memcpy:警告:取消引用"void *"指针 [英] memcpy: warning: dereferencing ‘void *’ pointer

问题描述

我使用read()函数从文件中读取40个字符，并且需要从10的偏移量复制20个字符的长度.换句话说，我需要将第10个字符至第30个字符进行memcpy转换为一个新的内存地址.但是，当我运行代码时(请参阅下文)，我收到警告消息:warning: dereferencing ‘void *’ pointer

I use the read() function to read in 40 characters from a file, and need to copy from the offset of 10 for the length of 20. In other words, I need to do memcpy from the 10th to 30th characters into a new memory address. When I run my code (see following), however, I got the warning message: warning: dereferencing ‘void *’ pointer

int main()
{
    void *buffer = malloc(40);
    int fd = open("example20.txt", O_RDONLY);
    printf("the value of fd is %d \n", fd);


/* read 40 characters from the file */ 
int bytes_read = read(fd, buffer, 40);

void *new_container = malloc(20);

/* copy from buffer, starting offset at 10 for length of 20 */
memcpy(new_container, &buffer[10], 20);
printf("new_container is %s \n", (char *) new_container);

return 0;
}

我想知道这个错误是什么意思，以及如何解决?

I am wondering what this error means, and how to fix it?

edit1:我找到了解决问题的方法:通过将缓冲区从void *转换为新的char *指针.

edit1: I found a way of solving the problem: by casting the buffer from void* to a new char* pointer.

char *buffer2 = (char *) buffer;
memcpy(new_container, &buffer2[10], 20);

edit2:我在memcpy中找到了使用void *指针的方法:memcpy(new_container, buffer+10, 20);这样的变量缓冲区"可以是void *类型

edit2: I found a way of using void* pointer in memcpy: memcpy(new_container, buffer+10, 20); the variable "buffer" in this way can be a void* type

推荐答案

更改行

memcpy(new_container, &buffer[10], 20);

到

memcpy(new_container, (char *)buffer + 10, 20);

这是因为&buffer[10]的计算结果为&(*(buffer + 10))，因为数组下标运算符的优先级高于address of运算符&.但是，buffer的类型为void *，并且由于没有大小信息，因此无法对void指针执行指针算术运算.在buffer上使用类型转换运算符(char *)提供必要的大小信息，以使
(char *)buffer + 10等效于buffer + 10 * sizeof(char)或变量11个元素的地址>.

That's because &buffer[10] evaluates to&(*(buffer + 10)) because the array subscript operator has higher precedence than the address of operator &. However, buffer is of type void * and pointer arithmetic cannot be done on void pointers because there is no size information. Using the typecast operator (char *) on buffer provides the necessary size information so that
(char *)buffer + 10 is equivalent to buffer + 10 * sizeof(char) or the address of the 11th element in the buffer pointed to by the variable buffer.

这篇关于memcpy:警告:取消引用"void *"指针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！