why do we need to put a & operator in scanf() for storing values in an integer array and not while storing a string in a char array?
i.e
int a[5];
for(i=0;i<5;i++)
scanf("%d",&a[i]);
but char s[5]; scanf("%s",s); ?
we need to pass in the address of the place we store the value,since array is a pointer to first element so in the case with int/float arrays it basically means (a+i) but what's in case of strings in C?
scanf("%d", a + i ) works too.
%d and %s just tell scanf what to expect but in both cases it expects an address
in C, arrays and pointers are related.
%s just says to scanf to expect a string which is \0 terminated, whether it will fit into the character array or not scanf doesn't care.
Because characters arrays are already pointers.
You can think of C arrays as pointers to a stack-allocated amount of RAM. You can even use pointer operations on them instead of array indexing. *a and a[0] both produce the same result (returning the first character in the array).
scanf accepts a pointer to whatever you are putting the value in. In the first instance, you are passing a reference to the specific int at position i in your integer array. In the second instance you are passing the entire array in to scanf. In C, arrays an pointers are synonymous and can be used interchangeably (sort of). The variable s is actually a pointer to memory that has contiguous space for 5 characters.
Please see my in-depth answer that will explain the purpose of 'C Strings Confusion'
When you use the name of an array in an expression (except as the operand of sizeof or the address-of operator &), it will evaluate to the address of the first item in that array -- i.e., a pointer value. That means no & is needed to get the address.
When you use an int (or short, long, char, float, double, etc.) in an expression (again, except as the operand of sizeof or &) it evaluates to the value of that object. To get the address (i.e., a pointer value) you need to use the & to take the address.