memmove


Does memmove actually “move” a chunk of memory and leave behind zeros at the source? [duplicate]

流过昼夜 提交于 2020-01-15 03:26:49
问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: memcpy vs memmove Does memmove actually "move" a chunk of memory? If so, does it leave the memory with zeros? Or, is it just like memcpy? I am looking at the man page, and I do not believe my assumption is correct. If I want to move a chunk of memory using memmove, would I have to manually zero out the chunk of memory where I did the move? 回答1: memmove has nothing to do with clearing the old memory, and in fact

memcpy vs assignment in C — should be memmove?

大城市里の小女人 提交于 2020-01-01 04:41:07
问题 As pointed out in an answer to this question, the compiler (in this case gcc-4.1.2, yes it's old, no I can't change it) can replace struct assignments with memcpy where it thinks it is appropriate. I'm running some code under valgrind and got a warning about memcpy source/destination overlap. When I look at the code, I see this (paraphrasing): struct outer { struct inner i; // lots of other stuff }; struct inner { int x; // lots of other stuff }; void frob(struct inner* i, struct outer* o) {

Bitwise memmove

℡╲_俬逩灬. 提交于 2019-12-30 18:06:07
问题 What is the best way to implement a bitwise memmove ? The method should take an additional destination and source bit-offset and the count should be in bits too. I saw that ARM provides a non-standard _membitmove, which does exactly what I need, but I couldn't find its source. Bind's bitset includes isc_bitstring_copy, but it's not efficient I'm aware that the C standard library doesn't provide such a method, but I also couldn't find any third-party code providing a similar method. 回答1:

What are real significant cases when memcpy() is faster than memmove()?

自闭症网瘾萝莉.ら 提交于 2019-12-18 12:54:15
问题 The key difference between memcpy() and memmove() is that memmove() will work fine when source and destination overlap. When buffers surely don't overlap memcpy() is preferable since it's potentially faster. What bothers me is this potentially . Is it a microoptimization or are there real significant examples when memcpy() is faster so that we really need to use memcpy() and not stick to memmove() everywhere? 回答1: At best, calling memcpy rather than memmove will save a pointer comparison and

Why is Linux memmove() implemented the way it is?

社会主义新天地 提交于 2019-12-18 03:11:51
问题 From the Linux manpage for memmove(3) The memmove() function copies n bytes from memory area src to memory area dest. The memory areas may overlap: copying takes place as though the bytes in src are first copied into a temporary array that does not overlap src or dest, and the bytes are then copied from the temporary array to dest. Instead of allocating a temporary array and copy the values twice we could just do the following: void *my_memmove(void *dest, const void *src, size_t n) { signed

Why is Linux memmove() implemented the way it is?

泄露秘密 提交于 2019-12-18 03:11:01
问题 From the Linux manpage for memmove(3) The memmove() function copies n bytes from memory area src to memory area dest. The memory areas may overlap: copying takes place as though the bytes in src are first copied into a temporary array that does not overlap src or dest, and the bytes are then copied from the temporary array to dest. Instead of allocating a temporary array and copy the values twice we could just do the following: void *my_memmove(void *dest, const void *src, size_t n) { signed

memmove implementation in C

╄→гoц情女王★ 提交于 2019-12-17 22:11:58
问题 Can some one help me to understand how memmove is implemented in C. I have only one special condition right ? if((src<dst)&&((src+sz) > dst)) copy from the back Also does it depend on the way stack grows ? 回答1: Mathematically, you don't have to worry about whether they overlap at all. If src is less than dst , just copy from the end. If src is greater than dst , just copy from the beginning. If src and dst are equal, just exit straight away. That's because your cases are one of: 1) <-----s---

Will memcpy or memmove cause problems copying classes?

点点圈 提交于 2019-12-17 13:06:51
问题 Suppose I have any kind of class or structure. No virtual functions or anything, just some custom constructors, as well as a few pointers that would require cleanup in the destructor. Would there be any adverse affects to using memcpy or memmove on this structure? Will deleting a moved structure cause problems? The question assumes that the memory alignment is also correct, and we are copying to safe memory. 回答1: In the general case, yes, there will be problems. Both memcpy and memmove are

memcpy的内存重叠问题

青春壹個敷衍的年華 提交于 2019-12-14 23:45:13
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 【注】改编自 memmove 和 memcpy的区别 。原作者如有不爽,请告知! memcpy是C语言中的库函数,在头文件string.h中,作用是拷贝一定长度的内存的内容,原型分别如下: void *memcpy(void *dest, const void *src, size_t count) 使用memcpy时,有可能会遇到内存重叠的问题: 第一种情况下,拷贝重叠的区域不会出现问题,内容均可以正确的被拷贝。 第二种情况下, 问题出现在右边的两个字节,这两个字节的原来的内容首先就被覆盖了,而且没有保存 。所以接下来拷贝的时候, 拷贝的是已经被覆盖的内容,显然这是有问题的 。 通过memmove可以避免这一问题。memmove和memcpy实现一样的功能:内存拷贝。原型如下: void *memmove(void *dest, const void *src, size_t count) 以下几点你需要了解: memove可以避免内存拷贝时的重叠问题。 实际上,memcpy只是memmove的一个子集。 memcpy比memmove的速度要快一些。 有兴趣的,可以看看 linux的源码 ,实现很简单,一看就明白。 /** * memcpy - Copy one area of memory to

malloc, free, and memmove inside a subfunction

≯℡__Kan透↙ 提交于 2019-12-11 10:37:39
问题 I want to use a subfunction to copy a char array. it is like this: void NSV_String_Copy (char *Source, char *Destination) { int len = strlen(Source); if (*Destination != NULL) free(Destination); Destination = malloc(len + 1); memmove(*Destination, Source, len); Destination[len] = '\0'; //null terminate } that way, I can call it from the main function and perform the operation this way: char *MySource = "abcd"; char *MyDestination; NSV_String_Copy (MySource, MyDestination); However, it does

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