一文带你玩转内存操作函数

港控/mmm° 2023-10-13 15:34 5阅读 0赞

> **作者主页：**paper jie的博客\_CSDN博客
> 
> **本文作者：**大家好，我是paper jie，感谢你阅读本文，欢迎一建三连哦。
> 
> **本文录入于**《系统解析C语言》专栏，本专栏是针对于大学生，编程小白精心打造的。笔者用重金(时间和精力)打造，将算法基础知识一网打尽，希望可以帮到读者们哦。
> 
> **其他专栏：**《算法详解》《C语言》《C语言-语法篇》等
> 
> **内容分享：**本期将对c语言中的内存操作函数进行详细的讲解，各位看官姥爷快搬好小板凳坐好叭。
> 
>  \-------- 不要998，不要98，只要一键三连，三连买不了吃亏，买不了上当

**目录**

memcpy

memcpy的介绍和使用

memcpy的模拟实现

memmove

memmove的介绍和使用

memmove的模拟实现

memcmp

memcmp的介绍和使用

memset

memset的介绍和使用

--------------------

# memcpy #

void * memcpy ( void * destination, const void * source, size_t num )

## ![d219e9c98b504ee3b9fdeecd009b1f44.png][] ##

## memcpy的介绍和使用 ##

Copies the values of *num* bytes from the location pointed to by *source* directly to the memory block pointed to by *destination*.The underlying type of the objects pointed to by both the *source* and *destination* pointers are irrelevant for this function; The result is a binary copy of the data.The function does not check for any terminating null character in *source* \- it always copies exactly *num* bytes.To avoid overflows, the size of the arrays pointed to by both the *destination* and *source* parameters, shall be at least *num* bytes, and should not overlap (for overlapping memory blocks, [memmove][] is a safer approach).

**函数memcpy从source这个位置开始向后复制num个字节的数据到destination的内存位置**

**这个函数在遇到'\\0'的时候并不会停止**

**如果source和destination有任何的重叠，复制的结果都是未定义的**

**解释：**这里的函数参数之所以是void\*，是为了适用于多种类型，可以接受多种类型的参数。且不能将一个数组的地址作为两个参数传给memcpy，不然复制的结果就是乱糟糟的，不可预测。

//memcpy的使用
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
    	int arr1[] = { 2, 3, 4, 5, 6, 7, 8, 9 };
    	int arr2[] = { 1,1,1,1,1 };
    	memcpy(arr1, arr2, 20);
    	for (int i = 0; i < 8; i++)
    	{
    		printf("%d ", arr1[i]);
    	}
    	return 0;
    }

## memcpy的模拟实现 ##

//memcpy的模拟实现
    #include <stdio.h>
    #include <string.h>
    #include <assert.h>
    
    void* my_memcpy(void* dest, const void* source, size_t num)
    {
    	void* ret = dest;
    	assert(dest && source);
    	while (num--)
    	{
    		*(char*)dest = *(char*)source;
    		dest = (char*)dest + 1;
    		source = (char*)source + 1;
    	}
    	return ret;
    }
    int main()
    {
    	int arr1[] = { 2, 3, 4, 5, 6, 7, 8, 9 };
    	int arr2[] = { 1,1,1,1,1 };
    	my_memcpy(arr1, arr2, 20);
    	for (int i = 0; i < 8; i++)
    	{
    		printf("%d ", arr1[i]);
    	}
    	return 0;
    }

# memmove #

void * memmove ( void * destination, const void * source, size_t num )

## ![b8c10e0d3a4c4868a036ac6fe62a7b4b.png][] ##

## memmove的介绍和使用 ##

Copies the values of *num* bytes from the location pointed by *source* to the memory block pointed by *destination*. Copying takes place as if an intermediate buffer were used, allowing the *destination* and *source* to overlap.The underlying type of the objects pointed by both the *source* and *destination* pointers are irrelevant for this function; The result is a binary copy of the data.The function does not check for any terminating null character in *source* \- it always copies exactly *num* bytes.To avoid overflows, the size of the arrays pointed by both the *destination* and *source* parameters, shall be at least *num* bytes.

**memmove是memcpy的优化，和memcpy的区别就是memmove函数处理的原内存块和目标内存块可以重叠**

**如果源空间和目标空间出现重叠，就得用memmove函数处理**

//memmove的使用
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
    	int arr1[] = { 1,2,3,4,5,6,7,8,9,10 };
    	memmove(arr1, arr1 + 3, 20);
    	for (int i = 0; i < 10; i++)
    	{
    		printf("%d ", arr1[i]);
    	}
    	return 0;
    }

## memmove的模拟实现 ##

//memmove的模拟实现
    #include <stdio.h>
    #include <string.h>
    #include <assert.h>
    
    void* my_memmove(void* dest, const void* source, size_t num)
    {
    	void* ret = dest;
    	assert(dest && source);
    	while (num--)
    	{
    		if (dest < source) //前->后
    		{
    			*(char*)dest = *(char*)source;
    			dest = (char*)dest + 1;
    			source = (char*)source + 1;
    		}
    		else if (dest > source) //后->前
    		{
    			*((char*)dest + num) = *((char*)source + num);
    		}
    	}
    	return ret;
    }
    int main()
    {
    	int arr1[] = { 1,2,3,4,5,6,7,8,9,10 };
    	my_memmove(arr1, arr1 + 3, 20);
    	for (int i = 0; i < 10; i++)
    	{
    		printf("%d ", arr1[i]);
    	}
    	return 0;
    }

# memcmp #

int memcmp ( const void * ptr1, const void * ptr2, size_t num )

![8b772b9985784b2997dd8072434be513.png][]

## memcmp的介绍和使用 ##

compares the first *num* bytes of the block of memory pointed by *ptr1* to the first *num* bytes pointed by *ptr2*, returning zero if they all match or a value different from zero representing which is greater if they do not.Notice that, unlike [strcmp][], the function does not stop comparing after finding a null character.

**比较它们从prt1和prt2指针开始的num个字节，相同返回0，prt1<prt2返回负数，prt1>prt2返回正数**

![1ced290c7c124bebb99dac9d3b82536f.png][]

//memcmp的使用
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
    	char arr1[] = {1,2,3,4,5,6,7};
    	char arr2[] = {1,2,3,4,2};
    	int ret = memcmp(arr1, arr2, 20);
    	printf("%d\n", ret);
    	return 0;
    }

# memset #

void * memset ( void * ptr, int value, size_t num )

![ae7a9ae1957a460b8eb1c1080343d358.png][]

## memset的介绍和使用 ##

Sets the first *num* bytes of the block of memory pointed by *ptr* to the specified *value* (interpreted as an unsigned char)

**从ptr指向的内存块开始向后num个字节，改为value的内容**

//memset的使用
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
    	char arr[] = "hello world";
    	memset(arr, '0', 3);
    	printf("%s\n", arr);
    	return 0;
    }

--------------------

[d219e9c98b504ee3b9fdeecd009b1f44.png]: https://img-blog.csdnimg.cn/d219e9c98b504ee3b9fdeecd009b1f44.png
[memmove]: https://legacy.cplusplus.com/memmove
[b8c10e0d3a4c4868a036ac6fe62a7b4b.png]: https://img-blog.csdnimg.cn/b8c10e0d3a4c4868a036ac6fe62a7b4b.png
[8b772b9985784b2997dd8072434be513.png]: https://img-blog.csdnimg.cn/8b772b9985784b2997dd8072434be513.png
[strcmp]: https://legacy.cplusplus.com/strcmp
[1ced290c7c124bebb99dac9d3b82536f.png]: https://img-blog.csdnimg.cn/1ced290c7c124bebb99dac9d3b82536f.png
[ae7a9ae1957a460b8eb1c1080343d358.png]: https://img-blog.csdnimg.cn/ae7a9ae1957a460b8eb1c1080343d358.png