problem

[samosato]

hi all I got following problem : I want to read and variable (age ) but when its not properly entered I want to write bad format or etc. example printf("enter age \n"); scanf ("%d,&age"); and I need to create a condition which is saying in case you enter string or non integer number (7,5 ) write bad format eter again thanks

[Nuri Ismail]

This is a common problem of scanf(). You can use this[^] technique to solve your problem. NOTE: Next time when you want to post a C question please use the C\C++\MFC forum[^].

[Lost User]

Use the features of the function to verify your input; see the MSDN documentation here[^]. A simple loop checking that you have a converted field and that its value is within the range you require should do the trick.

Just say 'NO' to evaluated arguments for diadic functions! Ash

[Alain Rist]

Hi, Supposing you search in this forum a C++ STL solution :)

#include<iostream>
#include<strstream>
#include<string>

int main()
{
using std::cout;
using std::endl;

// This is what you want
cout << "Enter age" << endl;
int age = -1;
std::string s;
while (age < 0)
{
	std::getline(std::cin, s);

	if (s.empty())
		cout << "Try again!" << endl;
	else if (s.find\_first\_not\_of("0123456789") != s.npos)
		cout << s << " is not a valid age. Try again!" << endl;
	else
		std::istrstream(s.c\_str()) >> age;

	const int HighestAgeEver = 122;
	if (age > HighestAgeEver)
	{
		cout << age << "!! You are kidding. Try again!" << endl;
		age = -1;
	}
}

cout << "Your age is " << age << endl;

return 0;

}

cheers, AR Edited to handle empty input and unrealistic age :) And for simpler code flow

When the wise (person) points at the moon the fool looks at the finger (Chinese proverb)

modified on Wednesday, October 13, 2010 1:37 PM