Need help with malloc in C programming. It is allocating more space than expected

Question

Let me preface this by saying that i am a newbie, and im in a entry level C class at school.

Im writing a program that required me to use malloc and malloc is allocating 8x the space i expect it to in all cases. Even when just to malloc(1), it is allocation 8 bytes instead of 1, and i am confused as to why.

Here is my code I tested with. This should only allow one character to be entered plus the escape character. Instead I can enter 8, so it is allocating 8 bytes instead of 1, this is the case even if I just use a integer in malloc(). Please ignore the x variable, it is used in the actual program, but not in this test. :

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


int main (int argc ,char* argv[]){

    int x = 0;
    char *A = NULL;
    A=(char*)malloc(sizeof(char)+1);
    scanf("%s",A);
    printf("%s", A);
    free(A);
    return 0;
}

Answer 1

 A=(char*)malloc(sizeof(char)+1);

is going to allocate at least 2 bytes (sizeof(char) is always 1). I don't understand how you are determining that it is allocating 8 bytes, however malloc is allowed to allocate more memory than you ask for, just never less.

The fact that you can use scanf to write a longer string to the memory pointed to by A does not mean that you have that memory allocated. It will overwrite whatever is there, which may result in your program crashing or producing unexpected results.