char array not assignable

Question

Okay, so I want to save a word in a char array but it gives me a error

Here's my code

#include <stdio.h>
#include <time.h> 
#include <stdlib.h>
#include <string.h>

int main(void)
{
    char yesno[30] = "n";        //yes/no answer
    char class[30] = "undefined";//choosen class 
    int classchoosen = 0;        

    /* initialize random seed: */
    srand ( time(NULL) );

    printf("Welcome, which class do you wanna play with? \n");
    printf("Type W for Warrior, M for Mage or R for Ranger. Then press Enter\n");
    while(classchoosen == 0)
    {
        scanf("%s", class);
        if(strcmp(class, "W") == 0)
        {
            classchoosen = 1; 
            class = "Warrior";
        }
        if(strcmp(class, "M") == 0)
        {
            classchoosen = 1; 
            class = "Mage";
        }
        if(strcmp(class, "R") == 0)
        {
            classchoosen = 1; 
            class = "Ranger";
        }
        if(classchoosen == 0)
        {
            class = "undefined";
        }
        printf("So you wanna play as a %s? Enter y/n", class);
        classchoosen = 0; //For testing, remove later 

    }
    while(1)
    {
        /* Irrelevant stuff */ 
    }
}

And it gives me following errors:

damagecalc.c:44:13: error: expected identifier
                        class -> "warrior";
                                 ^
damagecalc.c:49:10: error: array type 'char [30]' is not assignable
                        class = "mage";
                        ~~~~~ ^
damagecalc.c:54:10: error: array type 'char [30]' is not assignable
                        class = "ranger";
                        ~~~~~ ^
damagecalc.c:58:10: error: array type 'char [30]' is not assignable
                        class = "warlock";
                        ~~~~~ ^
4 errors generated.

I know I could just print out the class name just after the string comparison, but this thing is really bugging me and I want to know why it doesn't work.

PS: Please forgive me for any obvious mistakes I may do, I just got into PC programming recently after having worked on uC's for a couple of years.

why not char yesno='n'; instead of char yesno[30] = "n"; ? — Meninx - メネンックス
– Meninx - メネンックス, Commented Jan 9, 2015 at 19:11

Iharob Al Asimi · Accepted Answer · 2017-01-11 12:09:10Z

16

Yes char arrays are not assignable just as all arrays aren't. You should use strcpy for example

strcpy(class, "Warrior");

and so on.

Also, you don't need to declare char class[30]; the longest string I see in your code is "undefined" so

char class[10];

should be ok, 9 characters of "undefined" + 1 null terminating byte '\0'.

And you should prevent buffer overflow with scanf this way

scanf("%1s", class);

since you only want to read 1 character, also the comparison should be simpler just

if (class[0] == 'M')
    strcpy(class, "Mage");

or even thsi would be more readable

classchosen = 1;
switch (class[0])
{
case 'M':
    strcpy(class, "Mage");
    break;
case 'R':
    strcpy(class, "Ranger");
    break;
case 'W':
    strcpy(class, "Warrior");
    break;
default:
    classchosen = 0;
}

and finally check that scanf actually succeeded, it returns the number of arguments matched so in your case a check would be like

if (scanf("%1s", class) == 1) ...

edited Jan 11, 2017 at 12:09

answered Jan 9, 2015 at 19:05

Iharob Al Asimi

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3 Comments

Iharob Al Asimi Over a year ago

@Luxuspunch see some suggestions to improve your code.

Iharob Al Asimi Over a year ago

@Luxuspunch if you think any of the answers helped you, you can accept the answer by clicking the check mark see this. May be you already know this, but in case you don't...

Luxuspunch Over a year ago

Sorry I was eating. Anyways, thanks A LOT to everyone contributing here. Always makes my day when someone can help me or vice versa :) Also I do appreciate your improvements. It was an early version as stated in my post but I didn't notice that I could clean it up so much.

Anurag Bhakuni · Accepted Answer · 2016-05-03 04:42:50Z

6

see this is not only for character array , this implies for all type of arrays but the question aries why?,when we can assign other variable why not array the thing is:- suppose we have:

int a[size];
a = {2,3,4,5,6};

because here the name of an array mean address of the first location of an array

printf("%p",a); // let suppose the output is 0x7fff5fbff7f0

We are saying by that

0x7fff5fbff7f0 = something; which is not correct yes we can do a[0] = something , now it saying assign the value of array at 0th location.

edited May 3, 2016 at 4:42

answered Aug 18, 2015 at 2:11

Anurag Bhakuni

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Comments

Gopi · Accepted Answer · 2015-01-09 19:13:59Z

1

 class = "Ranger";

should be

strcpy(class,"Ranger");

Fix the same in all the places. char arrays are not assignable

edited Jan 9, 2015 at 19:13

answered Jan 9, 2015 at 19:06

Gopi

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4 Comments

Luxuspunch Over a year ago

Yep, thank you too for answering. Strangely I have never found this anywhere for about 1 hour of research. I'm really starting to doubt my googling skills. I will mark this question as solved as soon as I can

Gopi Over a year ago

@Luxuspunch no probs but this is basics of dealing with strings in c.

Luxuspunch Over a year ago

Well now I know. On the uC's I programmed we did never work with strings, that may be the reason I'm not too familiar with this.

Iharob Al Asimi Over a year ago

@Luxuspunch it's not easy to google for this things, but I assure you many similar questions are asked every day her on SO, i've ansewered a lot of questions where the problem was exactly this one.

Meninx - メネンックス · Accepted Answer · 2015-01-09 19:08:32Z

0

switch class = "Warrior"; to strcpy(class, "warrior");

switch class = "Mage"; to strcpy(class, "Mage");

switch class = "Ranger"; to strcpy(class, "Ranger");

switch class = "undefined"; to strcpy(class, "undefined");

and it should work !

answered Jan 9, 2015 at 19:08

Meninx - メネンックス

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