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Full Discussion: Please Explain me the output
Top Forums Programming Please Explain me the output Post 302155315 by shamrock on Thursday 3rd of January 2008 03:40:16 PM
Old 01-03-2008
Quote:
Originally Posted by vikashtulsiyan
#include<stdio.h>

char *def[5]={"pqrs","rstu","tuvw","vwxyz","xyzab"};
char abc[5][5]={"abc","def","ghi","jkl","mno"};

void main()
{
char *p=(char *)def;
p=p+40;
printf("%s\n",p);
}

the output of the abve code snippet is mno...
HOW??? beats me.. please help
There are some basic things that are wrong with the above code. That's why the output doesn't make sense. You can't equate p to def since p is a pointer to char while def is a pointer to pointer to char. In fact the proper declaration for p and the code should be...

Code:
char **p = def;
p=p+40;
printf("%s\n",*p);

The problem occurs probably because def is being cast to a unilevel pointer as in (char *)def and hence outputs garbage.
 

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