C ++中数学计算中的内部时间变量

Question

The following example is used to illustrate my question:

#include <iostream>
#include <string>

int main()
{

  signed char p;
  signed char temp=100;
  signed char t=4;
  p = (temp+temp+temp+temp)/t;
  std::cout << "Hello, " << int(p)<< "!\n";
}

In the above codes, variable p is defined as the average of four singed char variables. However, the sum of the signed char variable (temp+temp+temp+temp) will be larger than the range of signed char. So my question is how C++ handle this situation.

Answer 1

However, the sum of the signed char variable (temp+temp+temp+temp) will be larger than the range of signed char.

That does not matter as char will be promoted to int due to integral promotion . Details can be found here . So operations will be done over type int and you will get expected result.

Answer 2

Nothing happens, because of integral promotion

Prvalues of small integral types (such as char) may be converted to prvalues of larger integral types (such as int). In particular, arithmetic operators do not accept types smaller than int as arguments, and integral promotions are automatically applied after lvalue-to-rvalue conversion, if applicable. This conversion always preserves the value.

(temp+temp+temp+temp) will return an integer.

(temp+temp+temp+temp)/t will be inside of the char range.

so p == temp