replace the "*" in the above string with " ", a single space character using ANSI C??

ok, I am very new to C programming and i have a problem at hand, i wuld really really appreciate any help plz,



The string is defined as the following:

char * foo = " Baltimore *MD*is*a*really*good*place*to*work*and*play";

Using the smallest amount of code possible, replace the "*" in the above string with " ", a single space character using ANSI C language. The resulting string should be printed out at the end of the program. In addition to a manual examination, the resultant .C file will be compiled with GCC compiler with the "-ansi" flag set and executed to determine success

CAN SOMEBODY HELP PLEASE?

 

We'd be glad to help, however we don't do your work for you. Refer to your lessons, work out some come, and present it for help with any problems you encounter. Be sure to read the "Before you make a query" thread (upper right corner of this page) before you post code -- code snippets should be inside code tags. You can learn about them there.

 

You can also use standered function strtok( ) .Take a char pointer and allocate it memory enough to hold the string.call strtok for tokens and append space at the last of each token and copy first token to string then concadinate others.

 

also put terminating char in the last of string.

 

ok here is the code i made

Code:

#include <stning.h>
char *foo = " Charlotte *NC*is*a*really*cool*place*to*work*and*play";
{
     char delims[]="*";
     char *result = NULL;
     result = strtok ( str, delims );
     while ( result != NULL ) 
     {
           printf ("result is "\%s"\n", result );
           result = strtok (NULL, delims );
     }
}

can some one help me improve it?

 

He does not want to tokenize the string, he want to replace occurences of one character with other. There's no point in doing it in an unwieldy and inefficient way.

On most operating systems the current string, as defined, wil be const. That is, he will not be able to write to it. This will require copying the string, while replacing the characters on the fly. The terminating zero is already there, it just needs to be included in the copy.

 

We normally don't solve people's homework for them, Hardik. We help THEM solve it. It is how one learns. I'm making an exception in this case because you are complicating the problem unduly, and doing the original poster a huge disfavor.

Code:

#include <stdio.h>
#include <string.h>

int main (int argc, char* argv[])
{
    // If your OS makes a string defined as follows a const value (most do)
    char *foo = "Charlotte *NC*is*a*really*cool*place*to*work*and*play";
    // then you'll have to make a copy of the string, thusly:
    char *bar = strdup (foo);
    // or you may just define the string differently, thusly:
    char baz [] = "Charlotte *NC*is*a*really*cool*place*to*work*and*play";

    // In any even, you make the substitution like this:
    char *replace = baz;
    printf ("%s\n", baz);
    while (*replace)
    {
        if (*replace == '*') *replace = ' ';
        replace++;
    }

    printf ("%s\n", baz);
	return 0;
}

 

well, if it makes feel good at all, this problem is not a homework problem, nor i am a student at all, also i see how simply you have solved it when i spent almost half night reading about strtok and delims, i missed a simple thing of copying string and making a simple loop, i guess i complicated it too much, but in anycase your help is not used in some classroom to get a better grade, thank you very much for ur directions

 

i have a question though, i knw strcpy will not allocate the memory for the copy for you so u used strdup, but is there an easier way to it without strdup? i was just wondering