C Programming - Strings - Discussion

Discussion Forum : Strings - Find Output of Program (Q.No. 12)
12.
What will be the output of the program ?
#include<stdio.h>
#include<string.h>

int main()
{
    static char s[] = "Hello!";
    printf("%d\n", *(s+strlen(s)));
    return 0;
}
8
0
16
Error
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
38 comments Page 2 of 4.

Mohana Murali yk said:   9 years ago
S returns address where character 'H' is stored, it is added to length of string, which is unknown? Am I right?

Snehal said:   9 years ago
Thank you @Dipankar. Nice explanation.

Anant said:   10 years ago
It will print 0 because it points to null character.

Chandra vijay vishwakarma said:   1 decade ago
See, strlen(s)=6, and every string have a default null character, right?

So whenever we call *(s+strlen (s)) means, now char *s is pointing to null character and ASCII of null character is 0, So it is printing 0.

Lak said:   1 decade ago
Length of string = 8;

Vivek said:   1 decade ago
strlen(s) = 6.
*(s + 6) = s[6].
s[6] = null.
Therefore it is 0.

As @Meghana said.

Taj said:   1 decade ago
IT's ok that it will print "0" when "%d" is used.

Why it is printing blank when "%c" is used?

Ram bhawan said:   1 decade ago
Length of string = 6.

str+6 = \0(i.e. Ascii value = 0).
i.e. = 0;
Simply.

Subhankar Som said:   1 decade ago
strlen of Hello is 6. The expression *(s+strlen(s)) indicates *(s+6). According to pointers to array concept *(s+6)=s[6].

The last element contain '\0'. So the output gives 0 value as the value of \0 is 0(ASCII).

Meghana said:   1 decade ago
strlen of hello! is 6. So expression in printf statement reduces to *(s+6). *(s+6) is same as s[6], which is null character. ASCII value for null character is 0. So it prints 0.


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