错过了 scanf 和 function 没有它继续。 如果我添加一个空格仍然不起作用

Question

#include <stdio.h>


struct mychar {
    char value;
    struct mychar *nextPtr;
};

typedef struct mychar Mychar;


void instructions();
void append(Mychar **, char );
void printlist(Mychar *);


int main(){
    instructions();

    Mychar *startPtr = NULL;

    unsigned int choice;
    char newchar;
    do {
        scanf("%d",&choice);
        switch (choice) {
            case 1:
                printf("\nWrite the character you want to add.");
                printf("\n> ");
                scanf(" %c", &newchar);
                append(&startPtr, newchar);
                printlist(startPtr);
                break;
            case 2:
                break;
            default:
                printf("\nError, try again.\n");
                //main();
                instructions();
                break;
        }
    } while (choice!=3);
    printf("\n\nEnd of run.\n");
}


void instructions(){
    printf("\nSelect operation. 1 to add, 2 to remove, 3 to exit.");
    printf("\n> ");
}


void append(Mychar **sPtr, char newvalue){
    Mychar *newlinkPtr = calloc (1, sizeof(Mychar));
    newlinkPtr->value = newvalue;
    newlinkPtr->nextPtr = NULL;

    Mychar *previousPtr = NULL;
    Mychar *currentPtr = *sPtr;

    while(currentPtr!=NULL && newvalue > currentPtr->value){
        previousPtr = currentPtr;
        currentPtr = currentPtr->nextPtr;
    }

    if (previousPtr){
        previousPtr->nextPtr = newlinkPtr;
        newlinkPtr->nextPtr = currentPtr;
    } else {
        *sPtr = newlinkPtr;
    }

}


void printlist(Mychar *currentPtr){
    printf("\n\nCurrent list:\n");
    while (currentPtr!=NULL){
        printf("%c", currentPtr->value);
        currentPtr = currentPtr->nextPtr;
    }
}

为什么我会有这种行为？ 如果我运行程序，在我输入 1 后，它会打印“当前列表”并保持 scanf 输入处于打开状态，所以我只能在“当前列表”打印后输入值。 此外，只有在我使用 scanf 输入字符后才应调用“当前列表”，因为 function 打印列表位于 scanf 之后……但实际上是这样的：

Select operation. 1 to add, 2 to remove, 3 to exit.
> 1

Write the character you want to add.
> a


Current list:
ab

Write the character you want to add.
> 

Current list:
abc

Write the character you want to add.
> 

Current list:
abcd

Write the character you want to add.
> 

Current list:
abcd

Answer 1

从中吸取的教训是始终检查scanf是否返回0 ，至少还建议进行EOF检查，并采取相应的行动，至于代码的事件顺序，它并不完全存在，您可以进行一些调整有一个很好的，错误的输入证明，I/O 序列：

void clear_stdin() { //stdin buffer clearing function 
    int c;
    while ((c = getchar()) != '\n' && c != EOF){}
}

do {
    instructions(); //move inside the loop, user will be prompted in each cycle
    while (scanf("%d", &choice) == 0) {
        printf("\nError, try again.\n");
        instructions();
        clear_stdin(); // if input fails clear the buffer
    }
    clear_stdin(); // clear the buffer for 1hjh type input
    switch (choice) {
    case 1:

        printf("\nWrite the character you want to add.");
        printf("\n> ");
        while (scanf(" %c", &newchar) == 0) { //this can be a pattern
            clear_stdin();                    //see @ismick comment
        }                                     //
        clear_stdin();                        //
        append(&startPtr, newchar);
        printlist(startPtr);
        break;
    case 2:
        break;
    case 3:
        printf("\n\nEnd of run.\n"); //if you dont have a case default will catch 3
        break;
    default:
        printf("\nError, try again.\n");
        break;
    }
} while (choice != 3);