通过整数限定符打印字符 [英] Printing char by integer qualifier

问题描述

我正在尝试执行以下程序。

I am trying to execute the below program.

#‎include‬ "stdio.h" 
#include "string.h" 

void main()
{ 
    char c='\08'; 
    printf("%d",c); 
} 

我得到的输出为 56 。但是对于 8 以外的任何数字，输出本身就是数字，但是对于 8 ，答案是 56 。

I'm getting the output as 56 . But for any numbers other than 8 , the output is the number itself , but for 8 the answer is 56.

有人可以解释吗？

推荐答案

以 \0 代表总数字，是以8为底​​的数字系统，并使用数字 0到7 。因此 \08 是无效表示nofollow noreferrer>八进制数字，因为 8∉[0，7] ，因此您将获得实现定义的行为。

A characters that begins with \0 represents Octal number, is the base-8 number system, and uses the digits 0 to 7. So \08 is invalid representation of octal number because 8 ∉ [0, 7], hence you're getting implementation-defined behavior.

您的编译器可能识别出多字节字符 '\08'为'\0'一个字符和'8'作为另一个，并解释为'\08'为'\0' + '8'使其变为'8'。查看 ASCII 表后，您会注意到的十进制值8'是56。

Probably your compiler recognize a Multibyte Character '\08' as '\0' one character and '8' as another and interprets as '\08' as '\0' + '8' which makes it '8'. After looking at the ASCII table, you'll note that the decimal value of '8' is 56.

感谢@ DarkDust，@ GrijeshChauhan和@EricPostpischil。

Thanks to @DarkDust, @GrijeshChauhan and @EricPostpischil.

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