整数类型的指针大小与int *大小 [英] Size of pointer of integer type vs Size of int*

问题描述

我开始阅读Pointers并与此同时进行修改.我偶然发现了这个:

I started reading Pointers and while tinkering with them. I stumbled upon this :

#include<stdio.h>
int main()
{
    int *p,a;
    a=sizeof(*p);
    printf("%d",a);
}

输出:4

然后用sizeof(int*)代替sizeof(*p)，现在输出8.

Then in the place of sizeof(*p) I replaced it with sizeof(int*) Now it outputs 8 .

P是整数类型的指针，并且int *也是同一件事(我的假设正确吗?).然后为什么要打印两个不同的值.我正在64位gcc编译器上执行此操作.

P is a pointer of integer type and int* is also the same thing ( Is my assumption correct? ). Then why it is printing two different values. I am doing this on a 64bit gcc compiler.

推荐答案

每个初学者总是对指针声明和取消引用指针感到困惑，因为语法看起来是一样的.

Every beginner always gets confused with pointer declaration versus de-referencing the pointer, because the syntax looks the same.

• int *p;的意思是声明一个指向int的指针".您也可以将其写为int* p;(含义相同，个人喜好).
• *p，在声明中其他位置使用时，表示取p指向的内容".

• int *p; means "declare a pointer to int". You can also write it as int* p; (identical meaning, personal preference).
• *p, when used anywhere else but in the declaration, means "take the contents of what p points at".

因此，sizeof(*p)的意思是给我p指向的内容的大小"，但是sizeof(int*)的意思是给我指针类型本身的大小".在您的计算机上，int显然是4个字节，但指针是8个字节(通常是64位计算机).

Thus sizeof(*p) means "give me the size of the contents that p points at", but sizeof(int*) means "give me the size of the pointer type itself". On your machine, int is apparently 4 bytes but pointers are 8 bytes (typical 64 bit machine).

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