如何在错误条件下从头重启main()？

回答你的问题.

我不建议打电话给主.

您可以创建另一个包含所有代码的函数.

在main中,您可以调用该函数.

你可以在该函数内调用一个函数(称为递归)

但是,一个简单的循环可以完成这项工作.

do{
    printf("Hi!\nWelcome!\nThis is an expression based calculator\ndeveloped by Sankasuvra Bhattacharya\n");
    printf("that performs arithmetic operations on\ntwo numbers.\n");
    float num1;
    float num2;
    float ans = 0.0;
    char symbol;
    char ask;
    printf("Please type the expression you want to calculate: ");
    if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
    {
        printf("\nInvalid input! Please try again...\n\n");
    }
    else {
    switch(symbol) {
        case '+' : ans = num1 + num2;
                   break;
        case '-' : ans = num1 - num2;
                   break;
        case '*' :
        case 'x' :
                   ans = num1 * num2;
                   break;
        case '/' :
                   if (num2 == 0) {
                        printf("Division by zero is not possible!\nPlease try again...\n\n");
                        return main();
                    }
                    else {
                        ans = num1 / num2;
                        break;
                    }
        default :
                    printf("\nInvalid input! Please try again...\n\n");
                    return main();
    }
    printf("The answer is %g\n",ans);
    printf("\nTo use the calculator again, type 'Y'. ");
    printf("To exit, type any other character...\n");
    scanf("%s",&ask);
    printf("\n");
    }while(ask == 'y' || ask == 'Y') ;                        
    printf("Thank you for using the program. Please give full marks.");
}

编辑:要回答对此问题的评论,您要做的是:

while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
    printf("\nInvalid input! Please try again...\n\n");
}

并删除 else

EDIT2:完整代码.请注意,表达式不能超过99个字符.

#include 

int main()
{

float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
char string[100];
do{
    printf("Hi!\nWelcome!\nThis is an expression based calculator\ndeveloped by Sankasuvra Bhattacharya\n");
    printf("that performs arithmetic operations on\ntwo numbers.\n");

    printf("Please type the expression you want to calculate: ");

    while(1){
        fgets (string , 100 ,stdin);
        if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
            printf("\nInvalid input! Please try again...\n\n");
        else
            break;

    }

    switch(symbol) {
        case '+' : ans = num1 + num2;
                break;
        case '-' : ans = num1 - num2;
                break;
        case '*' :
        case 'x' :
                ans = num1 * num2;
                break;
        case '/' :
                if (num2 == 0) {
                        printf("Division by zero is not possible!\nPlease try again...\n\n");
                        return main();
                    }
                    else {
                        ans = num1 / num2;
                        break;
                    }
        default :
                    printf("\nInvalid input! Please try again...\n\n");
                    return main();
    }
    printf("The answer is %g\n",ans);
    printf("\nTo use the calculator again, type 'Y'. ");
    printf("To exit, type any other character...\n");
    scanf("%s",&ask);
    printf("\n");

}while(ask == 'y' || ask == 'Y') ;  

printf("Thank you for using the program. Please give full marks.");

return 0;

}