About Arrays

[sridevi101]

main()
{
char *str [] ={"Frogs","Do", "Not","Die.","T hey","Croak"};
printf("%d%d",s izeof(str),size of(str[0]));
}

Please let me know the answer ....

[Banfa]

Try compiling and running it yourself.

If (when really) you get compiler errors if you don't understand them please feel free to post back here.

[whodgson]

printf("%d\n",s izeof(str));//prints 20 because of '.' after Die and NULL char.
printf("%d\n",s izeof(str[0]));prints 4
printf("%d",siz eof(str[5]));prints 4
Remember array size is zero based and NULL will not print.

[Banfa]

No sizeof(str) is not 20, remember that str is an array of char pointers so size of str is dependent on the size of char pointer and the number of items in the array. What it is not dependent on is any of the contents of the string pointed to.

sizeof(str[0]) and sizeof(str[5]) might be 4 on a 32 bit platform that has 32 bit pointers.

[whodgson]

yes...I should have said that those byte sizes applied to the machine I was using.........t hanks for correcting.