Char str[]="DOLLAR";Here , you have the base address of array i.e. str.By using only one pointer variable, you have to replace the occurrence of 'L' with '$$'. (after replacement, it should be DO$$$$AR)can any body suggest efficient method?

Interview Candidate
Nov 22nd, 2006
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R.S.JAYASATHYANARAYANAN

Dec 27th, 2006

char str[]="DOLLAR"
char *ptr;
*(ptr+2)='$$';
*(ptr+3)='$$';
then it will display DO$$$$AR

mekala

Jan 9th, 2007

    #include<stdio.h>
#include<string.h>
void replace(void)
{
char *ptr="$$";
printf("%s",ptr);
}
int main()
{
char s[]="DOLLAR";
int i;
        for(i=0;i<=strlen(s);i++)
        {
                if (s[i]!='L')
                {
                printf("%c",s[i]);
                }
                else
                {
                replace();
                }
        }
}