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[] ) { } 

What is the difference between using [square brackets] vs *asterisk, and does anyone have a style guide as to which is preferrable, assuming they are equivalent to the compiler?

For completeness, an example

void readEmStar( int *arrayOfInt, int len ) {   for( int i = 0 ; i < len; i++ )     printf( "%d ", arrayOfInt[i] ) ;   puts(""); }   void readEmSquare( int arrayOfInt[], int len ) {   for( int i = 0 ; i < len; i++ )     printf( "%d ", arrayOfInt[i] ) ;   puts(""); }  int main() {   int r[] = { 2, 5, 8, 0, 22, 5 } ;    readEmStar( r, 6 ) ;   readEmSquare( r, 6 ) ; } 

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!