Confused between passing strings to a function (C)

user_19005139 0
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Why this works:

#include <stdio.h>
void slice(char *st, int m, int n)
{
    int i = 0;
    while ((i + m) < n)
    {
        st[i] = st[i + m];
        i++;
    }
    st[i-1] = '\0';
}

int main()
{
    char st[] = "Hello";
    slice(st, 1, 6);
    printf("The value of string is %s\n", st);
    return 0;
}

And this doesn't:

#include <stdio.h>
void slice(char *st, int m, int n)
{
    int i = 0;
    while ((i + m) < n)
    {
        st[i] = st[i + m];
        i++;
    }
    st[i-1] = '\0';
}

int main()
{
    char*st = "Hello";
    slice(st, 1, 6);
    printf("The value of string is %s\n", st);
    return 0;
}

In first I initialized my string using:

  • char st[]="Hello"; (using array)

And in latter I used:

  • char*st="Hello"; (using pointer)

I'm kind of getting confused between these 2 initialization types, what's the key difference between declaring a string by using char st[]="Hello"; and by using char*st = "Hello";.

pointers function c arrays string
3 Answers

With char st[] = "Hello";, st[] is a modifiable array of characters. The call slice(st, 1, 6); takes the array st and converts to a pointer to the first element of the array. slice() then receives that pointer, a pointer to modifiable characters.

With char *st = "Hello";, st is a pointer that points to a string literal "Hello". With the call slice(st, 1, 6);, the function receives a copy of the pointer - a pointer to the string literal. Inside slice(), code st[i] = ... is attempting to modify a string literal, that is undefined behavior (UB). It might work, it might fail, it might work today and fail tomorrow - it is not defined.

Do not attempt to modify a string literal.

... passing strings to a function ...

In both cases, code does not pass a string to slice(), but a pointer to a string. Knowing that subtle distinction helps in understanding what is truly happening.

user_24103590

This is an artifact of old syntax in C:

char * s = "Hello world!";

is a non-const character pointer to const memory. It is still permitted by syntax, but the string is still not a mutable object. To be pedantic it should really be written as:

const char * s = "Hello world!";

In contrast:

char s[] = "Hello world!";

allocates a local (on the stack), mutable array and copies the string data to it (from wherever the non-mutable copy is stored in memory). Your function can then do as it likes to your local copy of the string.

user_27067070

The type char [] is different from the type char* (char* is a variable - int. but char[] is an array which is not a variable). However, an array name can be used as a pointer to the array. So we can say that st[] is technically similar to *str .

the problem in the 2nd version of your code

If you have read-only strings then you can use const char* st = "hello"; or simply char* st = "hello"; . So the string is most probably be stored in a read-only memory location and you'll not be able to modify it. However, if you want to be able to modify it, use the malloc function:

char *st= (char*) malloc(n*sizeof(char)); /* n-The initial size that you need */
// ...
free(st);

**So to allocate memory for st, count the characters ("hello"-strlen(st)=5) and add 1 for this terminating null character , and functions like scanf and strcpy will add the null character;

so the code becomes :

#include <stdio.h>
void slice(char *st, int m, int n)
{
    int i = 0;
    while ((i + m) < n)
    {
        st[i] = st[i + m];
        i++;
    }
    st[i-1] = '\0';
}

int main()
{
    char *st =malloc(6*sizeof(char))  ;
    const char *cpy="hello";
    strcpy(st, cpy); /* copies the string pointed by cpy (including the null character) to the st. */
    slice(st, 1, 6);
    printf("The value of string is %s\n", st);
    return 0;
}

  1. you can fill your string also by a for loop or by scanf() .

  2. in the case of a large allocation you must end your code with free(st);

user_187761580
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