C char qsort in array of pointers to strings

Question

I have an array of pointers to string:

char *TAB[3] = { "dafafa", "alfkasf", "bafgr" };

I would like to sort characters in in each of those strings.

My compare function:

int cmp(const void *a, const void *b)
{
   return *(char *)a - *(char *)b; 
}

and while trying qsort on one of these:

qsort(TAB[0], 6, sizeof(char), cmp);

The program doesn't work. After many efforts I found that the reason of the problem is in delivering TAB[0] to qsort().

Can anyone explain why it doesn't work and how to fix that?

Literal strings in C are read only. Attempting to modify them leads to undefined behavior. — Some programmer dude
– Some programmer dude, Commented Aug 31, 2017 at 11:37
"I would like to sort characters in in each of those strings." --> and where would you like to put that sorted string given that writing to the source string, a string literal, is undefined behavior? — chux
– chux, Commented Aug 31, 2017 at 13:09

Sergey Kalinichenko · Accepted Answer · 2017-08-31 11:36:38Z

4

If you want to sort characters inside each string, the first thing you must ensure is that your strings can be written to. As it currently stands, your strings are read-only, so you cannot sort their characters without copying their content into memory that allows writing.

Next thing is that you need a loop. Since you are sorting each string individually, you need to loop through the array, and call qsort on each item. The initial item is TAB[i], and the length is strlen(TAB[i]). Your cmp function will work.

answered Aug 31, 2017 at 11:36

Sergey Kalinichenko

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4 Comments

Piotr Witkoś Over a year ago

How do you know that those strings are read-only? Could you show me code example which can fix that? I've checked that after declaring char tab[]="dafafa" instead of char* tab="dafafa" it worked. Whats the difference there?

Sergey Kalinichenko Over a year ago

@PiotrWitkoś All string literals are read-only, in the sense that you are not allowed to change their content. char tab[]="dafafa" is different, because you initialize an array of characters with a string literal. This array becomes writable. TAB[3] in your original code is an array of pointers, so it is very much like char* tab="dafafa" repeated three times. You can check here to see how to properly copy strings.

Piotr Witkoś Over a year ago

I have a question in my exam to sort characters in every string in an array of strings TAB[100]. How would you declare it to avoid copying the content?

Sergey Kalinichenko Over a year ago

@PiotrWitkoś char TAB[3][100]={"dafafa","alfkasf","bafgr"}; will do the trick (demo).

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