CODEWITHSUNDEEP
carrayspointers
sleighty
sleighty

Reputation: 1075

Passing char arrays to functions that take char pointers

In C, why is it that I'm able to pass character arrays to functions that take a char * as an argument, but I cannot pass the address of an array to functions that take a char **?

UPDATE: Interestingly, changing the argument type to char* qux[12] doesn't change the compiler warning at all

For example:

#include <stdio.h>

void foo(char* qux) { puts(qux); }
void bar(char** qux) { puts(*qux); }
void baz(char* qux[12]) { puts(*qux); }

int main() {
    char str[12] = "Hello there";

    foo(str);
    bar(&str); // Compiler warning
    baz(&str); // Same compiler warning

    return 0;
}

In the second case, I get a compiler warning:

warning: incompatible pointer types passing 'char (*)[12]' to
         parameter of type 'char **' [-Wincompatible-pointer-types]

What's going on here?

Upvotes: 4

Views: 3051

Answers (2)

ScottK
ScottK

Reputation: 1556

In C, char * represents a pointer to a contiguous sequence of characters. A contiguous sequence of characters with a null termination is what we call a string in C.

char ** is a pointer to a contiguous sequence of strings, and since each string is a contiguous sequence of characters terminated by a null ('\0') character, char ** represents a contiguous sequence to a contiguous sequence of null terminated characters.

Your declaration:

 char str[12] = "Hello there";

Declares str to be an array of characters of length 12, and it is initialized to the 12 characters {'H','e','l','l','o',' ','t','h','e','r','e','\0'}. This is compatible with the parameter in foo(), but not with bar and baz both of which expect a contiguous sequence of pointers to strings. That is why those two give you a compiler warning because the parameter is incompatible with the arguments passed in.

Upvotes: 2

Some programmer dude
Some programmer dude

Reputation: 409442

Arrays naturally decays to pointers to their first element. So in the call foo(str) it's really the same as foo(&str[0]). This is of type char * so it's all okay.

Now the second call, the one to bar, is a different matter. When you use &str you don't get a pointer to the arrays first element, you get a pointer to the array itself. And as the compiler noted, this is of type char (*)[12], which is very different from (and incompatible with) char **.

Lastly, when you declare baz you say that the argument is of type char *[12], that is you have an array or 12 pointers to char, not that you have a pointer to an array of 12 char. Furthermore, due to the array decay to pointer thing, char *[12] is actually the same as char **.

Upvotes: 5

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