如何打印指针指向的值？

Question

我试图向用户显示一系列问题，然后扫描答案。 我的代码构建没有错误，但是当我运行它时，我收到错误：期望指向 char 的指针，但发现指向聚合的指针。这里有什么错误？

#include <stdio.h>

int main ()

{

 char name[50] , lastname[50] , add[100], post[50], town[60], state[60], tel[50];

 printf ("**** PLEASE ENTER THIS CONFIDENTIAL INFORMATION****");

  printf ("\n=================================================\n=================================================");

printf ("\n\nFirst name:");
scanf ("%s",&name);

printf ("\nLast name:");
scanf  ("%s",&lastname);

printf ("\nAddress Please:");
scanf ("%s",&add);

printf ("\nPostcode:");
scanf("%s",&post);

printf ("\ntown:");
scanf ("%s",&town);

printf("\nTelephone number:");
scanf("%s",&tel);


printf ("\n\n****CONFIDENTIAL INFORMATION****");

printf ("\n=================================================\n=================================================");

printf ("\nName:%s %s \n" ,name ,lastname);
printf ("Address:%s\n",add);
printf ("postal code:%s\n",post);
printf ("Town:%s\n",town);
printf ("State:%s\n",state);
printf ("Tel:%s\n",tel);

}

Answer 1

尝试这个（&全部删除）请参阅我上面的评论以了解原因。

#include <stdio.h>

int main ()

{

 char name[50] , lastname[50] , add[100], post[50], town[60], state[60], tel[50];

 printf ("**** PLEASE ENTER THIS CONFIDENTIAL INFORMATION****");

  printf ("\n=================================================\n=================================================");

printf ("\n\nFirst name:");
scanf ("%s",name);

printf ("\nLast name:");
scanf  ("%s",lastname);

printf ("\nAddress Please:");
scanf ("%s",add);

printf ("\nPostcode:");
scanf("%s",post);

printf ("\ntown:");
scanf ("%s",town);

printf("\nTelephone number:");
scanf("%s",tel);


printf ("\n\n****CONFIDENTIAL INFORMATION****");

printf ("\n=================================================\n=================================================");

printf ("\nName:%s %s \n" ,name ,lastname);
printf ("Address:%s\n",add);
printf ("postal code:%s\n",post);
printf ("Town:%s\n",town);
printf ("State:%s\n",state);
printf ("Tel:%s\n",tel);

}

Answer 2

当你对字符串进行 scanf 时，你不需要

scanf("%s",&str);

而是直接做

scanf("%s",str).

Answer 3

#include <stdio.h>
int main()
{
    int i=3;
    int *j;
    j = &i;

    printf("i %d\n",i);//value of i
    printf("j %d",*j);//value of i in j
    printf("j %d",&j);// address of i in j
   return 0;
}

因此，这就是引用和取消引用指针的方式。我刚刚给出了线索。现在你用你的大脑......干杯！

Answer 4

问题出在 scanf 上：

printf ("\n\nFirst name:");
scanf ("%s",&name);

printf ("\nLast name:");
scanf  ("%s",&lastname);

printf ("\nAddress Please:");
scanf ("%s",&add);

printf ("\nPostcode:");
scanf("%s",&post);

printf ("\ntown:");
scanf ("%s",&town);

printf("\nTelephone number:");
scanf("%s",&tel);

您应该将这样的数组传递给 scanf （不带地址运算符 &）：

printf ("\n\nFirst name:");
scanf ("%s",name);

printf ("\nLast name:");
scanf  ("%s",lastname);

printf ("\nAddress Please:");
scanf ("%s",add);

printf ("\nPostcode:");
scanf("%s",post);

printf ("\ntown:");
scanf ("%s",town);

printf("\nTelephone number:");
scanf("%s",tel);

这是为什么呢？ 因为数组名称本身是指向其中元素首地址的指针，例如：

int x[2] = {0, 1};

x 指向第一个元素的地址，在本例中为 0 并且您还可以将数组名称视为普通指针（并非所有情况都允许） 例如：

int x[2] = {0, 1};
*(x + 0) = 55;
*(x + 1) = 105;

使用gdb调试器的输出：

5               int x[2] = {0, 1};
(gdb) s
7               *(x+0) = 55;
(gdb) p x
$1 = {0, 1}
(gdb) s
8               *(x+1) = 105;
(gdb) p x
$2 = {55, 1}
(gdb) s
10              return (0);
(gdb) p x
$3 = {55, 105}

但是这是违法的

int x[2] = {0, 1};
int y = 200;
int *p = NULL;

//x = &y;   wont work as x is like a const pointer only
p = &y;    //will work normally and p will point to address of y

希望这有帮助