CODEWITHSUNDEEP
cnullchar
daremkd
daremkd

Reputation: 8424

Confused about the Null Zero in C

Let's say you have:

char[5] = "March";

this won't work in C although:

char[0]='M'
char[1]='a'
char[2]='r'
char[3]='c'
char[4]='h'
char[5]='\0'

For some reason, I have to say to C that char[6]="March" and not char [5].

Why is this? What goes into char[6]?

Upvotes: 1

Views: 163

Answers (8)

paxdiablo
paxdiablo

Reputation: 882426

First things first, char is not a valid variable in C, it's a keyword. You probably meant 

char xyzzy[6];

or something similar, which would create a character array called xyzzy. But, once that's fixed up, nothing goes into xyyzy[6]. The statement char xyzzy[6]; means an array of six characters, the indexes of which are 0 through 5 inclusive: {0,1,2,3,4,5}, that's six elements.

In any case, unless you need the array to be bigger, you're usually better off letting the compiler choose the size with:

char xyzzy[] = "March";

Upvotes: 4

Siddarth
Siddarth

Reputation: 21

Char is a keyword not a variable name, so don't use it as one.

Array indexing: that is counting the number of characters in an array begining from 0, i.e a[0].

Suppose there are 5 characters in the array (for eg: tiger) then a[0] = 't' and a[4] = 'r' and all the string literals in this case an array of characters ends in a '\0' i.e a EOF
character. So to store an array of n characters use a[n] compiler will add '\0' at the end of that array.

Upvotes: 0

hmjd
hmjd

Reputation: 122001

A string literal, "March" in this case, has an implicit null terminator so a char[] array requires 6 elements to store it. To quote the relevant points from the section 6.4.5 String literals of the C99 standard:

A character string literal is a sequence of zero or more multibyte characters enclosed in double-quotes, as in "xyz".

In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals.

In both cases the code is overrunning the end of the array. The code currently has undefined behaviour.

Change to:

char a[6] = "March";

The second code snippet accesses beyond the end of the array as indexes run from 0 to N - 1, where N is the number of elements in the array. For an array char a[5]; the valid index are 0, 1, 2, 3, 4.

Upvotes: 1

user529758
user529758

Reputation: 

char array[n]

means an n elements long array, not an array whose last index is n. Because in C, arrays are zero-based, it means that the last valid index of an n-element array is n - 1. Since there's a zero terminator for a string, and "March" is 5 letters, that's a total of 6 characters, so you have to write char str[6] = "March";. If that's confusing for now, don't include the length of the array; for initialized arrays, the compiler will automagically fill it in, so you can write char str[] = "March"; instead.

Upvotes: 0

tomsv
tomsv

Reputation: 7277

Nothing goes into char[6], but you still need a char[] of size 6 to hold the 6 characters including the null-terminator byte. They will go into char[0] through char[5]. char[6] remains undefined.

Upvotes: 0

ckruse
ckruse

Reputation: 9740

char [5] means, you have memory for 5x sizeof(char). 0-5 are six times sizeof(char): 0 first, 1 second, 2 third, 3 fourth, 4 fifth, 5 sixth. So you need a segment with 6 times sizeof(char): char [6]. The first cell, zero, consumes space, too.

Upvotes: 0

Denys Séguret
Denys Séguret

Reputation: 382444

There are 6 elements if you count from 0 to 5 included.

So you must declare a char[6] in order to have str[0] to str[5].

Upvotes: 0

Minion91
Minion91

Reputation: 1929

The null terminator \0 goes in the last slot. It is there because otherwise there would be no way to check the length of the string.

Upvotes: 0

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