比特移位一个整数值 [英] bit-shifting by an integer value

问题描述

这code是用于高速缓存模拟器项目 - 我想从一个内存地址中提取一定位。当我尝试使用INT变量做位转移，我结束了不正确的结果，但是当我直接使用数字，我的结果是正确的。我一直在到处找一个答案，但我不能找到一个。这里有什么是我的问题？

This code is for a cache simulator project - I am trying to extract certain bits from a memory address. When I attempt to use int variables to do the bit shifting, I end up with an incorrect result, but when I use the numbers directly, my result is correct. I've been looking all over for an answer to this, but I can't find one. What is my issue here?

#include <stdio.h>

void main(){

    unsigned long long int mem_addr = 0x7fff5a8487c0;

    int byte_offset_bits = 2;
    int block_offset_bits = 5;
    int index_bits = 8;
    int tag_bits = 33;

    unsigned long long int tag1 = (mem_addr&(((1<<33)-1)<<(2+5+8)))>>(2+5+8);
    unsigned long long int tag2 = (mem_addr&(((1<<tag_bits)-1)<<(byte_offset_bits + block_offset_bits + index_bits)))>>(byte_offset_bits + block_offset_bits + index_bits);

    printf("%s %llx\n", "Tag 1:", tag1);
    printf("%s %llx\n", "Tag 2:", tag2);

}

输出是：

Tag 1: fffeb509
Tag 2: 1

我也越来越为正确计算TAG1的线，这是没有道理给我一个警告，因为Tag1是一个64位无符号长的长整型，而我只有48位转移：

I am also getting a warning for the line that computes tag1 correctly, which doesn't make sense to me, since tag1 is a 64-bit unsigned long long int, and I am only shifting by 48 bits:

 warning: left shift count >= width of type [enabled by default]

在此先感谢您的帮助。

Thanks in advance for the help.

推荐答案

所有整数文字值 INT 类型，除非你指定了preFIX诸如例如 1ULL 。

All integer literal values are int type, unless you specify a prefix such as e.g. 1ULL.

这意味着 1 LT;γ-33 移32位有符号值33的步骤。你需要做的 1ULL＆LT;＆LT; 33 。

That means that 1<<33 shifts a 32-bit signed value 33 steps. You need to do 1ULL << 33.

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