difference between scanf("%c" , ..) and getchar() - c

Question

I always think scanf("%c" , &addr); is equal to getchar() before I test this:
#include<stdio.h>
int main()
{
int i;
scanf("%c",&i);
printf("%d\n", i);
if(i == EOF)
printf("EOF int type and char input\n");
i =getchar();
printf("%d\n", i);
if(i == EOF)
printf("EOF int type and char input\n");
}
I got output when I use "Ctrl+D" twice:
-1217114112
-1
EOF int type and char input
Since EOF is -1 in int type ,I also try use scanf("%d",&i); replace scanf("%c",&i) , just get the same output.
I got confused. Can anybody explain this for me?
----------------------------------EDIT-----------------------------------------------
I want to know the behavior of scanf("%c",i) of Ctrl+D , I do test:
#include<stdio.h>
int main()
{
int i;
int j;
j = scanf("%c",&i);
printf("%c\n", i);
printf("%d\n", j);
if(i == EOF)
printf("EOF int type and char input");
i =getchar();
printf("%d\n", i);
if(i == EOF)
printf("EOF int type and char input");
}
OutPut:
k // If the scanf set 1 byte in i , why here print 'k' ?
-1
-1
EOF int type and char input

Answer 1

Your comparison does not fully set i as it involves Undefined Behavior (UB).
int i; // the value of i could be anything
scanf("%c",&i); // At most, only 1 byte of i is set, the remaining bytes are still unknown.
printf("%d\n", i);// Your are printing 'i' whose value is not fully determined.
Had you tried
char ch;
int y = scanf("%c",&ch);
printf("%d\n", ch);
if(ch == EOF)
You would potentially make a match even though the input was not EOF. Had you scanned in a char with the value of 255, the char would take on the 2s compliment 8-bit value of -1. The comparison would sign extend the 8-bit -1 to match the int size and you would match -1.
(Assumptions: 2s compliment integers, 8-bit byte, EOF == -1, char is signed).
The correct EOF test is
int y = scanf("%c",&ch);
if (y == EOF)
Note: getchar() & scanf() return EOF implies End-of-file or I/O error. A subsequent check of ferror(stdin) distinguishes this.

Answer 2

The first value is probably undefined behavior. You can't rely on i having a value unless scanf() returns 1.
With scanf() in particular, you seem to be confusing the scanned value (the conversion of characters according to a format specifier in the first argument) with the return value of the function call.
With getchar(), of course, this distinction doesn't exist since it only has a return value.