c++ - How does floating-point arithmetic work when one is added to a big number? -

if run code:

 #include <iostream>  int main ()  {       using namespace std;      float = 2.34e+22f;      float b = a+1.0f;        cout<<"a="<<a<<endl;      cout<<"b-a"<<b-a<<endl;      return 0;  } 

then result 0, because float number has 6 significant digits. float number 1.0 tries added 23 digit of number. so, how program realizes there no place number 1, algorithm?

step step:

ieee-754 32-bit binary floating-point format:

     sign         1 bit     significand 23 bits     exponent     8 bits 

i) float = 23400000000.f;

convert 23400000000.f float:

 23,400,000,000 = 101 0111 0010 1011 1111 1010 1010 0000 00002                = 1.01011100101011111110101010000000002 • 234. 

but significand can store 23 bits after point. must round:

   1.01011100101011111110101 010000000002 • 234 ≈ 1.010111001010111111101012 • 234 

so, after:

float = 23400000000.f;

a equal 23,399,991,808.

ii) float b = + 1;

 = 101011100101011111110101000000000002. b = 101011100101011111110101000000000012   = 1.01011100101011111110101000000000012 • 234. 

but, again, significand can store 23 binary digits after point. must round:

   1.01011100101011111110101 000000000012 • 234 ≈ 1.010111001010111111101012 • 234 

so, after:

float b = + 1; 

b equal 23,399,991,808.

iii) float c = b - a;

 101011100101011111110101000000000002 - 101011100101011111110101000000000002 = 0 

this value can stored in float without rounding.

so, after:

float c = b - a; 

с equal 0.