如何在C中将多字符常量转换为整数? [英] How to convert multi-character constant to integer in C?

问题描述

如何将x中的多字符常量转换为整数?

How to convert multi-character constant in x to integer?

例如，我尝试将 '13'设置为('3'+'1'<<<< 3)，但是它无法正常工作.我的意思不是"0123" ，而是'0123'.它可以编译，但是我不知道编译器在打印时如何获得八进制结果6014231063.我不是在寻找 atoi ，它只是将其转换为当前数字.例如， int x ='1'将以十进制数字系统显示49.现在，我对打印 int x ='0123'的内容很感兴趣.该任务来自编程竞赛，因此答案不应该是意外行为.

I tried for example '13' as ('3' + '1' << 3), but it doesn't work properly. I don't mean "0123", but '0123'. It compiles, but I don't how did compiler gets the octal result 6014231063 when printing it. I am not looking for atoi which just converts this to present number. For example int x = '1' would print 49 in decimal number system. Now I am interested what would print int x = '0123'. This task is from programming competition, so the answer shouldn't be unexpected behavior.

int main(void) { 
  int x = '0123';  
  printf("%o\n", x);
  printf("%d\n", x >> 24);
  printf("%d\n", x << 8 >> 24);
  printf("%d\n", x & 0xff);  
  return 0;
}

推荐答案

如何在C语言中将多字符常量转换为整数?

How to convert multi-character constant to integer in C?

'0123'在 int 中.

int x = '0123'; 

---

'0123'是字符常量.在C语言中，这是常量形式之一，其类型为 int .很少使用它，因为它的值是实现定义的.通常取决于字节顺序和字符填充(例如ASCII)，如下所示:

---

'0123' is a character-constant. In C, this is one of the forms of a constant and it has the type of int. It is rarely used as its value is implementation-defined. It's usually the following depending on endianness and character codding (e.g. ASCII):

(('0'*256 + '1')*256 + `2`)*256 + '3' = 858927408 = 0x33323130
(('3'*256 + '2')*256 + `1`)*256 + '0' = 808530483 = 0x30313233

进一步:用它编写有用的可移植代码是一个挑战.当使用超过1个字符时，许多编码样式都会禁止使用它.

Further: It is a challenge to write useful portable code with it. Many coding styles bar it when used with more than 1 character.

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