代码将永远不会执行C

-3

我有这个问题，它说，如果（knightATKC == 2 & & knightATKC == 4）我说，代码将永远不会执行。它将无法正常工作。如果有人知道如何解决这个问题，那会很好。不要担心你可能会看到的未使用的变量，我将尽快使用它，但现在我不知道如何修复代码将永远不会执行。对不起，我可怜的英语：/代码将永远不会执行C

#include <stdio.h> 
#include <stdlib.h> 
#include <stdbool.h> 
#include <unistd.h> 

int main() 
{ 
char enter = 0; 
char pNAME[30]; 
char pGENDER; 
int pAGE; 
int pATK; 
int pHP = 20; 
int pATKC; 
int knightATK = 3; 
int knightHP = 15; 
int knightATKC = 0; 
int kingATK = 5; 
int kingHP = 30; 
int kingATKC; 

printf("While you are enjoying your breakfast at your house, suddenly you get teleported to a different dimension and no one to be seen except an old man.\n"); 
printf("\nOld Man: What is your name warrior?\n"); 
printf("\nInsert name:\n"); 
scanf("%s",pNAME); 
fpurge(stdin); 

printf("\nInsert gender (m/f):\n"); 
scanf("%c",&pGENDER); 
fpurge(stdin); 
while (pGENDER != 'f' &&pGENDER != 'm') 
{ 
    printf("\nInvalid entry, please try again.\n"); 
    printf("\nInsert gender (m/f):\n"); 
    scanf("%c",&pGENDER); 
} 

printf("\nOld Man: Now how old are you young warrior?\n"); 
printf("\nInsert Age:\n"); 
scanf("%d",&pAGE); 
fpurge(stdin); 
if (pAGE < 18) 
{ 
    printf("\nOy mate! No stalker!\n"); 
    printf("\nTeleporting back to reality...\n"); 

    return 0; 
} 

if (pAGE > 80) 
{ 
    printf("\nNo oldies allowed!\n"); 
    printf("\nTeleporting back to reality...\n"); 

    return 0; 
} 

printf("\nOld Man: Welcome warrior I fear the knights has taken over our kingdom, and you, %s, are the only one who can save us.\n",pNAME); 
printf("\n%s: How am I suppose to do that?[Enter]\n",pNAME); 
if (enter != '\r' && enter != '\n') { enter = getchar(); } 
printf("HI"); 

while (pHP != 0 && knightHP != 0) 
{ 
    pATK = 5; 
    knightATK = 3; 

    knightATKC = (rand()%5); 

    if (knightATKC == 2 && knightATKC == 4) 
    { 
     printf("The knight swings his sword at %s",pNAME); 
     printf("T\nhe knight hits %s for 3HP\n",pNAME); 
     pHP = pHP - 3; 
    } 
    else if (knightATKC == 3) 
    { 
     printf("\nThe knight slashes his sword at %s\n",pNAME); 
     printf("\nThe knight critically hit %s for 5 HP\n",pNAME); 
     pHP = pHP - 5; 
    } 
    else 
    { 
     printf("\nThe knight swings his sword at %s\n",pNAME); 
     printf("\nThe knight fails to hit %s\n",pNAME); 
     printf("\nPress [Enter] to attack\n"); 
    } 

}}

来源

2015-08-30 S.P.

如果你想要执行代码，不要求一个永远不会成立的条件！ –

@DiegoBasch有时开发人员会犯简单的编码错误。有时候会有混乱。有时候我们会给这个人带来怀疑的好处。 – zaph

你的问题是不言自明的。 knightATKC == 2 && knightATKC == 4表示knightATKC同时是2和4，这是不可能的。也许你的意思是||（布尔或）而不是&&（布尔和）？

来源

2015-08-30 01:51:32

坏声明：

if (knightATKC == 2 && knightATKC == 4)

这整个如果因为&&运营商“和”声明绝不是真实的，因此第一比较和第二都必须是真实的。但是，如果knightATKC等于2那么knightATKC不能等于4。

也许你想要的是 “或”

if (knightATKC == 2 || knightATKC == 4)

如果以上任何一种情况。

来源

2015-08-30 01:52:13 zaph

代码将永远不会执行C

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