4=5 (a=1, t=4)

Thanks, please help me.

Upvotes: 0

Views: 119

Answers (5)

pmg
pmg

Reputation: 108988

Just do it with pencil and paper ...

                 | t | x | y | s | a |
-----------------+---+---+---+---+---+
before main      | 8 |#NA|#NA|#NA|#NA|
before x=dok(s)  | 8 | ? | ? | 6 |#NA|
inside dok       | 4 |#NA|#NA|#NA| 1 |
after dok        | 4 | 5 | ? | 6 |#NA|
before y=doky(s) | 4 | 5 | ? | 6 |#NA|
inside doky      | 5 |#NA|#NA|#NA| 1 |
after doky       | 5 | 5 | 6 | 6 |#NA|

Upvotes: 0

sergico
sergico

Reputation: 2621

y = a + t = a + t + a = 1 + 4 + 1 = 6 :)

Upvotes: 0

Sergio Morales
Sergio Morales

Reputation: 2650

The value of t is changed to 4 with the dok function, and the doky function increments that value by 1 (the value in a). Sum that (5 so far) to the value of a again (set to 1), and that's 4+1+1 = 6.

//t is 4, the value of a is irrelevant since it changes on the next instruction.

a=1;
t+=a;    // t is now 5

return(a+t);   // 1+5 = 6

Upvotes: 1

Mahesh
Mahesh

Reputation: 34625

tell me how y comes as 6 ...

Call to dok function modifies t to 4.

int doky(int a)
{
    a=1;
    t+=a;    // Previously t is 4 because of t-=4 in earlier function call
             // t = 4+1 = 5
    return(a+t);   // 1+5 = 6 retured
}

Upvotes: 5

fdermishin
fdermishin

Reputation: 3706

first t increases by a and then sum of a and t is returned

so, t was 4. then operator t += a is executed and t becomes 5. and a+t == 1+5 == 6 is returned

Upvotes: 1

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