C++ C指针地址打印

C pointer address printing

提问by Robert Rocha

I am learning about C pointers and addresses for the first time and how to use them on my tablet

我第一次学习 C 指针和地址以及如何在我的平板电脑上使用它们

Let's say:

让我们说：

int x = 1, y = 2;

int *ip; // declares *ip as an int type?

ip = &x; //assigns the address of x to ip?

y = *ip; //y is now 1 via dereferencing

Are all the comment explanations correct?

所有的评论解释都正确吗？

What happens if I print the result of ip? Will it print the address of variable x, something like

如果我打印 ip 的结果会发生什么？它会打印变量 x 的地址，比如

011001110

011001110

回答by haccks

Yes. All of your statements are correct. However in case of first

是的。你的所有陈述都是正确的。但是在第一个的情况下

int *ip;

it is better to say that ipis a pointer to an inttype.

最好说它ip是一个int类型的指针。

What happens if I print the result of ip?

如果我打印 ip 的结果会发生什么？

It will print the address of x.

它将打印 的地址x。

Will it print the address of variable x, something like

011001110  

它会打印变量 x 的地址，比如

011001110  

No. Addresses are generally represented in hexadecimal. You should use %pspecifier to print the address.

否。地址通常以十六进制表示。您应该使用%p说明符来打印地址。

printf("Address of x is %p\n", (void *)ip);  

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NOTE:
Note that in the above declaration *is not the indirection operator. Instead it specify the type of p, telling the compiler that pis a pointer to int. The *symbol performs indirection only when it appears in a statement.

注意：
请注意，在上面的声明*中不是间接运算符。相反，它指定 的类型p，告诉编译器这p是一个指向 的指针int。该*符号执行，只有当它出现在一份声明中间接。

回答by Andrew L

int x = 1, y = 2;

int *ip; // declares ip as a pointer to an int (holds an address of an int)

ip = &x; // ip now holds the address of x

y = *ip; // y now equals the value held at the address in ip

回答by Fiddling Bits

Consider the following as an example:

以以下为例：

Initializer       x        y        ip
Memory Value      [1]      [2]      [1000]
Memory Address    1000     1004     1008

As you can see:

如你看到的：

1. xhas the value 1and the address 1000
2. yhas the value 2and the address 1004
3. iphas the value 1000(the address of x) and the address 1008

1. x有值1和地址1000
2. y有值2和地址1004
3. ip具有值1000（的地址x）和地址1008

Consider the following:

考虑以下：

1. x == 1and &x == 1000
2. y == 2and &y == 1004
3. ip == 1000and &ip == 1008and *ip == 1(the value of x)

1. x == 1和 &x == 1000
2. y == 2和 &y == 1004
3. ip == 1000和&ip == 1008和*ip == 1（的值x）

Hope this helps you visualize what's happening.

希望这可以帮助您想象正在发生的事情。

回答by adrian

It's all correct.

都是对的。

1st line: you declare two variables

第一行：声明两个变量

2nd line: a memory pointer "ip" is defined

第二行：定义了内存指针“ip”

3rd line: The memory adress of X is given to the pointer ip

第 3 行：将 X 的内存地址赋予指针 ip

4th line: Y is now set to the value of the variable X at the address ip

第 4 行：Y 现在设置为地址 ip 处的变量 X 的值

The memory address, however, is in hexadecimal format. 011001110 is a byte of data and not a memory address. The address is more likely going to be something like 0x000000000022FE38 (it may be shorter). Consider this:

但是，内存地址是十六进制格式。011001110 是一个字节的数据，而不是一个内存地址。该地址更有可能类似于 0x000000000022FE38（可能更短）。考虑一下：

int x = 0;
int *ptr = &x; //the "&" character means "the address of"
printf("The variable X is at 0x%p and has the value of %i", (void *)ptr, x);

This would print the address of X, rather than *ptr. You'd need another pointer to print the address of *ptr. But that's rather pointless, as a pointer is defined to print the address of a variable. Think of it as an alias for another variable (it's value is the value at that memory address).

这将打印 X 的地址，而不是 *ptr。您需要另一个指针来打印 *ptr 的地址。但这毫无意义，因为定义了一个指针来打印变量的地址。将其视为另一个变量的别名（它的值是该内存地址处的值）。