c - printf("%d") doesn't display what I input -

my code:

printf("enter number : ");  scanf("%d", &number);  printf("%d entered\n", &number); 

i input 2,

expected output : 2 entered

actual output : 2293324 entered

what's problem here?

printf("%d entered\n", &number); 

is wrong because %d(in printf) expects argument of type int, not int*. invokes undefined behavior seen in draft (n1570) of c11 standard (emphasis mine):

7.21.6.1 fprintf function

[...]

9. if conversion specification invalid, behavior undefined. ²⁸²⁾ if argument not correct type corresponding conversion specification, behavior undefined.

fix using

printf("%d entered\n", number); 

---

then, why scanf require & before variable name?

keep in mind when use number, pass value of variable number , when use &number, pass address of number(& address-of operator).

so, scanf not need know value of number. needs address of (an int* in case) in order write input it.

printf, on other hand, not require address. needs know value (int, in case) printed. why don't use & before variable name in printf.