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C++ Discussion Thread
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Jan 11th, 2009

while loop issue

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Can I do this?

c++ Syntax (Toggle Plain Text)
while(variable=functionCall()!=0){
 
     /* ... */
 
}
while(variable=functionCall()!=0){

     /* ... */

}

the function I'm calling, is returning a 3, but the variable in the while loop is equal to 1...
I'm not sure what is happening in this statement, Originally I thought it would just always evaluate to true and loop forever since, variable would be successfully set to functionCall()'s return value, therefore evaluating to 1, and 1 does not equal zero, so that too would evaluate to 1 (true) and loop forever... but it sets variable to 1 and then exits the while loop--what's going on?

functionCall() gets input from the user so I can only call it once really, I was going for while(functionCall()!=0){ } , but then I wouldn't be able to check the return value within the loop. I could do something else but I wanted to use this method, so I'm just wondering what it's actually doing.

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winrawr

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Jan 11th, 2009

Re: while loop issue

Quote ...

variable would be successfully set to functionCall()'s return value

no way, variable is set to true, because functionCall()!=0 is true. in this case true means 1

da penguin

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Jan 11th, 2009

Re: while loop issue

p.s. And yes, this does loop forever if functionCall() really returns 3

da penguin

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Jan 11th, 2009

Re: while loop issue

you should try something like this:

C++ Syntax (Toggle Plain Text)
while ((variable = functionCall())!=0)
while ((variable = functionCall())!=0)

...
because it seems to me that variable is given the value of evaluation: functionCall()!=0, so you just need to use one more pair of brackets...
I hope this helps

Alibeg

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Junior Poster in Training

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Jan 11th, 2009

Re: while loop issue

@da penguin, oh right, I forgot about right->left order, and you're right it was looping forever--there are two loops working together and I was mistaken with which loop I was in at what time.

@Alibeg, it worked great, thank you!

winrawr

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