C语言 memcpy 和指针

memcpy and pointers

提问by Avinash

I am confuse on how to read the pointers copied in an array using memcpy. Following is what I have tried, but does not work.

我对如何使用 memcpy 读取在数组中复制的指针感到困惑。以下是我尝试过的，但不起作用。

Basically, I have allocated block of memory in which I am copying pointers similar to array fashion, but during retrial it is not working. While this works with basic data types properly

基本上，我已经分配了内存块，我在其中复制类似于数组方式的指针，但是在重试期间它不起作用。虽然这适用于基本数据类型

I want to store anything in the elementblock, It can be either integersor pointers.

我想在element块中存储任何东西，它可以是integers或pointers。

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

#define INDEX(x)  ((char *)elements + (sizeof(int*) * (x)))
int size = 10;

void f2_int_ptr() {    
    int i = 0;
    void *elements = (void*)malloc(size * sizeof(int*));
    for ( i = 0; i < size; i++ ) { 
       int *v = ( int *) malloc ( sizeof(int));
       memcpy ( v, &i, sizeof ( int ));


       memcpy ( INDEX(i) , v,  sizeof (int*));
    }
    for ( i = 0; i < size; i++ ) { 
        int *v = (int*)0;
        memcpy ( v, INDEX(i), sizeof(int *));
        printf ( "%d\n", *v );
    }
}
void f1_int() {
    int i = 0;
    void *elements = (void*)malloc(size * sizeof(int));
    for ( i = 0; i < size; i++ ) { 
       memcpy ( INDEX(i) , &i,  sizeof (int));
    }
    for ( i = 0; i < size; i++ ) { 
        int v;
        memcpy ( &v, INDEX(i), sizeof ( int ));
        printf ( "%d\n", v );
    }
}
int main(){
    f1_int();
    f2_int_ptr();
    return 0;
}

In the above code f1_intworks fine but f2_int_ptrdoes not work.

在上面的代码中f1_int工作正常但f2_int_ptr不起作用。

采纳答案by Oliver Charlesworth

f2_int_ptr()needs to become this:

f2_int_ptr()需要变成这样：

void f2_int_ptr() {
    int i = 0;
    void *elements = malloc(size * sizeof(int*));

    for ( i = 0; i < size; i++ ) { 
       int *v = malloc ( sizeof(int));
       memcpy ( v, &i, sizeof ( int ));
       memcpy ( INDEX(i) , &v,  sizeof (int*));
    }

    for ( i = 0; i < size; i++ ) {
        int *v;
        memcpy ( &v, INDEX(i), sizeof(int *));
        printf ( "%d\n", *v );
    }
}

Note the subtle changes to the memcpy()arguments.

注意memcpy()参数的细微变化。

Note: I really, really, really, wouldn't write code like this! It's incredibly difficult to follow.

注意：我真的，真的，真的，不会写这样的代码！遵循它是非常困难的。

回答by sickgemini

If you are storing the pointers in the elemnts, I think the final memcpy needs to use &v to copy the element to the pointer v. Similar to v=elements[i].

如果将指针存储在元素中，我认为最终的 memcpy 需要使用 &v 将元素复制到指针 v。类似于 v=elements[i]。

void f2_int_ptr() {    
    int i = 0;
    void *elements = (void*)malloc(size * sizeof(int*));
    for ( i = 0; i < size; i++ ) { 
       int *v = ( int *) malloc ( sizeof(int));
       memcpy ( v, &i, sizeof ( int *));


       memcpy ( INDEX(i) , &v,  sizeof (int*));
    }
    for ( i = 0; i < size; i++ ) { 
        int *v = (int *)0;
        memcpy ( &v, INDEX(i), sizeof(int *));
        printf ( "%d\n", *v );
    }
}

回答by Avinash

thanks Guys, it worked finally. i guess i have not allocated the space where to copy memcpy.

谢谢大家，终于成功了。我想我还没有分配复制 memcpy 的空间。

void f2_int_ptr() {    
    int i = 0;
    void *elements = (void*)malloc(size * sizeof(int*));
    for ( i = 0; i < size; i++ ) { 
       int *v = ( int *) malloc ( sizeof(int));
       memcpy ( v, &i, sizeof ( int ));

       memcpy ( INDEX(i) , v,  sizeof (int*));
    }
    for ( i = 0; i < size; i++ ) { 
        int *v = ( int *) malloc (sizeof(int*));
        memcpy ( v, INDEX(i), sizeof(int*));
        printf ( "%d\n", *v );
    }
}

回答by IProblemFactory

Code is very ugly, so I dont even know how it should works but: Why are u using &v here:

代码非常难看，所以我什至不知道它应该如何工作，但是：你为什么在这里使用 &v：

memcpy ( &v, INDEX(i), sizeof ( int ));

v is pointer itself:

v 是指针本身：

int *v = (int*)0;