Solution for Exercise 19-19

ex1919

#include <stdio.h>

int main()
{
    char *fruit[] = {
       "watermelon",
       "banana",
       "pear",
       "apple",
       "coconut",
       "grape",
       "blueberry"
    };
    int x;

    for(x=0;x<7;x++)
    {
        while( putchar(*(*(fruit+x))++))
            ;
        putchar('\n');
    }

    return(0);
}

Output

watermelon
banana
pear
apple
coconut
grape
blueberry

Notes

* Here's the translation for Line 18:

*(*(fruit+x))++

is quite similar to:

*pa++

Where pa holds the base address of a string. So while(*pa++) outputs a string just as while(*(*(fruit+x))++) because like pa, *(fruit+x) translates to an address. The parentheses are required.

* Here's how *(*(fruit+x))++ evaluates:

fruit+x is an element in the array of pointers, an address. As x increments, it points to each successive element in the array, the address of each string.

*(fruit+x) refers to the value at fruit+x, which is the string itself.

*(*(fruit+x)) refers to the first character of the string at address fruit+x. This expression is the same as **(fruit+x) shown in the book. The extra parentheses are required for the last part of the expression:

*(*(fruit+x))++ grabs the character at the address *(fruit+x). Then the ++ operator references the next address in the string, the next character. This is why (*(fruit+x)) is enclosed in parentheses.

* Don't fret if you didn't arrive at this solution. This exercise wasn't an easy one. Here's another way to work the while loop:

a=0; while(putchar(*(*(fruit+x)+a++))) ;

This method uses int variable a as the offset to the string. This solution looks more cumbersome, but it may help you understand what's going on — or if you devised a similar solution by yourself, so be it.

* This type of pointer construction doesn't often find itself into "real" C programs. It's just easier to use printf() to output the strings.

Exercise ex1920 →