指定一个数字字面8位？

问题描述

unsigned char ascii;
int a = 0;
char string[4] = ; "A1"
ascii = (string[a] - 'A' + 10) * 16;

警告：转换为无符号的字符
  从'诠释'可能会改变其值

似乎GCC考虑字符和数字面值为int默认。我知道我可以只投的前pression到（无符号字符），但我怎么可以指定字符的文字和数字文字为8位，但不强制转换？

It seems that gcc considers chars and number literals as int by default. I know I could just cast the expression to (unsigned char) but how can I specify char literals and number literals as 8 bit without casts ?

有一个类似的问题：

文字部分被认为是默认双，但他们可以指定浮动方式：

Literal fractions are considered double by default but they can be specified to float by:

3.1f

因此​​，3.1将被视为浮动，而不是双。

Therefore, 3.1 would be considered a float rather than a double.

推荐答案

在 C ，你不能做计算，在什么比 INT

In C, you cannot do calculations in anything shorter than int

char a = '8' - '0'; /* '8' is int */
char a = (char)'8' - '0'; /* (char)'8' is converted to `int` before the subtraction */
char a = (char)'8' - (char)'0'; /* both (char)'8' and (char)'0' are converted */

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