C中是，否答案

我正在尝试制作一个C程序，我可以询问以下用户是或否问题：

Do you like beer? Y or N Are you old enough? Y or N How old are you? 

如果用户说超过18岁打印消息： let's go for some beers

但如果第一个问题中的一个或年龄是N ，请打印： sorry maybe next time

我认为问题在于If语句，也许我没有封装条件。

 #include  int main(){ char answer; int age = 0 ; printf("Do you like beers Enter Y or N: \n"); scanf(" %c", &answer); printf("\n so your answer is %c\n", answer); printf("Are you old enough to have a beer?\n"); scanf(" %c", &answer); printf("\n so your answer is %c\n", answer); if (answer == 'Y') { printf("how old are you?\n"); scanf(" %d", &age); if (age >= 18) printf("\n Let's go for some beers , I will pay the first round \n"); }if else (age < 18 && answer == 'N') printf("\n sorry my friend , maybe next time \n"); // printf("You may NOT "); return 0; } 

Answer 1

问题是你的最后一个if语句只有在用户未满18岁时才是真的并且说不是你想要它是或者。

从中更改最后一个if语句

 if else (age < 18 && answer == 'N') 

至：

  if else (age < 18 || answer == 'N') 

Answer 2

从上面的代码片段看，你的else语句看起来格式不正确（ if else代替else if ）。 此外，由于您正在测试这两个问题是否都是错误的，您应该使用|| 操作数。 所以你想要做的事情如下：

 else if (age < 18 || answer == 'N') 

Answer 3

您正在使用相同的变量“answer”。 因此，当您输入第二个问题的答案时，它将替换第一个答案。

  int main(){ char answer; char answer2; int age = 0 ; printf("Do you like beers Enter Y or N: \n"); scanf(" %c", &answer); printf("\n so you answer is %c\n", answer); printf("Are you older enough to have a beer?\n"); scanf(" %c", &answer2); printf("\n so you answer is also %c\n", answer2); if (answer == 'Y' && answer2 == 'Y') { printf("how old are yo.\n"); scanf(" %d", &age); if (age >= 18) printf("\n lets go for some beers , I will paid the first round \n"); } else if (answer == 'N' || answer2 == 'N') printf("\n sorry my friend , maybe next time \n"); // printf("You may NOT "); return 0; } 

希望这是你需要的。 干杯。

Answer 4

我认为您可以认为程序的结构大致如下。

 REPLY = PROMPT(Q1_MESSAGE) IF(REPLY == YES){ //Narrow down the conditions REPLY = PROMPT(Q2_MESSAGE) IF(REPLY == YES){ //Narrow down the conditions REPLY = PROMPT(Q3_MESSAGE) IF(REPLY >= 18){ DISPLAY(GOOD_MESSAGE) } ELSE { DISPLAY(NO_GOOD_MESSAGE) } } ELSE { DISPLAY(NO_GOOD_MESSAGE) } } ELSE { DISPLAY(NO_GOOD_MESSAGE) } 

嵌套IF可以被视为AND条件作为条件。
因此，通过将问题消息及其响应部分汇总到函数中，可以如下编写。

 IF(FUNC1() == TRUE AND FUNC2() == TRUE AND FUNC3() == TRUE){//Function call proceeds from left to right (Shortcut evaluated) DISPLAY(GOOD_MESSAGE) } ELSE { DISPLAY(NO_GOOD_MESSAGE) } 

例如，您可以具体写如下。

 #include  int main(void){ char YN(const char *prompt); int enter_age(const char *prompt); if(YN("Do you like beers Enter Y or N: \n") == 'Y' && YN("Are you old enough to have a beer?\n") == 'Y' && enter_age("How old are you?\n") >= 20){ printf("\nLet's go for some beers, I will take the first round.\n"); } else { printf("\nSorry my friend, maybe next time.\n"); } return 0; } char YN(const char *prompt){ char ans[2], ret, ch; int ret_scnf; while(1){ fputs(prompt, stdout); if((ret_scnf = scanf(" %1[YNyn]%c", ans, &ch)) == 2 && ch == '\n'){ if(*ans == 'Y' || *ans == 'y'){ ret = 'Y'; break; } else if(*ans == 'N' || *ans == 'n'){ ret = 'N'; break; } } else if(ret_scnf == EOF){ ret = 'N'; break; } scanf("%*[^\n]");scanf("%*c");//clear input } return ret; } int enter_age(const char *prompt){ int ret_scnf, age; char ch; while(1){ fputs(prompt, stdout); if((ret_scnf = scanf("%d%c", &age, &ch)) == 2 && ch == '\n'){ if(age < 0)//Can I enter years old at the age of 0? continue; return age; } else if(ret_scnf == EOF){ age = -1; break; } scanf("%*[^\n]");scanf("%*c"); } return age; }