What does long * tempArray[10]; do?

Question

Can you please explain what this piece of C++ code does:

int main()
{
    long * tempArray[10];
}

Thanks for your help!

This is pretty basic C stuff. What instructional materials are you using to learn? — Greg Hewgill, Jul 23 '12 at 03:06
A great resource to help decipher C-style declarations is [cdecl.org](http://cdecl.ridiculousfish.com/?q=long+*+tempArray%5B10%5D%3B) — Blastfurnace, Jul 23 '12 at 03:07
@user1544818 http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list — jogojapan, Jul 23 '12 at 03:07
possible duplicate of [How do I create an array of pointers?](http://stackoverflow.com/questions/620843/how-do-i-create-an-array-of-pointers) — jogojapan, Jul 23 '12 at 03:07
As @GregHewgill said, maybe you should start from the beginning if you are asking what the '*' means, otherwise it might be a painful learning process. — Vincent B., Jul 23 '12 at 03:08
@user Please read [Clockwise Spiral Rule](http://c-faq.com/decl/spiral.anderson.html). — Mahesh, Jul 23 '12 at 03:08
@dreamlax: I hate to spoil the joke, but some novices might take you seriously. — Keith Thompson, Jul 23 '12 at 03:20

score 1 · Accepted Answer · answered Jul 23 '12 at 03:50

That particular snippet of code does nothing at all. If you compile that program, it will terminate and report to the hosting environment that it terminated successfully, and that's it.

The long * tempArray[10]; declares a variable called tempArray to have the type array 10 of pointer to long, which means that tempArray is capable of holding 10 long * objects.

For demonstrative purposes:

// declare some long integers
long myLong = 100;
long anotherLong = 400;
long thirdLong = 2100;

// declare array 3 of pointer to long
long *tempArray[3];

// remember that arrays in C and C++ are 0-based, meaning index 0
// refers to the first object in an array. Here we are using the `&'
// operator to obtain a long pointer to each of the longs we declared
// and storing these long pointers in our array.
tempArray[0] = &myLong;
tempArray[1] = &anotherLong;
tempArray[2] = &thirdLong;

score 0 · Answer 2 · edited Jul 23 '12 at 05:04

0

It creates an array of 10 long pointers.

edited Jul 23 '12 at 05:04

Anthony

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answered Jul 23 '12 at 03:05

Keith Nicholas

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what does the asterisk (*) mean? – Shivanshu Goyal Jul 23 '12 at 03:06

score 0 · Answer 3 · answered Jul 23 '12 at 03:06

0

The code doesn't actually do anything.

long * tempArray[10]; allocates an array capable of holding ten pointers to long integers.

answered Jul 23 '12 at 03:06

Russell Borogove

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What does long * tempArray[10]; do?

3 Answers3