在C中将单个字符（十六进制数字）转换为整数 [英] Convert single Character (hex number) to Integer in C

问题描述

因此，对于作业，我必须将字符（ 0 - F ）转换为整数（ 0 - 15 ）， 0 - 9 可以正常工作，但是如果给出任何字母，它会打印一个随机数：例如，对于 C ，它给出 19 ，对于 D 则返回 20 。

So for an assignment I have to convert a character (0-F) to an integer (0-15), 0-9 works fine, but if any letter is given, it prints a random number: For C for instance, it gives 19, for D is returns 20.

这是我的方法：

int char2int(char digit) {
    int i = 0;

    if (digit == 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9)
        i = digit - '0';
    else
    if (digit == 'A' || 'B' || 'C' || 'D' || 'E' || 'F')
        i = digit - '9';
    else
        i = -1;

    return i;
}

起初，我的if语句是这样的：

At first my if statements were like this:

if (digit => 0 && =< 9)

if (digit => A && =< F)

但这带来了许多错误。你可以说我不太了解C。我当前的If语句有效，但是我确定它的长度不必要。

But that gave a number of errors. You can tell I don't know C very well. My current If statement works but I'm sure it's unnecessarily long.

推荐答案

if（digit = = 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9）

if (digit == 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9)

条件表达式在C中不是这样。

This is not how conditional expressions work in C.

您需要将数字与每个数字进行比较单独的数字

You either need to compare digit against each of the numbers individually

if (digit == '0' || digit == '1' || digit == '2' ...

或巧妙地做到这一点：

if(digit >= '0' && digit <= '9')
                         ^^ not =<

请注意，由于要比较，我在数字周围加上了' 字母0 而不是数字的数字（这是不相同的，请参见此处）所有ASCII字符值）。

Notice that I put ' around the numbers because you want to compare the digit with the letter 0 and not the number (which is not the same see here for all the ASCII character values).

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