诠释为char铸造 [英] int to char casting

问题描述

int i = 259;       /* 03010000 in Little Endian ; 00000103 in Big Endian */
char c = (char)i;  /* returns 03 in both Little and Big Endian?? */

在我的电脑03分配到char C ，我有小端，但我不知道，如果焦炭铸造读取至少显著字节或读取由指出字节我变量。

In my computer it assigns 03 to char c and I have Little Endian, but I don't know if the char casting reads the least significant byte or reads the byte pointed by the i variable.

推荐答案

字节序实际上并没有改变任何东西。它不会尝试存储其中一个字节（MSB，LSB等等）。

Endianness doesn't actually change anything here. It doesn't try to store one of the bytes (MSB, LSB etc).

• 如果字符是无符号，将环绕。假设8位字符259％256 = 3


• 如果字符签署结果是实现默认。谢谢 PMG ：6.3.1.3/3在C99标准

• If char is unsigned it will wrap around. Assuming 8-bit char 259 % 256 = 3
• If char is signed the result is implementation default. Thank you pmg: 6.3.1.3/3 in the C99 Standard

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