How to restart main() from the beginning on an error condition? - c

Question

#include <stdio.h>
int main()
{
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");
/* want to restart main() again here */
}
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);
if (ask == 'y' || ask == 'Y') {
printf("\n");
main();
}
else {
printf("Thank you for using the program. Please give full marks.");
}
}
return 0;
}

Answer 1

To answer your question.
I would not recommend calling main.
You could create another function that has all you code.
Inside main, you call that function.
You can call a function inside that function (called recursion)
However, a simple loop could do the job.
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.");
}
Edit: To answer the comment to this question what you want to do is:
while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("\nInvalid input! Please try again...\n\n");
}
And remove the else
EDIT2: Full code. Note that the expression cannot be more than 99 characters.
#include <stdio.h>
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;
}

Answer 2

#Kristjan Kica answer is good. I think you are using spaces in your input like 1 + 2.
According to manual page of scanf
All conversions are introduced by the % (percent sign) character. The format string may also contain other characters. White space
(such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else
matches only itself. Scanning stops when an input character does not match such a format character. Scanning also stops when an input
conversion cannot be made.
Remove the spaces and try again.
Example:
1+2 should work by the changes mentioned in kristijan answer.
Edit:
Replace the line in #Kristjan Kica answer
while(ask == 'y' || ask == 'Y') ;
with
}while(ask == 'y' || ask == 'Y') ;
Edit 2:
Last closing } should be your main functions closing brace.

Answer 3

Finally solved it. All thanks to Kristjan Kica...
#include <stdio.h>
int main(void)
{
float num1;
float num2;
float ans;
char symbol;
char ask;
char string[100];
fflush(stdin);
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: ");
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
{
printf("\nInvalid input! Please try again...\n\n");
main();
}
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");
if (ask == 'y' || ask == 'Y')
{
main();
}
else {
return 0;
}
}
}