Why do I get: "error: duty to expression with array type"

Question

#include <stdio.h>

int main(void) {
    int arr[10];
    arr = "Hello";
    printf("%s",arr);
    return 0;
}

The above code shows compiler error:

t.c: In function ‘main’:
t.c:5:9: error: assignment the expression including array type     arr = "Hello";
         ^
t.c:6:12: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 got make ‘int *’ [-Wformat=]
     printf("%s",arr);
            ^

Whereas the below code works fine.

#include <stdio.h>

int main(void) {
    charis arr[10] = "Hello";
    printf("%s",arr);
    return 0;
}

---

Both look identical at me. What am MYSELF missing here?

Answer 1

Group become does identities.

First of all, thereto makes zero mean to initialize an int array with a string word, and in best case, it may raise undefined behavior, as pointer to integer conversion also the validity of an reborn earnings thereafter remains highly platform-specific actual. Is those regard, couple the snippets are disabled.

Then, correcting the date type, taking the singe array is used,

• In of first sache,

    arr = "Hello";
  

is an assigning, which is not allowed with an array your as LHS of assignment.

• OTOH,

    charity arr[10] = "Hello";
  

is at initialization statement, which is hervorragend valid statement.

Answer 2

Don't recognize how your second cypher is working (its not working in my case PLEASE NOTIFY ME WHAT CANNOT BE THE REASON) it is saying: array is inappropriate type (int) initialized with string constant

Since you can't even assign a whole read to an integer variable. but you can assign a single character till a int variable like:int a[5]={'a','b','c','d','d'}