C++ char pointer to char variable

Question

I'm given a method header as so:

char* thisMethod(char* input){}

is it possible to say,

char var = input[0];

? I know that "input" will be in the form of a char array.

I'm obviously new to C++ pointers are throwing me off. I tried researching how to work with char pointer function arguments but couldn't find anything specific enough to help. Thanks for the help.

Is that second line of code intended to go inside the function? If so, write it that way. And, yes, if you do that, it will copy the character that input points to into var. — Pete Becker
– Pete Becker, Commented Sep 14, 2016 at 14:21
Picking at nits, all you can guarantee is that there is one character that the pointer is pointing to. Anything else is an assumption. For passing more than one character, use std::string or std::vector<char> or some other container. — Thomas Matthews
– Thomas Matthews, Commented Sep 14, 2016 at 16:28

Emilio Garavaglia · Accepted Answer · 2016-09-14 19:01:52Z

There is a misconception that may lead you into further troubles:

I know that "input" will be in the form of a char array.

NOPE: By the scope of that function, input is a pointer to a character. The function has no way to know where such a pointer comes from.

If it has been taken from an array upon calling the function than it will be a pointer to the first element of that array.

Because pointer have an arithmetic that allows to add offsets and because the [] operator applied to pointers translates as a[b] = *(a+b) by definition, in whatever code, if a is a pointer, *a and a[0] are perfect synonymous.

Think to an array as a sequence of boxes and a pointer as your hand's index finger

adding an offset to a finger (like finger+2) means "re-point it aside" and de-referencing it (like *finger) means "look inside what it points to".

The [] operator on pointers is just a shortcut to do both operations at once.

Arrays are another distinct thing. Don't think to them when dealing with pointers, since -in more complex situations, like multidimensional array or multi-indirection pointers - the expression a[b][c] won't work anymore the same way.

Mark Ransom · Accepted Answer · 2016-09-14 14:22:24Z

0

There are two ways to get a value from a pointer, * and []. The following are equivalent:

char var1 = *input;
char var2 = input[0];

Using the brackets is more common when you know you were passed an array, since it allows you to supply an index. You need some way of knowing where the end of the array is so that you don't attempt any access past it, your function is missing that important detail.

answered Sep 14, 2016 at 14:22

Mark Ransom

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2 Comments

Brandon Walker Over a year ago

Now would it be ok to use input[1] input[2] etc to continue down the char array that I know is being looked at by the pointer? Or is there a way to place the char array into a string so it is easier to track the size/boundaries of the char[]/string?

Mark Ransom Over a year ago

@BrandonWalker The usual convention with strings is for every character to be valid until you reach a character with a value of zero. Not '0', but 0. If you have that you can easily copy it to a std::string with simple assignment.

Hatted Rooster · Accepted Answer · 2016-09-14 14:29:30Z

0

As long as it's inside the function and input points to something valid ( not NULL/nullptr and not a garbage location ) then doing char var = input[0]; is just fine. It's the same as char var = *input.

P.S If it's supposed to be a string I recommend using std::string.

edited Sep 14, 2016 at 14:29

answered Sep 14, 2016 at 14:21

Hatted Rooster

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