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Please Explain me the output

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Top Forums Programming Please Explain me the output
# 8  
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.
# 9  
Old 01-09-2008
actully what puzzled me (and why i posted this question) is that the output is consistent in several platforms i ran the code. again this code i got from a book which explains in some vague way why the outout should always be mno
# 10  
Old 01-11-2008
Quote:
Originally Posted by vikashtulsiyan
actully what puzzled me (and why i posted this question) is that the output is consistent in several platforms i ran the code. again this code i got from a book which explains in some vague way why the outout should always be mno
What compiler and on what platform is it giving this output? I am unable to duplicate your results.
# 11  
Old 01-11-2008
Quote:
Originally Posted by shamrock
What compiler and on what platform is it giving this output? I am unable to duplicate your results.


am using SUNOS 5.9 with cc compiler
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