CODEWITHSUNDEEP
ccommand-line-arguments
ABanerjee
ABanerjee

Reputation: 195

How can *++argv[0] refer to different command line arguments?

So I was reading The C Programming language and came across a section where programs were now allowed to have arguments... For example 

find -x -n pattern

Here, -x means except. -n means numbered lines... and pattern is what it will look for, in another few lines of input.

Now they regard find as *argv[0], -x and -n at *++argv[0], and pattern as *++argv[0]. How does a computer know one arg from the other?

If 3 things are all equal to *++argv[0], then they stay at argv[1], but all of them?? Could anyone please explain in depth?

Upvotes: 2

Views: 180

Answers (2)

Jim Lewis
Jim Lewis

Reputation: 45045

argv[0] = program name = "find"
argv[1] = first argument = "-x"
argv[2] = second argument = "-n"
argv[3] = third argument = "pattern"

argc = 4, so you know there are no other arguments to process.

Don't be confused by the use of the pre-increment operator in expressions like *++argv[0]. The arguments are passed in separate array elements.

When the shell executes your command, it uses whitespace to divide the command line up into the program name and arguments and pass them to your program. Sometimes you need to work around that by using double quotes, for example, if you need to deal with a file whose name contains embedded spaces:

mv some stupid filename sane_filename

This won't work because "some" "stupid" "filename" will be seen as separate arguments. But you can do this:

mv "some stupid filename" sane_filename

to get a single argument with embedded spaces.

Upvotes: 3

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798636

The ++n preincrement operator changes the variable it is applied to. The first time ++argv is executed, indexing it at 0 actually points to element 1 of the original value argv had, the second time it points to element 2, and so on.

Upvotes: 2

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