c 中 while 循环的基本问题，带有 or 运算符，用于从用户那里获取选择

Question

So the problem is very simple but I mess it up somehow. I create a menu with this code:

#include<stdio.h>
int main(){
    int n;
    int choice;
        printf("\nMENU:\n");
        printf("\n1- Add new student");
        printf("\n2- Print student List");
        printf("\n3- Find max average grade");
        printf("\n4- Find a student by name");
        printf("\n5- Delete student by ID");
        printf("\n6- Export data file");
        printf("\n0- Quit\n");
        printf("\nPlease enter your choice: ");
        scanf("%d", &choice);
        fflush(stdin);
        while(choice!=1 || choice!=2 || choice!=3 || choice!=4 || choice!=5 || choice!=6 || choice!=0){
            printf("Please enter again: ");
            scanf("%d", &choice);
            fflush(stdin);
        }
}

The while loop force user to only enter 1 to 6 or 0 to quit. However the loop will also execute when I enter those value. What have I messed up?

I know this is a basic concept in c and I have already using it many times but this is the first time I saw this.

Answer 1

Make an habit of reading C expressions out aloud/in your head, then apply common sense afterwards.

"if choice isn't 1 or it isn't 2..." Wait, that doesn't make sense.
"if choise isn't 1 AND it isn't 2..." That makes sense.

Also related and generally helpful for programmers: De Morgan's Laws

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Unrelated to your question, the source of learning that told you to use fflush(stdin) is bad and needs to be retired and replaced. See Using fflush(stdin)