Is there a way to use pointers + a variable to traverse an array?

Question

I'll admit the title may be confusing, but I really didn't know a better way to put this into words. Here's the code:

char * convertCase(char const * array){
  size_t i = 0;
  while(array[i] != '\0'){
    if(array[i] > 96 && array[i] < 123) *array+i -= 32; // fourth row
    else if(array[i] > 64 && array[i] < 91) array[i] += 32; // fifth row
    i++;
  }
  return(array);
}

What I want to do is something like what you can see in the fifth row, just in the "fourth row" fashion. I'd like to traverse an array in a similar way, but that gives me an l-value error. What's the correct way?

You got the precedence wrong, it should be *(array + i) -= 32. — August Karlstrom, Sep 02 '19 at 16:45
`char const * array` declares the pointer to const chars. You can't modify elements pointed by this `array`. — 273K, Sep 02 '19 at 16:47
`*array+i` == `(*array)+i` which isn't an l-value. It should be `*(array+i)` which is equivalent to `array[i]` — Ingo Leonhardt, Sep 02 '19 at 16:48
It would help with understanding your question and your problem if you#d describe what (by your understanding) fourth row and fifth row are doing (should be doing). Because of the operator precedences, you might be wrong and "what fourth/fifth row" actually IS doing is not what you WANT/THINK it to do. — Yunnosch, Sep 02 '19 at 16:49
This is not directly related to your question but you can do: if array[i] >= 'a' instead of using the ASCII value. This will make it easier to understand for anyone who reads your code in the future. — jmq, Sep 02 '19 at 16:53
@ S.M. this is actually something I was testing before that I forgot to delete, it's not meant to be there; @ jmq thanks for the advice! will do for sure; @ August Karlstrom thanks! clears a lot of stuff up; @ Yunnosch I wanted to increase / decrease the characters in the array to convert them to the opposite case — Impasse, Sep 02 '19 at 16:58
Related: https://stackoverflow.com/questions/1641957/is-an-array-name-a-pointer See second answer — S.S. Anne, Sep 02 '19 at 17:09

score 2 · Accepted Answer · edited Sep 02 '19 at 19:48

2

There are some things need to be changed in you code:

As Ingo said, *array + i is wrong; it should be *(array + i) which is equivalent to array[i]
Instead of checking if(array[i] > 96 && array[i] < 123), I would prefer using if(array[i] => 'a' && array[i] <= 'z') which adds readability. Same goes for second if
I would recommend you to not change original string, instead, create another string and collect new letters there:

char * convertCase(char const * array)
{
    size_t len = strlen(array);
    char *new_str = (char *) calloc(len + 1, sizeof(char));

    for (size_t i = 0; i < len; ++i)
    {
        if (array[i] >= 'a' && array[i] <= 'z') new_str[i] = array[i] - ('a' - 'A');
        else if (array[i] >= 'A' && array[i] <= 'Z') new_str[i] = array[i] + ('a' - 'A');
    }

    return new_str;
}

edited Sep 02 '19 at 19:48

chux - Reinstate Monica

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answered Sep 02 '19 at 17:22

Miradil Zeynalli

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Technically using `char` literals is not equivalent behavior, because the C standard does not guarantee that the character set is ASCII. So, while your code would always perform a case inversion in the implementation's character set, the numeric literals would always perform case inversion in ASCII, so be mindful of the intention. – Patrick Roberts Sep 02 '19 at 17:47
@chux You are right, I forgot about that. Already edited. Also added `('a' - 'A')` Thanks for mentioning. – Miradil Zeynalli Sep 02 '19 at 17:59
Why is changing the original string not advised? I think I heard something similar before but never got an answer to that question – Impasse Sep 03 '19 at 07:59
@Impasse like I said, either could be correct, or incorrect, depending on your intent, but my point was just to make the answerer aware they were not equivalent behavior according to the specification. – Patrick Roberts Sep 03 '19 at 13:21

Is there a way to use pointers + a variable to traverse an array?

1 Answers1