I need help with this simple calculation!!!

Quote:

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Originally Posted by Lindley

It can; just don't get confused between the while which *opens* a while loop and the while which *closes* a do/while loop. The latter can't have a block after it.

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oh ok ok i got you didn't know the closing while loop could not show a message that's how you learn though by mistakes thanks alot.

I have another question...this is prob my last one for a while,

how would i go about

Code:

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if ( a!= A NUMBER )
{
cout<<" you did not enter a number, you cannot input character's that are not numbers " <<endl;

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so i want to make sure they entered a number could be decimal doesn't matter as long as it is a number but if they eneter for example

Code:

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@ or T or  "

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get it?


im not sure it's NAN

Quote:

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Originally Posted by J-C

so i want to make sure they entered a number could be decimal doesn't matter as long as it is a number but if they eneter for example @ or T or "
get it?
im not sure it's NAN

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if you're comparing a single character a to see if it is not a digit, you can use the same function Lindlay recommended earlier:

Code:

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if (strchr("0123456789",b) == NULL)

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Adding support for decimals makes it slightly harder -- you need to add '.' to the list, and keep track of if you have already seen a decimal point inside this particular number. This can be simply done by using a boolean flag that you set to true the first time you see a decimal, and you reset to false when you're done with that number.

If you just want to convert a character / string to a number without verifying that it is actually a number, you can use atoi()... if they enter something that is not a number, atoi returns 0.

There's also an isdigit() function which operates on individual characters.