如何修复"return 0/ exit(0)在for循环内不能正常工作“？
我在做一个科学计算器。并考虑循环语句，直到用户通过使用0作为其中一个选项来请求停止。但即使在输入0之后。它最后一次询问以下语句：printf("Enter two numbers (For only one no. required you can just enter other number anything)\n");
我尝试过使用goto, exit(0)和return 0语句。甚至while(1)和for(;;)也会循环。
#include <stdio.h>
#include<math.h>
int main()
{
int a;
float b,c;
float d=3.14159/180;
while(1)
{
printf("\nScientific Calculator :\n");
printf("Enter option:\n 0- Exit, 1-Add, 2-Sub, 3-Multiply, 4-Divide,\n 5-sin(x), 6-cos(x), 7-tan(x), 8-sinh(x), 9-cosh(x), 10-tanh(x),\n11-log10(x),12-exponent,13-power of x w.r.t y \n");
scanf("%d",&a);
printf("Enter two numbers (For only one no. required you can just enter other number anything)\n"); //Here is where it starts even after return 0
scanf("%f%f",&b,&c); //Here after inputting value it ends.
switch(a)
{
case 0:return 0; //Here is the return 0;
case 1:printf("%d",(int)(b+c)); break;
case 2:printf("%d",(int)(b-c)); break;
case 3:printf("%d",(int)(b*c)); break;
case 4:printf("%f",b/c); break;
case 5:printf("%f",sin(b*d)); break;
case 6:printf("%f",cos(b*d)); break;
case 7:printf("%f",tan(b*d)); break;
case 8:printf("%f",sinh(b*d)); break;
case 9:printf("%f",cosh(b*d)); break;
case 10:printf("%f",cosh(b*d)); break;
case 11:printf("%f",tanh(b*d)); break;
case 12:printf("%f",log10(b)); break;
case 13:printf("%f",exp(b)); break;
case 14:printf("%f",pow(b,c)); break;
default:printf("Enter correct option\n");
}
}
return 0;
}我想让它退出程序，但它要求输入printf("Enter two numbers ---\n");，在输入值之后，它退出。
回答 1
Stack Overflow用户
发布于 2019-04-29 00:53:31
，但即使在输入0之后。它最后一次询问以下语句：
printf("Enter two numbers (For only one no. required you can just enter other number anything)\n");
您需要立即检查大小写“0”，因此
printf("Enter option:\n 0-退出，1-加，2-除，3-乘，4除，\n 5-正弦(X)，6-cos(x)，7-tanh(X)，8-sinh(x)，9-cosh(x)，10-tanh(x)，\n11-log10(X)，12-指数，13-x的幂w.r.t y \n")；scanf("%d"，&a)；printf(“输入两个数字(只有一个否。required您只需输入其他数字)\n“)；//即使在返回0 scanf("%f%f"，&b，&c)之后，它也会从这里开始；//在这里输入值之后，它就会结束。开关(A)情况0:返回0；//这里是返回0；情况1:printf("%d"，(int)(b+c))；break；...
必须是
printf("Enter option:\n 0- Exit, 1-Add, 2-Sub, 3-Multiply, 4-Divide,\n 5-sin(x), 6-cos(x), 7-tan(x), 8-sinh(x), 9-cosh(x), 10-tanh(x),\n11-log10(x),12-exponent,13-power of x w.r.t y \n");
scanf("%d",&a);
if (a == 0)
return 0;
printf("Enter two numbers (For only one no. required you can just enter other number anything)\n"); //Here is where it starts even after return 0
scanf("%f%f",&b,&c); //Here after inputting value it ends.
switch(a)
case 1:printf("%d",(int)(b+c)); break;
...注检查输入的int是否有效也是一个好主意，因此要检查scanf("%d",&a)是否返回1，对于scanf("%f%f",&b,&c)也是如此，因此检查它将返回2……
正如@OznOg对于案例12和13所说的，你只需要读一个数字，而不是两个。另请注意，如果代码无效，则询问号码是无用的，因此：
if (scanf("%d",&a) != 1)
return -1;
if (a == 0)
return 0;
if ((a < 0) || (a > 14))
printf("Enter correct option\n");
else if ((a == 12) || (a == 13)) {
printf("Enter one number\n");
if (scanf("%f",&b) != 1)
return -1;
if (a == 12)
printf("%f",log10(b));
else
printf("%f",exp(b)); break;
}
else {
printf("Enter two numbers\n");
if (scanf("%f%f",&b,&c) != 2)
return -1;
switch(a)
{
case 1:printf("%d",(int)(b+c)); break;
case 2:printf("%d",(int)(b-c)); break;
case 3:printf("%d",(int)(b*c)); break;
case 4:printf("%f",b/c); break;
case 5:printf("%f",sin(b*d)); break;
case 6:printf("%f",cos(b*d)); break;
case 7:printf("%f",tan(b*d)); break;
case 8:printf("%f",sinh(b*d)); break;
case 9:printf("%f",cosh(b*d)); break;
case 10:printf("%f",cosh(b*d)); break;
case 11:printf("%f",tanh(b*d)); break;
case 14:printf("%f",pow(b,c)); break;
}
}https://stackoverflow.com/questions/55892502
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