I have trouble using sscanf and fgets to check for overflow.

I will post the assignment specification so I could help whoever would kindly like to offer his/her help.

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1) The program should expect 2 decimal integers per line. Let's call these n1 and n2.2) For each such input line it should output the value of n1 + 2 * n2 (followed by a newline character '\n').

3) If there are not 2 integers on the line or there is something other than exactly 2 integers other than whitespace, the corresponding output should be the line

**"illegal input"**.

4) If the input is legal but the entered

__n1__or

__n2__or

__the calculated output value__would be greater than the maximum long int value, then the corresponding output should be

**"overflow"**.

5) If it would be less than the minimum (i.e. most negative) long int value, then the corresponding output should be

**"underflow"**.

• You can assume that the input line does not contain the null char '\0';

• You can also assume that no legal input line is longer than

__50 characters__(not counting the '\n' character). If such a line is encountered, the output should be

**"illegal input"**.

__n1 and n2 should be declared to be "long int".__

**RESTRICTION:****Do not**try to use a "larger type" to detect overflow/underflow. Instead, use the return values, etc. to detect overflow/underflow.

You are expected to use appropriate functions in your assignments. For example, reading the input one character at a time here is not appropriate.

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Now to my question. I have trouble checking if the input line is longer than 50 characters using fgets and sscanf.

Same with overflow. How do I check if n1 or n2 is overflow/underflow? As for the calculated result I think I could manage that.

I guess I would check if n1 > LONG_MAX - n2.

I think I will be able to do the program by reading character by character but the point of this assignment is to use tools such as sscanf fgets strlen and so on...

I have this assignment due in 2 days and I will appreciate any comments, hints, help before then?

Thank you very much and I will wait for your replies.