Passing char pointer as argument to function

Question

This porgram prints different permutations of a string.It works correctly if I declare string as char array in main and pass array name in printAnagram function.But if I declare string as char* s = "hello" and pass 's' then it crashes. Why?

#include <stdio.h>
#include <conio.h>

void printAnagram(char *str, int b, int e);

int main()
{
    char *s = "ABC"; // works fine when char s[] = "ABC" is used.
    printAnagram(s, 0, 2);
    return 0;
}

void swap(char *a, char* b)
{
    char temp = *a;
    *a = *b;
    *b = temp;
}
void printAnagram(char *str, int b, int e)
{
    int i = 0;
    if(b==e)
        printf("%s\n", str);
    else
    {
        for(i=b;i<=e;i++)
        {
            swap((str+b),(str+i));
            printAnagram(str, b+1, e);
            swap((str+b), (str+i));
        }
    }
}

because a string literal is a constant. no tutorial/reference/language book told you about that? this has been asked literally hundreds of times. — The Paramagnetic Croissant, Jan 23 '15 at 11:18

score 3 · Accepted Answer · answered Jan 23 '15 at 11:17

3

In char *s = "ABC"; - "ABC" is a string literal modifying it is UB.

Pointers are usually used to point to data that already exists, so you can use it like this

char arr[] = "C__";

char* t = &arr[0];

Also modifiable,

t[1] = 'p';

t[2] = 'p';

Here

char *t= "C__";

points to a string constant.

There is a better way of writing the above:

const char* t= "C__";

answered Jan 23 '15 at 11:17

Sadique

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@Sadique I try this approach and when I send the char* t to the function, the arr[] does not change. Any ideas? – JC203 Apr 06 '18 at 18:39

score 2 · Answer 2 · answered Jan 23 '15 at 11:17

The crash has to do with the kind of memory where the characters of your string are stored, not with whether or not you declare it as a character array or a pointer. In other words, you can keep the pointer, and copy the string into it, like this:

int main()
{
    char *s = strdup("ABC");
    printAnagram(s, 0, 2);
    free(s);
    return 0;
}

C standard considers memory allocated to string literals, such as "ABC", non-writable. Hence, any changes you do to them cause undefined behavior. When you change declaration to an array, C copies the literal into writable memory, so there is no undefined behavior (and no crash).

score 1 · Answer 3 · edited May 23 '17 at 12:20

1

char *s = "ABC"; is a string literal stored in read only memory location. When you try modifying it, yes it will crash. In fact it is of const char *s. Instead you should be using as mentioned already as,

char s[] = "ABC";

Look at this previous SO answer Modifying a string literal.

edited May 23 '17 at 12:20

Community

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answered Jan 23 '15 at 11:17

Sunil Bojanapally

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score 0 · Answer 4 · answered Jan 23 '15 at 11:17

0

You are accessing the read only memory.

char *a="ABC";// It will stored in the read only memory.

If you are trying to access this you cannot do. You can use like this.

char array[]="ABC";
char *a=array;

answered Jan 23 '15 at 11:17

Karthikeyan.R.S

3,991
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1

no he doesn't have to allocate dynamically. Initializing an automatic array is perfectly fine. – The Paramagnetic Croissant Jan 23 '15 at 11:19

capagira87 · Answer 5 · 2015-01-23T11:31:52.730

0

You could try to declare a char array, for example char [6] str="ABC"; and then use a char pointer, to manage that array: char *p=&str; The crash is due to the fact that your pointer to a string that is not present in the stack memory.

edited Jan 23 '15 at 11:31

answered Jan 23 '15 at 11:21

capagira87

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Passing char pointer as argument to function

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