gcc（TDM-GCC）中的无符号整数溢出错误？

Question

#include <iostream>
#include <climits>
#include <cinttypes>
using namespace std;

int main()
{
    uint16_t i = 0;
    cout << USHRT_MAX << '\n' << i - 1 << '\n';
    return 0;
}

Output

65535
-1

I expected two equal outputs, but it wasn't. Isn't this a non-standard-compliant behaviour?

*System: Windows7

*Compile Option: g++ -o $(FileNameNoExt) $(FileName) -std=c++11 -Wall -Wextra

Answer 1

When C++ sees the expression

i - 1

it automatically promotes i and 1 to int types, so the result of the expression is an int , hence the output of -1.

To fix this, either cast the overall result of the expression back to uint16_t , or do something like

i--;

to modify i in-place, then print i .

Hope this helps!

Answer 2

i is promoted to an int before the evaluation of i - 1 , so the expression i - 1 is itself evaluated as a signed integer ( int ), try :

cout << USHRT_MAX << '\n' << (uint16_t)(i - 1) << '\n';

Live Demo