inialisation for the while loop

Hi folks.

Is there a problem with nbr in the below sample code from my C book not being initialised? I am thinking that because of this, it could actually contain 99, so the program could actually stop before the user enters any values?

Thanks in advance.

Code:

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int value[5];
int ctr = 0;
int nbr;

while(ctr < 5 && nbr != 99)
{
    puts("Enter a number, 99 to quit");
    scanf("%d", &nbr);
    value[ctr] = nbr;
    ctr++;
}

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When in doubt zero it out.

Yes, the loop acts on uninitialised output. This would lead to undefined behaviour (even if, most of the time, it would "work"). You wouldn't want to be on the end of that code if the loop did something interesting, like controlling a nuclear reactor.

When a variable isn't initialised immediately on declaration, always be suspicious and audit its use to be sure that it gets initialised SOMEHOW before you do anything with it, no matter what code-path you end up taking.

And most compilers will pick things like this up when they have warnings enabled.

When you always want the loop body to execute at least once, then the appropriate type of loop to use is a do..while loop:

Code:

---

int value[5];
 int ctr = 0;
 int nbr = 0;
 do {
    puts("Enter a number, 99 to quit");
    scanf("%d ", &nbr);
    value[ctr++] = nbr;
 } while (ctr < sizeof(value)/sizeof(value[0]) && nbr != 99);

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Often this means that the initialisation is not necessary because it will be set to a value during the loop body. However in this case nbr might still not be given a value if the data entered by the user cannot be converted to a number. Thus initialising it to zero is still a good idea, or you could just check the return result of the scanf.

Quote:

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Originally Posted by whiteflags

When in doubt zero it out.

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Nah, when in doubt, find out what needs to be done, and do it.