I understand the concept of pointers and arrays in C. My problem is the syntax of them. What is this:
char* var;
char var*;
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3 Answers
Your premise is wrong. The first is the pointer type and the second is well... nothing, it is not valid C++ (neither is C for that matter). Static arrays are more like this:
char var[15];
Dynamic arrays are on the other hand usually done with vectors.
The confusion between arrays and pointers is a legacy from the C programming language, where *(a+n) and a[n] mean the same thing, so we can use the a[n] even if a is a pointer, not an array. This is not very idiomatic in C++ though.
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char* var; declares var to be a "pointer to char". char var*; is not a valid declaration.
An array would be declared something like: char var[10];.
answered May 4, 2012 at 20:37
6 Comments
Qaz
Or something like:
std::array<char, 10> var;Qaz
@classdaknok_t, it's the ideal static array to be using if you have the resources available to do so.
leftaroundabout
@classdaknok_t of course it's an array. Just not a C-style array, but this question is about C++.
Qaz
@SethCarnegie, of course, but the whole purpose of it is to replace the use of an array with a better alternative. We say
std::string objects are strings, but strings in the base sense are still C-style character arrays. Hopefully std::array will take over in the same way one day. |
This is a pointer:
char* var;
This is syntactically incorrect, and will not compile:
char var*;
Arrays and pointers in C++ are not the same thing, but the syntax might make it seem like they are. You can declare an array in C++ like this:
char ary[] = {'a','b','c'};
...and this is valid -- there will be an array created with 3 elements.
You can also create a function which takes a pointer:
void foo(char* bar)
{
// ...
}
...and pass it ary:
foo(ary);
...and it will compile and run fine. This might lead you to believe that pointers and arrays are the same thing; but they aren't. bar in foo above isn't an array -- it's a pointer that points to the first element of the array.
answered May 4, 2012 at 20:42
1 Comment
Qaz
It might be worth mentioning that in the call
foo(ary);, ary decays into the needed pointer.