How does the string populate an array

Question

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?

no, you only have one string of length 100. so your puts will issue a kind of shifted version of your sentence, or garbage if your sentence is shorter than 15 chars. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Feb 16, 2017 at 10:13
A string is several characters, string1[0] is a single character. How can it be stored there in its entirety? — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Feb 16, 2017 at 10:14
Yes, you've misunderstood. C doesn't really have strings, it has arrays of chars with a null byte to terminate. Yu can manipulate the chars as small integers using the [] syntax, and string[0] is the first letter, string[1] the second letters, and string[N] is nul. — Malcolm McLean
– Malcolm McLean, Commented Feb 16, 2017 at 10:18
@StoryTeller If I get rid of the for loop and use puts(&string1[0]); It will output the original sentence instead of just outputting the first character, which is causing me confusion. — Project Skeith
– Project Skeith, Commented Feb 16, 2017 at 10:21

StoryTeller - Unslander Monica · Accepted Answer · 2017-02-16 10:19:36Z

2

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.

answered Feb 16, 2017 at 10:19

StoryTeller - Unslander Monica

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Comments

myaut · Accepted Answer · 2017-02-16 10:20:36Z

1

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])

answered Feb 16, 2017 at 10:20

myaut

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1 Comment

Project Skeith Over a year ago

Thank you for this. I was misinterpreting the other answers, because I was misreading as if only one letter should have been outputted per loop based on their reasoning. However, showing the &string1[5] example help cement what was going on when I use puts.

sameerkn · Accepted Answer · 2017-02-16 10:24:09Z

1

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)...

edited Feb 16, 2017 at 10:24

answered Feb 16, 2017 at 10:19

sameerkn

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