fopen不兼容的类型char（*）[9]而不是C中的char（*）

Question

My program in C reads data in a file in order to initialise variables but it won't open the file. It fails when it reaches fopen. Xcode outputs the following warning for both fopen and printf. I understand the error but I don't know how to correct it, I tried many tricks but it won't do, can anyone help me out ? I just want my program to open Try1.txt .

Incompatible pointer types passing 'char (*)[9]' to parameter of type const char*'

So this is the code inside my main function :

FILE *infile = NULL;
const char infilename [] = "Try1.txt";

infile = fopen(&infilename, "r");
if (infile == NULL)
{
    printf("Failed to open file: %s\n", &infilename);
    return 1;
}

Note that the program stops before reaching the if loop because it never prints. I tried to initialise the size and to add '\\0' at the end of my string too.

Answer 1

It's because of &infilename , which gives you a pointer to the array of characters. Drop the address-of operator and it will work.

Answer 2

Problem

&infilename is a pointer to the array, which means it is a pointer to an object of type char (*)[9] .

But, infilename will be a pointer of type char (*) , which points to the first element of the array.

Fix

infile = fopen(infilename, "r");