ASCII TABLE - 负值 [英] ASCII TABLE - negative value

问题描述

可能重复：

负数ASCII值

int main() {
    char b = 8-'3';
    printf("%c\n",b);

    return 0;
}

我运行这个程序，我得到一个看起来像问号？）。

I run this program and I get a sign which looks like a question mark (?).

我的问题是，为什么是打印，不打印什么，只要我知道b的值由ASCII表减去43

My question to you is why is it prints that and not printing nothing, beacause as far as I know the value of b by the ASCII table is minus 43 which is not exist.

by the way, when I compile this code: 

c $ c> int main（）{
char b = -16;
printf（％c \\\
，b）;

return 0;
}

int main() { char b = -16; printf("%c\n",b); return 0; }

我什么都没有。

推荐答案

以下是 C 2011标准

7.21.6.1 fprintf函数

...

8转换规范及其含义如下：

...

c 如果不存在 l 长度修改，则 int 参数被转换为
unsigned char ，并写入结果字符。


如果存在 l 长度修改符， wint_t 参数被转换为
和 ls 转换规范，没有精度，参数指向
到 wchar_t 的两个元素数组的初始元素，第一个元素
包含 wint_t cc的参数 lc 转换规范和
第二个零宽字符。

7.21.6.1 The fprintf function

...
8 The conversion specifiers and their meanings are:
...
c If no l length modifier is present, the int argument is converted to an unsigned char, and the resulting character is written.

If an l length modifier is present, the wint_t argument is converted as if by an ls conversion specification with no precision and an argument that points to the initial element of a two-element array of wchar_t, the first element containing the wint_t argument to the lc conversion specification and the second a null wide character.

该-43被转换为无符号值（213），所以它打印一个扩展的ASCII字符。

That -43 is being converted to an unsigned value (213), so it's printing an extended ASCII character.

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