assigning a string to another string

Question

Why this code is not running? Why str1 is not assigned to str2 ?? I know i have an option of using strcpy but i wish to know the reason why this is not working??

#include<stdio.h>
int main()
{
char str1[]="hello";
char str2[10];
str2=str1;
printf("%s",str2);
return 0;
}

Whereas if I use pointers than it works like here..

#include<stdio.h>
int main()
(
char *s="good morning";
char *q;
q=s;
while(*q!='\0')
{
printf("%c",*q);
q++;
}
return 0;
}

This works. Now the string has been copied via pointers so why such difference??

possible duplicate of What is the difference between char s[] and char *s in C? — Bo Persson
– Bo Persson, Commented Jul 8, 2012 at 8:48

chrisaycock · Accepted Answer · 2012-07-08 14:43:07Z

1

str2 is a static array. You can't just reassign it like a pointer.

As for your pointer example, q=s reassigns q to point to the same space that s is pointing to. However, the pointer reassignment does not copy the value.

For static arrays, use strcpy() to copy the value. For pointers, use strdup().

edited Jul 8, 2012 at 14:43

answered Jul 8, 2012 at 7:43

chrisaycock

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Comments

unkulunkulu · Accepted Answer · 2012-07-08 09:09:57Z

0

You need to do

strcpy (str2, str1)

you need to copy each element of the str1 to str2 character by character.

What you are doing is to attempt assign the address of the string in str1 assign to the array var str2 which is not allowed. Static arrays couldn't be used to be assigned values.

Even if you had char *str2; and then str2 = str1 although your code would work, but in such a case the string is not copied, instead the address of the string in str1 is copied in str2 so now dereferencing str2 will point to the string

Also note that when you copy string into a char *str2 always allocate enough memory before copy. one possibility is:

str2 = malloc (sizeof (char) * (strlen (str1) + 1));
strcpy (str2, str1);

edited Jul 8, 2012 at 9:09

unkulunkulu

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answered Jul 8, 2012 at 7:41

phoxis

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1 Comment

unkulunkulu Over a year ago

strlen of memory is one byte too small, oops

akash · Accepted Answer · 2012-07-08 07:44:50Z

0

str2 and str1 refers to the address of their first elments so assignment of arrays is not done in this way use either str[] to assign char by char or use inbuilt strcpy

answered Jul 8, 2012 at 7:44

akash

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Comments

user313885 · Accepted Answer · 2012-07-08 07:45:59Z

0

You may also use strdup() which allocates exact amount of space needed to store a copy.

answered Jul 8, 2012 at 7:45

user313885

1 Comment

Ghasan غسان Over a year ago

Is this function found in Windows? I think it is only for POSIX-compliant operating systems. Am I wrong?

Clifford · Accepted Answer · 2012-07-08 09:02:31Z

In C you can only assign objects that have type; and in C a string is not a data type. Instead it is a convention - the convention being that strings are represented by null-terminated character arrays; and because arrays are not considered as types, neither are strings.

In your second code fragment, you have simply made q point to s; this is not a copy of the string at s, but merely a copy of the pointer 's'. They both refer to the same data in memory. So if for example you changed the string at 's', and then printed the string at 'q', you will see that the data at 'q' has also changed. More directly if you did printf( "s:%p, q:%p" (void*)s, (void*)q ) ; you will see that they both hold the same pointer value.

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