如何检查变量是否等于多个值 [英] How to check if variable equal to multiple values

问题描述

在C中我想检查变量是否等于多个值，并且不完全分离也不知道如何编码.

In C I want to check if variable equal to multiple values and I don't know how to code it without separating it fully.

if(str [i] =='u'||'o'||'i'|| e'||'a')始终使我为真，而我没有不明白为什么，我需要解释.

if (str[i]=='u'||'o'||'i'||'e'||'a') giving me always true and I don't understand why, I need explanation.

if(str [i] ==('u'||'o'||'i'||'e'||'a'))给我总是假，我不明白为什么，我需要解释.

if (str[i]==('u'||'o'||'i'||'e'||'a')) giving me always false and I don't understand why, I need explanation.

谢谢.

推荐答案

不同的结果与运算符优先级有关.

The different results has to do with operator precedence.

x == y || z

与

(x == y) || z

与

x == (y || z)    

您的表达式为'u'||'o'||'i'||'e'||'a'，因此在我们的情况下为 y 将为'u'和 z 为'o'||'i'||'e'||'a'. z 的计算结果为true，因为至少一个操作数(在本例中为所有操作数)为非零.因此，第一行等效于(str [i] =='u')||1 当然会一直取值为1，这是正确的.另一方面， str [i] ==('u'|| 1)与 str [i] == 1 相同，因为'u'||1 等于1.

You have the expression 'u'||'o'||'i'||'e'||'a' so in our case, y will be 'u' and z will be 'o'||'i'||'e'||'a'. z will evaluate to true, because at least one of the operands (all of them in this case) is non-zero. So the first line will be equivalent to (str[i] == 'u') || 1 which of course always will evaluate to 1, which is true. On the other hand, str[i] == ('u' || 1) is the same as str[i] == 1 because 'u' || 1 will evaluate to 1.

在C语言中没有很好的内置方法来完成这样的事情.您可以做的，很容易概括的是编写这样的自定义函数:

There is no good built in way to do such a thing in C. What you could do, that is pretty easy to generalize is to write a custom function like this:

bool isMember(char e, char*s, size_t size)
{
    for(size_t i; i<size; i++) {
        if(s[i] == e)
            return true;
    }
    return false;
}

上述功能很容易针对不同类型进行修改.但是在您的情况下，它可以像这样使用:

The above function is easy to modify for different types. But in your case it can be used like this:

char characters[] = {'u','o','i','e','a'};
if (isMember(str[i], characters, sizeof(characters)) {

在处理 char 时，有一些更简单的方法，但我选择此解决方案是因为它不仅限于 char .

When dealing with char there are somewhat easier methods, but I chose this solution because it is not restricted to char.

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