如何memset的初始化-1整数数组？ [英] How memset initializes an array of integers by -1?

问题描述

借助手册页说，关于 memset的：

 的#include＆LT;＆string.h中GT;
无效* memset的（无效* S，INT C，为size_t N）
 

的 memset的（）函数填充第一个 N 字节的内存区通过指出为s 随着不断的字节 C 。

显然， memset的不能用于初始化 INT 阵列，如下图所示：

  int类型的[10];
memset的（一，1，sizeof的（一））;
 

这是因为 INT 重新由4个字节psented（说）$ P $，没有人可以得到在阵列在。结果
但我经常看到程序员使用 memset的到 INT 数组元素设置为 0 或 1 。

  int类型的[10];
INT B〔10〕;
memset的（一，0，sizeof的（一））;
memset的（二，-1，sizeof的（B））;
 

按我的理解，与整数初始化 0 是可以的，因为 0 可重新$ P $在psented 1个字节（可能是我错了，在这种情况下）。但怎么可能初始化 B 与 1 （一个4字节的值）？

解决方案

奇怪的是，这个工程的原因 1 是完全一样的，这与工作的原因零点：在二进制补码再presentation ， 1 的 1 S IN的所有位，无论整数的大小，在一个地区做馅用充满了 1 <字节/ code>取值产生 1 签署 INT S，长的区域 s和短 S于补硬件。

在来自两个互补的不同硬件的结果将是不同的。在 1 整型常量将被转换为 unsigned char型所有的人的，因为标准是具体就如何转换必须执行。然而，与设置为其所有比特字节区域1 将PTED如根据平台的规则积分值间$ P $。例如，标志和硬件级阵列中的所有元素将包含相应类型的最小负值。

The manpage says about memset:

#include <string.h>
void *memset(void *s, int c, size_t n)

The memset() function fills the first n bytes of the memory area pointed to by s with the constant byte c.

It is obvious that memset can't be used to initialize int array as shown below:

int a[10];
memset(a, 1, sizeof(a));  

it is because int is represented by 4 bytes (say) and one can not get the desired value for the integers in array a.
But I often see the programmers use memset to set the int array elements to either 0 or -1.

int a[10];
int b[10];
memset(a, 0, sizeof(a));  
memset(b, -1, sizeof(b));  

As per my understanding, initializing with integer 0 is OK because 0 can be represented in 1 byte (may be I am wrong in this context). But how is it possible to initialize b with -1 (a 4 bytes value)?

解决方案

Oddly, the reason this works with -1 is exactly the same as the reason that this works with zeros: in two's complement binary representation, -1 has 1s in all its bits, regardless of the size of the integer, so filling in a region with bytes filled with all 1s produces a region of -1 signed ints, longs, and shorts on two's complement hardware.

On hardware that differs from two's complement the result will be different. The -1 integer constant would be converted to an unsigned char of all ones, because the standard is specific on how the conversion has to be performed. However, a region of bytes with all their bits set to 1 would be interpreted as integral values in accordance with the rules of the platform. For example, on sign-and-magnitude hardware all elements of your array would contain the smallest negative value of the corresponding type.

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