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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?

  
Total Answers and Comments: 2 Last Update: January 09, 2007     Asked by: pavan kumar reddy 
  
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December 27, 2006 03:41:31   #1  
R.S.JAYASATHYANARAYANAN        

RE: char str[]="DOLLAR";Here , you have the base...

char str[] DOLLAR

char *ptr;

*(ptr+2) '$$';

*(ptr+3) '$$';

then it will display DO$$$$AR


 
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January 09, 2007 01:12:17   #2  
mekala        

RE: char str[]="DOLLAR";Here , you have the base...
#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();
}
}
}


 
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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?
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