Why does scanf() need & operator (address-of) in some cases, and not others?

pranay picture pranay · Aug 9, 2010 · Viewed 13k times · Source

Why do we need to put a & operator in scanf() for storing values in an integer array but not while storing a string in a char array?

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 whats the case with strings?

Answer

Scott M. picture Scott M. · Aug 9, 2010

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.