char array declaration in C

Question

I have been reading C book by K&R. I came across this words

Character arrays are a special case of initialization; a string may be used instead of the braces and commas notation:

char pattern = "ould";

is a shorthand for the longer but equivalent

char pattern[] = { 'o', 'u', 'l', 'd', '\0' };

In this case, the array size is five (four characters plus the terminating '\0').

I coded a simple program to test it as

#include<stdio.h>
int main()
{
    char c= "Hello";
    printf("%s",c);
    return 0;
}

According to that book there should be no error but it returns an error

`cannot convert char * to char`

As per my previous knowledge of arrays,the array statement should be like char *c="Hello"; Can you explain the meaning of the words in that book please. Did i Misunderstand the meaning of that words??

Is there really a line in your book telling " char pattern = "ould"; " ? — Kabulan0lak
– Kabulan0lak, Commented Aug 4, 2014 at 9:58
Page 86 of my copy of K&R (2nd Edition) reads char pattern[] = "ould";. If you have a digital copy (you did say you copied and pasted it), it may have been an oversight in the transcription or automated scanning process. — user539810
– user539810, Commented Aug 4, 2014 at 10:06

Quentin · Accepted Answer · 2014-08-04 10:00:49Z

2

Well it's missing two brackets. char pattern[] = "ould"; will work fine. Time to trash that book !

answered Aug 4, 2014 at 10:00

Quentin

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Comments

Baj Mile · Accepted Answer · 2014-08-04 10:24:17Z

It should be:

char pattern[] =  "ould" ;

Also be careful with :

char* pattern = "ould";

Please note that with "" you define a const characters array. It is supposed not to change anymore during app execution. If you do:

pattern[2] = '9';

your app may crash.

Correct is:

const char* pattern = "ould";

Why is that? Your compiler probably will allocate the "ould" in a protected read only memory block and pattern is just a pointer to it. If you however define :

char pattern[] = { 'o', 'u', 'l', 'd', '\0' };

It is totally different. It is a modifiable buffer in your local program stack. You can write pattern[2] = '9'; in this case. It is the same as to define char buffer with size 5 and copy text in it. Please read from somewhere else about string initializations and string literals.

it should be: {thecode} that's wrong. It wouldn't fit the context, since the one you've provided is not equivalent to the other.
The string literal is not const h, however it causes undefined behaviour to modify it.

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