是否将负数打印为C中定义良好的字符？ [英] Is printing negative numbers as characters well-defined in C?

问题描述

例如，以下代码：

#include <stdio.h>
#include <limits.h>

int main(void)
{
    char a;
    signed char b;
    for(a = CHAR_MIN, b = CHAR_MIN; a < CHAR_MAX ; a++, b++ )
    printf("%c %c\n", a, b);
}

输出：

! !
" "
# #
$ $
% %
& &
' '
( (
……

当 a 和 b 为负，仍然在屏幕上打印字符。我想知道这种行为是否定义明确吗？

When a and b are negative, characters are still printed on the screen. I wonder whether this behaviour is well-defined?

如果是，是由标准定义还是由特定实现定义？定义这种行为的意义何在？

If so, is it defined by the standard or an specific implementation? And what's the point of defining such behaviour?

推荐答案

是

传递值以匹配％c 是由于 int 或 unsigned 作为通常的整数促销的一部分> ... 自变量 printf（）。

When a value is passed to match "%c" is it converted to an int or unsigned as part of the usual integer promotions due to the ... of the printf() arguments.

当 printf（）看到 int 的值，它将转换为 unsigned char

When printf() see that int value, it converts to to an unsigned char. Then the corresponding character is printed.

c 如果没有 l 长度修饰符存在， int 参数转换为 unsigned char ，并生成c写作。 C11dr§7.21.6.18

c If no l length modifier is present, the int argument is converted to an unsigned char, and the resulting character is written. C11dr §7.21.6.1 8

因此，传递任何提升为 int 甚至在 INT_MAX 范围内的 unsigned 都不是问题。

So passing any narrow type integer that is promoted to int or even an unsigned within the range of INT_MAX is not a problem.

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