[英]Why does this code work to convert hexadecimal to decimal
This code will convert one hexadecimal digit into a decimal value. 此代码将一个十六进制数字转换为十进制数值。
int value;
// ch is a char variable holding a hexadecimal digit
if (isxdigit(ch))
if (isdigit(ch))
value = ch - '0';
else
value = tolower(ch) - 'a' + 10;
else
fprintf(stderr, "%c is not a valid hex digit", ch);
I don't fully understand how it works though. 我不完全理解它是如何工作的。 I can see that different things are subtracted from the char variable depending on whether it is a number or a letter. 我可以看到从char变量中减去不同的东西,具体取决于它是数字还是字母。 I can understand the part where a number gets converted, but I don't understand why 10 has to be added to the value when the character is a letter. 我可以理解数字转换的部分,但我不明白为什么当字符是字母时必须将10添加到数值中。
The subtraction of tolower(ch) - 'a'
will map the character to a number in the range 0..5 for the letters a..f. tolower(ch) - 'a'
的减法会将字符映射到字母a..f的0..5范围内的数字。 However, the (decimal) value of the hexadecimal digit a 16 is 10 10 , so to move the range back up to 10..15 where it needs to be, 10 is added. 但是,十六进制数字a 16的(十进制)值是10 10 ,因此要将范围移回到需要的10..15,添加10。
Perhaps this helps: 也许这有助于:
+---------+------------+-----------------+-------------+
Character | Subtracted | Resulting value | Digit value |
+---------+------------+-----------------+-------------+
| '0' | '0' | 0 | 0 |
| '1' | '0' | 1 | 1 |
| '2' | '0' | 2 | 2 |
| '3' | '0' | 3 | 3 |
| '4' | '0' | 4 | 4 |
| '5' | '0' | 5 | 5 |
| '6' | '0' | 6 | 6 |
| '7' | '0' | 7 | 7 |
| '8' | '0' | 8 | 8 |
| '9' | '0' | 9 | 9 |
| 'a' | 'a' | 0 | 10 |
| 'b' | 'a' | 1 | 11 |
| 'c' | 'a' | 2 | 12 |
| 'd' | 'a' | 3 | 13 |
| 'e' | 'a' | 4 | 14 |
| 'f' | 'a' | 5 | 15 |
+---------+------------+-----------------+-------------+
Notice how the "resulting value" column resets back to 0 at 'a', which is not where it needs to be according to the final "digit value" column, which shows each hexadecimal digit's value in decimal. 注意“结果值”列如何在'a'处重置为0,这不是根据最终“数字值”列所需的位置,该列显示每个十六进制数字的十进制数值。
The expression ch - '0'
works because in C "the value of each character after 0 ... shall be one greater than the value of the previous" ( C99 section 5.2.1). 表达式ch - '0'
起作用,因为在C“中,0之后的每个字符的值应比前一个值大1”( C99第5.2.1节)。
So, for example, the value of the character '3'
is 3 greater than the value of '0'
, so when you subtract these two values, you get the integer 3. 因此,例如,字符'3'
的值比'0'
的值大3,因此当您减去这两个值时,您将得到整数3。
The expression tolower(ch) - 'a' + 10
works by luck, because, except for the above constraint for digits, all values of characters are implementation-defined. 表达式tolower(ch) - 'a' + 10
运气很好,因为除了上面的数字约束外,所有字符值都是实现定义的。
So when you subtract 'c' - 'a'
you get 2 (and, adding 10, you get 12 – the correct value of that digit), because most computers work in ASCII or EBCDIC. 因此,当你减去'c' - 'a'
你得到2(并且,加上10,你得到12 - 该数字的正确值),因为大多数计算机都使用ASCII或EBCDIC。 But when you run this program on the DS9K, you might get −42. 但是当你在DS9K上运行这个程序时,你可能得到-42。
To be truly portable you would need to compare ch
to each of the six letters in turn. 要真正便携,您需要依次将ch
与六个字母中的每个字母进行比较。 That's why some systems provide a digittoint()
function. 这就是为什么有些系统提供了digittoint()
函数。
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