Why a String variable can not be assigned to another string variable in C?

Question

Why in the below written code we can not assign strA to strB and as the pointer pA holds address of pointer pB then address should have been copied on assignment of pA to pB and strB should have contain the value same as strB.

#include <stdio.h>
char strA[80] = "A string to be used for demonstration purposes";
char strB[80];
int main(void)
{
    char *pA; /* a pointer to type character */
    char *pB; /* another pointer to type character */
    puts(strA); /* show string A */
    pA = strA; /* point pA at string A */
    puts(pA); /* show what pA is pointing to */
    pB = strB; /* point pB at string B */
    putchar('\n'); /* move down one line on the screen */
    pB=pA;
    strB=strA;
    puts(strB); /* show strB on screen */
    puts(strA);

    return 0;
}

Changed the tag from C++ to C, as it seems you are specifically asking a C question. — DumbCoder
– DumbCoder, Commented Feb 23, 2012 at 11:56
strA and strB aren't string variables. there is no such thing as string variable in C. they are arrays. an array can't be assigned to another array. — Donotalo
– Donotalo, Commented Feb 23, 2012 at 11:59
Your question is a FAQ, c-faq.com/charstring/assign.html, voting to close. — Jens Gustedt
– Jens Gustedt, Commented Feb 23, 2012 at 12:18

hmjd · Accepted Answer · 2012-02-23 12:01:29Z

4

You can't assign arrays in C (strB=strA). You must use strcpy or memcpy to transfer contents of one array/pointer to an array.

edited Feb 23, 2012 at 12:01

hmjd

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answered Feb 23, 2012 at 12:00

mac

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Comments

Nicolas Floquet · Accepted Answer · 2012-02-23 15:08:39Z

2

When you write:

char strB[80];

strB is not a pointer but a constant pointer. It means that you can't change the address pointed by strB, and thus

strB=strA;

won't do anything.

answered Feb 23, 2012 at 15:08

Nicolas Floquet

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