I can't understand the problem I am having in this simple code with datatypes?

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I think it is related to the bit width of long in the particular machine.

int a;
	long b;
	char c;
	float d;
	double e;
	
	std::cin >> a >> b >> c >> d >> e;
	std::cout << a << std::endl;
	std::cout << b << std::endl;
	std::cout << c << std::endl;
	std::cout << d << std::endl;
	std::cout << e << std::endl;

What I have tried:

It is not working on these inputs.

211916801 97592151379235457 p 19856.992 -5279235.721231465

as output i am getting is:

211916801
2147483647
╠
-1.07374e+08
-9.25596e+61

Posted 21-Jun-19 2:42am

Member NFOC

Updated 21-Jun-19 10:02am

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jsc42 · Answer 1 · 2019-06-21T03:05:00

Solution 1

I'm not a C++ programmer but ...
2147483647 is 2^31 - 1 which suggests long is 4 bytes. If you Google (for example) cpp long bit width you will find

c++ faq - What does the C++ standard state the size of int, long type to be? - Stack Overflow[^]

which I'll let you read for yourself - it gives all the clues that you need.

Posted 21-Jun-19 3:05am

jsc42

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[no name] 21-Jun-19 16:05pm

+5

CPallini · Answer 2 · 2019-06-21T10:02:00

Quote:
97592151379235457

is too big for a 32-bit machine.
It is worth nothing cin provides some hint about, try:

C++

#include <iostream>
using namespace std;

int main()
{

  int a;
  long b;
  char c;
  float d;
  double e;

  cout << "size_of long = " << sizeof(long) << endl;

  cin >> a;
  cout << boolalpha;
  cout << cin.good() << endl;

  cin >> b;
  cout << cin.good() << endl;

  cin >> c;
  cout << cin.good() << endl;

  cin >> d;
  cout << cin.good() << endl;

  cin >> e;
  cout << cin.good() << endl;

  cout << a << endl;
  cout << b << endl;
  cout << c << endl;
  cout << d << endl;
  cout << e << endl;
}