Static variable in C++ value is not getting updated even when incrementing in function

Asked Jul 09 '16 at 17:30

Active Jul 09 '16 at 17:35

Viewed 43 times

/* why is the static variable value getting displayed as 0 even when it is getting incremented in display()*/

#include "iostream"
using namespace std;
int count = 0;

int display(int n)
{
    count++;
    if(n)
        return 1+display(--n);
    else
        return 0;
}
int main()
{

    cout << display(10) << " and count is " << count << endl;   
    return 0;
}

edited Jul 09 '16 at 17:35

Jesper Juhl

30,449
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asked Jul 09 '16 at 17:30

Bupa

5

C++ isn't evaluated from left to right. – melpomene Jul 09 '16 at 17:31
Try `cout << display(10); cout << " and count is " << count << "\n";` – melpomene Jul 09 '16 at 17:34
@melpomene What about side effects? I doubt std::cout is really made for non-left-to-right evaluation of the arguments. – Paul Stelian Jul 09 '16 at 17:41
Also see: http://stackoverflow.com/questions/2129230/cout-order-of-call-to-functions-it-prints, http://stackoverflow.com/questions/18093758/reverse-order-when-using-stdcout, http://stackoverflow.com/questions/376278/parameter-evaluation-order-before-a-function-calling-in-c – Cody Gray - on strike Jul 09 '16 at 17:41
@PaulStelian I have no idea what you're asking. – melpomene Jul 09 '16 at 17:43
@paul The `<<` operator does not create a sequence point, so the order of evaluation is unspecified. std::cout is not "made" for either left-to-right or right-to-left. – Cody Gray - on strike Jul 09 '16 at 17:43
@CodyGray Oh, so the variable may be read before the function call? Of course, the operators themselves on cout must be called left-to-right for syntactical reasons. – Paul Stelian Jul 09 '16 at 17:44
@CodyGray The duplicate is actually my question though, and perhaps the OP's too. – Paul Stelian Jul 09 '16 at 17:44

Static variable in C++ value is not getting updated even when incrementing in function

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