I am writing this simple code for checking if the user has entered the key or not. If not I want to return 1 and show an error msg. Below is the snapshot of my code which I am using. I am getting the following error. I will appreciate your help on the same. Thank You.
It's quite literally what it says. First, understand that all warnings are treated as errors because of one of the flags in the compile statement. This would otherwise just be a warning.
Now to the real issue. It literally means that the var argv is not being used anywhere in the program. If you were to do anything with the input vars, like print them for instance, this error would disappear.
It would be the same as declaring
int foo; in the program and never doing anything with foo after that.
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I concur with Cliff B explanation to the problem. In order to resolve the error,there are two ways forward :
1.You can either remove the parameter from the main (which will not fulfill the specification of the problem set)
2.Use the argv array or some of its element somewhere in the program inside the main code.For that you should know the values in the argv array. That is enough hint ,i guess.