strings of characters in C - c

Question

I am trying to solve a C problem where I have to sort n strings of characters using pointers.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void sort (char *s, int n)
{
int i,j; char aux[]="";
for (i=1;i<=n-1;i++)
{
for (j=i+1;j<=n;j++) //if s[i]> s[j] switch
{
if (strcmp(*(s+i),*(s+j))==1)
{
strcpy(aux,*(s+i);
strcpy(*(s+i),*(s+j));
strcpy(*(s+j),*(s+i));
}
}
}
}
void show(char *s, int n)
{
int i;
for (i=1;i<=n;i++)
{
printf("%s",*(s+i));
}
}
int main()
{
int i,n; char *s;
printf("give the number of strings:\n");
scanf("%d",&n);
s=(char*)calloc(n,sizeof(char));
for (i=1;i<=n;i++)
{
printf("s[%d]= ",i);
scanf("%s",s+i);
}
sort(s,n);
show(s,n);
return 0;
}
The warnings that I get are when I use strcmp to compare and when I use strcpy to switch *(s+i) and *(s+j) values.
"passing argument 2 of strcmp makes pointer from integer without a cast"

Answer 1

The warnings that I get are when I use strcmp to compare and when I use strcpy to switch *(s+i) and *(s+j) values.
"passing argument 2 of strcmp makes pointer from integer without a cast"
You are wrong argument to strcmp and strcpy. Signature of strcmp and strcpy are
int strcmp(char *string1, char *string2);
char *strcpy(char *dest, const char *src)
But you are passing it the arguments of char type. Removefrom(s+i)and*(s+j)`. It should be
if (strcmp((s+i), (s+j))==1)
{
strcpy(aux, (s+i));
strcpy((s+i), (s+j));
strcpy((s+j), (s+i));
}
Another problem is you have not allocated memory for the pointer s. For characters, you can declare it as
s = malloc ( (n + 1)*sizeof(char) );
or simply
s = malloc ( n + 1 ); // +1 is for string terminator '\0'

Answer 2

You seem to have mis-understood the return value of strcmp(); it's an integer that will be 0 (zero) if and only if the two string arguments are equal. You're testing for 1, which it will only be rarely; it has no specific meaning.
Consider just using qsort().

Answer 3

You're allocating space for n strings of 1 byte each. The easiest approach is to assume each string will be less than some set length, like 40 bytes. Then you would allocate the memory like this:
s=(char*)calloc(n*(40),sizeof(char));
Then your scanf needs to be modified:
scanf("%s",s+(i*40));
Now, string 1 will be at *s, string 2 will be at *(s+40), etc. Keep in mind that a string ends with a null character (0x00), so the string can only contain 39 chars. Any unused data will also be 0x00.
Do the same s+(i*40) for the sorting algorithm, compare to >0, not ==1, and strcmp, strcpy expect pointers. Then you should be good.

Answer 4

In your code there are a lots of mistakes: in variable declarations, memory allocation and pointer usage.
First remember that you should always reserve enough space to store your string values, and be aware to don't try to store a string with a length greater then the reserved space.
Also you should be very careful managing pointers (s+1 doesn't point to the first string but to the second one, s+n points out of the allocated memory)
So, as mentioned in other answers you should decide the maximum size of your strings an then allocate for them the right amount of memory.
Then I suggest you to use a pointer to strings, i.e. a char** to access your strings and manage the sort, in this way the code is more readable and the sort is faster (you don't need to copy strings, but only to switch pointers)
So your main function should be:
int main()
{
int i, n;
// Declare the pointer to the memory where allocate the space for the strings
char* a; // points to a char ( == is a string)
// Declare the pointer to the memory where store the pointer to the strings
char** s; // points to a char* ( == is a pointer to a string)
printf("give the number of strings:\n");
scanf("%d", &n);
// Allocate the memory for n strings of maximum 40 chars + one more for the null terminator
a=(char*)calloc(n*41, sizeof(char));
// Allocate memory for n pointers to a string
s=(char**)calloc(n, sizeof(char*));
// Notice the 0-based loop
for (i=0; i<n; i++)
{
s[i] = a + i*41; // set the i-th element of s to point the i*41-th char in a
printf("s[%d]= ", i);
scanf("%40s", s[i]); // read at least 40 characters in s[i],
// otherwise it will overflow the allocated size
// and could generate errors or dangerous side effects
}
sort(s,n);
show(s,n);
// Remember always to free the allocated memory
free(s);
free(a);
return 0;
}
the sort function sorts an array of pointers, without copying strings:
void sort (char** s, int n)
{
int i, j;
char* aux;
// Notice the 0-based loop
for (i=0; i<n; i++)
{
for (j=i+1; j<n; j++) //if s[i]> s[j] switch
{
if (strcmp(s[i], s[j])>0)
{
// You don't need to copy strings because you simply copy pointers
aux = s[i];
s[i] = s[j];
s[j] = aux;
}
}
}
}
finally the correct show function:
void show(char** s, int n)
{
int i;
// Notice the 0-based loop
for (i=0; i<n; i++)
{
printf("%s", s[i]);
}
}
I can't test the code but it should works
Added other options
1) If you have troubles with dynamic memory allocation you can use instead static allocation, your main function then will be:
int main()
{
int i, n;
// Declare an array of fixed lenght strings
char s[100][41]; // You manage max 100 strings
printf("give the number of strings:\n");
scanf("%d", &n);
// Ensure that n is less or equal maximum
if (n>100)
{
printf("you'll be asked for a maximum of 100 strings\n");
n=100;
}
// Notice the 0-based loop
for (i=0; i<n; i++)
{
printf("s[%d]= ", i);
scanf("%40s", s[i]); // read at least 40 characters in s[i],
// otherwise it will overflow the allocated size
// and could generate errors or dangerous side effects
}
sort(s,n);
show(s,n);
// You don't need to free any memory
return 0;
}
all other functions remain unchanged.
2) Whereas, if you want the maximum freedom in the memory you need you can choose a two allocate memory separately for each string only for the needed size, in this case your main funcion will be:
int main()
{
int i, n;
char buffer[1000]; // You need a buffer to read input, before known its actual size
// Declare the pointer to the memory where store the pointer to the strings
char** s; // points to a char* ( == is a pointer to a string)
printf("give the number of strings:\n");
scanf("%d", &n);
// Allocate memory for n pointers to a string
s=(char**)calloc(n, sizeof(char*));
// Notice the 0-based loop
for (i=0; i<n; i++)
{
printf("s[%d]= ", i);
scanf("%999s", buffer); // read at least 999 characters in buffer,
// otherwise it will overflow the allocated size
// and could generate errors or dangerous side effects
// Allocate only the memory you need to store the actual size of the string
s[i] = (char*)calloc(strlen(buffer)+1, sizeof(char)); // Remember always 1 more char for the null terminator
// Copy the buffer into the newly allocated string
strcpy(s[i], buffer);
}
sort(s,n);
show(s,n);
// Remember always to free the allocated memory
// Now you have first to free the memory allocated for each string
for (i=0; i<n; i++)
{
free(s[i]);
}
// Then you can free the memory allocated for the array of strings
free(s);
return 0;
}
all other functions remains unchanged too.