CODEWITHSUNDEEP
carraysstring
Project Skeith
Project Skeith

Reputation: 25

How does the string populate an array

I'm creating a simple program to see how a string populates an array.

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

int main(void)
{
    char string1[100];
    int i;

    printf("Enter sentence.\n");
    fgets(string1, 100, stdin);

    for (i=0; i<=15; i++)
       puts(&string1[i]);
    return 0;
}

I'm having a bit of a problem understanding how the string is populating an array. My expectation is that the string will be completely stored in string1[0] and any further indexes will come up blank. However, when I throw the loop to see if my assumption is true, it turns out that every index has been filled in by the string. Am I misunderstanding how the string is filling the array?

Upvotes: 1

Views: 71

Answers (3)

sameerkn
sameerkn

Reputation: 2259

string is not stored in string1[0] but string's first character is stored at string1[0] or string starts at (string1+0). Here, &string1[0] or (string1+0) can be seen as a pointer, a pointer to C String string1.

In that sense, every valid index i of string1 will give you a valid pointer (string1 + i) which will point to some part of C String string1.

In the last for loop you are printing the suffixes of string string1 which are pointed by (string1 + 0), (string1 + 1), (string1 + 2)...

Upvotes: 1

myaut
myaut

Reputation: 11514

Your problem is not string1 layout per se but how puts interprets it. Strings are represented by char arrays in C while their end is marked as null terminator (character with code 0):

 S e n t e n c e \0
 ^         ^
 string1   &string1[5]

&string1[5] is a pointer to a one character, but since the following character is not null terminator, following memory is interpreted as a string and nce gets printed.

You'll need to use putc and access individual characters:

putc(string1[i])

Upvotes: 1

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170202

For the string "Hello!", the memory representation would be something like this

+-------+-------+-------+-------+-------+-------+-------+
|  'H'  |  'e'  |  'l'  |  'l'  |  'o'  |  '!'  | '\0'  |
+-------+-------+-------+-------+-------+-------+-------+

The first cell, at index 0, contains the first character. And each subsequent character is contained in a cell with an increasing index.

Library functions like puts expect you to pass the address of the first character, and then they read the string up to \0.

So if you pass simply string1 or &string1[0], it will resolve to the address of 'H'.

If you pass &string[1], it will resolve to the address of 'e', and the library function will think that is the first character, because that's the contract C strings are designed with.

Upvotes: 2

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