working with char pointer and integer pointer

Question

My question is about dereferencing a char pointer

Here is my code -

#define MAX 10

char s[80]="Hello";

int main(){

    char *stackValue;

    stackValue=&s;//here I assined the address of s to stackValue

    if(!stackValue){

        printf("No place for Value");

        exit(1);

    }

    else{

        printf("\n%s",*stackValue);//This doesn't work with * before it 

        printf("\n%s",stackValue);//This works properly
    }
    return 0;
}

In the above code I have assigned the address of S[] to stackValue and when I am printing *stackValue it doesn't work ,

But If I print only 'stackValue' That works.

When I do same thing with Integer

int main(){
    int i=10, *a; 
    a=&i;
    printf("%d",*a);//this gives the value 
    printf("%d",a)//this gives the address
    return 0;
}

Is printing char pointer and integer pointer is different. When I use * in int value it gives the value but gives an error when I use it as a char pointer.

Help me out?

artm · Accepted Answer · 2016-11-29T07:17:20.080

1

With the first code snippet:

stackValue=&s; is incorrect given s is already an array to char. If you write like that then stackValue becomes pointer to pointer to char (not pointer to char).

Fix that by changing to stackValue=s;

Also, again %s expect a pointer to char (NOT pointer to pointer to char) - that explains why this doesn't work printf("\n%s",*stackValue); // this doesn't work

You need printf("\n%s",stackValue); instead.

With the second code snippet.

a=&i; is ok because i is a single int, NOT an array.

edited Nov 29 '16 at 07:17

answered Nov 29 '16 at 06:59

artm

17,291
6
38
54

yeah it is , but my question still remains, when I use *stackValue it doesnt print the value of s[] but print an int value, and when I print 'stackValue ' only it prints the value – tiwarinitin94 Nov 29 '16 at 07:04
it doesn't matter what it prints - it's undefined behaviour (that means anything can happen). Your code is wrong from the start so there is not much point to elaborate in the wrong printf. – artm Nov 29 '16 at 07:07
I have amend my code again still it not giving the right output – tiwarinitin94 Nov 29 '16 at 07:12
@NitinTiwari accepting an answer is ok, but you should vote first before accept – artm Nov 29 '16 at 07:41
Sorry now I did – tiwarinitin94 Dec 10 '16 at 05:02

Mohan · Answer 2 · 2016-11-29T07:23:52.137

0

The "%s" format specifier for printf always expects a char* argument. so this is working and correct statement

printf("\n%s",stackValue);

and in first statement you are passing value so it will give you undefined behaviour.

edited Nov 29 '16 at 07:23

answered Nov 29 '16 at 07:16

Mohan

1,871
21
34

score 0 · Answer 3 · answered Nov 29 '16 at 07:21

0

What you are trying to do is this:

int main(void)
{
    char a_data = "Hello, this is example";
    char *pa_stack[] = {a_data};

    printf("We have: %s\n", *pa_stack);
}

answered Nov 29 '16 at 07:21

aeb-dev

313
5
16

working with char pointer and integer pointer

3 Answers3