如何使指针指向2D数组的任何数组元素?

Coming straight to the point,

直截了当地说，

I want the character pointer p to point to the only array element that contains the character 'T'.

我希望字符指针p指向唯一包含字符'T'的数组元素。

char a[100][100];
char *p;

for(int i=0;i<4;i++)
  for(int j=0;j<4;j++)
    if(a[i][j] == 'T')
      p = a[i][j];

P.S. I tried with various combinations of *, **, etc but nothing seems to work.

附注:我尝试了各种组合的*，**等，但似乎没有任何效果。

3 个解决方案

#1

Use its address:

使用它的地址:

char a[100][100];
char *p;

for(int i=0;i<4;i++)
  for(int j=0;j<4;j++)
    if(a[i][j] == 'T')
      p = &a[i][j];

a[i][j] is of type char and p is of type char *, which holds an address. To get the address of a variable, prepend it with &.

[i][j]是char类型的，p类型是char *，它持有一个地址。要获取变量的地址，请在其前面加上&。

The * operator on a pointer works the other way round. If you would want to get the 'T' back, you'd use:

指针上的*操作符反过来工作。如果你想把T拿回来，你可以用:

 char theT = *p;

#2

there is another way to get it

还有另一种方法。

char a[100][100];
char *p;

for(int i=0;i<4;i++)
   for(int j=0;j<4;j++)
       if(a[i][j] == 'T')
           p = a[i]+j;

By writing p = a[i]+j; you actually say, We have a pointer at the begging of an array called a[i] and you point to the position that is j times away from the begging of that array!

写p = a[i]+j;你会说，我们有一个指针指向一个叫a[i]的数组，你指向那个位置，这个位置是j的距离，而不是那个数组的请求!

#3

-1

change the if part as follows

更改if部分如下所示

  if(a[i][j] == 'T' ) {
       p = (char *) &a[i][j];
       i = 4; break;
  }

#1