Incorrect variable value incrementing

Question

I have the following code:

#include <stdio.h>
#include <ctype.h>

int main(int argc, char **argv)
{
        int ch, lower, upper = 0;
        printf("Enter a line of text: \n");
        while ((ch = getchar()) != EOF) {
        if (islower(ch)) {
                ch = toupper(ch);
                ++upper;
        } else if (isupper(ch)) {
                ch = tolower(ch);

                printf("Looking at lower: %d\n", lower);
                ++lower;
                printf("Looking at lower: %d\n", lower);
        }
        putchar(ch);
        }
        printf("Hello\n");
        printf("\nRead %d characters in total. %d converted to upper-case, %d to lower-case.", upper+lower, upper, lower);
}

For some reason the upper variable is being set correctly, but can't work out why lower is giving an erroneous value. E.g. If I type in 'Football' it says 4195825 converted to lower-case, where the actual output should be 1.

I can't see where I'm going wrong here.

haccks · Accepted Answer · 2014-04-07 21:16:21Z

3

You haven't initialized lower. It's value is indeterminate.

C11: 6.7.9 Initialization (p10):

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

Initialize it to 0.

int ch, lower = 0, upper = 0;

answered Apr 7, 2014 at 21:16

haccks

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6 Comments

AkshaiShah Over a year ago

I don't think that's the issue. If I do int ch, lower, upper; It gives the same output.

haccks Over a year ago

Yes this is the issue. I quoted the standard.

AkshaiShah Over a year ago

Oh sorry yeah you are right, but how come the upper variable works correctly if I do int ch, lower, upper; but the lower variable does not?

haccks Over a year ago

Indirectly, your program invokes undefined behavior. You may either get expected or unexpected output. Nothing can be said.

AkshaiShah Over a year ago

Interesting.. is there anywhere I can read more about this random behaviour?

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