C: I have one confusion in the following C program related to the string and character array.

Question

#include<stdio.h>
#include<string.h>

int main()
{
    char ptr[]= {'a','b','c','0','e'};
    char str[]= "abc0e";
    printf("\nptr = %s\t len = %d\n",ptr, strlen(ptr));
    printf("\nstr = %s\t len = %d\n",str, strlen(str));
    return 0;
}

Output : ptr = abc0e len = 6
str = abc0e len = 5

Why the length for ptr is 6 ? Can someone please explain it ?

Answer 1

char ptr[]= {'a','b','c','0','e'};
in above declaration of ptr array.You are not giving '\0' character at last of ptr array so when you will print the length of ptr.You may got length of ptr 5 or other than 5.
But in below of str array declaration
char str[]= "abc0e";
the compiler adjust NULL character at the last of e element.So always you will get length of str array is 5.

Answer 2

Both are giving the length as 5 see the codepad link of your program http://codepad.org/rHdVgNMe

However explanation provided by rajan is correct in one case of str compiler added the '\0' in the end of the string so valid pointer is str[0] - str[5] where as in case of ptr valid ptr is ptr[0] to ptr[4].

Comments

I have checked many compilers e.g. gnu c,turbo c,gnu c++ also.In all cases every compiler giving different different length of ptr array.So check the program on other compilers also.

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Yes you are right, codepad may be different. In nutshell there is a difference in both :)