what is the output of the following program code?

Question

void myfunc(int X){
    if(X > 0)
        myfunc( --X );
    printf("%d", X);
}

int main(){
    myfunc(5);
    return 0;
}

• A. 0,0,1,2,3,4
• B. 4,3,2,1,0
• C. 4,3,2,1,0,0
• D. 0,1,2,3,4

Answer 1

A.0,0,1,2,3,4 --- 0 printed twice because of --x statement @ if condition for fun(0) 0 is printed and for fun(1) 0is printed because value x is decremented before fun(0) fun call starts to execute!

Comments

how two zero's are coming before 1?? pls explain this part...

Answer 2

Key point which u have to note here is that there is no braces after the if statement therefore the first statement written after it would only fall into the if portion so myfunc(0) when called would simply print 0 ,so 2  0's would be printed. it won't enter into the if check condition and rest explanation is already given by others ,My intention was to just highlight the braces portion.

Answer 3

A. 0,0,1,2,3,4 

recursive call sequence -myfun(5)--> myfun(4)--> myfun(3)--> myfun(2)--> myfun(1)-->myfun(0)

so,myfun(0) prints a 0 then myfun(1) prints  0 again as 1 in decremented first in  similiar way then 2 ,3 and lastly 4 get printed.

Comments

ok got it thank u..

Answer 4

Answer is Option A

Answer 5

It's tail recursion problem..

so the output is 0,0,1,2,3,4