Difference between [square brackets] and *asterisk

Question

If you write a C++ function like

void readEmStar( int *arrayOfInt )
{
}

vs a C++ function like:

void readEmSquare( int arrayOfInt[] )
{
}
         


        

Answer 1

When you use the type char x[] instead of char *x without initialization, you can consider them the same. You cannot declare a new type as char x[] without initialization, but you can accept them as parameters to functions. In which case they are the same as pointers.

When you use the type char x[] instead of char *x with initialization, they are completely 100% different.

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Example of how char x[] is different from char *x:

char sz[] = "hello";
char *p = "hello";

sz is actually an array, not a pointer.

assert(sizeof(sz) == 6);
assert(sizeof(sz) != sizeof(char*)); 
assert(sizeof(p) == sizeof(char*));

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Example of how char x[] is the same as char *x:

void test1(char *p)
{
  assert(sizeof(p) == sizeof(char*));
}

void test2(char p[])
{
  assert(sizeof(p) == sizeof(char*));
}

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Coding style for passing to functions:

It really doesn't matter which one you do. Some people prefer char x[] because it is clear that you want an array passed in, and not the address of a single element.

Usually this is already clear though because you would have another parameter for the length of the array.

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Further reading:

Please see this post entitled Arrays are not the same as pointers!