c - 为什么我需要按三次 '\n' 才能运行？

Question

嗨，有人知道为什么这段代码让我在运行前按 Enter 三次吗？我需要它才能在第一个 '\n' 运行学校作业这是代码:非常感谢!!

int main()
{
    char  L1=0, L2=0, L3=0 ;
    int count_hya=0, countLA=0, countLC=0, countLH=0, countLL=0 ;
    printf("give me some letters\n");

    while ((L1!='\n')&&(L2!='\n')&&(L3!='\n'))
    {
        L1=getchar();
        if (L1=='A')
            countLA++;
        if (L1=='C')
            countLC++;
        if (L1=='L')
            countLL++;
        if (L1=='H')
            countLH++;
        if (L1=='H'&&L2=='Y'&&L3=='A')
            count_hya++;
    }
    printf("number of words begning with: A=%d  C=%d  L=%d  H=%d Hydra=%d", 
           countLA, countLC, countLL, countLH, count_hya);
    return 0;
}

编辑将最后几行更改为:

 if (L1=='H')
  {   
  countLH++;
  L2=getchar();
  L3=getchar();
  if (L1=='H'&&L2=='Y'&&L3=='A')
   count_hya++;
  }

非常感谢您的帮助!

Answer 1

当你使用的时候:

while ((L1!='\n')||(L2!='\n')||(L3!='\n'))

应该是

while ((L1!='\n')&&(L2!='\n')&&(L3!='\n'))

使用下面的代码

#define SCANF_CHK(L) \
     scanf("%c", &L); \
     if(L=='\n') continue;

int main()
{
    char  L1=0, L2=0, L3=0 ;
    int count_hya=0, countLA=0, countLC=0, countLH=0, countLL=0 ;
    printf("give me some letters\n");
    while ((L1!='\n')&&(L2!='\n')&&(L3!='\n'))
    {
        SCANF_CHK(L1);
        SCANF_CHK(L2);
        SCANF_CHK(L3);
        if (L1=='A')
            countLA++;
        if (L1=='C')
            countLC++;
        if (L1=='L')
            countLL++;
        if (L1=='H')
            countLH++;
        if (L1=='H'&&L2=='Y'&&L3=='A')
            count_hya++;
    }
    printf("number of words begning with: A=%d  C=%d  L=%d  H=%d Hydra=%d\n", 
           countLA, countLC, countLL, countLH, count_hya);
    return 0;
}