Pass a char array to a function that takes a char pointer

Bruno Ely

In C, why can I pass a character array to a function that takes a char *as a parameter, but not the address of the array to a function that takes a char **?

UPDATE: Interestingly, changing the parameter type char* qux[12]to does n't completely change the compiler warning

E.g:

#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?

some programmer buddies

Arrays decay naturally to a pointer to the first element. So in a call, foo(str)it's effectively the same as foo(&str[0]). It's type, char *so it's ok.

Now, the second call is a barproblem. When used &str, instead of getting a pointer to the first element of the array, you get a pointer to the array itself . And as the compiler points out, this is the type , which is very different from (and is not compatible with) .char (*)[12]char **

Finally, when declaring, bazyou say the parameter is type char *[12], i.e. you have an array or 12 pointers to char , not that you have an array of 12 pointerschar . Also, since the array decays to the pointer stuff, it 's char *[12]actually the same .char **

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