c-memcpy和指针.仍在工作.为什么? [英] c - memcpy and pointers. Still work. Why?

问题描述

那我该怎么做才能获得正确的代码?

What should I do to have a correct code then ?

好的，我更正下面的代码

Ok, I correct the code below

摆弄memcpy. Linux，64位. gcc 4.8.x

Fiddling with memcpy. Linux, 64 bits. gcc 4.8.x

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void d(char ** restrict a, char ** restrict b){

    char r[20];
    memcpy(r,*a, strlen(*a)+1);

    // it is the same thing as before since *c is equivalent to &r (or put simply r).
    char *restrict c = malloc(20);
    memcpy(c,*a, strlen(*a)+1);

    // that one bugs me. why b alone and not *b ??
    // EDIT : this is incorrect
    memcpy(b,*a, strlen(*a)+1);

    // EDIT : this is correct
    memcpy(*b,*a, strlen(*a)+1);


    printf("pointer c -> hey %s\n",c);
    printf("array r -> hey %s\n",r);

    // EDIT : this is incorrect
    printf("double pointer -> hey %s\n",b);

    // EDIT : this is correct
    printf("double pointer -> hey %s\n",*b);
}

int main(void)
{

    char a[] = "YOU";
    char * b = a;
    char * c = malloc(20);

    d(&b, &c);


    return 0;
}

问题

我想了解为什么memcpy不会抱怨我将双指针传递给它，而只需要一个指针.

Question

I would like to undertsand why memcpy doesn't complain about me passing double pointer to it, while it needs a pointer only.

我知道使用chars * b ==& a == a，并且数组的第一个成员引用的数组最多为'\ 0'.问题确实在于将双指针传递给memcpy.

I know that with chars *b == &a == a and that an array is referenced by its first member up to '\0'. The problem really is with passing a double pointer to memcpy.

我为什么不必做

memcpy(*b, *a, strlen(*a)+1);

因为memcpy签名是

since memcpy signature is

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

，根据请问这里的渔获物"是什么?

What is the "catch" here please ?

非常感谢

推荐答案

双指针是指向单个指针的单个指针，因此可以将其传递给需要空指针的函数.

Well, a double pointer is a single pointer to single pointer, so it can be passed to a function that expects a void pointer.

您的代码是否正确，当然是另一回事...并非如此，并且仅出于巧合.请注意，不仅您使用memcpy()写入了错误的位置，而且还打印了与printf()中的字符串相同的错误内存位置.这里的巧合"是这两个错误"位置都相同，因此您错误地认为它可以正常工作.

It is of course another thing whether or not your code is correct... It's not and works only by coincidence. Note that not only you use memcpy() to write to a wrong location, but you also print the same wrong location of memory as a string in your printf(). The "coincidence" here is that both of these "wrong" locations are the same, so you falsely assumed that it works fine.

尝试真正打印正确的东西，您会看到混乱的地方:

Try to really print the right thing and you'll see the mayhem:

printf("double pointer -> hey %s\n",*b);

这篇关于c-memcpy和指针.仍在工作.为什么?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！