按位移位将无符号字符提升为int [英] bitwise shift promotes unsigned char to int

问题描述

以下代码:

unsigned char result;
result = (result << 4 );

使用gcc版本4.6.4(Debian 4.6.4-2)进行编译，带有-Wconversion标志会导致警告

Compiled with gcc version 4.6.4 (Debian 4.6.4-2), with the -Wconversion flag results in the warning

警告:从'int'转换为'unsigned char'可能会更改其值[-Wconversion]

warning: conversion to 'unsigned char' from 'int' may alter its value [-Wconversion]

那是为什么?

推荐答案

因为该标准如此规定.二进制运算符的操作数 进行整体提升，其中任何小于 int升级为int;手术的结果有 还要键入int.如果说原始价值是 0x12，结果将是0x120，并将其分配给 unsigned char将导致值更改. (分配 值将为0x20.)发出警告.

Because the standard says so. The operands to binary operators undergo integral promotion, in which anything smaller than an int is promoted to int; the results of the operation have type int as well. And if the original value were, say, 0x12, the results would be 0x120, and assigning this to an unsigned char will cause a change in value. (The assigned value will be 0x20.) Whence the warning.

从标准(第5.8节Shift运算符): 属于整数或无范围枚举类型且为整数 进行促销.结果的类型是 提升为左操作数."与其他二进制运算符不同的是， no 努力从两个运算符中找到一个通用类型: 结果类型是左操作数句点的类型.但不可或缺 促销确实仍然存在:unsigned char将是 提升为int(如果int的大小为，则提升为unsigned int 1).

From the standard (§5.8 Shift operators): " The operands shall be of integral or unscoped enumeration type and integral promotions are performed. The type of the result is that of the promoted left operand." Unlike other binary operators, there is no effort to find a common type from the two operators: the result type is that of the left operand, period. But integral promotion does still occur: the unsigned char will be promoted to int (or to unsigned int if int has a size of 1).

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