为何向char类型赋值256会触发编译警告，赋值255却不会，但二者均存在整数溢出？

Let's break this down step by step, focusing on C++'s type rules and compiler warning logic:

Core Background #

First, remember that C++ leaves the signedness of char as implementation-defined—it can behave like either signed char (range: -128 to 127 on 8-bit systems) or unsigned char (range: 0 to 255). The -Woverflow warning your compiler uses considers this flexibility when deciding whether to flag an assignment.

Example 2: char c = 255; #

• 255 is exactly the maximum value for unsigned char, which is a valid possible type for char. Even if your compiler treats char as signed char, 255 will wrap around to -1 (via 8-bit modulo 256 arithmetic), and -1 is within the signed char range.
• Since there exists a valid char type (unsigned) that can hold 255 without overflow, the compiler doesn't trigger a -Woverflow warning—it can't be certain the assignment is problematic across all valid C++ implementations.

Example 1: char c = 256; #

• 256 exceeds the range of both signed char (max 127) and unsigned char (max 255). No matter how char is configured on your system, this value will overflow and wrap around to 0 (256 modulo 256 = 0).
• The compiler can definitively confirm this assignment will cause an overflow regardless of implementation, so it emits the -Woverflow warning to alert you to the value change.

Why the output isn't 255 or 256? #

This ties back to your compiler's default char type (most mainstream compilers like GCC/Clang on x86 use signed char):

• For char c = 255;: The 8-bit binary 11111111 represents -1 in two's complement (the standard for signed integers), so static_cast<int>(c) outputs -1.
• For char c = 256;: 256 wraps to 0 modulo 256, so static_cast<int>(c) outputs 0.

If your compiler used unsigned char by default, Example 2 would output 255, but Example 1 would still output 0.

内容的提问来源于stack exchange，提问作者I'mNotAModerator