Variable incorrect in my code

Question

It compiles correctly, but when I run it and input a number it outputs some random string of seven numbers. I have no idea what to do with this. For example I put in 5 and it gives me 2751724 then I do it again and get 3537324. I don't know if I have some settings off or what but this seems pretty simple to me.

#include "stdafx.h"
#include <stdio.h>

int _tmain(int argc, _TCHAR* argv[])
{
    int number;

    printf("Input number: ");
    scanf_s("%d", &number);
    printf("Number is: %d \n", &number);

    return 0;
}

just print number not &number, &number is the adresss in memory where number is — chris
– chris, Commented Feb 16, 2015 at 1:15
You can printing the wrong thing in printf("Number is: %d \n", &number);. Hint: what does & do here? — Yu Hao
– Yu Hao, Commented Feb 16, 2015 at 1:15
Oh thanks to the both of you. I got the syntax messed up in my head. — Xandr
– Xandr, Commented Feb 16, 2015 at 1:18
I'm sure this is a duplicate but search doesn't find anything since they have all been downvoted to death :) — M.M
– M.M, Commented Feb 16, 2015 at 1:29
when calling any of the scanf family of functions, always check the returned value to assure the input/conversion operation was successful — user3629249
– user3629249, Commented Feb 16, 2015 at 1:31

chris · Accepted Answer · 2015-02-16 01:28:40Z

4

Using the & operator gets the address of a variable in memory (aka a pointer). By using this functions can alter the value of that variable directly without having to return a value. You should read up about pointers and how to use them.

In short, printf("Number is: %d \n", number).

answered Feb 16, 2015 at 1:28

chris

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Mr.CodeMonkey · Accepted Answer · 2015-02-16 02:22:23Z

1

when you do like this:

printf("Number is: %d \n", &number);

you print the address of number instead of the value of number

answered Feb 16, 2015 at 2:22

Mr.CodeMonkey

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