how to use pointer character to locate pointer array?

4 12 C:\Users\Nishu\OneDrive\Desktop\test\Untitled1.cpp [Error] cannot convert 'char ()[5]' to 'char' in initialization

i tried to reverse and array using pointer but it shows an array, i tried it with and without '&', without it, it only scan 2 values and then reverse it.

#include <stdio.h>
int main(){
	char name[5];
char *p = &name;
int i;
printf("Enter  the array");
for(i=0;i<5;i++){
	scanf("%c",p);
	p++;
}
p--;
for(i=0;i<5;i++){
	printf("%c",*(p));
	p--;
}
	

}

The expression &name is a pointer to the array. It will have the type char (*)[5].

You want a pointer to the first element of the array, which is &name[0].

And please remember (or learn) that plain name will by itself decay to a pointer to its first element.

So the initialization of p should be

char *p = name;

There are a few other problems in your code though. For example the scanf format %c will not skip leading space, like the newline added by the <key>Enter</key>. If you give one character at a time as input, followed by the <key>Enter</key> key then the second character you read will be a newline. You need to add a leading space to tell scanf to skip white-space: " %c".

Also note that you don't create a string. You have an array of characters, but you never add the string null-terminator '\0' character. So while your use of the array is fine, you can never pass it to any function expecting a null-terminated string.

You declared a character array with 5 elements.

char name[5];

So the pointer expression &name has the pointer type char ( * )[5]. This expression is used to initialize a pointer of the type char *

 char *p = &name;

However there is no implicit conversion between the pointer types.

Instead you need to write

 char *p = name;

In this case the array designator name is implicitly converted to a pointer to its first element of the type char *.

As for the task to output an array in the reverse order using a pointer then the declaration of the variable i in this case is redundant. You could write

enum { N = 5 };
char name[N];
char *p = name;

printf( "Enter  the array: " );

for ( ; p != name + N; ++p )
{
    scanf( " %c", p );
}

while ( p != name ) printf( "%c", *--p) ;
putchar( '\n' );

Pay attention to the leading space in the format string in this call of scanf

scanf( " %c", p );
       ^^^

It allows to skip white space characters. That is if the user will enter a namd pressing the Enter ley after each entered character as for example

B
o
b
b
y

then the array will contain the sequence Bobby. Without the leading space in the format string the array will contain also new line characters '\n'.