Head First C string.h related questions

Question

#include <stdio.h>
#include <string.h>

void print_reverse(char *s)
{
    size_t len = strlen(s);
    char *t = s + len - 1;
    while (t >= s)
    {
        printf("%c", *t);
        t = t - 1;
    }
    puts("");
}

Above is a function that will display a string backward on the screen. But I don't understand the 7th line (char *t = s+ len-1;). Could anybody explain this is spoken English please?

`char *t = s+ len-1;` points to the last character of the string. — Weather Vane, Jun 05 '17 at 13:02
You might also want to [find a good beginners book](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) and read about *pointers* and *pointer arithmetic*. — Some programmer dude, Jun 05 '17 at 13:03
....or if s is not nul-terminated.... but is not constructive here... — LPs, Jun 05 '17 at 13:04
@Stargateur Of if `s` is not a valid pointer, or points to an unterminated string, or many other possible things. — Some programmer dude, Jun 05 '17 at 13:04
But technically, the code is illegal since `t` will be left pointing to 1 `char` before the array passed to it, and the comparison `t >= s` may not work. — Weather Vane, Jun 05 '17 at 13:04
See also [fputc](http://pubs.opengroup.org/onlinepubs/9699919799/functions/fputc.html) and [putc](http://pubs.opengroup.org/onlinepubs/9699919799/functions/putc.html). — Sinan Ünür, Jun 05 '17 at 13:05
`char *t = s + len - 1;` is undefined behavior if the input string is empty, and `t = t - 1;` is always undefined behavior when `t` points to the beginning of `s`, since it is UB to even form this address with pointer arithmetic. — ad absurdum, Jun 05 '17 at 13:23

Vlad from Moscow · Answer 1 · 2017-06-05T13:27:12.290

For starters this function

void print_reverse(char *s)
{
    size_t len = strlen(s);
    char *t = s + len - 1;
    while (t >= s)
    {
        printf("%c", *t);
        t = t - 1;
    }
    puts("");
}

is wrong and has undefined behavior.:)

There are two problems.

The first one is that the passed string as the argument can have a zero-length. In this case this declaration

char *t = s + len - 1;

will look like

char *t = s - 1;

and the pointer t can be wrong.

The second problem is that this expression statement

t = t - 1;

has undefined behavior in case when the pointer t is equal to s.

From the C Standard (6.5.6 Additive operators)

...If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined.

A correct function implementation can look the following way

void print_reverse( const char *s)
                    ^^^^^
{
    size_t len = strlen(s);
    const char *t = s + len;
                    ^^^^^^^
    while (t != s)
           ^^^^^^
    {
        printf("%c", *--t);
                     ^^^^
    }
    puts("");
}

As for your question then in this declaration

char *t = s + len - 1;

the pointer t is tried to be initialized by the address of the last character of the string before the terminating zero.

Made my comment under OP, then saw your answer. If this is example code from the book, it does not speak well of "Head First C". — ad absurdum, Jun 05 '17 at 13:26

score 2 · Answer 2 · answered Jun 05 '17 at 13:06

Main logic behind this functions is that this code:

char *t = s+ len-1;

return a pointer to the address of the last char in the char pointer you are passing to the function. The loop prints it by decrementing it:

t = t - 1;

So in simple words it prints the char pointer from backwards.

Head First C string.h related questions

2 Answers2