Difference between char* and char[] in C [duplicate]

Question

Possible Duplicate:
C - Difference between “char var[]” and “char *var”?

I have written the following C code

#include<stdio.h>
int main()
{
    char name[31];
    char *temp;
    int i ;
    scanf("%s",name);
    temp  = name;
    name = temp;

}

I got the following error when compiling

incompatible types when assigning to type 'char[31]' from type 'char *'

Array name is a pointer to first element(here char pointer ..right?). right?The above code means that character array and char* are different types ..Is it true? Why type of name != char * why i cannot assign another char pointer to a char pointer(array name)

Answer 1

Array name is a pointer to first element(here char pointer ..right?). right?

Wrong. Arrays decay to a pointer to the first element in most contexts, but they certainly aren't pointers. There's a good explanation in the C FAQ and an invaluable picture (a is an array and p is a pointer):

incompatible types when assigning to type 'char[31]' from type 'char *'

In C arrays are non-modifiable lvalues, you can't change what they point to since they don't point anywhere in the first place.

Answer 2

"Array name is a pointer to first element(here char pointer ..right?). right?"

char name[31];
char *temp;
/* ... */
name = temp;

In the name = temp assignment, the value of name is converted to a pointer to char. The value is converted, not the object. The object is still an array and arrays are not modifiable lvalues. As the constraints of the assignment operand require the left operand of the assignment operator to be a modifiable lvalue, you got an error.