C++中int转char时123456为何被转换为ASCII值64？

When you assign an integer value larger than char's range to a char variable, the conversion follows these clear rules:

• Char size context: On most systems, char occupies 8 bits. If unsigned, it holds values from 0–255; if signed, from -128–127. Either way, 123456 is far outside this range.
• Modulo arithmetic is the key: For unsigned char, the C++ standard requires converting larger integers using modulo 2^8 (256). For signed char, pre-C20 this behavior was implementation-defined, but nearly all compilers used the same modulo logic—and C20 standardized this wrapping behavior for signed types too.

Let's break down the math for 123456:

• Calculate how many times 256 fits into 123456: 256 × 482 = 123392
• Subtract that product from the original value: 123456 - 123392 = 64
  This remainder is the value stored in your char variable. Since 64 falls within both signed and unsigned char ranges, the result is consistent either way.

Why only a warning, not an error? #

C++ allows implicit narrowing conversions (like int to char when the value is out of range) but flags them as risky because they discard data. Compilers issue a warning (like -Wconstant-conversion here) to alert you to the potential problem, but don't block compilation by default. If you want to enforce errors for such cases, use flags like -Werror to turn warnings into errors, or use an explicit cast to make your intent clear.

The ASCII value 64 maps to the character '@', which is what your cout statement outputs.

内容的提问来源于stack exchange，提问作者Tarun Malviya