Struct with pointer and char in c

Question

My code is working with integer's value, but when I tried to add the character value ,then I have error.

Here's my code:

  1 #include <stdio.h>
  2 
  3 struct _pointer{
  4 
  5     int x;
  6     int y;
  7     char Q[200];
  8 
  9 }address,*pointer;
 10 
 11 
 12 main()
 13 {
 14     pointer = &address;  // here we give the pointer the address.
 15     pointer->x = 10;    // here we give the pointer the value to   variable x.
 16     pointer->y = 30;   // here we give the pointer the value to variable y.
 17     (*pointer).Q = "BANGO!";

 18     printf("The x variable is %d\nThe y variable is %d\nTheText\n",pointer->    x,pointer->y,pointer->Q);
 19 
 20 }
 21

So where is my mistake ?

Thanks

What're u trying to do?

ForceBru
– ForceBru

2015-02-10 16:20:51 +00:00
Commented Feb 10, 2015 at 16:20 — ForceBru
– ForceBru, Commented Feb 10, 2015 at 16:20

Raghu Srikanth Reddy · Accepted Answer · 2015-02-10 16:24:48Z

2

copying a string is done by strcpy(char *dst, const char *src)

Copy the string like this

strcpy(pointer->Q,"BANGO!");

edited Feb 10, 2015 at 16:24

answered Feb 10, 2015 at 16:22

Raghu Srikanth Reddy

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1 Comment

VolAnd Over a year ago

in C language strings (char * type) do not support copying with operator =.

justanothercoder · Accepted Answer · 2015-02-10 17:58:52Z

2

You're passing pointer->Q to printf but there is no %s in the format string.

And also you should copy string with strcpy(pointer->Q, "mystring");

edited Feb 10, 2015 at 17:58

answered Feb 10, 2015 at 16:23

justanothercoder

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Comments

Iharob Al Asimi · Accepted Answer · 2015-02-10 16:25:57Z

1

I can see a few errors, the most important one is that you can't assign to arrays in c, to set the contents of the array to those of the string you need to copy the contents, you can use strcpy() for that, in your case you need

 strcpy((*pointer).Q, "BANGO!");

also, the rest of your code doesn't seem to be a good Idea, I recommend this instead

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

struct MyStruct
{
    int x;
    int y;
    char Q[200]; 
};

int
main()
{
    struct MyStruct  instance;
    struct MyStruct *pointer;

    pointer = &instance;

    pointer->x = 10;    // here we give the pointer the value to   variable x.
    pointer->y = 30;   // here we give the pointer the value to variable y.

    /* copy the contents of "BANGO!" into the array Q */
    strcpy(pointer->Q, "BANGO!");

    printf("x = %d\ny = %d\nQ = %s\n", pointer->x, pointer->y, pointer->Q);
    /*                           ^ you need this for ---------------^ this */

    /* or even */
    printf("x = %d\ny = %d\nQ = %s\n", instance.x, instance.y, instance.Q);
    /* which will print the same, since you modified it through the pointer */
    return 0;
}

you should also note, that main() returns and int.

There is no good reason to use global variables in a general case, there are situations where they are needed or useful, but in general you should avoid them.

answered Feb 10, 2015 at 16:25

Iharob Al Asimi

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1 Comment

Iharob Al Asimi Over a year ago

@pythonlover Just to illustrate how pointers work. Did you know

ForceBru · Accepted Answer · 2015-02-10 16:32:41Z

1

To my mind, you didn't even try to compile your code.

First of all, you cannot just assign a string to char[] you need to use strcpy(char *to, char *from);.

Then, you have three arguments to printf but only two formatting %'s.

Correct way:

 printf("The x variable is %d\nThe y variable is %d\nTheText variable is %s\n",pointer->x,pointer->y,pointer->Q);

and

strcpy(pointer->Q,"Your text");

edited Feb 10, 2015 at 16:32

answered Feb 10, 2015 at 16:26

ForceBru

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1 Comment

nullptr Over a year ago

yea! , I think the error is in the (*pointer).Q = "BANGO!"; .

Gopi · Accepted Answer · 2015-02-10 16:28:19Z

0

I would suggest to use

strlcpy(pointer->Q,"BANGO!",sizeof(pointer->Q));

which is less error prone and guarantees a null termination of your string.

answered Feb 10, 2015 at 16:28

Gopi

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3 Comments

Iharob Al Asimi Over a year ago

strlcpy is not standard.

Gopi Over a year ago

@iharob Yes you are right but no harm to using it right? Since it is safer

Iharob Al Asimi Over a year ago

It's safer if it's available, which is not guaranteed since it's not standard.

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