我试图写我的第一个计算器,并发现了一些在线的例子,然后我改变,使他们更容易在流动的条款。然而,当我改变这个流程:简单的计算器()
#include <stdio.h>
main()
{
char operator;
float num1,num2;
printf("Enter an operator (+, -, *, /): ");
scanf("%c" ,&operator);
printf("Enter first operand: ");
scanf("%f" ,&num1);
printf("Enter second operand: ");
scanf("%f" ,&num2);
switch(operator)
{
case '+':
printf("num1+num2=%.2f\n" ,num1+num2);
break;
case '-':
printf("num1-num2=%.2f\n" ,num1-num2);
break;
case '*':
printf("num1*num2=%.2f\n" ,num1*num2);
break;
case '/':
printf("num1/num2=%.2f\n" ,num1/num2);
break;
default: //of operator is other than +, -, *, /, erros message shown
printf("Error! Invalid operator, this is basic math only.\n");
}
return 0;
}
这样:
#include <stdio.h>
main()
{
char operator;
float num1,num2;
printf("Enter first operand: ");
scanf("%f" ,&num1);
printf("Enter an operator (+, -, *, /): ");
scanf("%c" ,&operator);
printf("Enter second operand: ");
scanf("%f" ,&num2);
switch(operator)
{
case '+':
printf("num1+num2=%.2f\n" ,num1+num2);
break;
case '-':
printf("num1-num2=%.2f\n" ,num1-num2);
break;
case '*':
printf("num1*num2=%.2f\n" ,num1*num2);
break;
case '/':
printf("num1/num2=%.2f\n" ,num1/num2);
break;
default: //of operator is other than +, -, *, /, erros message shown
printf("Error! Invalid operator, this is basic math only.\n");
}
return 0;
}
根本改变流从:进入运营商,然后输入第一个数字,然后第二个数字。要:输入第一个号码,然后输入运营商,然后输入第二个号码。 我的问题是,当我这样做时,我看到了Enter运算符,但程序跳过了输入运算符的选项并要求输入第一个数字,然后输入第二个数字。响应是默认开关。
你看过[这个](http://stackoverflow.com/questions/14484431/scanf-getting-skipped?rq=1)的问题吗? – charmlessCoin