是什么导致“警告:条件表达式中的指针/整数类型不匹配"? [英] What is causing "warning: pointer/integer type mismatch in conditional expression"?

问题描述

我有一个枚举，以及一​​个全部使用该枚举的宏定义和方法.我无法编译它.请考虑以下代码.

I have an enum, and a macro definition and a method that all use the enum. I can't get it to compile. Consider the following pieces of code.

typedef enum fruits_t
{
    APPLE,
    ORANGE,
    BANANA
} fruits_t;

#define KEY_TO_VALUE(x) ((x == APPLE) ? 0 :  \
                         (x == ORANGE) ? 1 :  \
                         (x == BANANA) ? 2 : \
                         "Undefined")

static void foo(char fruit) {
    if (fruit == KEY_TO_VALUE(APPLE)) {
        /* do something */
    }
}

这可以编译，但是我收到以下警告.

This compiles, but I get the following warnings.

warning: pointer/integer type mismatch in conditional expression

warning: comparison between pointer and integer

为什么?我对C非常陌生，因此，如果您能解释一些对经验丰富的C开发人员而言显而易见的事情，我将不胜感激.我大部分的编程知识都是基于Java的.

Why? I am very new to C, so if you could explain things that may seem obvious to an experienced C developer, I'd appreciate it. Most of my programming knowledge is Java based.

推荐答案

编译器试图找出程序中每个表达式的类型.

The compiler is trying to figure out the type of each expression in the program.

诸如x > 0 ? 5 : "no"的表达式使编译器抓狂.如果x大于零，则类型为int，但如果不大于0，则类型为const char *.这是一个问题，因为没有从指针到int的自动转换(反之亦然).因此，编译器会对此发出警告.

An expression such as x > 0 ? 5 : "no" makes the compiler scratch its head. If x is greater than zero, the type is int, but if it isn't then the type is const char *. This is a problem, because there is no automatic conversion from pointer to int (and vice versa). So the compiler warns about it.

解决方案是确保无论fruit的值是什么，KEY_TO_VALUE的值都具有单一类型.例如，可以使用诸如-1之类的特殊值来代替"Undefined"(由于它是文字字符串，其类型为const char *).

The solution is to make sure that no matter what the value of fruit is, the value of KEY_TO_VALUE has a single type. For example, instead of "Undefined" (which is of type const char *, because it is a literal string), you can use a special value such as -1.

此外，请注意，APPLE是值为0的常数，ORANGE是值为1的常数，而BANANA是值为2的常数(这是enum的工作方式).因此，您不需要KEY_TO_VALUE，因为常量已经具有所需的值.您可以直接将fruit与APPLE直接进行比较:

Also, note that APPLE is a constant with the value 0, ORANGE is a constant with the value 1 and BANANA is a constant with the value 2 (this is how enum works). So you don't need KEY_TO_VALUE, as the constants already have the desired values. You can simply compare fruit to APPLE directly:

if (fruit == APPLE) { ... }

这篇关于是什么导致“警告:条件表达式中的指针/整数类型不匹配"?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！