c - 为什么这个 C 代码没有生成预期的输出？

Question

我在 Microsoft Visual C++ 2010 中编写了这个简单的 C 代码。

   #include<stdio.h>
    #include<conio.h>
    void main()
    {
    char title[20], artist[30];
    int numtrack, price;
    char type;

    printf("Enter the title of CD \n");
    scanf("%s",title);
    printf("\nName of the artist \n");
    scanf("%s",artist);
    printf("\nEnter the type of CD(enter a for album and s for single)\n");
    scanf("%c",&type);
    printf("\n Enter the number of tracks \n");
    scanf("%d", &numtrack);
    printf("\n Enter the price of the cd \n");
    scanf("%d", &price);
    printf("%s\n%s\n%c\n%d\n%d\n",title, artist, type, numtrack, price);
    getch();
    }

输出是

Enter the title of CD
ranjit

Name of the artist
mahanti

Enter the type of CD(enter a for album and s for single)

 Enter the number of tracks

4

 Enter the price of the cd
4
ranjit
mahanti


4
4

我不明白为什么它不等待类型变量的输入？有人可以解释一下吗？提前致谢。

Answer 1

代替

scanf("%c",&type);

你想要

scanf(" %c",&type);

否则，前一个字符串中的换行符之一将被用作类型。