你能帮我解决下面的代码吗？我正在尝试制作一个程序，用户需要选择以下选项：
1 - 乘法表
2-测试
当用户选择其中任何一个并且它已完成运行时，程序将提示用户选择以下内容：
1 - 另一个乘法表
2 - 另一个测试
3 - 退出
选择第二个选项时出现问题。每当我尝试退出（第3个选项）时，它都会返回第1个选项。当我选择第一个选项然后选择第三个选项时，一切正常。我该如何解决这个问题？
这是我的代码：
int c1;
int num;
int multi;
int num1, num2;
int answer;
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("Choice: ");
scanf("%d", &c1);
while (c1 != 1 && c1 != 2)
{
printf("\nInvalid selection. Please choose again.\n");
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("Choice: ");
scanf("%d", &c1);
}
if (c1 == 1)
{
do
{
printf("\nChoose a number ranging from 1 to 12.\n");
printf("Choice: ");
scanf("%d", &num);
while (num > 12)
{
printf("\nInvalid selection. Please choose again.\n");
printf("Choose a number ranging from 1 to 12.\n");
printf("Choice: ");
scanf("%d", &num);
}
printf("\nMultiply of %d\n", num);
for (multi = 1; multi <= 12; multi++)
{
printf("%d x %d = %d\n", num, multi, num * multi);
}
printf("\nNext program?\n");
printf("1 - Another multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d", &c1);
while (c1 != 1 && c1 != 2 && c1 != 3)
{
printf("\nInvalid selection. Please choose again.\n");
printf("1 - Another multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d", &c1);
}
} while (c1 == 1);
}
if (c1 == 2)
{
do
{
printf("\nEnter 1st number: ");
scanf("%d", &num1);
printf("Enter 2nd number: ");
scanf("%d", &num2);
printf("Your answer is: ");
scanf("%d", &answer);
if (answer == num1 * num2)
{
printf("\nYou are correct. The answer for %d times %d is %d.", num1, num2, num1 * num2);
}
else if (answer != num1 * num2)
{
printf("\nYou are incorrect. The answer for %d times %d is %d, not %d.", num1, num2,
num1 * num2, answer);
}
printf("\nNext program?\n");
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d", &c1);
while (c1 != 1 && c1 != 2 && c1 != 3)
{
printf("\nInvalid selection. Please choose again.\n");
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d", &c1);
}
do
{
printf("\nChoose a number ranging from 1 to 12.\n");
printf("Choice: ");
scanf("%d", &num);
while (num > 12)
{
printf("\nInvalid selection. Please choose again.\n");
printf("Choose a number ranging from 1 to 12.\n");
printf("Choice: ");
scanf("%d", &num);
}
printf("\nMultiply of %d\n", num);
for (multi = 1; multi <= 12; multi++)
{
printf("%d x %d = %d\n", num, multi, num * multi);
}
printf("\nNext program?\n");
printf("1 - Another multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d", &c1);
while (c1 != 1 && c1 != 2 && c1 != 3)
{
printf("\nInvalid selection. Please choose again.\n");
printf("1 - Another multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d", &c1);
}
} while (c1 == 1);
} while (c1 == 2);
}
答案 0 :(得分：1)
C中的 do..while 循环在检查前执行一次。在第二个 if ：“if（c1 == 2）”中，您的代码首先执行测试，然后再次要求选择。即使您的答案为“3”，它也会执行以下 do..while 部分中的代码，这部分实际上是“乘法表”的代码。
您可以使用 while 循环代替 do..while 。但是代码看起来过于复杂，有些部分会重复我建议您查看逻辑（还考虑使用开关语句）。
答案 1 :(得分：0)
找到解决方案。显然，while while循环不能很好地工作。不得不改为while循环。这是代码：
int c1;
int num;
int multi;
int num1,num2;
int answer;
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("Choice: ");
scanf("%d",&c1);
while(c1!=1 && c1!=2){
printf("\nInvalid selection. Please choose again.\n");
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("Choice: ");
scanf("%d",&c1);
}
while(c1==1 || c1==2){
if(c1==1){
printf("\nChoose a number ranging from 1 to 12.\n");
printf("Choice: ");
scanf("%d",&num);
while(num>12){
printf("\nInvalid selection. Please choose again.\n");
printf("Choose a number ranging from 1 to 12.\n");
printf("Choice: ");
scanf("%d",&num);
}
printf("\nMultiply of %d\n",num);
for(multi=1;multi<=12;multi++){
printf("%d x %d = %d\n",num,multi,num*multi);
}
}else if(c1==2){
printf("\nEnter 1st number: ");
scanf("%d",&num1);
printf("Enter 2nd number: ");
scanf("%d",&num2);
printf("%d times %d equals to: ",num1,num2);
scanf("%d",&answer);
if(answer==num1*num2){
printf("Your answer is correct! %d times %d equals to %d.",num1,num2,num1*num2);
}else if(answer!=num1*num2){
printf("Your answer is incorrect! %d times %d equals to %d, not %d.",num1,num2,num1*num2,answer);
}
}
printf("\n\nNext program?\n");
printf("1 - Another multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d",&c1);
while(c1!=1 && c1!=2 && c1!=3){
printf("\nInvalid selection. Please choose again.\n");
printf("1 - Another multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
scanf("%d",&c1);
}
}
答案 2 :(得分：0)
您不应该依赖于从scanf()获取正确的输入，更安全的选择是使用fgets()读取输入并使用atoi()将值转换为int。
int getValidNumberInput(void)
{
char buf[BUFSIZ];
int value = 0;
if (fgets(buf, sizeof(buf), stdin) != NULL)
value = atoi(buf);
return value;
}
atoi()函数返回0表示非数值，因此非常适合您的需要。
正如评论中所提到的，使用while()循环和switch()语句可以轻松简化程序的逻辑。
int main(void)
{
bool hasLooped = false;
bool invalidChoice = false;
int choice;
while (1)
{
/* Display the relevant menu */
displayProgramMenu(hasLooped, invalidChoice);
/* Get the selected menu choice */
choice = getValidNumberInput();
if (choice < 1 || choice > 3)
invalidChoice = true;
else
invalidChoice = false;
/* Perform the chosen action */
switch (choice)
{
case 1:
doFirstOption();
hasLooped = true;
break;
case 2:
doSecondOption();
hasLooped = true;
break;
case 3:
exit(0);
break;
default:
break;
}
}
}
使用布尔变量可以将复杂性转移到易于推理的函数中。
void displayMenu(bool hasLooped, bool showInvalidMsg)
{
if (showInvalidMsg)
printf("\nInvalid selection. Please choose again.\n");
else if (hasLooped)
printf("\nNext program?\n");
if (hasLooped)
printf("1 - Another multiplication table\n");
else
printf("1 - Multiplication table\n");
printf("2 - Test\n");
printf("3 - Exit\n");
printf("Choice: ");
}