Parsing pointer parameter in C

Question

This code take 5 strings and sort them in ascending way.

 void swap (char data[5][255], int i, int j) {
     char temp[255];
     strcpy(temp,data[i]);
     strcpy(data[i],data[j]);
     strcpy(data[j],temp);
 }

 void sort (char data[5][255], int n) {
     // * : first address contact
     int i, j;
     for(i = 0; i < n-1; i++)
         for( j = i+1; j > 0; j--)
             if(strcmp(data[j-1],data[j])>0)
             {
                 printf("%s",data[j-1]);
                 swap(data, j-1, j);
             }
 }

 int main() {
     char strings[5][255];
     char comp[255];
     int i, n;
     n = sizeof(strings)/sizeof(comp);
     printf("Enter 5 strings, max 255 chars each:\n");
     for(i=0; i < n; i++)
         scanf("%s",strings[i]);
     sort(strings, n);
     printf("Sorted data:\n");
     for(i=0; i < n-1; i++)
         printf("%s, ",strings[i]);
     printf("%s.\n",strings[i]);
     return 0;
 }

In addition of that, how can I possibly parse my static array string[5][255] to function by using pointer? I tried that, for example,

void sort ( char **data, int i ) { ... }

but it throws out error like this.

incompatible pointer types passing 'char [5][255]' to parameter of type 'char **'

Is there anything I can parse my array using pointer?

Since array parsed to function its first address(pointer), I thought function will accept those expression. Please give me some advice to understand.

http://stackoverflow.com/questions/12674094/array-to-pointer-decay-and-passing-multidimensional-arrays-to-functions << see here — Jekyll, Dec 17 '13 at 23:38

Olaf Dietsche · Accepted Answer · 2013-12-18T23:11:47.550

4

The parameter you must pass is not a pointer to a pointer char**, but a pointer to a char array char(*)[255]

void sort (char (*data)[255], int n)

When you pass an array, you can omit the size of the first dimension

void sort (char data[][255], int n)

which is equivalent to char(*)[255].

char** is a pointer, which points to another pointer. Whereas char(*)[255] is a pointer, which points to an array.

edited Dec 18 '13 at 23:11

answered Dec 17 '13 at 23:33

Olaf Dietsche

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Thanks Mr. Olaf, but I can not understand why it is impossible. Is that because of array saving priority to do not allocate parsing parameters as an infinite way? – sogo Dec 17 '13 at 23:38
It is not possible, because these are different types. Please see the updated answer. – Olaf Dietsche Dec 17 '13 at 23:41
Thanks a lot Mr. Olaf, to sum up, function can not be point a pointer, but can point any other type (like array), right? – sogo Dec 17 '13 at 23:46
1

You can use pointer to a pointer as @Jekyll shows. In your case however, you just don't have one, but a pointer to array. – Olaf Dietsche Dec 17 '13 at 23:49
it's not `char *[255]`, but `char (*)[255]`. They are very different. – newacct Dec 18 '13 at 22:54

Jekyll · Answer 2 · 2013-12-19T09:29:00.353

1

As I wrote you in the comment char[][] doesn't decay to char** but it decays to char(*)[] ( char (*data)[255] in your case) as the first element decay in a pointer which is not a "pointer to a pointer" but a pointer to an array of 255 characters. It is possible to use a char** pointer if you do something like this (c++):

char **array = new char *[N];

for(int i = 0; i<N; i++)
    array[i] = new char[N];

or using the malloc (c).

As @newact suggests it is important to distinguish between

char *data[255] -> char *[255]data = => "data is an array of 255 pointers to char" And

char (*data)[255] -> char [255](*data) => "data is a pointer to an array of 255 chars"

edited Dec 19 '13 at 09:29

answered Dec 17 '13 at 23:44

Jekyll

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I got it, Mr. Jekyll! Thanks. It helped me a lot! – sogo Dec 17 '13 at 23:55
1

`char*[]` is an array of pointers. very different from `char (*)[255]` – newacct Dec 18 '13 at 22:55
@newacct corrected the char*[] in char(*)[] in the description :). – Jekyll Dec 19 '13 at 09:48

Darkhan · Answer 3 · 2013-12-19T01:45:34.293

1

You have to understand the difference between: char **data vs char (*data)[255]

Those are two different types because the allocation of memory may be different:

When char **data is used pointer arithmetic may not work properly, meaning data could be scattered all over the memory

When char (*data)[255] is used pointer arithmetic works perfectly because all elements of array are adjacent to each other

edited Dec 19 '13 at 01:45

answered Dec 17 '13 at 23:46

Darkhan

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Thanks Mr. Daveel. Then if I assign array by using memory allocation like 'malloc', then is it possible to parse to double pointing function like, 'sort(char **data, int n)'? Because I remember I have tried parse dynamic array to pointer. – sogo Dec 17 '13 at 23:51
1

@Sogo Yes, if you will allocate your strings on the heap, then your sort(char **data, int n) will work – Darkhan Dec 18 '13 at 00:03
`char *data[255]` is not `char (*data)[255]` – newacct Dec 18 '13 at 22:55

score -1 · Answer 4 · edited Dec 18 '13 at 00:04

-1

Try to make the following changes:

void swap ( char *data, int i, int j)

*(data + index1*255 + index2)

edited Dec 18 '13 at 00:04

Rubens

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answered Dec 17 '13 at 23:36

SliceSort

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Parsing pointer parameter in C

4 Answers4