CODEWITHSUNDEEP
cif-statement
fiorenza2
fiorenza2

Reputation: 15

'if' statements not behaving as expected without 'else'

Just a quick question; I've been working through K&R and the code for the digit/whitespace/other counter works fine. However, whilst trying to get my head round the functionality of else I've encountered something which doesn't work as expected.

The code from the book is as follows:

#include <stdio.h>

/* count digits, white space, others */
main()
{
    int c, i, nwhite, nother;
    int ndigit[10];

    nwhite = nother = 0;
    for (i = 0; i < 10; ++i)
        ndigit[i] = 0;

    while ((c = getchar()) != EOF)
        if (c >= '0' && c <= '9')
            ++ndigit[c-'0'];
        else if (c == ' ' || c == '\n' || c == '\t')
            ++nwhite;
        else
            ++nother;

     printf("digits =");
    for (i = 0; i < 10; ++i)
        printf(" %d", ndigit[i]);
    printf(", white space = %d, other = %d\n", nwhite, nother);
}

If I then modify the while loop so it reads:

 while ((c = getchar()) != EOF)
            if (c >= '0' && c <= '9')
                ++ndigit[c-'0'];
            if (c == ' ' || c == '\n' || c == '\t')
                ++nwhite;

It should still have the same functionality as the original code except for the fact it won't count 'other' characters. However what I actually get is in fact just the 'digit' part working, with 'nwhite' returning zero no matter what the input. I feel the disparity is perhaps due to a fundamental misunderstanding of how if statements function.

Upvotes: 1

Views: 172

Answers (2)

LogicG8
LogicG8

Reputation: 1757

while ((c = getchar()) != EOF)
        if (c >= '0' && c <= '9')
            ++ndigit[c-'0'];
        if (c == ' ' || c == '\n' || c == '\t')
            ++nwhite;

Is equivalent to

while ((c = getchar()) != EOF) {
        if (c >= '0' && c <= '9')
            ++ndigit[c-'0'];
}
if (c == ' ' || c == '\n' || c == '\t')
        ++nwhite;

Only the first statement that follows a looping or branch construct "belongs" to that construct. That is why the original if-else if-else chain works without braces. Each statement chains to the previous with the first if/else statement belonging to the while loop and the second if/else belonging to the first if/else. It is idiomatic to express the logic this way to avoid unnecessary indenting.

It may help to visualize the code with braces 

while ((c = getchar()) != EOF) {
    if (c >= '0' && c <= '9') {
        ++ndigit[c-'0'];
    }
    else {
        if (c == ' ' || c == '\n' || c == '\t') {
            ++nwhite;
        }
        else {
            ++nother;
        }
    }
}

Upvotes: 2

ouah
ouah

Reputation: 145829

 while ((c = getchar()) != EOF)
        if (c >= '0' && c <= '9')
            ++ndigit[c-'0'];
        if (c == ' ' || c == '\n' || c == '\t')
            ++nwhite;

The second if statement is no longer in the loop. Use { and } to enclose the loop statements.

Upvotes: 10

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