Char array in c & pointer addition

Question

My instructors give me the following function. So don't blame me for ambiguity lol

void step_step_step(char *first, char *second, char *third)
{
    if (third[3] == second[2] + 8 && second[2] == first[1] + 8)
        printf("8: Illinois\n");
    else
                printf("8: ERROR\n");
}

I try to call the function in the this way:

char *p8_1 = (char*) malloc(sizeof(char)*11);
    char *p8_2 = (char*) malloc(sizeof(char)*11);
    char *p8_3 = (char*) malloc(sizeof(char)*11);
    p8_1[9] = 'u';
    p8_2[2] = p8_1[9];
    p8_2[10] = p8_1[9];
    p8_3[3] = p8_2[10];
    step_step_step(p8_1, p8_2, p8_3);

And it keeps printing error. What am I doing wrong here? I don't understand why that doesn't work when my other solution does:

p8_2[2] = p8_1[1] + 8;
p8_3[3] = p8_2[2] + 8;
step_step_step(p8_1, p8_2, p8_3);

Your + operators are adding characters, not pointers, so there's no "pointer addition" here. Instead you're comparing 'u' to 'u' + 8 and getting non-equal (not suprisingly) — Chris Dodd
– Chris Dodd, Commented Sep 6, 2013 at 20:37

Morten Jensen · Accepted Answer · 2013-09-07 11:50:57Z

5

Possible operator precedence bug:

Change to: if ((third[3] == second[2] + 8) && (second[2] == first[1] + 8))

The wiki article on 'Order of operations' tells more about the evaluation precedence.

If you can't explain to someone in the blink of an eye, in which order some expression will be evaluated - you should use enough parentheses that it won't matter (because no order precedence to care of) or that it's instantly obvious how the clauses are separated.

EDIT:

A closer look makes me realize something fishy is going on. You're not incrementing any pointers. If you were, you would do something like if ((third[3] == second[2 + 8]) && (second[2] == first[1 + 8]))

edited Sep 7, 2013 at 11:50

answered Sep 6, 2013 at 20:12

Morten Jensen

5,9573 gold badges46 silver badges57 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

jxh Over a year ago

There is no operator precedence issue per se, although parentheses may add clarity.

jxh · Accepted Answer · 2013-09-06 20:48:05Z

3

You believe the if condition should pass, since all the elements you are intending to check is initialized to 'u'. In your code, this is true for p8_3[3], p8_2[10], p8_2[2], and p8_1[9]. Then, there is the call:

step_step_step(p8_1, p8_2, p8_3);

As third takes the p8_3 value, indexing its 3^rd element is fine, and it has value 'u'. However, second[2] has the value 'u', and then 8 is added to it before the == comparison. This results in the == to result in a false evaluation. The && is short-circuited, and the else case of the if is taken.

If short-circuiting had not taken place, the right side expression of the && would compare second[2] with first[1] + 8. As first[1] is uninitialized, this would have resulted in undefined behavior.

The indexing operator ptr[index] behaves like *(ptr + index). So, the expression:

second[2] + 8

translates into:

*(second + 2) + 8

which is definitely NOT the same as *(second + 2 + 8).

edited Sep 6, 2013 at 20:48

answered Sep 6, 2013 at 20:16

jxh

70.8k8 gold badges116 silver badges204 bronze badges

8 Comments

Michael Pryor Over a year ago

How can you even figure out what he's trying to do? :) I'm trying really hard and failing.

pm100 Over a year ago

this ones not clear - why wouldnt he say second[10]; maybe he means 'the ascii char 8 steps higher'; although his test doesnt set it up that way

jxh Over a year ago

I am inferring the intended outcome by how he initialized the input.

Collin Over a year ago

Again, my instructors wrote the function. I took it as the the third char arrays index [3] should hold the same character as the second char arrays index [10].. since it's [2] + 8. And the second char arrays index [2] should store the same character as the first char arrays index [9]. since it's [1] + 8

jxh Over a year ago

The [] operator does not distribute over +.

|

user2694688 · Accepted Answer · 2013-09-06 20:22:10Z

1

Looks like p8_1[1] has not been initialized. In your "working" example you assign the random value that's in p8_1[1] to p8_2[2], resulting in their equality. If you're trying to do pointer addition, you need to use a different syntax:

(pointer + (8*sizeof(char))[<index>]

answered Sep 6, 2013 at 20:22

user2694688

Collectives™ on Stack Overflow

Char array in c & pointer addition

3 Answers 3

1 Comment

8 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

8 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related