将一个指针内容复制到另一个 [英] Copy one pointer content to another

问题描述

我想我在某处读到过，当使用指针时，我们想将一个指针的内容复制到另一个指针，有两个选项:

I thought I've read somewhere that when using pointers and we want to copy the content of one to another that there are two options:

• 使用 memcpy 或
• 只是用 = 分配它们?

然而，在下面的例子中，我只是通过为两个指针分配内存来测试它，然后分配第二个，首先改变......但是我的第二个指针的条目也在改变......我做错了什么:/.

However in the lower example, I just tested it by allocating memory for two pointers, then assigning the second, changing first..but then the entry of my second pointer is also changing.. What am I doing wrong :/.

typedef struct {

    int a;
    int b;
    int c;
} my_struct;


int main(int argc, char** argv) {

    my_struct* first = malloc(sizeof(my_struct));   
    first->a = 100; first->b = 101; first->c = 1000;

    my_struct* bb = malloc(sizeof(my_struct));  

    printf("first %d %d %d\n", first->a, first->b, first->c);
    bb = first;
    printf("second %d %d %d\n", bb->a, first->b, bb->c);


    first->a = 55; first->b = 55; first->c = 89;
    printf("second %d %d %d\n", bb->a, first->b, bb->c);
}

推荐答案

你做的那一刻 bb = first;, bb and first指向相同的内存位置.first->a = 55;首先-> b = 55；first->c = 89; 将更改该位置中 a、b 和 c 的值.first的原始值，还在内存中徘徊，却再也无法访问了.

The moment you do bb = first;, bb and first are pointing to the same location of memory. first->a = 55; first->b = 55; first->c = 89; will change the values for a, b, and c in that location. The original value of first, is still lingering in memory but no way to access it anymore.

我认为您可能想要做的是*bb = *first;.

I think what you may want to do is *bb = *first;.

这篇关于将一个指针内容复制到另一个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！