c - 使用指针时 C 中的奇怪代码行为，其中注释代码的某个不相关部分会影响输出

Question

我编写的以下代码显示了奇怪的行为，具体取决于我是否在 main() 中“注释”第三行和第四行。功能。如果我“评论”printf和scanf语句，代码的行为符合预期，否则会产生意外的结果/输出。 请帮助我了解我缺少什么。基本上，我只是一个试图理解指针的 C 新手。

#include <stdio.h>
#include <conio.h>

void main() {
    int n, *ptr1;
    char ch, *ptr2;
    printf("Enter an integer\n");
    scanf("%d", &n);
    printf("Enter any single alphabetic character\n");
    scanf("%c", &ch);
    ptr1 = &n;
    ptr2 = &ch;
    printf("\nThe integer is %d and its pointer is %d in the address %d\n", n, *ptr1, ptr1);
    printf("\nThe character is %c and its pointer is %c in the address %d\n", ch, *ptr2, ptr2);
}

Answer 1

您的问题是由于当您插入整数时，您输入了一个换行符，该换行符被视为字符并插入到 ch 变量中。将前导空格添加到第二个 scanf 调用中以忽略前一个换行符。

#include<stdio.h>

int main()
{
    int n,*ptr1;
    char ch,*ptr2;
    printf("Enter an integer\n");
    scanf("%d",&n);

    printf("Enter any single alphabetic character\n");
    scanf(" %c",&ch);// <- space added

    ptr1=&n;
    ptr2=&ch;
    printf("\nThe integer is %d and its pointer is %d in the address %p\n", n, *ptr1, (void*) ptr1);
    printf("\nThe character is %c and its pointer is %c in the address %p\n",ch, *ptr2, (void*) ptr2);

    return 0;
}

工作 session :

Enter an integer
6
Enter any single alphabetic character
b

The integer is 6 and its pointer is 6 in the address 0x22cc84

The character is b and its pointer is b in the address 0x22cc83

PS。不要使用 conio.h - 它是非标准的，并且某些编译器不支持它。另外，您不会从中调用任何函数。