Why doesn't C need '&' in scanf() while dealing with strings/char

Question

My code is very simple just basic IO. Now when I use this code it works perfectly.

#include <stdio.h>

int main()
{
    int age = 0;
    char name[100]; //Unlike c++ We need to specify char limit instead of "Name", so name can't go above 99 char + 1 terinator
    printf("Enter your age: ");
    scanf("%d", &age);
    printf("Enter your name: ");
    scanf("%s", name);// dont need & for char
    printf("Your name is %s and your age is %d\n", name, age);
    return 0;
}

Now

#include <stdio.h>

int main()
{
    int age = 0;
    char name[100]; //Unlike c++ We need to specify char limit instead of "Name", so name can't go above 99 char + 1 terinator
    printf("Enter your age: ");
    scanf("%d", &age);
    printf("Enter your name: ");
    scanf("%s", &name);// dont need & for char
    printf("Your name is %s and your age is %d\n", name, age);
    return 0;
}

when I make changes in line 10. and add &name. compiler throws this error. Why is that?

p2.c:10:17: error: format specifies type 'char *' but the argument has type 'char (*)[100]' [-Werror,-Wformat]
    scanf("%s", &name);// dont need & for char
           ~~   ^~~~~

I don't know much about C.

Because in second code snippet ```%s``` need starting address of char array. So, ```name``` stores starting address of array. Where as in char ```%c``` needs address as well. In ```char ch;``` ```ch``` won't return address you need to use ```&ch```. — Sathvik, Sep 12 '20 at 18:05
@Sathvik: `name` does not store an address - name is *converted* to an address as necessary. — John Bode, Sep 12 '20 at 23:52

score 3 · Accepted Answer · answered Sep 12 '20 at 18:02

Strings in C are simply sequences of characters. The %s format specifier thus expects a pointer to char such that it can write whatever scanf reads into the character sequence at the memory location pointed to by that pointer. In your case name is a character array and not a pointer, but in C you can often use arrays in contexts where pointers are expected, we say that the array decays to a pointer to its first member.

Deepak Patankar · Answer 2 · 2020-09-13T05:59:18.910

1

The scanf function takes a pointer of the variable whose value is to be set.

For other types like int, we use & operator which specifies the address of the variable whereas for char[] the variable name is converted to the pointer to the first element of the array so we don't need a &.

edited Sep 13 '20 at 05:59

answered Sep 12 '20 at 18:02

Deepak Patankar

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`char name[100];` is an array rather than a pointer. It's automatically converted to the pointer to its first element in this and many other cases. – HolyBlackCat Sep 12 '20 at 18:03
Thanks a lot, @HolyBlackCat!!. I understood the mistake and had fixed it. Can you have a look ? – Deepak Patankar Sep 12 '20 at 18:08
2

the name designates the array and is not an address. It is converted to a pointer to the first element. – Antti Haapala -- Слава Україні Sep 12 '20 at 19:20

score 0 · Answer 3 · answered Sep 12 '20 at 18:19

0

Strings are arrays of chars. Arrays are passed by reference in C.

So when you pass array to the function you actually pass the pointer.

answered Sep 12 '20 at 18:19

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Why doesn't C need '&' in scanf() while dealing with strings/char

3 Answers3