CODEWITHSUNDEEP
carrayspointers
TimmayB
TimmayB

Reputation: 55

pointer to value in array?

So I need to have a pointer to a value in a const char array. But I can't quite get it to work without errors. Here's the code.

int main (void)
{
    const char *alp = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    char *ptr = &alp[3];

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

    return 0;
}

Edit- Sorry for not mentioning the errors. The thing is I get tons of different errors depending on where I put different asterisks and ampersands. There is no one particular error. One of the more frequent ones I get says "incompatible integer to pointer conversion assigning to 'char *' from 'const char';"

In the end I just want "ptr" to be equal to a pointer pointing to "D" in the array "alp".

Upvotes: 1

Views: 754

Answers (3)

Gavin H
Gavin H

Reputation: 10502

alp is a pointer to constant char.

ptr is a pointer to a non-const char.

If you want that to work out you would need to change ptr to be defined as:

char const * ptr = &alp[3];

Upvotes: 2

asc99c
asc99c

Reputation: 3905

The const is saying that you're not going to change it, but then you're trying to copy the reference to a variable without that limitation. Instead, try:

const char *ptr = &alp[3];

But note that as other posters have already suggested, this code will give you a pointer to a string beginning with the D (e.g. DEFGHI...), not just the character D.

Upvotes: 0

axblount
axblount

Reputation: 2662

If you only want one character to print, change the %s to %c and dereference the pointer

printf("%c\n", *ptr);

It's true that you had a character pointer but %s tells printf to print from that pointer until it reads a null character. So we switch to %c which will print one character but it expects a value rather than a pointer to a value.

Upvotes: 4

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