CODEWITHSUNDEEP
csortingqsortdouble-pointer
bodacydo
bodacydo

Reputation: 79339

How to qsort an array of pointers to char in C?

Suppose I have an array of pointers to char in C:

char *data[5] = { "boda", "cydo", "washington", "dc", "obama" };

And I wish to sort this array using qsort:

qsort(data, 5, sizeof(char *), compare_function);

I am unable to come up with the compare function. For some reason this doesn't work:

int compare_function(const void *name1, const void *name2)
{
    const char *name1_ = (const char *)name1;
    const char *name2_ = (const char *)name2;
    return strcmp(name1_, name2_);
}

I did a lot of searching and found that I had to use ** inside of qsort:

int compare_function(const void *name1, const void *name2)
{
    const char *name1_ = *(const char **)name1;
    const char *name2_ = *(const char **)name2;
    return strcmp(name1_, name2_);
}

And this works.

Can anyone explain the use of *(const char **)name1 in this function? I don't understand it at all. Why the double pointer? Why didn't my original function work?

Thanks, Boda Cydo.

Upvotes: 29

Views: 23410

Answers (9)

Roshan Tiwari
Roshan Tiwari

Reputation: 1

qsort() passes a pointer to the user-defined comparison function and as you have a char * (pointer to char array) hence your comparison function should dereference from pointer to pointer hence char **.

Upvotes: 0

Kemin Zhou
Kemin Zhou

Reputation: 6891

Maybe it is easier to give you an code example from me. I am trying to sort an array of TreeNodes and the first few lines of my comparator looks like:

int compareTreeNode(const void* tt1, const void* tt2) {
   const TreeNode *t1, *t2;
   t1=*(const TreeNode**)tt1;
   t2=*(const TreeNode**)tt2;

After that you do your comparison using t1 and t2.

Upvotes: 1

linuxstack
linuxstack

Reputation: 777

@bodacydo here is a program that may explain what other programmers are trying to convey but this would be in context of "integers"

#include <stdio.h>


int main()
{
    int i , j;
    int *x[2] = {&i, &j};

    i = 10; j = 20;

    printf("in main() address of i = %p, address of j = %p \r\n", &i, &j);

    fun(x);
    fun(x + 1);

    return 0;
}


void fun(int **ptr)
{
    printf("value(it would be an address) of decayed element received = %p, double dereferenced value is %d \r\n",*ptr, **ptr);
    printf("the decayed value can also be printed as *(int **)ptr = %p \r\n", *(int **)ptr );
}

Upvotes: 0

R.. GitHub STOP HELPING ICE
R.. GitHub STOP HELPING ICE

Reputation: 215257

The comparison function takes pointers to the type of object that's in the array you want to sort. Since the array contains char *, your comparison function takes pointers to char *, aka char **.

Upvotes: 2

Steve Jessop
Steve Jessop

Reputation: 279255

If it helps keep things straight in your head, the type that you should cast the pointers to in your comparator is the same as the original type of the data pointer you pass into qsort (that the qsort docs call base). But for qsort to be generic, it just handles everything as void*, regardless of what it "really" is.

So, if you're sorting an array of ints, then you will pass in an int* (converted to void*). qsort will give you back two void* pointers to the comparator, which you convert to int*, and dereference to get the int values that you actually compare.

Now replace int with char*:

if you're sorting an array of char*, then you will pass in a char** (converted to void*). qsort will give you back two void* pointers to the comparator, which you convert to char**, and dereference to get the char* values you actually compare.

In your example, because you're using an array, the char** that you pass in is the result of the array of char* "decaying" to a pointer to its first element. Since the first element is a char*, a pointer to it is a char**.

Upvotes: 26

Billy ONeal
Billy ONeal

Reputation: 106549

char *data[5] = { "boda", "cydo", "washington", "dc", "obama" };

is a statement asking the compiler for an array of size 5 of character pointers. You have initialized those pointers to string literals, but to the compiler, it's still an array of five pointers.

When you pass that array into qsort, the array of pointers decays into a pointer pointing to the first element, in accordance with C array parameter passing rules.

Therefore you must process one level of indirection before you can get to the actual character arrays containing the constants.

Upvotes: 0

Andre Holzner
Andre Holzner

Reputation: 18675

from man qsort:

The  contents of the array are sorted in ascending 
order according to a comparison function pointed to by
compar, which is called with two arguments that **point**
to the objects being compared.

So it sounds like the comparison function gets pointers to the array elements. Now a pointer to a char * is a char ** (i.e. a pointer to a pointer to a character).

Upvotes: 0

jilles
jilles

Reputation: 11232

qsort is general enough to sort arrays consisting of other things than pointers. That's why the size parameter is there. It cannot pass the array elements to the comparison function directly, as it does not know at compile time how large they are. Therefore it passes pointers. In your case you get pointers to char *, char **.

Upvotes: 2

Henk Holterman
Henk Holterman

Reputation: 273264

Imagine your data was double data[5] .

Your compare method would receive pointers (double*, passed as void*) to the elements (double).
Now replace double with char* again.

Upvotes: 4

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