为什么指针没有开关？ [英] Why no switch on pointers?

问题描述

例如：

#include <stdio.h>

void why_cant_we_switch_him(void *ptr)
{
    switch (ptr) {
        case NULL:
            printf("NULL!\n");
            break;
        default:
            printf("%p!\n", ptr);
            break;
    }
}

int main(void)
{
    void *foo = "toast";
    why_cant_we_switch_him(foo);
    return 0;
}

gcc test.c -o test
test.c: In function 'why_cant_we_switch_him':
test.c:5: error: switch quantity not an integer
test.c:6: error: pointers are not permitted as case values

只是好奇。这是一个技术上的限制？

Just curious. Is this a technical limitation?

人们似乎认为，只有一个常量指针前pression。是真的有关系吗？例如，下面是Objective-C中一种常见的范例（真的是C仅除了的NSString ， ID 和无，它仅仅是一个指针，所以它仍然是相关的 - 我只是想指出，有是的，其实是一个普遍使用它，尽管这是唯一一个技术问题）：

People seem to think there is only one constant pointer expression. Is that is really true, though? For instance, here is a common paradigm in Objective-C (it is really only C aside from NSString, id and nil, which are merely a pointers, so it is still relevant — I just wanted to point out that there is, in fact, a common use for it, despite this being only a technical question):

#include <stdio.h>
#include <Foundation/Foundation.h>

static NSString * const kMyConstantObject = @"Foo";

void why_cant_we_switch_him(id ptr)
{
    switch (ptr) {
        case kMyConstantObject: // (Note that we are comparing pointers, not string values.)
            printf("We found him!\n");
            break;
        case nil:
            printf("He appears to be nil (or NULL, whichever you prefer).\n");
            break;
        default:
            printf("%p!\n", ptr);
            break;
    }
}

int main(void)
{
    NSString *foo = @"toast";
    why_cant_we_switch_him(foo);
    foo = kMyConstantObject;
    why_cant_we_switch_him(foo);

    return 0;
}

gcc test.c -o test -framework Foundation
test.c: In function 'why_cant_we_switch_him':
test.c:5: error: switch quantity not an integer
test.c:6: error: pointers are not permitted as case values

似乎原因是开关只允许积分值（作为编译器警告所述）。所以我想一个更好的问题是要问，为什么是这样？ （虽然它可能是现在太晚了。）

It appears that the reason is that switch only allows integral values (as the compiler warning said). So I suppose a better question would be to ask why this is the case? (though it is probably too late now.)

推荐答案

switch语句只整数值进行操作。这就是为什么该错误信息是开关量不是一个整数。我不认为这是一个技术上的限制这么多，因为它是语言的语法之外。

Switch statements operate on integral values only. That's why the error message is "switch quantity not an integer." I don't think it's a technical limitation so much as it's outside the language syntax.

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