I need some help with the following C code.?

The real problem is printf(temp); This needs to be printf("%s\n", temp);

I've also added an "x" so you can tell when the original array is printed.

I've also added a prototype to get rid of a compiler warning.

I've also added explicit braces so that it's easier to edit later.

#include <stdio.h>

#include <stdlib.h>

#include <errno.h>

#include <math.h>

#include <string.h>

void reorder (int n, char *x[]);

int

main (void)

{

static char *x[2] = { "Hello",

"Bye"

};

int i;

for (i = 0; i < 2; ++i) {

printf ("%s\n", x[i]);

}

printf ("x\n");

int n = 2;

reorder (n, x);

return 0;

}

void

reorder (int n, char *x[])

{

char *temp;

int i, item;

for (item = 0; item < n - 1; ++item) {

for (i = item + 1; i < n; ++i) {

if (strcmp (x[item], x[i]) > 0) {

temp = x[item];

x[item] = x[i];

x[i] = temp;

printf ("%s\n", temp);

}

}

}

return;

}

It's amazing how wrong the other two answers are.

One said char * x[2] is an array of two characters. That's not right. It's an array of two pointers to a character. So you're fine there.

Another said your printf() statement was wrong but didn't point out that you're not really printing the reordered array, you're printing only when the values are exchanged, and then only the greater of the two strings.

So what you need to do is repeat the little loop that prints the array after you call reorder(). And remove the printf() statement in reorder(). Otherwise it looks fine.

The problem is that you're treating Strings as objects, which they are not in C.

To C, a String is an array of chars terminated by null.

In you're code you say:

static char *x[2] = {"Hello", "Bye" };

which is an array with only enough room for two chars.

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