*日期=“星期日”；vs int *number = 7；

Question

我有这样一段代码：

#include <stdio.h>
int main(void)
{
    char *date = "Sunday";
    //int *number = 7;

    printf("Today is %s, the 7 days of this week", date);
}

它按预期工作并打印。

$ ./a.out
Today is Sunday, the 7 days of this week

尽管如此，当我评论

#include <stdio.h>
int main(void)
{
    char *date = "Sunday";
    int *number = 7;

    printf("Today is %s, the %d days of this week", date, *number);
}

它报告错误：

$ cc draft.c
draft.c:5:10: warning: incompatible integer to pointer conversion initializing 'int *' with an expression of type
      'int' [-Wint-conversion]
    int *number = 7;
         ^        ~
1 warning generated.

我的代码有什么问题？

Answer 1

它应该是

int number = 7;
printf("Today is %s, the %d days of this week", date, number);

而不是

int *number = 7;
printf("Today is %s, the %d days of this week", date, *number);

在第一个片段中，用7初始化一个整数变量，并打印其值。在第二段中，使用地址7初始化指针，然后打印内存地址7处的整数值。