changing from array to pointer format?

[mmaria21]

Ive tried changing it to pointer format but when i do, the new string without spaces doesnt appear. i appreciate the help!

#include <iostream>
int stripspaces(char str[]);
using namespace std;
[LEFT][CENTER][RIGHT][CODE]

[/RIGHT][/CENTER][/LEFT]

int main()
{
char ar[50];
cout << "enter a sentence: " <<endl;
cin.getline(ar, 49);
cout<<" your sentence has " <<stripspaces(a r)<<" spaces "<<endl;

return 0;
}

int stripspaces (char str[])
{
int spacecounter = 0;
char * p =str;

while (*p)
{
if (*p == ' ')
spacecounter++;
p++;
}
int i =0;
char temp[50];
int count= 0;

while (str[i])
{
if(str[i]!= ' ')
{
temp[count] = str[i];
count++;
}
i++;
}
temp[count] = '\0';
cout <<" The new string without spaces is:" <<temp<<endl;
return spacecounter;
}[/CODE]

[newb16]

What doesn't work in the program above and what is the expected result?

[mmaria21]

the program above works but its written in array notation but i have to change it to pointer notation and so i tried writing something like char * t= temp; after line 27 so that later i could refer to the values in temp as *(t +i) and such so that i could change to pointer notation. but when i did that the program doesnt output the new string without the spaces(which is what is supposed to output). it does output the second part correctly, which is the part that tells you how many spaces are in the entered string.

[whodgson]

Apart from not understanding how your code keeps the screen open I would have thought that line 7 should be char* ar[50]; etc

[newb16]

I tried writing char *t=temp and replacing temp[count] with *(temp+count) after line 27 and it worked for me - it counts spaces and outputs trimmed string.

[mmaria21]

hm.. I guess ill try again then! Thank you! would that be sufficient for the program to be considered in pointer notation?

[whodgson]

Yes.....because thats what it would be using.