无法理解涉及指针C程序中的一小部分 [英] Can't understand small part of a C program involving pointers

问题描述

  INT X = 5;
INT Y = 6;
为int * p =＆放大器; X;
* P = X + Y;
P =＆安培; Y;
* P = X + Y;
的printf（％D，％D，X，Y）;
 

我能理解第3行。我知道，为int * p =＆放大器; X; 表示指针 P 是在内存地址指向 X 。但我无法理解接下来的4行code的。此code的输出是：

  11,17
 

但我无法理解怎么样？有人可以解释给我。

解决方案

  

指针是一个变量，其值是另一个变量的地址，即内存位置的直接地址。

当你分配这样的，

 为int * p =＆放大器; X; //这将指向x变量的地址。
 

在 * P做任何更改，它会影响记忆力。所以当你访问一个与 X 它具有的价值。

  * P = X + Y; //这相当于X = X + Y;
 

然后是也是这样。

现在 X 的值为 11 。 P =＆安培; Y; ​​

  * P = X + Y; //它等同于Y = X + Y;
 

所以现在x的值 11 和y值 5 。所以结果是 17 。

int x = 5;
int y = 6;
int *p = &x;
*p = x + y;
p = &y;
*p = x + y;
printf("%d,%d", x, y);

I can understand the first 3 lines. I know that int *p = &x; means that the pointer p is pointing at the memory address of x. But I can't understand the next 4 lines of the code. The output for this code is:

 11,17

But I unable to understand how? Can someone explain it to me.

解决方案

A pointer is a variable whose value is the address of another variable, i.e., direct address of the memory location.

When you are assigning like this,

 int *p=&x;// It will point to the address of x variable. 

Any change done in the *p, It will affect the memory. So while you are accessing that with the x it have the value.

 *p=x+y; // It's equivalent x=x+y;

Then y also like this.

Now x have the value 11. p=&y;

  *p=x+y;// It's equivalent to y=x+y;

so now x have the value 11 and y value 5. So the result is 17.

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