如何乘以c中的32位整数 [英] How to multiply 32-bit integers in c

问题描述

执行：

  #define HIGH32（V64）（（uint32_t）（（V64>> 32）& amp ; 0xffFFFFFF））
 #define LOW32（V64）（（uint32_t）（V64& 0xffFFffFF））
 
 uint32_t a = 0xffFFffFF; 
 uint32_t b = 0xffFFffFF; 
 uint64_t res = a * b; 
 printf（res =％08X％08X \\\
，HIGH32（res），LOW32（res））; 
  

给出：

 res = 00000000 00000001
  

但我预计：fffffffe00000001。我做错了什么？单一作业：

  res = 0x0123456789ABCDEF; 
 printf（res =％08X％08X \\\
，HIGH32（res），LOW32（res））; 
  

给出

  res = 01234567 89ABCDEF 
  

环境：

  $ gcc --version 
 gcc（GCC）4.8.3 
版权所有（C）2013自由软件基金会，Inc. 
 
 $ gcc -v 
 COLLECT_GCC = gcc 
 COLLECT_LTO_WRAPPER = / usr / lib / gcc / i686-pc-cygwin / 4.8.3 / lto-wrapper.exe 
 dest arch：i686- pc-cygwin 
 
 $文件a.exe 
 a.exe：PE32可执行文件（控制台）Intel 80386，用于MS Windows 
  

解决方案

您目前有：

  uint64_t res =（uint32_t）a *（uint32_t）b; 
  

您需要在乘法之前将参数提升为64位数字。因此：

  uint64_t res =（uint64_t）a * b; 
  

Execution of:

#define HIGH32(V64) ((uint32_t)((V64 >> 32)&0xffFFffFF))
#define LOW32(V64) ((uint32_t)(V64&0xffFFffFF))

uint32_t a = 0xffFFffFF;
uint32_t b = 0xffFFffFF;
uint64_t res = a * b;
printf("res = %08X %08X\n", HIGH32(res), LOW32(res));

Gives:

"res = 00000000 00000001"

But I expect: fffffffe00000001. What did I do wrong? Single assignment:

res = 0x0123456789ABCDEF;
printf("res = %08X %08X\n", HIGH32(res), LOW32(res));

Gives

res = 01234567 89ABCDEF

Environment:

$gcc --version
gcc (GCC) 4.8.3
Copyright (C) 2013 Free Software Foundation, Inc.

$ gcc -v
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=/usr/lib/gcc/i686-pc-cygwin/4.8.3/lto-wrapper.exe
dest arch: i686-pc-cygwin

$ file a.exe
a.exe: PE32 executable (console) Intel 80386, for MS Windows

解决方案

You currently have:

uint64_t res = (uint32_t) a * (uint32_t) b;

You will want to promote the arguments to 64bit numbers before the multiplication. Therefore:

uint64_t res = (uint64_t) a * b;

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