C语言 我可以用 printf() 显示枚举的值吗？
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原文地址: http://stackoverflow.com/questions/2161790/
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Can I display the value of an enum with printf()?
提问by Pieter
Is there a one-liner that lets me output the current value of an enum?
是否有一个单行可以让我输出枚举的当前值？
采纳答案by bmargulies
As a string, no. As an integer, %d.
作为一个字符串，不。作为整数，%d。
Unless you count:
除非你算：
static char* enumStrings[] = { /* filler 0's to get to the first value, */
"enum0", "enum1",
/* filler for hole in the middle: ,0 */
"enum2", "enum3", .... };
...
printf("The value is %s\n", enumStrings[thevalue]);
This won't work for something like an enum of bit masks. At that point, you need a hash table or some other more elaborate data structure.
这不适用于位掩码枚举之类的东西。此时，您需要一个哈希表或其他一些更复杂的数据结构。
回答by Matthieu
enum MyEnum
{ A_ENUM_VALUE=0,
B_ENUM_VALUE,
C_ENUM_VALUE
};
int main()
{
printf("My enum Value : %d\n", (int)C_ENUM_VALUE);
return 0;
}
You have just to cast enum to int !
Output : My enum Value : 2
您只需将 enum 转换为 int ！
输出：我的枚举值：2
回答by DrAl
The correct answer to this has already been given: no, you can't give the name of an enum, only it's value.
已经给出了正确答案：不，您不能给出枚举的名称，只能给出它的值。
Nevertheless, just for fun, this will give you an enum and a lookup-table all in one and give you a means of printing it by name:
尽管如此，只是为了好玩，这将为您提供一个枚举和一个查找表，并为您提供一种按名称打印它的方法：
main.c:
主文件：
#include "Enum.h"
CreateEnum(
EnumerationName,
ENUMValue1,
ENUMValue2,
ENUMValue3);
int main(void)
{
int i;
EnumerationName EnumInstance = ENUMValue1;
/* Prints "ENUMValue1" */
PrintEnumValue(EnumerationName, EnumInstance);
/* Prints:
* ENUMValue1
* ENUMValue2
* ENUMValue3
*/
for (i=0;i<3;i++)
{
PrintEnumValue(EnumerationName, i);
}
return 0;
}
Enum.h:
枚举.h：
#include <stdio.h>
#include <string.h>
#ifdef NDEBUG
#define CreateEnum(name,...) \
typedef enum \
{ \
__VA_ARGS__ \
} name;
#define PrintEnumValue(name,value)
#else
#define CreateEnum(name,...) \
typedef enum \
{ \
__VA_ARGS__ \
} name; \
const char Lookup##name[] = \
#__VA_ARGS__;
#define PrintEnumValue(name, value) print_enum_value(Lookup##name, value)
void print_enum_value(const char *lookup, int value);
#endif
Enum.c
枚举文件
#include "Enum.h"
#ifndef NDEBUG
void print_enum_value(const char *lookup, int value)
{
char *lookup_copy;
int lookup_length;
char *pch;
lookup_length = strlen(lookup);
lookup_copy = malloc((1+lookup_length)*sizeof(char));
strcpy(lookup_copy, lookup);
pch = strtok(lookup_copy," ,");
while (pch != NULL)
{
if (value == 0)
{
printf("%s\n",pch);
break;
}
else
{
pch = strtok(NULL, " ,.-");
value--;
}
}
free(lookup_copy);
}
#endif
Disclaimer: don't do this.
免责声明：不要这样做。
回答by BlueTrin
Some dude has come up with a smart preprocessor idea in this post
一些家伙在这篇文章中提出了一个聪明的预处理器想法
回答by BlueTrin
enum A { foo, bar } a;
a = foo;
printf( "%d", a ); // see comments below
回答by Alex Bondor
I had the same problem.
我有同样的问题。
I had to print the color of the nodes where the color was: enum col { WHITE, GRAY, BLACK };and the node: typedef struct Node { col color; };
我必须打印颜色所在节点的颜色：enum col { WHITE, GRAY, BLACK };和节点：typedef struct Node { col color; };
I tried to print node->colorwith printf("%s\n", node->color);but all I got on the screen was (null)\n.
我试图打印node->color，printf("%s\n", node->color);但我在屏幕上看到的只有(null)\n.
The answer bmarguliesgave almost solved the problem.
bmargulies给出的答案几乎解决了问题。
So my final solution is:
所以我的最终解决方案是：
static char *enumStrings[] = {"WHITE", "GRAY", "BLACK"};
printf("%s\n", enumStrings[node->color]);
