Thread: sscanf replacing chars

does anyone know how i can use sscanf to replace characters in a string?


say i have string

abcd

and i want to replace a and d with a space, how coul di do that?


thanks




also does anyone know why im getting this warning?

Code:

char * func1() { 

char blah[30];

return blah;

}

warning C4172: returning address of local variable or temporary

basically, you were returning the address of the variable which goes out of scope after the function returns. The memory pointer that your are returning to calling function is in "dangling" state after that. If its good till the time you use it in calling function, you are lucky. Otherwise , if its currpted or allocated to some other process, your calling function accessing it might produce crash.

Static solves the problem by making its scope persistant even when the function declaring it goes out of scope. These variables are stored in a seperate locations in memory and are valid throughout the program scope.

Originally Posted by paperbox005

does anyone know how i can use sscanf to replace characters in a string?

First, I wouldn't use sscanf.

Originally Posted by paperbox005

say i have string

abcd

and i want to replace a and d with a space, how coul di do that?

I'd just loop through the source string looking for the character to find and replace it with the desired string.

Code:

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

char *foo(char *dst, const char *src, char find, const char *replace)
{
   char *start = dst;
   while ( *src )
   {
      if ( *src == find )
      {
         strcpy(dst, replace);
         dst += strlen(replace);
      }
      else
      {
         *dst++ = *src;
      }
      ++src;
   }
   *dst = '\0';
   return start;
}

int main()
{
   static const char before[] = "abcdabcdabcd";
   char after [ sizeof before * 2 ];
   puts(foo(after, before, 'a', "d "));
   return 0;
}

/* my output
d bcdd bcdd bcd
*/