在unix中采用unsigned long int的默认值 [英] taking default value of unsigned long int in unix

问题描述

#include每个头文件(我认为)

#included every header file (i think)

int main(int argc, char* argv[])
{
  unsigned long long x=atoi(argv[1]);
  unsigned long long y=atoi(argv[2]);
  if(x>0 && x<=4294967295 && y>0 && y<=4294967295)
  {
    cout<<"done";
  }
  else
  {
    cout<<"not done";
  }
  return 0;
}

//但是当我插入x = 6543848485或大于2254354363(10位数)的任何值时...
//它会自动采用默认值x = 2147483647(自动)...我不y ...并没有进入else条件...

请帮助

// But when i insert x=6543848485 or any value greater than 2254354363(10 digit)...
// It is automatically taking the default value of x=2147483647(automatically)... I donot y... and not entering in the else condition...

please help

推荐答案

是否可能是因为atoi返回了integer?如带符号的32位数字? 2147483647恰好是十六进制的7FFFFFFF吗? IE.可以容纳带符号的32位整数的最大数字吗?

Could it possibly be because atoi returns an integer? As in a signed, 32 bit quantity? And 2147483647 just happens to be 7FFFFFFF in hex? I.e. The largest number that fits in a signed, 32 bit integer?

atoi [ ^ ]是不合适的处理这么大的数字.使用 strtol [

atoi[^] is not appropriate for dealing with such big numbers. Use strtol[^] instead.

但是当我在Windows环境中运行该代码时.那么它运行得很好...

But when i run that code in Windows environment. then its running perfectly fine...

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